Ratio & Proportion 1.1 GOPAL BHOOT CHAPTER RATIO AND … · 2019-09-28 · Ratio & Proportion 1.1...

Ratio & Proportion 1.1 GOPAL BHOOT

CHAPTER

RATIO AND PROPORTION PAST EXAMINATION QUESTIONS

RATIO AND PROPORTION 1. Two numbers are in the ratio 2 : 3 and the difference of their squares is 320. The numbers are : Nov-

2006 Ans:(b) (a) 12, 18 (b) 16, 24 (c) 14, 21 (d) None. Solution : use E Method or =(3x)2- (3x)2= 3200 16,24864320532049 2222 xxxxx

2. If p : q is the sub-duplicate ratio of p – x2: q – x2, then x2 is : Nov - 2006Ans:(d) (a) (b) (c) (d) None

Solution: 2

2

2

2

q

p

xq

xp

q2(p-x2) = p2(q-x2) pqqpxqxpxpqpxqpqxqpxpq 2222222222222222 )()(

)(

)())(()( 2222222

qp

pqxqppqqpqpxpqqpqpx

3. An alloy is to contain copper and zinc in the ratio 9 : 4. The zinc required to melt with 24kg of copper is: Nov-2006 Ans:(a)

(a) 10 (b) 10 (c) 9 (d) 9 kg

Solution: px

x 24

4

9

9

424 P = Kg

3

210

3

32

4. Two numbers are in the ratio 7 : 8. If 3 is added to each of them, their ratio becomes 8 : 9. The

numbers are : Feb - 2007 Ans:(c) (a) 14,16 (b) 24,27 (c) 21,24 (d) 16,18

Solution: 9

8

38

37

x

x

4083575127632464 xxx

= 21, 24 (dividing by 17 we get) 5. The third proportional between (a2 – b2) and (a+b)2 is : Feb - 2008Ans:(d)

(a) (b) (c) (d)

Solution:

ba

baP

baba

babaP

p

ba

ba

ba

3222

2

22 )(

))((

)()()(

)(

6. If then the value of is : June - 2009Ans:(c)

(a) 1 (b) -1/7 (c) 1/7 (d) 7 Solution:

3

2

3

3 qP

q

p

= qp

qp

2

2

qq

qq

qq

qq

3

43

4

3

22

3

22


q

qq

q

qq

qq

7

3

33

73

3

343

34

7

1

7. Fourth proportional to x, 2x, (x+1) is : Ans:(c) (a)(x+2) (b) (x-2) (c) (2x+2) (d) (2x-2) Solution:

221

2

xPx

p

x

x

x

8. If A : B = 2 : 5, then (10A + 3B): (5A + 2B) is equal to Dec - 2010 Ans:(a)

(a) 7:4 (b) 7:3 (c) 6:5 (d) 7:9Solution: A=2 B = 5

20

35

5.22.5

5.32.10

25

310

BA

BA

9. The ratio Compounded of 4 : 5 and sub-duplicate ratio of “a” : 9 is 8 : 15 Then Value of “a” is : Dec-2011 Ans. (c) a) 2 b) 3 c) 4 d) 5

Solution:

42154

358

95

4

15

8

aaa

10. If X Varies inversely as square of Y and given that Y = 2 for X = 1, then the Value of X for Y = 6 will be : Dec - 2011Ans. (d) a) 6 b) 9 c) 1/3 d) 1/9

Solution: 22

1

y

kx

yx

11. Which of the numbers are not in proportion ? June-2012 Ans. (a) (a) 6, 8, 5, 7 (b) 7, 14, 6, 12 (c) 18, 27, 12, 18 (d) 8, 6, 12, 9

Solution:22 12

6

14

7

7

5

8

6Proportion

d

c

b

a

18

12

27

18

39

26

3

2

3

2

3

4

3

24

9

12

6

8

3

4

3

4

12. Find two numbers such that mean proportional between them is 18 and third proportional between them is 144 June-2012 Ans. (a)

(a) 9,36 (b) 8,32 (c) 7,28 (d) 6,24

Solution : let the two numbers be x & y x : 18:18:y y

xy

x 32418

18 ……. (i)

x : y : y : 144 n

yxyx

y

y

x

14144.

144

22 ………….. (ii)

Comparing (i) and (ii)

3646656144

3243

2

yyy

y 13. The mean proportional between 24 and 54 is : June-2013 Ans. (d)

a) 33 b) 34 c) 35 d) 36

Solution: 24 : x : x : 54 3654

24 x

x

x


14. The triplicate ratio of 4 : 5 is: Ans. (c) a) 125 : 64 b) 16 : 25 c) 64 : 125 d) 120 : 46

Solution: Triplicate ratio of 4 : 5 is 64 : 12 15. If x : y = 2: 3 then (5x + 2y) : (3x – y) = June - 2014Ans. (b)

a) 19:3 b) 16:3 c) 7:2 d) 7:3. Solution:

3:163

1616

3

610

32.3

3.22.5

3

25

3,2,,3

2

4

k

k

k

kk

kk

kk

yx

yx

kykxletx

16. If ,7215 22 pqqp where p and q are positive, then p : q will be: Dec-2015 Ans. (a)

a) 5 : 6 b) 5 : 7 c) 3 : 5 d) 8 : 3 Solution: 15(2p2-q2) = 7pq 30p2-15q2 =7pq30p2-7pq-15q2 = 0 30p2-(25-18)pq-15q2=0

30p2-25pq+18pq-15q2=0 5p(6p-5q)+3q(6p-5a)=0 (6p-5q)(5p+3q)=0 6p=5q 5p= - 3q

17. The ratio of third proportion of 12, 30 to the mean proportion of 9, 25 is : Dec-2015 Ans. (b) a) 2 : 1 b) 5 : 1 c) 7 : 15 d) 3 : 5 Solution :

12:30::30: x mean proportion = 259

x

30

30

12 =15

x = 75 ratio=75 : 15 =5 : 1

18. What number must be added to each of the numbers 10, 18, 22, 38 to make the numbers is proportion? Dec-2015 Ans. (a) a) 2 b) 4 c) 8 d) None of these Solution : let, the number is x (10+x) : (18+x) : (22+x) : (38+x)

x

x

x

x

38

22

18

10

(10+x)(38+x)=(22+x)(18+x) 380+10x+38x+x2=396+22x+18x+x2 8x=16

19. The value of

22

22

22

22

22

22

xzy

yxz

zyx

zxy

yzx

zyxis June-2016 Ans. (b)

a) 0 b) 1 c) – 1 d) ∞

Solution: 22

22

22

22

22

22

)(

)(

)(

)(

)(

)(

xzy

yxz

zyx

zxy

yzx

zyx


=))((

))((

))((

))((

))((

))((

xzyxzy

yxzyxz

zyxzyx

zxyzxy

yzxyzx

zyxzyx

zyx

yxzzxyzyx

zyx

zyx

=1

20. If a:b = 2:3, b:c = 4:5 and c;d = 6:7, then a:d is: June - 2017 (Ans: c)

a) 24:35 b) 8:15 c) 16:35 d) 7:15

43

2

b

a 3

5

4

c

b

7

6

d

c

12

8

b

a

15

12

c

b

15

8

c

a

15

7

6

d

c

90

48

c

a

105

90

d

c

35:1635

16

105

48

d

a

21. If a = 56

56

and b =56

56

then the value of 22

11

ba is equal to: June - 2017 (Ans: b)

a) 480 b) 482 c) 484 d) 486

Solution: a+b=56

56

56

56

5656

565622

=22

ab= 156

56

56

56

2

2

22

22

22

211

ab

abba

ba

ab

ba

1

1.22 2 =482

CONCEPTUAL

22. A box contains ` 56 in the form of coins of one rupee, 50 paise and 25 paise. The number of 50 paise double the number of 25 paise coins and four times the numbers of one rupee coins. The numbers of 50 paise in the box is : : Feb-2007 Ans:(a) (a) 64 (b) 32 (c) 16 (d) 14 Solution: Let, no. of one rupee coin is x. Fifty paise 2x25 paise 4X

564

22

4

xxx

56

2

656

2356

22

xxxx

xxx =7x =112x =16

no. of 50paise coin is = 4×x =69 23. Eight people are planning to share equally the cost of a rental car. If one person withdraws from the

arrangement and the others share equally entire cost of the car, then the share of each of the remaining persons increased by : : May - 2007Ans:(c) (a) 1/9 (b) 1/8 (c) 1/7 (d) 7/8

Solution: 8

x

7

x iner =

7

x-

8

x=

5656

78 xxx


remaining 8

1 cost born by 7 persons.

8

1

7

1ofth

Cost increase by

7

1

24. A bag contains `187 in the form of 1 rupee, 50 paise and 10 paise coins in the ratio 3:4:5 Find the number of each type of coins : : May - 2007 Ans:(a) (a) 102, 136, 170 (b) 136, 102, 170 (c) 170,102,136 (d) None Solution:

18710

52030187

10

5

2

43

xxxxxx x = 34 =102,136,170

25. Ratio of earnings of A and B is 4 : 7. If the earnings of A increase by 50% and those of B decrease by 25%, the new ratio of their earning becomes 8:7. What is A's earning ? : Aug - 2007Ans:(d)

(a) ` 21,000 (b) ` 26,000 (c) ` 28,000 (d) Data inadequate Solution:

7

8

7.17

24

xxx

xx

xxxx 14286.1356

26. P, Q and R are three cities. The ratio of average temperature between P and Q is 11 : 12 and that between P and R is 9 : 8. The ratio between the average temperature of Q and R is : Aug-2007 Ans:(b)

(a) 22:27 (b) 27 :22 (c) 32 :33 (d) None.

27. ` 407 are to be divided among A, B and C so that their shares are in the ratio : : . The respective

shares of A, B, C are : Nov-2007 Ans:(a) (a) ` 165, ` 132, ` 110 (b) ` 165, ` 110, ` 132 (c) ` 132, ` 110, ` 165 (d) ` 110, ` 132, ` 165

Solution:

407654

xxx

110,132,1656

660,

5

660,

4

660660407

120

202480

x

xxx

28. The incomes of A and B are in the ratio 3 : 2 and their expenditures in the ratio 5:3. If each saves ` 1,500, then B's income is : Nov-2007 Ans:(a)

(a) ` 6,000 (b) ` 4,500 (c) ` 3,000 (d) ` 7,500 Solution : 3x-5y=1500 (i) 2x-3y=1500 (ii)

(i) ×2 & (ii) × 3 6x-104=3000 6x- 94=4500 -y = -1500 y=1500

3x - 5.1500 = 1500

300090003

2xx

29. In 40 litres mixture of glycerine and water, the ratio of glycerine and water is 3:1. The quantity of water added in the mixture in order to make this ratio 2 is: Feb-2008 Ans:(d) (a) 15 litres (b) 10 litres (c) 8 litres (d) 5 litres

Solution: Assume x

5362201

2

10

3010

4

14030

4

340

xx

x

30. In what ratio should tea worth ` 10 per kg be mixed with tea worth ` 14 per kg, so that the average price of the mixture may be ` 11 per kg? June-2008 Ans:(b) (a) 2:1 (b) 3:1 (c) 3:2 (d) 4:3 Solution: Let the ratio is =x:y

31. The ages of two persons are in the ratio 5:7. Eighteen years ago their ages were in the ratio of 8:13, their present ages (in years) are: June-2008 Ans:(a) (a) 50,70 (b) 70,50 (c) 40,56 (d) Non Solution: Let, this ages 5x,7x

13

8

187

185

x

x 56x-144=65x-234 x990 10 x


32. If A, B and C started a business by investing ` 1,26,000, ` 84,000 and ` 2,10,000. If at the end of the

year profit is ` 2,42,000 then the share of each is: Dec-2008 Ans:(a) (a) 72,600, 48,400, 1,21,000 (b) 48,400, 1,21,000, 72,600 (c) 72,000, 49,000, 1,21,000 (d) 48,000, 1,21,400, 72,600

Solution: 84

126

84000

126000

B

A

210

84

210000

84000

C

B

A : B : C = 126 : 84 : 210 33. What must be added to each term of the ratio 49 : 68, so that it becomes 3:4? June-2010 Ans:(c)

(a) 3 (b) 5 (c) 8 (d) 9

Solution: 4

3

68

49

x

x

34. The students of two classes are in the ratio 5 : 7, if 10 students left from each class, the remaining students are in the ratio of 4 :6 then the number of students in each class is : June-2010 Ans:(c) (a) 30,40 (b) 25,24 (c) 40,60 (d) 50,70

Solution :

6

4

107

105

x

x 60304028 xx x220

B = 5

20

35

5.22.5

5.32.10

25

310

BA

BA

35. In a film shooting, A and B received money in a certain ratio and B and C also received the money in the same ratio. If A gets ` 1,60,000 and C gets ` 2,50,000. Find the amount received by B.: June-2011 Ans. (a) (a) ` 2,00,000 (b) ` 2,50,000 (c) ` 1,00,000 (d) ` 1,50,000

Solution :

∴ √160000 250000 = ` 2,00,000

36. Find three numbers in the ratio 1 : 2 : 3, so that the sum of their squares is equal to 504 Dec-2013 Ans. (a) (a) 6, 12, 18 (b) 3, 6, 9 (c) 4, 8, 12 (d) 5, 10, 15 Solution: Let, the no's x, 2x, 3x x2 + (2x)2 + (3x)2 = 504 x2 + 4x2 + 9x2 = 504 14x2 = 504

38. If the salary of P is 25% lower than that Q and the salary of R is 20% higher than that of Q, the Ratio of the salary of R and P will be : June-2014 Ans. (b) a) 5:8 b) 8:5 c) 5:3 d) 3:5 Solution: let, the salary of Q=100 P's Salary =100 - 25: = 75 R's Salary =100 + 20: =120 x2+y2+2x = 9xy (x + y)2 = 9xy log (x + y)2 = log 9xy 2 log(x+y) = 2 log3 + log x + log y 2log(x+y)-2log3 = logx + logy

2(log(3

yx )= logx + logy

log3

1(x+4)=

2

1(log x + log y)


39. A person has assets worth `1,48,200. He wish to divide it amongst his wife, son and daughter in the ratio 3:2:1 respectively. From this assets, the share of his son will be: June-2014 Ans. (b) a) `24,700 b) `49,400 c) `47,100 d) `37,050 Solution: Total Worth of Assets =1,48,200

40. For three months, the salary of a person are in the ratio 2:4:5. If the difference between the product of salaries of the first two months and last two months is ` 4,80,00,000; then the salary of the second month will be : Dec-2014 Ans. (c) a) `4,000 b) `6,000 c) `8,000 d) `12,000 Solution: let, the salary of three months 2x, 4x, 5x (4x × 5x) - (4x × 2x) = 4,80,00,000 20x2 - 8x2 = 4,80,00,000 12x2 = 4,80,00,000 x = 2000

41. A dealer mixes rise costing ` 13.84 per Kg. with rice costing ` 15.54 and sells the mixture at ` 17.60 per kg. So, he earns a profit of 14.6% on his sale price. The proportion in which he mixes the two qualities of rice is June-2015 Ans. (a) a) 3 : 7 b) 5 : 7 c) 7 : 9 d) 9 : 11

Solution: 17.60 - 14.6% = yx

yx

54.1584.13

0304.15yx

yx

54.1584.13

15.0304x +15.0304y=13.84x+15.54y 1.1904x=0.5096y

428.01904.1

5096.0

y

x

log 2log

log

y

x

log10+logx-2logy 1+ m + n -2m+2n

42. X, Y, Z together starts a business, If X invests 3 times as much as Y invests and Y invests two third of what Z invests, then the ratio of capitals of X, Y, Z is June-2016 Ans. (d) a) 3:9:2 b) 6:3:2 c) 3:6:2 d) 6:2:3

Solution: X=3y y= z3

2

1

3

y

x

3

2

z

y

x : y = 3 : 1 y : z = 2 : 3 x : y = 6 : 2 x : y : z =6:2

43. There are total 23 coins of `1,`2 and`5 in a bag. If their value is `43 and the ratio of coins of `1 and `2 is 3:2. Then the number of coins of `1 is : Dec-2016 Ans.(a) (a)12 (b)5 (c)10 (d)14 Solution: let, one rupee coin is 3x 2 rupee coin is +2x 5 rupee coin= 23- 2x-3x = 23-5x 3x×1+2x×2+(23-5x) ×5=43 3x+4x+115-25x=43 - 18x= -72 x=4 no. of one rupee coin is 3x= 3×4=12


Answer Key 1. b 2. d 3. a 4. c 5. d 6. c 7. c 8. a 9. c 10. d 11. a 12. a 13. d 14. c 15. b 16. a 17. b 18. a 19. b 20. c 21. b 22. a 23. c 24. a 25. d 26. b 27. a 28. a 29. d 30. b 31. a 32. a 33. 34. c 35. c 36. a 37. a 38. c 39. b 40. b 41. c 42. a 43. d 44. 45. 46.

Indices & Log [Solution] 2.6 GOPAL BHOOT

CHAPTER

INDICES AND LOGARITHM PAST EXAMINATION QUESTIONS

INDICES 1. Value of ( a1/8 + a-1/8) (a1/8 – a-1/8) (a1/4 + a-1/4) (a1/2 + a-1/2) is : : Feb-2007 Ans:(b)

(a) (b) (c) (d)

Solution:

aaaaaaaaaaaaaa

aaaaaaaa1

)())(())()((

))()()((112

12

12

12

12

12

14

14

14

14

1

21

21

41

41

81

81

81

81

2. Simplification of n6m6

n9m4n3m

x

x.x

is : : May-2007 Ans:(b)

(a) mx (b)

mx (c) nx (d)

nx

Solution: nm

nmnm

x

x

66

943

mx

3. On simplificationbcacabcbcaba zz1

1

zz1

1

zz1

1

reduces to :Aug-2007Ans.(c)

(a) (b) (c) 1 (d) 0

Solution :

b

c

a

c

a

a

b

c

c

a

b

a

z

z

z

z

z

z

z

z

z

z

z

z

1

1

1

1

1

1

c

b

c

a

b

a

b

c

a

c

a

b

z

z

z

z

z

z

z

z

z

z

z

z

1

1

1

1

1

1

bac

c

acb

b

cba

a

zzz

z

zzz

z

zzz

z

1

cba

cba

zzz

zzz

4. √/

√

/ is equal to: : Nov-2007Ans:(b)

(a) 1 (b) 9

1 (c) 33 (d) √

Solution :

2

7

21

225

2

21

3.3

3

3

3

2

72322

522

133 )3()3( 2

3225

23 2

7

9

133

4

71533333 24

84

7

4

15

27

21

25

23

5. If ,422 1xx then the value of xx is : : Feb-2008 Ans:(d)

(a) 2 (b) 1 (c) 64 (d) 27 Solution:

122 xx = 4

4)211(2

42

22

x

xx

273

22242

33

3

3

xx

x

x


6. If ba zy,yx and

cxz then abc is : : June-2008 Ans:(b) (a) 2 (b) 1 (c) 3 (d) 4 Solution: z=xc =(y2)c =[(zb)]c =z=zabc =abc=1

7. If 3/133 31 x then find value of x9x3 3 : June-2009 Ans:(d)

(a) 3 (b) 9 (c) 12 (d)10 Solution:

x =3 31

31

3

3x = 33133

13

1093

3

2933

13333

13

.3

13.3

3

133333.333

2333

31

3133

13

13

13

113

nnxxxxxx

xxx

8. Find the value of: 1 1 1 / : June-2009 Ans:(b) (a) 1/x (b) x (c) 1 (d) None of these

Solution: [1-{1-(1-x2)-1] 21

=

21

1

21

111

x

21

1

2

2

1

111

x

x

21

2

211

x

x

2

22 1

x

xx2

1

x

xx

x

11

1

11

2

21

2

9. : Dec-2009 Ans:(b)

(a) 1/2 (b) 3/2 (c) 2/3 (d) 1/3

Solution:

2312

3

)12(22

112

22.22

122

nnn

nn n

10. If 360532 zyx Then what is the value of x, y, z.? : Dec-2009 Ans:(a)

(a) 3, 2, 1 (b) 1, 2, 3 (c) 2, 3, 1 (d) 1, 3, 2 Solution: Use E factors of 360

23 32 51 x = 3,y = 2, z = 1

11. If 2 2 4then is equal to : : June-2010 Ans:(c) (a) 7 (b) 3 (c) 27 (d) 9 Solution: 2x-2x-1=4

2x-2x 42

1

3228242

22.23

xxx

xx

=xx = 33 =27

12. The recurring decimal 2.7777………… can be expressed as : Dec-2010 Ans:(d) (a) 24/9 (b) 22/9 (c) 26/9 (d) 25/9

13. The value of

is equal to : June-2012 Ans. (b)

(a) 1/5 (b) 1/6 (c) 1/4 (d) 1/9

Solution:13

1

33

33

nn

nn

6

1

24

4

)33(3

)13(3

3.33.3

33.3313

1

n

n

nn

nn

14. Find the value of x, if xxxx 313

1

: Dec-2012 Ans. (b)


(a) 3 (b) 4 (c) 2 (d) 6

Solution: 34

31

xxx 334 x

xx

433

4 x

x

15. If ,0333 cba then the value of : Dec-2013 Ans. (a)

(a) abc (b) 9abc (c) (d)

Solution:

0333 cba

31

31

31

cba 3313

31

31

cba

= 331

31

31

31

313

313

31

.3 cbababa ccbaba ).(.3 31

31

31

31

31

31

3 cbacba

31

31

31

3cba

cba

abc

cba

3

3 17. The value of : June-2014 Ans. (c)

a) y b) -1 c) 1 d) None

Solution: 222222 acacaccbcbcbbababa yyy 333333 accbba yyy 1333333 accbbay

18. If rqqp yx , and ,6pr z then the value of xyz will be: : June-2015 Ans. (d) a) 0 b) 1 c) 3 d) 6 Solution: px=q, qy=r, r2=p6 =(qy)2=p6 =((px)y)z=p6 pxyz=p6 on comparing, xyz=6

19. The value of nn

nn

22

221

1

is : : Dec-2015 Ans. (b)

a) 1/2 b) 3/2 c) 2/3 d) 2

Solution:nn

nn

22

221

1

= )12(2

)2

11(2

1

n

n

= 12

3

=2

3

20. If 3 5 75 , then : Dec-2016 Ans.(c)

(a)x + y - z =0 (b) zyx

112

(c)

zyx

121

(d)

yzx

112

Solution : 3x=5y=75z=k

3x=k

3=k x1

5y=k

5=k y1

75z=k

75=k z1

3×5×5=75 =k x1

×k y1

×k y1

=k z1

= k zyyx k1111

zyyx

1111

zyx

121

LOGARITHM

29. 7 5 3 is equal to : : Nov-2006 Ans:(c)

(a) 0 (b) 1 (c) log 2 (d) log 3

Solution: 7 log

15

16 + 5 log

24

25+ 3 log

80

81

)80log81(log3)24log25(log5)15log16(log7

)loglog4(log43)log3(loglog25)log(loglog47 523235532 523235532 log3log12log12log15log5log10log7log7log28 2log

30. The value of the expression: tdlog.d

clog.cblog.balog

a : Feb-2007Ans:(a)

(a) t (b) abcdt (c) (a + b + c + d) (d) None


Solution: tdlog.d

clog.cblog.balog

a

taaaaa

t

d

t

c

d

b

c

a

b

loglog

log

log

log

log

log

log

log

log

log

31. If , then x is given by : : Feb-2007 Ans:(b)

(a) 1/100 (b) 1/10 (c) 1/20 (b) None of these

Solution:

44

41

41

10

1

)10000(

11000

xx

10

1 x =10%

32. If log (2a – 3b) = log a – log b, then a = : : May-2007Ans:(a)

(a) (b) (c) (d)

Solution: log (2a-3b)=loga-logb

log (2a- 3b)= log ( )b

a

12

3)32(

2

b

ba

b

aba

33. )abc(log

1

)abc(log

1

)abc(log

1

cabcab

is equal to : : Aug-2007Ans:(c)

(a) 0 (b) 1 (c) 2 (d) -1

Solution=abc

ca

abc

bc

abc

ab

log

log

log

log

log

log

abc

cabcab

log

logloglog

2log

log2

log

..log

abc

abc

abc

cabcab

34. Number of digits in the numeral for 264. [Given log 2 = 0.30103] : : Aug-2007Ans:(c) (a) 18 digits (b) 19 digits (c) 20 digits (d) 21 digits Solution: 264 taking log = 64 log2 = 64×0.30103 = 19.26 = 20digits.

35. If zyx 2054 then z is equal to : : Nov-2007Ans:(d)

(a) xy (b) (c) (d)

Solution: 4x = 5y = 20z = k 4x= k

4= k x

1

5y= k

5=k y

1

20z

20= k z

1

4×5=20 =k x

1

× k y

1

= k z

1

=k xy

yx

= k z

1

= yx

xyz

36. The value

. is : : Nov-2007Ans:(a)

(a) 2log3 10 (b) 3log7 10 (c) zlog3 e (d) None

Solution:

4

10

9

16

3

8

log

log

log

loglog

log

=

2

10

3

2

3

2

log2

log

log2

log4log

log3

= 10

2

log4

log223 =

10

2

log4

log223

3 log10 2

37. If ,ee

eex

nn

nn

then the value of n is : : Feb-2008Ans:(a)

(a) (b) (c)

(d)


Solution:

nn

n

nnnn

nnnn

nn

nn

nn

n

ex

x

e

e

x

x

eeee

eeee

x

x

ee

ee

xee

ex2

1

1

2

2

1

1

1

11

1

Take log

nx

xn

x

xene

x

xn

1

1log

2

12

1

1loglog2log

1

12

38. log 144 is equal to : : Feb-2008 Ans:(b) (a) 2 1og 4 + 2 1og 2 (b) 4 1og 2 + 2 1og 3 (c) 3 log 2 + 4 log 3 (d) 3 1og 2 – 4 1og3

Solution: log144

= log(12 × 12) 1212 loglog log(3×4)+log(3×4) log3+2log2+log3+2log2

4log2+2log3

39. If ,1xlogloglog 232 then x equals : : June-2008Ans:(c)

(a) 128 (b) 256 (c) 512 (d) None Solution: log2[log3(log2x)]-1

92

22

123 29log3log2)(loglog xxxx =512

40. If log log log then : : Dec-2008Ans:(b)

(a) 12 (b) 14 (c) 16 (d) 8 Solution:

abba

abbaba

ba

4)log(

4log)log(log

2

1

4log 2

1

abbaabbabaabbaabba 1416216)(4 22222

14a

b

b

a

41. log (m + n) = log m + log n, m can be expressed as: : June-2009Ans:(a)

(a) (b) (c) (d)

Solution: log (m+n)=log m+ log n 42. 1 2.Find x : June-2009Ans:(a)

(a)16 (b) 0 (c) – 1 (d) None of these

Solution: log 4 (x2+x)-log4(x+1)=2

2

log

)1log

log

)log(4

2

xxxn

0)1)(16(0)16(1)16(01616016)116(01615161622

21

2log1

loglog2.21

log2log

)1log)log(

2

222244

42

42

22

4

2

xxxxxxxxxxxxxxxxxx

x

xx

x

xx

x

xxxxx

x=16 m-1

43. Find the value of √25 2 4 : Dec-2009Ans:(c) (a) x (b) 10 (c) 1 (d) None Solution:

43) X2101010 4log23log25log

x

10

2

10

2

10

5

log

log4

log

log3

log

log

x

10

225

log

log4log3log

x

10

25

log

loglog


1log

log10

10

x

44. If logab + loga c = 0 then : June-2010Ans:(d) (a) b = c (b) b = -c (c) b = c = 1 (d) b and c are reciprocals. Solution:

0log

loglog0

log

log

log

log

a

cb

a

c

a

b =log b + log c = 0 = log bc = 0 = log1 = bc =1 b =

c

1

45. The value of n32 xlog2xlog2xlog2xlog2 will be :- : Dec-2010Ans:(b)

(a)

2

xlog1nn (b) n(n+1) log x (c) n2 log x (d) None of these

Solution : 2 log x+2log x2+2 log x3+…….2log xn 2[log x+ 2log x+3log x+……+n log n]

2 log x[1+2+3+…….+n] 2 log n 2

)1( nn

n(n+1) log n

46. Solve : 2 : Dec-2010Ans:(a)

(a) 110 (b) 210 (c) 10 (d) 310

Solution:

23

)log11(

2

3log 1010

xx

2

6

log2229log3 1010 xx 1

1010 101log1213log xxx

47. If n = m! where (‘m’ is a positive integer > 2) then the value of : ⋯ :

June-2011 Ans. (a) (a) 1 (b) 0 (c) –1 (d) 2

Solution: m is an(+) ve integer, n= m

nnnn mlog

1......

log

1

log

1

log

1

432

mnnnn log.............logloglog 432

).........432(log mn mn log 1log nn

48. If 6loglog 42 xx , then the value of x is : Dec-2011Ans. (a) a) 16 b) 32 c) 64 d) 128

Solution: log 2 x+log 4 x=6

6log2

3

2log2

log36

2log2

loglog26

2log2

log

2log

log2

xxxxxx

1624log2 xxxx

49. If yxlog = 100 and 10,xlog2 then the value of ‘y’ is : : June-2012Ans. (c)

(a) 102 (b)

1002 (c) 000,12 (d)

000,102

Solution: log xy =100 log 2 x=10

y= x100 x=210

=(2 )10 10

=2 1000

50. Which of the following is true. If + = : Dec-2012Ans. (d)

(a) log (ab + bc + ca) = abc (b) log

(c) lob (abc) = 0 (d) lob (a + b + c) = 0 Solution:


abccabcab

1111

abcabc

bac 1

1log)log( cba 0)log( cba

51. For what value of x, the equation 22 log2log xx is true? : June-2013Ans. (a)

a) 16 b) 32 c) 8 d) 4

Solution: xx2

22 loglog xx log

log

log

log 22

2

xxxxxx log

log

log

log4

log

log

log

log2

log

log

log2

1log 2

222

222

2

2

xx log

log

log

log4

22

2

162loglogloglog41

log

log4 422

24 xxx

x

52. The value of log4 9. log3 2 is : Dec-2013Ans. (d) (a) 3 (b) 9 (c) 2 (d) 1

Solution: 3

2

2

2

log

log.

2log

3log

13log

2log

log2

log22

3

53. The value of (logy x. logz y. logx z)3 is : Dec-2013 Ans. (c) (a) 0 (b) -1 (c) 1 (d) 3

Solution: 3log.log.log zx

yz

xy =

3

log

log.

log

log.

log

log

x

z

z

y

y

x

54. If x2 + y2 = 7xy, then log = __________ : June-2014Ans. (b)

a) (log x + log y) b) (log x + log y) c) (log x / log y) d) (log x + log y)

Solution : x2 + y2 = 7xy x2+y2+2x=9xy (x + y)2=9xy log (x + y)2 = log 9xy 2 log(x+y) = 2 log3 + log x + log y 2log(x+y)-2log3 = logx + logy

2(log(3

yx )= logx + logy

log3

1(x+4)=

2

1(log x + log y)

55. If x = log24 12, y = log36 24 and z = log48 36, then xyz + 1 = ______ : June-2014Ans. (c) a) 2xy b) 2zx c) 2yz d) 2 Solution: x = log24

12, y = log3624, z = log48

36

xyz+1 148log

36log

36log

24log

24log

12log 1

48log

12log

48log

48log12log

48log

576log

48log

242log

2 48log

24log 2 48log

36log

36log

24log

2 log36

24 log4836 x y z


56. If log x = a+b ; log y = a-b then 2

10log

y

x ___________. : Dec-2014Ans. (a)

a) 1 – a + 3b b) a – 1 + 3b c) a + 3b + 1 d) 1 – b + 3a

Solution: log x= a + b, log y = a - b log 2

10

y

x

= log 10x - log y2 log 10 + log x - 2 log y 1+ a + b - 2a + 2b

1- a+ 3b

57. If rpyqrx qp log1,log1 and pqz rlog1 then the value of ____:Dec-2014

Ans.(b) a) 0 b) 1 c) 2 d) –1 Solution :

x= 1+ logpqr

x=1+ p

qr

log

log

x= p

qpp

log

loglog

pqr

p

x log

log1

y = 1+ rpqlog

y= 1+q

rp

log

log

y= q

rpq

log

loglog

y

1=

qrp

q

log

log

pqrlog1

r

pq

zlog

log1

r

pqr

zlog

loglog

rpq

r

z log

log1

1log

log

log

log

log

log

log

log111

pqr

pqr

pqr

r

pqr

q

pqr

p

zyx

58. If log x = m + n and log y = m – n, then 2/10log yx = : June-2015Ans. (a) a) 3n – m + 1 b) 3m – n + 1 c) 3n + n + 1 d) 3m + n + 1 Solution: logx =m+n, log y=m-n

log 2log

log

y

x

log10+logx-2logy 1+ m + n -2m+2n 1- m +3n

59. The value of 5log4log3log 235 . : Dec-2015 Ans. (c)

a) 0 b) 1 c) 2 d) 1/2

Solution: log53×log3

4×log25

2

5

3

4

5

3

log

log

log

log

log

log 2

log

log2

log

log

2

2

2

4

60. The integral part of a logarithm is called _____ and the decimal part of a logarithm is called ____ : : June-2016 Ans. (b) a) Mantissa, Characteristic b) Characteristic, Mantissa c) Whole, Decimal d) None of these.

61. If 21loglog 42

4 xxx , then the value of X is : June-2016Ans. (c) a) 2 b) 3 c) 16 d) 8 Solution : log4(x

2+x)-log4(x+1)=2

21log

)log( 2

x

xx =log 4

2

1

1

x

xx =log4 x = 2 =x=42=16


62. Value of 60log

1

60log

1

60log

1

543

is : : June-2016Ans. (b)

a) 0 b) 1 c) 5 d) 60

Solution : 60log

1

60log

1

60log

1

543

=log 360 + 5

604

60 loglog =log 54360

= 6060log =1

63. If log 2 = 0.3010 and log 3 = 0.4771, then the value of log 24 is: Dec-2016 Ans. (c) (a)1.0791 (b)1.7323 (c)1.3801 (d)1.8301

Solution : log24 = 1.380211 formula times19 227695227695

64. The value of 5log4log3log 235 . : Dec-2016 Ans. (d)

a) 0 b) 1 c) 2 d) 1/2

Solution : log 35 5

243 loglog =

2

5

3

4

5

3

log

log

log

log

log

log

= 2

log

log22

2

65. The value of log (13+23+33+…….n3) is equal to: : June-2017 Ans: (b) a) 3 log 1+ 3 log 2 +…….+3 log n b) 2 log n + 2 log (n+1) - 2 log 2 c) log n + log (n+1) + log (2n +1) - log 6 d) 1 Solution: log(13+23+33+……n3)

=log

2

2

)1(

nn

=2 log 2

)1( nn

=2 log n + 2 log(n+1)-2log 2

Answer Key 1. b 2. b 3. c 4. b 5. d 6. b 7. d 8. b 9. b 10. a 11. c 12. d 13. b 14. b 15. a 16. c 17. c 18. d 19. b 20. c 21. a 22. c 23. a 24. c 25. b 26. c 27. b 28. c 29. c 30. a 31. b 32. a 33. c 34. c 35. d 36. a 37. a 38. b 39. c 40. b 41. a 42. a 43. c 44. d 45. b 46. a 47. a 48. a 49. c 50. d 51. a 52. d 53. c 54. b 55. c 56. a 57. b 58. a 59. c 60. b 61. c 62. b 63. c 64. d 65. b

Time Value of Money GOPAL BHOOT

6.1

CHAPTER TIME VALUE OF MONEY

b d

Past Exam Questions Time Value of Money

1. ` 8,000 becomes ` 10,000 in two years at simple interest. The amount that will become ` 6,875 in 3 years at the same rate of interest is : Nov-2006 Ans:(b) (a) ` 4,850 (b) ` 5,000 : (c) ` 5,500 (d) ` 5,275 Solution : Simple Interest = P × n × r 2,000 = 8000 × 2 × r r = 0.125 = 12.5% simple Let required principal is x Amount = principal + P × n × r 6875 = x + x×3×0.125 Solving = x = 5,000

2. The difference between the simple and compound interest on a certain sum for 3 years at 5% p.a. is ` 228.75 The compound interest on the sum for 2 years at 5% p.a. is : Nov-2006 Ans:(b) (a) ` 3,175 (b) ` 3,075 (c) ` 3,275 (d) ` 2,975 Solution :

Let principal = x Difference between CI and SI is 228.75 = P(1+r)n – P – Pnr 228.75 = x(1+0.05)3 – x – x × 3 × 0.05 Principal x = 30,000 CI for 2 years = FV – PV

= 30,000(1+0.05)2 – 30,000 = 33,075 – 30,000 = 3,075 3. Mr. X Invests ` 10,000 every year starting from today for next 10 years suppose interest rate is 8%

per annum compounded annually. Calculate future value of the annuity : Nov-2006 Ans:(a) Given that (1 + 0.08)10 = 2.15892500] (a) ` 156,454.88 (b) ` 1,448,656.25 (c) ` 1,56,554.88 (d)None of these

Solution :

FV of Annuity = A

= 10,000 A.

. = 14,4865.625 × (1 + 0.08) = 15,6454.875

4. The present value of an annuity of ` 3,000 for 15 years at 4.5% p.a. C.I. is : Nov-2006 Ans:(b) [ Given that (1.045)15 = 1.935282] (a) ` 23,809.67 (b) ` 32,218.67 (c) ` 32,908.67 (d) None of these

Solution :

PV of Annuity = A

A = 3,000 n = 15 r = 0.045 = 3,000 .

. . = 32218.67

5. The rate of simple interest on a sum of money is 6% p.a. for first 3 years, 8% p.a. for the next five years and 10% p.a. for the period beyond 8 years. If the simple interest accrued by the sum for a period for 10 years is ` 1,560. The sum is : Feb-2007 Ans:(b) (a) ` 1,500 (b) ` 2,000 (c) ` 3,000 (d) ` 5,000 Solution : Total Simple Interest = P × t × r 1560 = (P × 3 × 0.06 + P × 5 × 0.08 = P × 2 × 0.10) 1560 = 0.78P P = 2,000

6. A sum of money doubles itself in 10 years. The number of years it would treble itself is : Feb-2007 Ans:(c) (a) 25 years (b) 15 years (c) 20 years (d) None Solution :

If S.I assumed 100 --------200 --------300 10 years 10 years

Total 20 years If C.I. assumed Rule of 72 Doubles in years


6.2

= 10 years r = 7.2%

Assume CI rate 7.2% Rule of 114

Triples in = .

= 15.83 years.

7. In what time will ` 3,90,625 amount to ` 4,56,976 at 8% per annum, when the interest is compounded semiannually? Feb-2007 Ans:(a)

[Given :(1.04)4= 1.16986] (a) 2 years (b) 4 years (c) 5 years (d) 7 years Solution : Semi-annually 2 times in a year.

456976 = 390625.

= (1 + 0.04)2n 1.16986 = (1.04)2n (1.04)4 = (1.04)2n 2n = 4 n = 2 years

8. A machine can be purchased for ` 50,000. Machine will contribute ` 12,000 per year for the next five years. Assume borrowing cost is 10% per annum. Determine whether machine should be purchased or not: Feb-2007 Ans:(b) (a) Should be purchased (b) Should not be purchased

(c) Can't say about purchase (d) None of the above Solution :

Inflow Outflow 12,000 12,000 12,000 12,000 12,000 = 50,000 One time PV of Inflow

= PV of annuity = A

= 12,000 .

. . = 45,489.60

Net PV = PV of Inflow – PV of outflow = 45,489.60 – 50,000 = – 4510.40

Ans:- (b) Machine should not be purchased. 9. How much amount is required to be invested every year so as to accumulate ` 3,00,000 at the end of

10 years. If interest is compounded annually at 10%? Feb-2007 Ans:(a) [Give (1.1)10 = 2.5937] (a) ` 18,823.65 (b) ` 18,828.65 (c) ` 18,832.65 (d) ` 18,882.65

Solution :

FV of Annuity = A

30,000 = A .

. Solving A = 18823.65

10. A certain sum of money amounts to ` 6,300 in two years and ` 7,875 in three years nine months at simple interest. Find the rate of interest per annum : May-2007 Ans:(a) (a) 20% (b) 18% (c) 15% (d) 10% 6300 = P + 900 × 2 Solution : In 2 years amount = 6300 In 3.75 years Amount = 7875

Amount Principal + P × n × r In 2 Years 6,300 = p + p × n × r In 3.75 Years 7875 = p + p ×1.75 × r 1575 1.75 × r

Pr = 900 6300 = P + P × r × 2 6300 = P + 900 × 2 P = 4500 R = 20%

11. How long will ` 12,000 take to amount to ` 14,000 at 5% p.a. converted quarterly? May-2007 Ans:(b)

[Given: (1.0125)12.4 = 1.1666] (a) 3 years (b) 3.1 years (c) 13.5 years (d) 12.4 years.

Solution : FV = PV 1


6.3

14000 = 12000 1 . 1.1666 = (1+0.0125)4n

(1.0125)12.4 = (1.0125)4n 4n = 12.4 n = 3.1 years 12. A company is considering proposal of purchasing a machine either by making full payment of `

4,000 or by leasing it for four years at an annual rate of ` 1,250. Which course of action is preferable, if the company can borrow money at 14% compounded annually? May-2007 Ans:(a)

[Given : ]68896.114.1 4 (a) Leasing is preferable (b) Should be purchased (c) No difference (d) None of these

Solution : Buy Lease

Outflow = 4000

Outflow 4000

Outflow 1250 1250 1250 1250 (4 years)

PV of annuity= A = 1250 .

. .

Outflow= 3642 13. Vipul purchases a car for ` 5,50,000. He gets a loan of ` 5,00,000 at 15% p.a. from a Bank and

balance ` 50,000 he pays at the time of purchase. He has to pay the whole amount of loan in equal monthly instilments with interest starting from the end of the first month The money he has to pay at the end of every month is : May-2007 Ans:(a)

[Given (1.0125)12 = 1.16075452] (a) ` 45,130.43 (b) ` 45,230.43 (c) ` 45,330.43 (d) None of these

Solution : Cash price of car = Down Payment + PV of all installments forming an Annuity

5,50,000 = 50,000 + A A = ? r/k = 0.15/12 = 0.0125

nk = 1×12 = 12 5,00,000 = A .

. . A = 45130.43

14. If ` 1,000 be invested at interest rate of 5% and the interest be added to the principal every 10 years, then the number of years in which it will amount to ` 2,000 is : Aug-2007 Ans:(a)

(a) 16 years (b) 6

years (c) 16 years (d) 6

years

Solution : 1000-----------------1500 -------------------2000 10 years at 5% S.I. Principal 1000 Simple Interest of 10 years 500 (100 × ×5% ×10) Revised principal 1500 Required Interest 500 (2000 – 1500)

P × t × r = 500 1500 × t × 0.05 = 500 T = 6.667 years Total 10 + 6.67 = 16.67 years 15. The annual birth and death rates 1000 are 39.4 and 19.4 respectively. The number of years in which

the population will be doubled assuming there is no immigration or emigration is :Aug-2007Ans:(a) (a) 35 years (b) 30 years (c) 25 years (d) None of thee. Solution :

Growth rate . . 100= 2%

Population will double in = = 36 years approx

Exactly no of years = 02.01log

2log

1log

2log

r

16. The effective rate equivalent to nominal rate of 6% compounded monthly is : Aug-2007Ans:(b) (a) 6.05 (b) 6.16 (c) 6.26 (d) 6.07 Solution :

Effective ROI = 1 1 100 = 1 . 1 100 = 6.1678%


6.4

17. A company establishes a sinking fund to provide for the payment of `2,00,000 debt maturing in 20 years. Contributions to the fund are to be made at the end of every year. Find the amount of each annual deposit if interest is 5% per annum : Aug-2007Ans:(b) (a) ` 6,142 (b) ` 6,049 (c) ` 6,052 (d) ` 6,159 Solution : In sinking fund we use FV of Annuity FV of Annuity 2,00,000 n = 20 r = 0.05

FV of Annuity = A 2,00,000 = A .

. A = 6049

18. A person borrows ` 5,000 for 2 years at 4% p.a. simple interest. He immediately lends to another

person at 6 %p.a. for 2 years. Find his gain in the transaction per year : Nov-2007Ans:(a)

(a) ` 112.50 (b) ` 125 (c) ` 225 (d) ` 167.50 Solution : Gain = 6.25 – 4 =2.25% Gain every year = 5000 × 2.25% = 112.50

19. A person deposited ` 5,000 in a bank. The deposit was left to accumulate at 6% compounded quarterly for the first five years and at 8% compounded semiannually for the next eight years. The compound amount at the end of 13 years is : Nov-2007Ans:(b)

(a) ` 12,621.50 (b) ` 12,613.10 (c) ` 13,613.10 (d) None. Solution : 5000 -------------------------------- --------------------------------- 6% compounded for 5 years 8% compounded semi-annually for 8 years

FV after 5 years = PV 45

4

06.0150001

nk

k

r= 6734.275

Invested for another 8 years

FV after another 8 years = PVnk

k

r

1 = 6734.275

28

2

08.01

= 12613.10

20. Raja aged 40 wishes his wife Rani to have ` 40 lakhs at his death. If his expectation of life is another 30 years and he starts making equal annual investments commencing now at 3% compound interest p.a. How much should he invest annually ? Nov-2007Ans:(b) (a) ` 84,077 (b) ` 81,628 (c) ` 84,449 (d) ` 84,247 Solution : He requires 40 lakh after 30 years.Commencing now

FV of Annuity immediate = A 1

40,00,000 = A .

. 1 0.03

A = .= 81628

21. Two equal sums of money were lent at simple interest at 11% p.a. for 3 years and 4 years

respectively. If the difference in interests for two periods was ` 412.50, then each sum is: Feb-2008 Ans:(c) (a) ` 3,250 (b) ` 3,500 (c) ` 3,750 (d) ` 4,350 Solution : Interest of 4.5 – 3.5 = 1 year is 412.50 P × n × r = 4125.5 P × 1× 0.11 = 412.50 P = 3750

22. Anshul's father wishes to have ` 75,000 in a bank account when his first college expenses begin. How much amount his father should deposit now at 6.5% compounded annually if Anshul is to start college in 8 years hence from now ? Feb-2008 Ans:(a)

(a) ` 45,317 (b) ` 46,317 (c) ` 55,317 (d) ` 48,317. Solution : He requires 75,000 after 8 years FV = PV 1 75,000 = PV 1 0.065 PV =45317


6.5

23. A company may obtain a machine either by leasing it for 5 years (useful life) at an annual rent of ` 2,000 or by purchasing the machine for ` 8,100. If the company can borrow money at 18% . per annum, which alternative is preferable ? Feb-2008 Ans:(a) (a) Leasing (b) Purchasing (c) Can't say (d) None of these Solution :

Buy Lease Outflow = 8100

Outflow 8100

Outflow 2000 2000 2000 2000 2000 (5 years) PV of outflow

= A

= 2000 .

. .

Outflow= 6254 Lease is preferable as lower outflow

24. In how much time would the simple interest on a certain sum be 0.125 times the principal at 10% per annum? June-2008 Ans:(a)

(a) 1 years (b) 1

years (c) 2

years (d) 2 years

Solution : S.I = P × t × r 0.125 × P = P × t × 0.10 t = 1.25 25. The difference between compound interest and simple interest on a certain sum for 2 years @ 10%

p.a. is ` 10. Find the sum : Ans:(c) (a) ` 1,010 (b) ` 1,095 (c) ` 1,000 (d) ` 990 Solution : Difference = 10 = P(1 + r)n – P – Pnr = P(1 + 0.10)2 – P – P × 2 × 0.10 0.01P = 10 P = 1000

26. A machine worth ` 4,90,740 is depreciated at 15% on its opening value each year. When its value would reduce to ` 2,00,000 : June-2008 Ans:(a) (a) 5 years 6 months (b) 5years 7monyhs (c) 5years 5months (d) None. Solution :

Its WDV method WDV = Historical cost 1 2,00,000 = 4,90,740 1 0.15 0.4075 = (0.85)n Taking log log 0.4075 = log(0.85)n log 0.4075 = n log 0.85

n = 52.585.0log

4075.0log years n = 5.52 years = 5 + 0.52 years = 5 years + (0.52×12) months

= 5 yrs 6.24 months 27. A sinking fund is created for redeeming debentures worth ` 5 lakhs at the end of 25 years. How

much provision needs to be made out of profits each year provided sinking fund investments car earn interest at 4% p.a.? June-2008 Ans:(a) (a) 12,006 (b) 12,040 (c) 12,039 (d) 12,035 Solution :

In sinking Fund we use FV of Annuity FV of Annuity = A 5,00,000

= A [ .

.] A = 12006

28. If the difference between simple interest and compound interest is ` 11 at the rate of 10% for two years, then find the - sum. Dec-2008 Ans:(b) (a) ` 1,200 (b) ` 1,100 (c) ` 1,000 (d) None of these Solution : Difference 11 = P(1 + r)n – P – Pnr 11 = P(1+ 0.10)2 – P – P × 2 × 0.10 Solving P= 1000

29. Future value of an ordinary annuity Dec-2008 Ans:(a)

(a) A (n, i) =

i

1)i1(A

n

(b) A (n, i) =

i

1)i1(A

n

(c) A (n, i) =

i

)i1(1A

n

(d) A (n, i) =

n

n

)i1(i

1)i1(A

Solution :(a) FV of Annuity = A


6.6

30. Find the numbers of years in which a sum doubles itself at the rate of 8% per annum. Dec-2008 Ans:(b)

(a) 11 years (b) 12

years (c) 9

years (d) 13 years

Solution : Assuming simple Interest Doubles in = 12.5 years

31. In how many years, a sum will become double at 5% p.a. compound interest. June - 2009Ans:(c) (a) 14.0 years (b) 14.1 years (c) 14.2 years (d) 14.3 years

Solution : Doubles in 05.01log

2log

1log

2log

r = 14.2 years

32. The time by which a sum of money is 8 times of itself if it doubles itself in 15 years. June-2009 Ans:(c) (a) 42 years (b) 43 years (c) 45 years (d) 46 years Solution : 1 ------------- 2 ------------- 4 -------------8 15 years 15 years 15 years Assuming Compound Interest Required time 15 + 15 + 15 = 45 years

33. What is the rate of simple interest if a sum of money amounts to ` 2,784 in 4 years and ` 2,688 in 3 years ? June - 2009 Ans:(b) (a) l% p.a. (b) 4% p.a. (c) 5% p.a. (d) 8% p.a. Solution : Amount = P + Interest = P + P × t × r In 4 years 2784 = P + P × 4 × r In 3 years 2688 = P + P × 3 × r __________________________________________ + 96 = + Pr Pr = 96 2784 = P + (Pr) ×4 2784 = P + (96 × 4) P = 2400 Solving r = 4%

34. A sum amount to ` 1,331 at a principal of ` 1,000 at 10 % compounded annually. Find the time. June - 2009 Ans:(c) (a) 3.31 years (b) 4 years (c) 3 years (d) 2 years Solution : FV = PV(1 + r)n 1331 = 100(1 + 0.10)n 1.331 = (1.10)n (1.10)3 = (1.10)n n = 3 years

35. Paul borrows ` 20,000 on condition to repay it with compound interest at 5% p.a. in annual instalment of ` 2,000 each. Find the number of years in which the debt would be paid off.June-2009 Ans:(c) (a) 10 years (b) 12 years (c) 14 years (d) 15 years Solution : Loan Amount = PV of annuity

20,000 = A

20,000 = 2000.

.

10× 0.05 = .

.

n = 15 years 36. In how many years, a sum of ` 1000 compounded annually @ 10%, will amount to ` 1331? Dec-

2009 Ans:(d) (a) 6 years (b) 5 years (c) 4 years (d) 3 years Solution : FV = PV(1+r)n 1331 = 1000(1 + 0.10)n n = 3 years


6.7

37. The compound interest for a certain sum @ 5% p.a. for first year is ` 25 The S-I for the same money @ 5% p.a. for 2 years will be. Dec-2009 Ans:(b) (a) ` 40 (b) ` 50 (c) ` 60 (d) ` 70 Solution : Compound Interest for 1st year = SI for 1st year = 25 SI for 2nd years = 25 Total SI for 2 years = 50

38. At what % rate of compound interest (C.I) will a sum of money become 16 times in four years, if interest is being calculated compounding annually: June-2010 Ans:(a) (a) r =100% (b) r = 10% (c) r = 200% (d) r = 20% Solution : 1-----------2-------------------4 --------------- 8 ---------------16 1 year 1 year 1 year 1 year FV = PV(1 + r)n 16x = x(1 + r)4 By Trial and error r = 100%

39. Find the present value of an annuity of ` 1,000 payable at the end of each year for 10 years. If rate of interest is 6% compounding per annum (given (1.06)-10= 0.5584) : June-2010 Ans:(a) (a) ` 7,360 (b) ` 8,360 (c) ` 12,000 (d) None of these

Solution : PV of Annuity = A = 1000 .

. . = 7360

40. If the simple interest on a sum of money at 12% p.a. for two years is ` 3,600. The compound interest on the same sum for two years at the same rate is : June-2010 Ans:(a) (a) ` 3,816 (b) ` 3,806 (c) ` 3,861 (d) ` 3,860 Solution : Simple Interest = P.t.r = 3600 P × 2 × 0.12 = 3600 P = 15,000 2 years compound Interest = FV – PV = PV (1 + r)n – PV = 15,000 (1 + 0.12)2 – 15,000 = 3816

41. of an annuity of ` 5,000 is made annually for 8 years at interest rate of 9% compounded annually [Given that (1.09)8 =1.99256] is __________ Dec-2010 Ans:(a) (a) ` 55,142.22 (b) ` 65,142.22 (b) ` 65,532.22 (c) ` 57,425.22

Solution : FV of Annuity = A = 5000 .

. = 55142.2

42. The effective annual rate of interest corresponding to nominal rate 6% p.a. payable half yearly is Dec-2010 Ans: (d)

(a) 6.06% (b) 6.07% (c) 6.08% (d) 6.09% Solution :

Effective ROI = 10011

k

k

r = 1001

2

01

2

= 6.09%

43. The cost of Machinery is ` 1,25,000 If its useful life is estimated to be 20 years and the rate of depreciation of its cost is 10% p.a., then the scrap value of the Machinery is Dec-2010 Ans:(a)

(Given that (0.9)20=0.1215) (a) 15,187 (b) 15,400 (c) 15,300 (d) 15,250 Solution : WDV = HC (1 – r)n = 125000(1 – 0.10)20 = 15187.5 = 125000 × 0.1215 = 15187.5

44. Mr. X invests 'P' amount at Simple Interest rate 10% and Mr. Y invests 'Q' amount at Compound Interest rate 5% compounded annually. At the end of two years both get the same amount of interest, then the relation between two amounts P and Q is Dec-2010 Ans:(a)

(a) (b) (c) (d)

Solution : Mr X SI of 2 years = P × t × r = P × 2 × 0.10 = 0.20 P Mr. Y CI for 2 years = FV – PV = PV(1 + r)n – PV = Q (1 + 0.05)2 – Q = 0.1025Q

2 Interests are equal 0.20P = 0.1025Q P = .

. =

45. If the difference of S.I and C.I is ` 72 at 12% for 2 years. Calculated the amount.June-2011 Ans. (c) (a) ` 8,000 (b) ` 6,000 (c) ` 5,000 (d) ` 7,750. Solution : Difference = 72 = P(1+r)n – P – Pnr P(1+0.12)2 – P – P × 2 × 0.12 P = 5000


6.8

46. If a simple interest on a sum of money at 6% p.a. for 7 years is equal to twice of simple interest on another sum for 9 years at 5% p.a.. The ratio will be : June-2011 Ans. (c) (a) 2 : 15 (b) 7 : 15 (c) 15 : 7 (d) 1 : 7

Solution : SI on 1st principal =2×SI on 2nd principal P ×7 × 0.06 = 2×Q×9×0.05 = = 15:7

47. By mistake a clerk, calculated the simple interest on principal for 5 months at 6.5% p.a. instead of 6 months at 5.5% p.a. if the error in calculation was ` 25.40. The original sum of principal was _______. June-2011 Ans. (b) (a) ` 60,690 (b) ` 60,960 (c) ` 90,660 (d) ` 90,690 Solution :

Wrong SI = P × ×0.065 Correct SI = P × ×0.055

______________________________

Difference (0.325 – 0.33) = 25.40 P = 60960

48. If the Simple Interest on ` 1,400 for 3 years is less than the simple interest on ` 1,800 for the same period by ` 80, then the rate of interest is Dec-2011 Ans. (b) a) 567% b) 6.67% c) 7.20% d) 500% Solution :

SI on 1400 =1400 × 3 × r = 4200r SI on 1800 = 1800 × 3 × r = 5400r

Difference = 5400r – 4200r = 80 R = 0.0667 = 6.67% 49. Nominal rate of interest is 9.9% p.a. If interest is Compounded monthly. What will be the effective

rate of interest Dec-2011 Ans. (a)

1.1036 ?

a) 10.36% b) 9.36% c) 11.36% d) 9.9%

Solution : Effective ROI = 1 1 100 = 1 . 1 100 = 10.36%

50. The S.I. on a sum of money is of the principal and the no. of years is equal to the rate of interest

per annum. Find the rate of interest per annum ? June-2012 Ans. (b) (a) 5% (b) 20/3% (c) 22/7% (d) 6% Solution :

Simple Interest = P × t × = × P

here t = r r × = r2 = r = %

51. Simple interest on ` 2,000 for 5 months at 16% p.a. is ____________ June-2012 Ans. (a) (a) ` 133.33 (b) ` 133.26 (c) ` 134.00 (d) ` 132.09

Solution : 2000 × ×16% = 133.33

52. How much investment is required to yield an Annual income of ` 420 at 7% p.a. simple interest Dec-2012 Ans. (a)

(a) ` 6,000 (b) ` 6,420 (c) ` 5,580 (d) ` 5,000 Solution : P × rate = 420 P × 0.07 = 420 P = 6000.

53. Mr. X invests ` 90,500 in post office at 7.5 % p.a. simple interest. While calculating the rate was wrongly taken as 5.7% p.a. The difference in amounts at maturity is ` 9,774. Find the period for which the sum was invested Dec-2012 Ans. (c)

(a) 7 years (b) 5.8 years (c) 6 years (d) 8 years Solution : Let no of years = n

Interest = P × r × n 90500 × 0.075 × n – 90500 × 0.057 × n = 9774 n = 6 years 54. The difference between compound and simple interest on a certain sum of money for 2 years at 4%

p.a. is ` 1. The sum (in `) is: June-2013 Ans. (a) a) 625 b) 630 c) 640 d) 635 Solution : Difference = P(1+r)n – P – Pnr 1 = P(1+0.04)2 – P – P × 2 × 0.04 P = 625

55. A sum of money compounded annually becomes ` 1,140 in two years and ` 1,710 in three years. June-2013 Ans. (c)


6.9

Find the rate of interest per annum. a) 30% b) 40% c) 50% d) 60% Solution :

FV = PV(1+r)n 1140 = P(1+r)2 1710 = P(1+r)3 Divide 1140

1710

)1(

)1(2

3

rP

rP

1+r = 1.5 r = 50% 56. On what sum difference between compound interest and simple interest for two years at 7% p.a.

interest is ` 29.4 Dec-2013 Ans. (c) (a) ` 5,000 (b) ` 55,000 (c) ` 6,000 (d) ` 6,500 Solution :

Rate = 7% p.a., n = 2 years Difference between S.I and C.I = 29.4

4.2911

nrnrp

4.2907.021207.01

p

p [1.1449 – 1 – 0.14] = 29.4 p [0.0049] = 29.4 p = 0049.0

4.29 p = 6000 Ans.

57. In what time will a sum of money double itself at 6.25% p.a. simple interest? Dec-2013Ans. (d) (a) 5 years (b) 8 years (c) 12 years (d) 16 years Solution :

Rate = 6.25% p.a.

Money double itself = r

72.0 =

625.0

72.0 = 11.52 years or 12 years (Ans.)

58. What principal will amount to ` 370 in 6 years at 8% p.a. at simple interest? Dec-2013Ans. (b) (a) ` 10 (b) ` 250 (c) ` 310 (d) ` 350 Solution :

S.I = P × R × T S.I = 370 Time = 6 years Rate = 8% p.a. Amount = Principal + Interest 370 = Principal + P × R × T 370 = P[1+ RT]

⇒ 370 = p[1+0.08 × 6] .

= principal ∴ Principal = 250 (Ans.)

60. If a sum triples in 15 yrs at Simple rate of interest, the rate of interest per annum will be : June-2014 Ans. (b) a) 13.0% b) 13.3% c) 13.5% d) 18.0% Solution :

300 = 100 + 100

15100 R 30,000 = 10,000 + 1500 R R = 13.3%.

61. How much amount is required to be invested every year as to accumulate ` 6,00,000 at the end of 10th year, if interest is compounded annually at 10% rate of interest? June-2014Ans. (c) [Given : (1.1)10 = 2.59374] a) ` 37,467 b) ` 37,476 c) ` 37,647 d) ` 37,674

Solution : FV =

r

rA

n 11

60,00,000

=

1.

11.01 10 A = A × 15.937

937.15

000,00,6

= A A = 37,647 (Approx.)

62. The future value of an annuity of ` 1,000 made annually for 5 years at the interest of 14% compounded annually is : Dec-2014 Ans. (b) a) ` 5,610 b) ` 6,610 c) ` 6,160 d) ` 5,160

Solution : FV =

r

rA

n 11 =

6610

14.0

114.011000

5

Future value of annuity = ` 6610. 63. A sum of money invested of compound interest doubles itself in four years. It becomes 32 times of

itself at the same rate of compound interest in Dec-2014 Ans. (c)


6.10

a) 12 years b) 16 years c) 20 years d) 24 years Solution : 2 times = 4 years

4 times = 8 years [+4] 8 times = 12 years [+4] 16 times = 16 years [+4] 32 times = 20 years [+4]

When Double 65. A sum of money doubles itself in 8 years at simple interest. The number of years it would triple is

June-2015 Ans. (c) a) 20 years b) 12 years c) 16 years d) None of these Solution :

A = 200 P = 100 (Lets) S.I. = 100 100 = 100

8100 R 100 = 8R R = 12.5

When Triple A = 300

Let, P = 100 S. I. = 200 200 = 100

5.12100 T T = 16 years (Ans.)

66. A sum of ` 44,000 is divided into three parts such that the corresponding interest earned after 2 years, 3 years and 6 years may be equal. If the rates of simple interest are 6% p.a., 8% p.a. and 6% p.a. respectively, then the smallest part of the sum will be June-2015 Ans.(b) a) ` 4,000 b) ` 8,000 c) ` 10,000 d) ` 12,000 Solution : Let three parts x, y, z x + y + z = 44000 x × 2 × 0.06 = y × 3 × 0.08 = z × 6 × 0.06 0.12x = 0.24y = 0.36z 0.12x = 0.24y 0.24y = 0.36z

1

2

y

x

2

3

24.0

36.0

z

y

times12:3

z:ytimes3

1:2

y:x

2:3

:

3:6

: zyyx

2:3:6

:: zyxDivide 44000 in 6 : 3 : 2 = 24000, 12000, 8000

Answer Key

1. b 2. b 3. a 4. b 5. b 6. c 7. a 8. b 9. a 10. a 11. b 12. a 13. a 14. a 15. a 16. b 17. b 18. a 19. b 20. b 21. c 22. a 23. a 24. a 25. c 26. a 27. a 28. b 29. a 30. b 31. c 32. c 33. b 34. c 35. c 36. d 37. b 38. a 39. a 40. a 41. a 42. d 43. a 44. a 45. c 46. c 47. b 48. b 49. a 50. b 51. a 52. a 53. c 54. a 55. c 56. c 57. d 58. b 59. c 60. b 61. c 62. b 63. c 64. c 65. c 66. b 67. c 68. c 69. c 70. b 71. c 72. b 73. c 74. a 75. b 76. c 77. d 78. c 79. b 80. 81. 82. 83. 84. 85. a 86. a 87. c 88. a 89. b 90. b 91. a 92. c 93. a 94. a 95. a 96. d 97. a 98. a

Permutations & Combinations GOPAL BHOOT

7.1

CHAPTER PERMUTATIONS & COMBINATIONS

Past Exam Question Permutations And Combinations1. The number of triangles that can be formed by choosing the vertices from a set of 12 points, seven of

which lie on the same straight line, is: Nov-2006 Ans:(a) (a) 185 (b) 175 (c) 115 (d) 105 Solution : n = 12 p = 7

∴ no. of triangles 37

312

33 CCCC pn =

= 220 – 35 = 185

2. A code word is to consist of two distinct English alphabets followed by two distinct numbers between 1 and 9. How many such code words are there? Nov-2006 Ans:(b) (a) 6,15,800 (b) 46,800 (c) 7,19,500 (d) 4,10,800 Solution : Code Number = Two English alphabets / Two distinct number 1 and 9 ∴ Two English alphabet → any 26 × 25 Two distinct number → any 9 × 8 ∴ Code Words = 26 × 25 × 9 × 8 = 46800

3. A boy has 3 library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow Mathematics part II unless Mathematics part -1 is also borrowed? In how many ways can he choose the three books to be borrowed? Nov-2006 Ans:(c) (a) 41 (b) 51 (c) 47 (d) 71 Solution : Either 3 books out of balance 6 in 3

6 C (I) (II) (6 Others) = 20 Way

Part I only 26

11 CC = 15

both I & II in 16

22 CC = 6

= 41 4. An examination paper consists of 12 questions divided into two parts A and B. Part A contains 7

questions and part B contains 5 questions. A candidate is required to attempt 8 questions selecting at least 3 from each part. In how many maximum ways can the candidate select the questions? Feb-2007 Ans:(d) (a) 35 (b) 175 (c) 210 (d) 420 Solution : Attempt → 8 question selecting at least 3 from each part. Possible combinations 355

53

7 CC

17545

47 CC

21035

57 CC

Total = 420 5. A Supreme Court Bench consists of 5 judges. In how many ways, the bench can give a majority

decision? Feb-2007 Ans:(d) (a) 10 (b) 5 (c) 15 (d) 16 Solution : The bench can give majority division when 3 judges, 4 judges and 5 judges Required combination = 5 5 5 10 + 5 + 1 = 16

6. Given : P (7, k) = 60 P(7, k - 3). Then : Feb-2007 Ans:(c) (a) k = 9 (b) k = 8 (c) k = 5 (d) k = 0 Solution :

377 60 kk PP

Or, !

!60. !

! Or, (10 – k)! = 60.(7 – k)!

Or, (10 – k) (9 – k) (8 –k) (7 – k)! = 60.(7 – k)!


7.2

Or, (10 – k) (9 – k) (8 – k) = 5 × 4 × 3 ∴ 10 – k = 5

k = 5 9 – k = 4

k = 5 8 – k = 3

k = 5 7. The number of ways in which n books can be arranged on a shelf so that two particular books are not

together is : Feb-2007 Ans:(a) (a) (n-2)x(n-l)! (b) (n-2)x(n+l)! (c) (n-l)x(n+l)! (d) (n-2)x(n+2)! Solution : Total number of books = n When two particular books are not together then remaining number of books = (n – 2) ∴ the number of ways when two particular books are not together is (n – 2)! × 1

= (n – 2)! !

! =

!

! = (n – 2) (n – 1)!

8. In how many ways can the letters of the word FAILURE be arranged so that the consonants may occupy only odd positions? May-2007 Ans:(a) (a) 576 (b) 476 (c) 376 (d) 276 Solution : 'FAILURE' F __ L __ R __ __ (1) (2) (3) (4) (5) (6) 7 ∴ There are 3 consonants out of 4 ∴ possible permutation = 3

4 P = 24 ways

Remaining place = 4 Total vowels = 4 Possible permutation = 4

4 P = 4! = 24 ways ∴ total permutation = (24 × 24) ways = 576

9. Five bulbs of which three are defective are to be tried in two lights - points in a dark-room. In how many trials the room shall be lighted? May-2007 Ans:(b) (a) 10 (b) 7 (c) 3 (d) None of these Solution :

3 defective 2 good Select 1 1 or 0 2 Possible combination 6231

21

3 CC Required ways = (6 × 24) ways = 144 ways 2

20

3 CC = 1

Total = 7

10. In how many ways can a party of 4 men and 4 women be seated at a circular table, so that no two woman are adjacent? May-2007 Ans:(c) (a) 164 (b) 174 (c) 144 (b) 154 Solution :

4 men can be seated in a circular table. So no, of ways = (n – 1)! = 3! = 6 ways ∴ There are 4 places for 4 women ∴ No of ways = 4! = 24 ways

11. The value of

5

1rr

5C is : May-2007 Ans:(b)

(a) 29 (b) 31 (c) 35 (d) 26 Solution :

5

1

5

rrC 31125

55

45

35

25

15 CCCCC

12. If ,C24P r6

r6 then find r : Aug-2007 Ans:(a)


7.3

(a) 4 (b) 6 (c) 2 (d) 1 Solution : rr PP 66 24

Or, !

!24. !

! ! Or, r! = 24 Or, r! = 4! Or, r = 4

13. Find the number of combinations of the letters of the word COLLEGE taken four together:Aug-2007 Ans:(a) (a) 18 (b) 16 (c) 20 (d) 26 Solution :

14. How many words can be formed with the letters of the word 'ORIENTAL' so that A and E always occupy odd places: Aug-2007 Ans:(b) (a) 540 (b) 8640 (c) 8460 (d) 8450 Solution :

A __ E __ __ __ __ __ (1) (2) (3) (4) (5) (6) (7) (8)

Possible permutation when A and E always Occupy odd places = 24 P = 12 ways

And possible permutation for remaining letters 72066 P ways

∴ total permutation = (720 × 12) ways = 8640 ways 15. If 1000C98 = 999C97 + XC901, find x : Nov-2007 Ans:(a)

(a) 999 (b) 998 (c) 997 (d) 1000 Solution : 90197

99998

100 CCC x 901902999

902100 CCC x x = 999

16. How many numbers greater than a million can be formed with the digits 4, 5, 5, 0, 4, 5, 3? Nov-2007 Ans:(b) (a) 260 (d) 360 (c) 280 (d) 380 Solution :

(except o)

16 P 1

6 P 15 P 1

4 P 13 P 1

2 P 11 P

(1) (2) (3) (4) (5) (6) (7) = 4320

∴ There 5 → 3 4 – 2 ∴ they can be arranged In 3! × 2!

∴ Number = ! !

= 360

17. A building contractor needs three helpers and ten men apply. In how many ways can these selections take place? Nov-2007 Ans:(d) (a) 36 (b) 15 (c) 150 (d) 120

Solution : 310C !

! = 120 ways

18. There are three blue balls, four red balls and five green balls. In how many ways can they be arranged in a row? Feb-2008 Ans:(b) (a)26720 (b) 27720 (c) 27820 (d) 26,620

Solution :∴ Required arrangement = !

! ! ! = 27720 ways

19. If C(n, r): C(n, r +1) = 1 : 2 and C (n, r + 1): C (n, r + 2) = 2:3, determine the value of n and r : Feb-

2008 Ans:(a) (a) (14, 4) (b) (12, 4) (c) (14, 6) (d) None Solution :

or, 2 or, 2

or, 2 or, n – r = 2r + 2

or, n – 3r = 2 ………….(i) × 2


7.4

or, or,

or, or, 2n – 2r – 2 = 3r + 6

or, 2n – 5r = 8 ………….(ii) 2n – 6r = 4 ∴ 2n – 5r = 8

2n + 5r or, 2n – 20 = 8

r = 4 or, 2n = 28

r = 4 or, n = 14

20. Six seats of articled clerks are vacant in a 'Chartered Accountant Firm'. How many different batches of candidates can be chosen out of ten candidates ? June-2008 Ans:(b) (a) 216 (b) 210 (c) 220 (d) None Solution : Different batches of candidates 6

10 C ways = 210 ways

21. Six persons A, B, C, D, E and F are to be seated at a circular table. In how many ways can this be done, if A must always have either B or C on his right and B must always have either C or D on his right? June-2008 Ans:(d) (a) 3 (b) 6 (c) 12 (d) 36 Solution : (ABC) DEF ∠ 4 1 6 (ABD) CEF ∠ 4 1 6 (ACBD) EF ∠ 3 1 = 2 14

22. If n Pr = n Pr+1 and n Cr = n Cr-1 then find the value of ‘n’ Dec-2008 Ans:(b)

(a) 2 (b) 3 (c) 4 (d) 5 Solution :

=

Or, !

! !

! Or, (n – r – 1)! = (n – r)! Or, n – r = 1 Or, n = 1 + r

∴ n + r – 1 = n Or, n = 2r - 1 Or, r + 1 = 2r – 1 Or, r = 2 ∴ n = 2r – 1 n = 4 – 1 = 3

24. In how many ways a committee of 6 members can be formed from a group of 7 boys and 4 girls having at least 2 girls in the committee. Dec-2008Ans:(c) (a) 731 (b) 137 (c) 371 (d) 351 Solution : 7 Boys 4 Girls Select 4 2 in

24

47 CC

3 3 in 3

43

7 CC

2 4 in 4

42

7 CC Total 371

25. Number of ways of painting a face of a cube by 6 colours is ________ June-2009 Ans:(a) (a) 36 (b) 6 (c) 24 (d) 1 Solution : ∟6 = 720

26. If 21818 rr CC find the value of 5Cr June-2009 Ans:(c)

(b) 50 (b) 50 (c) 56 (d) None of these Solution :

21818 rr CC ∴ r + r + 2 = 18 r = 8 ∴ the value of 563

85

85 CCCr


7.5

27. 7 books are to be arranged in such a way so that two particular books are always at first and last place. Final the number of arrangements. June-2009 Ans:(c) (a) 60 (b) 120 (c) 240 (d) 480 Solution : 2 particular books are always at first and last place so arranged in 2! ways So, remaining books i.e. 5 books can be arranged in 5 ways i.e. 5! Finally the number of arrangements = 5! × 2! = 240 ways

28. Find the number of arrangements in which the letters of the word 'MONDAY' be arranged so that the words thus formed begin with 'M' and do not end with 'N'. June-2009 Ans:(c) (a) 720 (b) 120 (c) 96 (d) None. Solution : begin with M arrangements are (6 -1)! i.e. 5! = 120 ways begin with ‘M’ and end with ‘N’ arrangements is (6 – 2)! = 4! = 24 ways. ∴ Required ways are (120 – 24) ways = 96 ways

29. In how many ways can 17 billiard balls be arranged if 7 of them are black, 6 red and 4 white ? (Ans: c)

(a) 4084080 (b) 1 (c) 8048040 (d) None of these

Solution : Possible ways = !

! ! ! = 4084080

30. (n + l)!=20(n-l)!,find n Dec-2009 Ans:(c) (a) 6 (b) 5 (c) 4 (d) 10 Solution : (n + 1)! = 20(n – 1)! or, (n + 1) (n) (n – 1)! = 20 (n – 1) ! or, n(n + 1) = 4 × 5 n = 4 n + 1 = 5 n = 4

31. Out of 4 gents and 6 ladies, a committee is to be formed find the number of ways the committee can be formed such that it comprises of at least 2 gents and at least the number of ladies should be double of gents. Dec-2009 Ans:(c) (a) 94 (b) 132 (c) 136 (d) 104 Solution : 4 Gents 6 Ladies Select 2 4 in 904

62

4 CC 2 5 in 365

62

4 CC

2 6 in 666

24 CC

3 6 in 466

34 CC

Total 136 32. In a bag, there were 5 white, 3 red, and 2 black balls. Three balls are drawn at a time what is the

probability that the three balls drawn are white? Dec-2009Ans:(a) (a) 1/12 (b) 1/24 (c) 1/120 (d) None of these Solution :

Total cases =10 =

= 120

∴ favorable cases = 5 white 3 red 2 black 100

20

33

5 CCC

∴ the probability that the three balls drawn are white = =

33. What is the probability that when the letters of ‘REGULATION’ be arranged so that the vowels come at odd places? Dec-2009 Ans:(a) (a) 1/252 (b) 1/144 (c) 144/252 (d) None of these Solution : Vowels = 5 Consonants = 5

E U A I O (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) 5 vowels in 5 places in 1205

5 P 5 consonants in 5 places 12055 P


7.6

Required ways (cases favourable) = 120 × 120 = 14400 ∴ total cases = 10!

Probability = !

=

34. Six points are on a circle. The number of quadrilaterals that can be formed are: June-2010 Ans:(c) (a) 30 (b) 360 (c) 15 (d) None of the above

Solution : 152

564

64

CCn

35. The number of ways of arranging 6 boys and 4 girls in a row so that all 4 girls are together is : June-2010 Ans:(c) (a) 6!. 4! (b) 2 (7!. 4!) (c) 7!. 4! (d) 2. (6!. 4!) Solution : Total number of boys and girls = 6 + 4 = 10 ∴ possible ways when 4 girls are together = 7! [ by taking 4 girls as 1] And in which 4 girls can be arranged in 4! ∴ total possible ways = 7! × 4!

36. How many numbers not exceeding 1000 can be made from the digits 1,2,3,4,5, 6, 7, 8, 9 if repetition is not allowed. June-2010 Ans:(b) (a) 364 (b) 585 (c) 728 (d) 819 Solution : - in 9P1 = 9 - - in 9P1.

8P1 = 72 - - - in 9P1.

8P1. 7P1 = 504 Total = 585

37. A garden having 6 tall trees in a row. In how many ways 5 children stand, one in a gap between the trees in order to pose for a photograph ? Dec-2010 Ans:(b) (a) 24 (b) 120 (c) 720 (d) 30 Solution :

* (1) * (2) * (3) * (4) * (5) * → Total children = 5 ∴ no of ways in which 5 children can stand one in a gap between trees in order to pose for a photograph 5 places for 5 children = 1205

5 P .

39. How many ways a team of 11 players can be made out of 15 players if one particular player is not to

be selected in the team. Dec-2010 Ans:(a) (a) 364 (b) 728 (c) 1,001 (d) 1,234 Solution : 1 particular Player is not selected Out of Remaining 14 players 11 can be Selected in =

3641114 C

40. Find the number of arrangements of 5 things taken out of 12 things, in which one particular thing must always be included. June-2011 Ans. (c) (a) 39,000 (b) 37,600 (c) 39,600 (d) 36,000 Solution : 12 things

1 particular

Thing must include 5

Remaining 11 thing Can be arranged

11

∴ no of arrangements = 5 11 = 5 × 11 × 10 × 9 × 8 = 39600

41. Exactly 3 girls are to be selected from 5 Girls and 3 Boys. The probability of selecting 3 Girls will be _______. June-2010 Ans. (a) (a) 5/28 (b) 1/56 (c) 15/28 (d) None. Solution :


7.7

Total cases = 8 = !

! ! =

!

! = 56

∴ favorable cases 5 girls 3 boys 5 3 = 10 × 1 = 10

Required probability =

42. In how many ways 3 prizes out of 5 can be distributed amongst 3 brothers Equally? Dec-2011 Ans. (c) a) 10 b) 45 c) 60 d) 120 Solution : Number of ways in which 3 prizes out of 5 can be distributed amongst

3 brothers equally = 5 = !

! =

!

!= 60 ways

43. There are 12 questions to be Answered Yes or No. How many ways can these be Answered? Dec-2011 Ans. (c) a) 1024 b) 2048 c) 4096 d) None Solution : 2 ways

yes Or no

2 ways yes

Or no

2 ways yes

Or no

2 ways yes

Or no

2 ways yes

Or no

2 ways yes

Or no

2 ways yes

Or no

2 ways yes

Or no

2 ways yes

Or no

2 ways yes

Or no

2 ways yes

Or no

2 ways yes

Or no = 212 = 4096

44. A team of 5 is to be selected from 8 boys and three girls. Find the probability that it includes two particulars girls. Dec-2011 Ans. (c) a) 2/30 b) 1/5 c) 2/11 d) 8/9 Solution :

Total cases = 11 = !

! ! =

!

! = 462

Cases favourable = 1.2.3

7.8.992

2 BCC

Required Probability =

45. The letters of the word “VIOLENT” are arranged so that the vowels occupy even place only. The number of permutations is __________. June-2012 Ans. (a) (a) 144 (b) 120 (c) 24 (d) 72 Solution :

I O E (1) (2) (3) (4) (5) (6) (7)

There are 3 vowels and 3 place. So they arranged in 3! = 6 ways Remaining 4 consonants and 4 place. So they arranged in 4! = 24 ways ∴ Total permutation = (6 × 24) ways = 144 ways.

46. If 2n

4n P20P then the value of ‘n’ is _________. June-2012 Ans. (b)

(a) – 2 (b) 7 (c) – 2 and 7 both (d) None of these. Solution : n(n – 1) (n – 2) (n – 3) = 20 n(n – 1) (n – 2) (n – 3) = 20 = 5 × 4 n – 2 = 5 n – 3 = 4 n = 7 n = 7

47. A man has 3 sons and 6 schools within his reach. In how many ways he can send them to school, if no two of his sons are to read in the same schools? Dec-2012 Ans. (b)

(a)6P2 (b) 6P3 (c) 63 (d) 36

Solution : 1st son in any of 6 school 2nd son in any of 5 schools 3rd son in any of 4 schools Required ways = 6 × 5 × 4 = 6P3

48. How many permutations can be formed form the letters of the word “DRAUGHT”, if both vowels may not be separated? Dec-2012 Ans. (b)

(a) 720 (b) 1,440 (c) 140 (d) 1,000


7.8

Solution : Taking A∪ together we have 6 letters (A∪) DRG HT in 6! Ways = 720 ways And two vowels can be arranged in 2! Ways = 2 ways ∴ Required permutation = (720 × 2) ways = 1440 ways 49. If 13C6 + 2 13C5 + 13C4 = 15Cx then, x = ___________ Dec-2012 Ans. (a) (a) 6 (b) 7 (c) 8 (d) 9 Solution :

13 2. 13 13 15

or, 13 13 13 13 15 or, 14 14 15 or, 15 15 x = 6 50. A polygon has 44 diagonals then the number of its sides are : June-2013 Ans. (d)

a) 8 b) 9 c) 10 d) 11 Solution :

No. of diagonals = – n 44 = – n 44 = nnn

2

)1(

44 = 2

22 nnn

n2 – 3n = 88 n(n – 3) = 88 = 11 × 8 n = 11

51. The number of words that can be formed out of the letters of the word “ARTICLE” so that vowels occupy even place is: June-2013 Ans. (b) a) 36 b) 144 c) 574 d) 754 Solution :

A I E (1) (2) (3) (4) (5) (6) (7)

3 vowels in 3 places = 33 P = 6 ways 4 consonants in 4 places = 4

4 P = 24 ways

required permutation = (6 × 24) ways = 144 ways 52. Number of ways of shaking hands in a group of 10 persons shaking hands to each other are: June-

2013 Ans. (a) a) 45 b) 54 c) 90 d) 10 Solution : Number of ways of shaking hands = = 10 = 45

53. If ,315

315

rr CC then ‘r’ is equal is . Dec-2013 Ans. (b)

(a) 2 (b) 3 (c) 4 (d) 5 Solution : 3r + r + 3 = 15 Solving r = 3

54. How many different words can be formed with the letters of the word “LIBERTY” Dec-2013Ans. (b) (a) 4050 (b) 5040 (c) 5400 (d) 4500 Solution : 7 letters in 50407

7 P ways

55. In how many ways can a family consist of three children have different birthdays in a leap year Dec-2013 Ans.(c) (a) 3

365 C (b) 33365 C (c) 366 × 365 × 364 (d) 3

366 C Solution :

(1) (2) (3) any of 366 days × any of 365 days × any of 364 days

58. If 3606 rP , then the value of ‘r’ is : Dec-2014 Ans. (c) a) 5 b) 3 c) 4 d) None of these Solution :

3606 rP Using hit & Trial Taking r = 4

!2

23456

!2

!64

6 P = 360 4r (Ans.)


7.9

59. There are 5 books on English, 4 Books on Tamil and 3 books on Hindi. In how many ways can these books be placed on a shelf if the books on the same subjects are to be together? Dec-2014 Ans. (c) a) 1,36,800 b) 1,83,600 c) 1,03,680 d) 1,63,800 Solution :

Take x Take y Take z

5 Books English , 4 Books Tamil , 3 Books Hindi Books on the same subject are to be together. x, y, z 3 things 3 ways to arrange = 63

3 P

Now, 5 English Books = x 5 ways to arrange = 1205

5 P

Now, 4 Tamil Books = y 4 ways to arrange = 244

4 P Now, 3 Hindi Books = z 3 ways to arrange = 63

3 P

Required ways = 6 × 120 × 24 × 6 = 1, 03, 680 (Ans.)

60. 5 Men and 4 Women to sit in a row in such a manner that the woman always occupy the even places. The number of such arrangement will be : Dec-2014 Ans. (d) a) 126 b) 1056 c) 2080 d) 2880 Solution : 5 Men 4 Women ______ ______ ______ ______ ______ _____ _____ _____ _____

| | | | W W W W

15 P

14 P

14 P 1

3 P 13 P 1

2 P 12 P 1

1 P 11 P

Required ways = 5 × 4 × 4 × 3 × 3 × 2 × 2 × 1 × 1 = 2880 (Ans.) 61. The four digit numbers that can be formed out of the seven digits 1, 2, 3, 5, 7, 8, 9 such that no digit

is repeated in any number and are greater than 3000 are June-2015 Ans. (c) a) 120 b) 480 c) 600 d) 840 Solution : 1, 2, 3, 4, 5, 7, 8, 9

3 ------ ------ ------ Any of the 6 digit = 36 P

5 ------ ------ ------ Any of the 6 digit = 36 P

7 ------ ------ ------ Any of the 6 digit = 36 P

8 ------ ------ ------ Any of the 6 digit = 36 P

9 ------ ------ ------ Any of the 6 digit

Required ways = 536 P = 5

!3

!6

= 6 × 5 × 4 × 5 = 600 (Ans.) 62. A person has ten friends of whom six are relatives. If he invites five guests such that three of them

are his relatives, then the total number of ways in which he can invite them are: June-2015 Ans. (c) a) 30 b) 60 c) 120 d) 75 Solution : Total 10 friends 6 Relative

Total No. of ways he can invite 5 guest in which 3 are relative =

120!2!2

!4

!3!3

!62

43

6

CC (Ans.)


7.10

63. A student has three books on computer, three books on Economics and five books on Commerce. If these books are to be arranged subject wise, then these can be placed on a shelf in the number of ways June-2015 Ans. (b) a) 25290 b) 25920 c) 4230 d) 4320 Solution :

3 Books on Computer 3 Books on Economics 5 Books on Commerce Total ways it can be arranged. 43205

53

33

3 PPP

If subject wise arrangement is to be done, No. of ways

waysTotal

Economics

Commerce

Economics

Computer

Computer

Commerce

Computer

Computer

Commerce

Commerce

Economics

Economics

Commerce

Economics

Computer

Economics

Commerce

Computer

6

Required ways = 6 × 4,320 = 25,920 (Ans.) 64. An examination paper with 10 questions consists of 6 questions in statistic part. At least one

question from each part is to be attempted in how many ways can this be done? Dec-2015 Ans. (b) a) 1024 b) 945 c) 1005 d) 1022 Solution : Out of 6 statistic questions, atleast one is to be done.

66

56

46

36

26

16 CCCCCC

63126 Out of remaining 4 questions, atleast one is to be done.

15124 44

43

42

41 CCCC

∴ Total questions = 63 × 15 = 945 65. If 720r

n p and 120rn c , then value of ‘r’ is: Dec-2015 Ans. (d)

a) 4 b) 5 c) 6 d) 3 Solution :

720rn p

120rn c

r

pc r

n

rn

r

720120

36 rr 66. There are 6 men and 4 women in a group, then the number of ways in which a committee of 5

persons can be formed of them, if the committee is to include at least 2 women are : Dec-2015 Ans.(b) a) 180 b) 186 c) 120 d) 105 Solution :

6 Men 4 Women 3 2 ----------- 5 2 3 ----------- 5 1 4 ----------- 5

∴ The no. of ways = 24

36 CC

34

26 CC

44

16 CC

= 186 67. In how many ways can a selection of 6 out of 4 teachers and 8 students be done so as to include at

least two teachers? June-2016 Ans. (b)


7.11

a) 220 b) 672 c) 596 d) 968 Solution :

4 Teachers 8 Students 6 - 2 4 - 4

82

4 CC (+)

6 - 3 4 - 38

34 CC (+)

6 - 4 4 - 28

44 CC (+)

672

68. There are 10 students in a class including 3 girls. The number of ways to arrange them in a row

when any two girls out of three never comes together : June-2016 Ans. (a) a) 3

8 P ∟7 b) 33 P ∟7 c) a) 3

8 P ∟10 d) None of these.

Solution : No. of arrange to arrange them in a row = 73

8 P

69. The maximum number of points of inter section of 10 circles will be June-2016 Ans. (c) a) 2 b) 20 c) 90 d) 180 Solution :

Points of intersection of 10 circles = .909108

102

102

PPn

71. An examination paper with 10 questions consists of 6 questions in statistic part. At least one question from each part is to be attempted in how many ways can this be done? Dec -2016 Ans. (b) a) 1024 b) 945 c) 1005 d) 1022 Solution : Same as (64).

72. The number of numbers between 1,000 and 10,000, which can be formed by the digits 1,2,3,4,5,6 without repetition is Dec -2016 Ans.(c) (a)720 (b)180 (c)360 (d)540 Solution : Number of numbers between 1000 and 10000 which can be formed by digits 1, 2, 3, 4, 5, 6 without repetition is

36013

14

15

16 PPPP

73. The number of ways in which 4 persons can occupy 9 vacant seats is: Dec -2016 Ans.(b) (a) 6048 (b) 3024 (c)1512 (d) 4536 Solution : No. of ways in which 4 persons can occupy 9 vacant places = 30241

61

71

81

9 PPPP 74. If 10C3 + 2. 10C4 + 10C5 = nC5 then value of n is: June-2017 (Ans: c)

a) 10 b) 11 c) 12 d) 13 Solution :

510

4210

310 CCC

510

410

410

310 CCCC

512

511

410 CCC

125512 nCC n

75. The number of parallelograms, formed from a set of six parallel lines intersecting another set of four parallel lines is: June-2017 (Ans: b) a) 360 b) 90 c) 180 d) 45 Solution : No. of parallelograms formed from a set of six parallel lines intersecting another set of four parallel lines _______

6 52 1

4 32 1

15 6 90.


7.12

76. The number of words which can be formed by letters of the word "ALLAHABAD' is: June-2017 (Ans: a) a) 7560 b) 3780 c) 30240 d) 15120 Solution : The no. of words which can be formed with the letters of "ALLAHABAD"

= 7560

Answer Key

1. a 2. b 3. c 4. d 5. d 6. c 7. a 8. a 9. b 10. c 11. b 12. a 13. a 14. b 15. a 16. b 17. d 18. b 19. a 20. b 21. d 22. b 23. d 24. c 25. a 26. c 27. c 28. c 29. c 30. c 31. c 32. a 33. a 34. c 35. c 36. b 37. b 38. a 39. a 40. c 41. a 42. c 43. c 44. c 45. a 46. b 47. b 48. b 49. a 50. d 51. a 52. a 53. b 54. b 55. c 56. a 57. d 58. c 59. c 60. d 61. c 62. c 63. b 64. b 65. d 66. b 67. b 68. a 69. c 70. d 71. b 72. c 73. b 74. c 75. b 76. a 77. b 78. c 79. a 80. a 81. 82. a 83. a 84. a 85. a 86. b 87. b

A.P. & G.P. GOPAL BHOOT

8.1

CHAPTER A.P. AND G.P.

Past Exam Questions Arithmetic And Geometric Progression1. The sum of all natural numbers from 100 and 1000 which are multiple of 5 is: Nov-2006

(a) 99,550 (b) 96,450 (c) 97,450 (d) 95,450 Solution : The series is 100, 105, 110, 115 .................. 1000 a = 100 ℓ 1000 d = 105 - 100 = 5. The sum of all natural numbers between 100 and 1000 which are multiple of 5 is

ℓ 100 1000

1 = 99,550 ⇒1000 = 100 + (n - 1)5 ⇒900 = (n - 1)5 ⇒n = 181 Answer ⇒99,550

2. Find n such that may be the geometric mean between a and b : Nov-2006

(a) 1/2 (b) 1 (c) -1/2 (d) 0 Solution : Geometric mean between a and b = √

∴ √

⇒ √ ⇒ . . . √ . √ ⇒ . √ √ . ⇒ √ √ √ √ √ √

⇒√

√

√ √

√ √

⇒√

√1

⇒ √ √

⇒√

√

⇒

⇒

⇒ 1 2⁄ 3. The first and the last terms of an AP are - 4 and 146 The sum of the terms is 7171 The number

of terms is : Nov-2006 (a) 101 (b) 100 (c) 99 (d) None Solution : a = - 4 ℓ = 146 a + (n - 1)d = 146

2ℓ

71712

4 146


8.2

⇒ 71712142

⇒7171 2142

101

4. If the first term of a GP exceeds the second term by 2 and the sum to infinity is 50, the series is :

(a) 10, 8, …… (b) 10, 8, …… (c) 10, , … (d) None

Solution : Let first term be a then second term be a - 2

∴

50

⇒ 50 ∴ 2 10 2 8

⇒ 100 ∴ 10, 8, . . . . ..

⇒ 10

5. The sum of square of first n natural numbers is : Feb-2007

(a)

2

1nn

(b)

6

1n21nn (c)

6

2n1nn

(d)

6

2n1nn

6. Divide 30 into five parts in AP such that the first and last parts are in the ratio 2 :3 : Feb-2007

(a) 5

36,

5

33,6,

5

27,

5

24

(b) 5

27,

5

24,

5

33,

5

36,6

(c) 6,

5

33,

5

36,

4

24,

5

27

(d) 5

36,

5

33,

5

27,

5

24,6

Solution :

Let the series be a, a + d, a + 2d, a+3d, 4 + 4d sum of the series = 30 ⇒ a + a + d + a + 2d + a + 3d + a + 4d = 30 ⇒ 5a + 10d = 30 or 1

⇒a + 2d = 6 or or 30 = 4

or 30 2 4

⇒d = -------------- (i) or 12 = 2a + 4d

or a + 2d = 6 or d =

The ratio of first and last part be 2 : 3

∴4

23

⇒

⇒2 6

23

⇒3a = 2a + 4(6-a) ⇒3a 2a + 24 - 4a ⇒3a = 2a - 2a

⇒5a = 24 d =

⇒a = =

3/5 = 6


8.3

3/5

27/5

2 2

6

∴ The series is , , 6, , , . . . . . . ..

7. If ⁄ ⁄ ⁄ and a, b, c are in G.P; the x, y, z are in : Feb-2007 (a) A. P. (b) G. P. (c) Both (a) & (b) (d) None Solution :

Let

∴ ⇒ Similarly, c ∴a,b,&careG.P , , areinG.P ∴ ⇒ . ⇒ ⇒ 2 ∴InA.Pseries,2b a c ∴ThisisinA.P.

8. Find the sum to n terms of the series: 7+77+777+__________ to n terms: Feb-2007

(a) 9

n71010

9

7 1n

(b)

9

n71010

9

7 1n

(c)

9

n71010

81

7 1n

(d)

9

n71010

81

7 1n

Solution : (Navratri Special) 7 + 77 + 777 + ................... to n terms = 7 (1+11+111 + ................. to n terms)

= (9+99+999 + .................. to n terms)

= [(10-1)+(100-1)+(1000-1)+ .................. to n terms)

= [(10+100+1000+ .................. to n terms) - n]

=

=

=

= 10 10

9. Find the sum of all natural numbers: between 250 and 1,000 which exactly divisible by 3: Feb-2007 (a) 1,56,375 (b) 1,56,357 (c) 1,65,375 (d) 1,65,357 Solution : The numbers divisible by 3 between 250 and 1000 are 252, 255, ............ 999 ∴ a = 252 d = 255 - 252 = 3

999 ⇒ a+(n-1)d = 999 ⇒ 252 + (n-1)3 = 999 ⇒ (n-1)3 = 747


8.4

⇒ 17473

⇒ n = 250 ∴ ℓ

252 999

1251 1,56,375

10. If the pth term of a GP is x and the qth term is y, then find the nth term : Feb-2007

(a)

pn

qn

y

x

(b)

qp

pn

qn

y

x

(c) 1 (d)

qp

1

pn

qn

y

x

Solution :

⇒ - (1) ⇒ - (11)

⇒

⇒

⇒

∴

⇒

⇒ . p - q - p + 1

⇒ .

. ∴

.

11. A person pays ` 975 in monthly instalments, each instalment is less than former by ` 5 The amount of first instalment is `100. In what time will the entire amount be paid? Feb-2007 (a) 26 months (b) 15 months (c) Both (a) & (b) (d) 18 months Solution : Let the entire amount be paid in n months First installment (a) = 100 d = - 5 ∴ 100 5 95 Similarly, The series is 100, 95, 90, 80, ................................ upto n months Sum = 2 1


8.5

⇒ 975 2 100 1 5

⇒ 975 100 . 5 1

⇒ 975

⇒ 1950 = 200n - 5 5 ⇒ 5 205 1950 0 ⇒ 41 390 0 ⇒ 15 26 390 0 ⇒ n(n - 15) - 26(n - 15) = 0 ∴ 15, 26 But we take n = 15 months because after twenty one installments, the installments become negative. So ignored n = 26.

12. If the sum of n terms of an AP is (3n2 -n) and its common difference is 6, then its first term is: Aug-2007 (a) 3 (b) 2 (c) 4 (d) 1 Solution :

3 Similarly 3 1 1 2 3 2 2 10

13. Find the sum of the series: 2 + 7+ 12+----------- 297 Aug-2007 (a) 8970 (b) 8870 (c) 7630 (d) 9875 Solution : a = 2 d = 5 297 ⇒ a + (n - 1)d = 297 ⇒ 2 + (n - 1)5 = 297 ⇒ (n - 1)5 = 295 ℓ = last term = 297 ⇒ (n - 1) = 59 ⇒ n = 60 ∴ Sum = ℓ

= 2 297

= 8970 14. A certain ball when dropped to the ground rebounds to 4th/5 of the height from which it falls; it is

dropped from a height of 100 metres find the total distance it travels before finally coming to rest: (a) 600m (b) 700m (c) 900m (d) 200m Solution : i.e. 100 + (80×2) + (64×2) + (51.2×2) + ............. ∞ = 100 + 160 + 128 + 102.4 + ......... ∞ GP Series So, r = 0.8 a = 160

= 100 + .

= 100 + 800 = 900m

15. In a GP if the (p + q)th term is m and (p - q)th term is n, then the pth term is : Aug-2007 (a) mn (b) √ (c) m2 (d) n2 Solution : ⇒ ----(i) ----(ii) Divide (i) by (ii)

⇒


8.6

⇒

⇒

⇒

Putting the value of r in ----- (i) ∴

⇒

⇒

⇒

⇒

⇒

⇒

∴

. .

.

.

.

√ 16. The sum of the series : 0.5 + 0.55 + 0.555 + ___________ to n term is Nov-2007

(a) ])1.0(1[9

5

9

n5 n

(b) ])1.0(1[

81

5

9

n5 n

(c) )1.0(1[

81

5

9

n5 n

(d)

])1.0(1[81

5

9

n5 n

Solution : 0.5 + 0.55 + 0.555 + ......... to n terms = 5 (0.1 + 0.11 + 0.111 + ......... to n terms)

= 0.9 0.99 0.999 ⋯ . .

= 1 0.1 1 0.01 1 0.001 ⋯ . .

= 0.1 0.01 0.001 ⋯ . .

= . .

.5 1

= .

.1 0.1

= 1 0.1

= 1 0.1

17. A contractor who fails to complete a building in a certain specified time is compelled to forfeit ` 200 for the first day of extra time required and thereafter forfeited amount is increased by `25 for every day If he loses ` 9,450, for how many days did he over-run the contract time? Nov-2007 (a) 19 days (b) 21 days (c) 23 days (d) 25 days


8.7

Solution : a = 200 d = 25 s = 9450 n = ?

∴22 1

⇒ 9450 200252

1

⇒ 18900 400 25 25 ⇒ 25 375 18900 0 ⇒ 15 756 0 ⇒ 36 21 756 0 ⇒ 36 21 36 0 ⇒ 36 21 0

∴ 36 & n = 21 The no. of days cannot be negative. So, we consider n = 21

18. The first, second and seventh term of AP are in GP and the common difference is 2, the 2nd term of AP is : Nov-2007 (a) 5/2 (b) 2 (c) 3/2 (d)1/2 Solution :

d = 2 = a + 2 6 = a + 12

, 2, 12 are in G.P. ∴ 2 12 ⇒ 4 4 12 ⇒ 4 1 12 ⇒ 1 3 ⇒ 3 1

⇒ 2 1 ⇒12

∴ 2

2

19. A man employed in a company is promised a salary of ` 3,000 every month for the first year and an increment of ` 1,000 in his monthly salary every succeeding year How much does the man earn from the company in 20 years? Feb-2008 (a) ` 30,00,000 (b) ` 27,50,000 (c) ` 19,10,000 (d) ` 7,90,000 Solution : a = 3000 d = 1000 n = 20 years = 20 × 12 months = 240 months S = 2 1

S = 2 3000 20 1 1000

S = 120 (6000 + 19000) S = 120 (25000) = 30,00,000

20. If a, b, c are in AP and x, y, z are in GP, then the value of x(b-c) y(c-a) z(a-b) is: Feb-2008 (a) 1 (b) 0 (c) b(c-a) (d) None Solution : Let a = 1 b = 2 ∴1, 2, 3 are in A. P. c = 3 and x = 1 ∴1, 2, 4 are in G. P.


8.8

y = 2 The value of

1 2 4 1 4 1

21. Insert 4 AM's between 3 and 18 : Feb-2008 (a) 12, 15, 9, 6 (b) 6, 9, 12, 15 (c) 9, 6, 12, 15 (d) 15, 12, 9, 6 Solution :

3, ______, ______, ______, ______, 18 a = 3

5 ⇒ 18 = 3 + 5d ⇒ 15 = 5d ⇒ d = 3

3 3 6 2 3 6 9 3 3 9 12 4 3 12 15

23. On 1st January every year a person buys National Saving Certificates of value exceeding that of his last year's purchase by ` 100 After 10 years, he finds that the total value of the certificates purchased by him is ` 54,500 Find the value of certificates purchased by him in the first year : June-2008 (a) ` 6,000 (b) ` 4,000 (c) ` 5,000 (d) ` 5,500 Solution : Let the value of certificates purchased by him in the first year be a n = 10 d = 100 2 1

⇒ 54500 102

2 9

⇒ 54500 5 2 9 100 ⇒ 54500 5 2 900 ⇒ 10900 2 900 ⇒ 10,000 2 ⇒ 2 10,000 ⇒ 5000

26. The sum of how many terms of the sequence 256, 128, 64,________is 511 Dec-2008 (a) 8 (b) 9 (c) 7 (d) None of these Solution :

a = 256 128256

0.5 1

11

511256 1 0.5

1 0.5

⇒ 511 0.5 256 1 0.5 ⇒ 255.5 256 1 0.5 ⇒ 1 0.5 0.998047 ⇒ 1 0.998047 0.5 ⇒ 0.5 0.001953 ⇒ 0.5 0.5 ⇒ 9

27. (x + 1), 3x, (4x + 2)are in AP Find the value of x Dec-2008 (a) 2 (b) 3 (c) 4 (d) 5


8.9

Solution : In A. P. 2b = a + c 2 × 3x = 4x + 2 x + 1 ⇒ ⇒

28. Find two numbers whose AM is 10 and GM is 8 Dec-2008 (a) [10,10] (b) [16,4] (c) [18,2] (d) [14,6] Solution : By Trial & Error Method

AM =

GM = Not possible. Again

AM =

GM = √ ∴ Answer (b)

30. The sum of terms of an infinite GP is 15 And the sum of the squares of the term is 45 Find the common ratio (a) 3/2 (b) 1 (c) -2/3 (d) 2/3 Solution :

⇒

Sum of the squares of the term i.e. sum of , , ∞

∴

⇒

⇒

⇒

⇒ ⇒ ⇒

⇒

⇒

31. If in an AP, Tn represents nth term If t7 :t10 = 5:7 then t8 : t11 = _________

(a) 13 : 16 (b) 17:23 (c) 14 : 17 (d) 15:19 Solution :

⇒

⇒ ⇒

⇒

∴

∶ ∶


8.10

32. The sum of an A P, whose first term is -4 and last term is 146 is 7171 Find the value of n Dec-2009 (a) 99 (b) 100 (c) 101 (d) 102 Solution :

2

1

⇒ 7171 2

4 146

⇒ 7171 2 142

⇒7171 2142

101

33. Find the sum to infinity of the following series: 1 – 1 + 1 – 1 + 1 – 1 + ……… ∞ Dec-2009 (a) 1 (b) ∞ (c) ½ (d) Does not exist Solution : 1 - 1 + 1 - 1 + 1 - 1 + ............... ∞ = (1+1+1+......∞) - (1+1+1+......∞)

=

= Does not exist. 34. If , , . , represents first, second and third terms of an AP respectively, the first term is 2 and

is minimum, then the common difference is equal to June-2010 (a) 5/2 (b) -5/2 (c) 2/5 (d) -2/5

Solution :

∴

⇒

2 2d 2 4 d 2 2d 8 2d 10 4d

⇒ 0Minimum∴Formaximumandminimumvalue, ⇒ ⇒

⇒

35. Divide 144 into three parts which are in AP and such that the largest is twice the smallest, the smallest of three numbers will be : June-2010 (a) 48 (b) 36 (c) 13 (d) 32 Solution : Let a - d, a, a+d are in A.P. ∴ 3a = 144 a = 48 ∴ By given question a + d = 2 (a-d) ⇒ a + d = 2a - 2d ⇒ 3d - a ⇒ d = 16 ∴ The smallest of three nos. is a - d = 48 - 16 = 32


8.11

36. Sum of Series ⋯… . .∞ is June-2010

(a) 15/36 (b) 35/36 (c) 35/16 (d) 15/16 Solution :

⋯………… .∞

⋯………… .∞

⋯…… .………… .∞

⋯………… .∞

37. If G be Geometric Mean between numbers a and b, then the value of is equal to Dec-

2010 (a) G2 (b) 3G2 (c) 1/G2 (d) 2/G2

Solution : √

∴

38. If Sum of ‘n’ terms of an Arithmetic Progression is What is the difference of its 10th and 1st term ? June-2011 (a) 207 (b) 36 (c) 90 (d) 63 Solution :

∴

∴ = 3 + 36 = 39 ∴

39. Find the product of : (243), (243)1/6, ⁄ …… . .∞, June-2011


8.12

(a) 1,024 (b) 27 (c) 729 (d) 246 Solution : (243), ⁄ , ⁄ …… . .∞,

⋯……..

a = 1 r = 1/6

.

40. Insert two Arithmetic means between 68 and 260 June-2011

(a) 132, 196 (b) 130, 194 (c) 70, 258 (d) None of the above Solution : 68, _______, _______, 260 a = 68

⇒ ⇒ ⇒ ⇒

41. Geometric Mean of P, P2, P3………Pn will be : June-2011

(a) (b) (c) (d) None of the above Solution :

, , …… . ⋯…………….

.

42. Find the numbers whose arithmetic mean is 12.5 geometric mean is 10 Dec-2011

a) 20 and 5 b) 10 and 5 c) 5 and 4 d) None of these Solution : By Trial and Error Method,

AM = 12.5

GM = √20 5 10 43. If sum of 3 arithmetic means between “a” and 22 is 42, then “a” = _______ Dec-2011

a) 14 b) 11 c) 10 d) 6 Solution : a, ______, ______,______, 22

4 22 = a + 4d ----- (i) 2

2ℓ

⇒

⇒ .

⇒ ⇒ ------ (ii) a + 2d = 14 × 2 ---------- (iii) a + 4d = 22 --------------- (iv) ∴ 2a + 4d = 28 ----------------------


8.13

a = 6 44. If each month ` 100 increases in any sum then find out the total sum after 10 months, if the sum of

first month is ` 2,000 Dec-2011 a) ` 24,500 b) ` 24,000 c) ` 50,000 d) ` 60,000 Solution :

a = 2000 d = 100 n = 10

22 1

102

4000 900

S = 24500 45. The sum of all two Digit odd numbers is Dec-2011

a) 2475 b) 2575 c) 4950 d) 5049 Solution : 11, 13, 15, 17 ............ 99 a = 11

d = 2 ℓ

1 11 99

⇒

⇒ ⇒n = 45 = 2475

46. If 5th term of a GP is 3 3 , then the product of first nine terms is Dec-2011 a) 8 b) 27 c) 243 d) 9 Solution :

√3 √3

∴ The product 06 first nine terms = ∴

√3

3 3 9 47. The sum of the third and ninth term of an AP is 8. Find the sum of the first 11 terms of the

progression Dec-2011 a) 44 b) 22 c) 19 d) 11 Solution : 8 ⇒ ⇒ ⇒

∴

48. If 8th term of an AP is 15, then sum of its 15 terms is June-2012

(a) 15 (b) 0 (c) 225 (d) 225/2 Solution :

7 14 15 = a + 7d

ℓ


8.14

14

2 14 15 7 15 15 225

49. Find the sum of the infinite terms June-2012

2,4,

8,

16…… . . ; 2,

(a) (b) (c) (d) None of these

Solution : a = 2

d = . 2

1

50. The 4th term of an AP is three times the first and the 7th term exceeds twice the third term by 1 Find the first term ‘a’ and common difference ‘d’ June-2012 (a) a = 3, d = 2 (b) a = 4, d = 3 (c) a = 5, d = 4 (d) a = 6, d = 5 Solution :

3 a = 3d = 3a

⇒ 3d = 2a ⇒

∴ ⇒ = a ⇒ ⇒ ⇒ ∴

⇒

⇒ [a = 3] ∴ a = 3 d = 2

51. In an AP , if common difference is 2, Sum of n terms is 49,7th term is 13 then n = __________ Dec-2012 (a) 0 (b) 5 (c) 7 (d) 13 Solution : d = 2 49

6 =13 ⇒ ⇒ 49

⇒

⇒ ⇒ ⇒ ⇒

52. The first term of a GP where second term is 2 and sum of infinite term is 8 will be Dec-2012 (a) 6 (b) 3 (c) 4 (d) 1 Solution : 2 ⇒ ar = 2 ⇒8(1 - r)r = 2 ⇒ 4(1-r)r = 2

8 ⇒

⇒


8.15

∴ ⇒

⇒ ⇒

⇒ ⇒ ⇒

53. If the sum of n terms of an AP be 2n2 + 5n, then its nth terms is Dec-2012 (a) 4n - 2 (b) 3n - 4 (c) 4n + 3 (d) 3n + 4 Solution :

2 5 2 1 5 1 7 2 2 5 2 18

18 ⇒ 18 7 ⇒ 11 ⇒ 11 ⇒ 11 7 7 ⇒ 4

∴ 1 = 7 + (n - 1)4 = 7 + 4n - 4 = 3 + 4n

54. If the sum of n terms of an AP be nn 23 and its common difference is 6, then its first term is : June-2013 a) 2 b) 3 c) 4 d) 5 Solution :

3 d = 6 3 1 1 2 3 2 2 10

2 ⇒ 2

55. If the sum of the 4th term and the 12th term of an AP is 8, what is the sum of the first 15 terms of the progression? June-2013 a) 60 b) 120 c) 110 d) 150 Solution :

8 ⇒ 3 11 8 ⇒ 14 8 ⇒ 7 4 152

2 14

= 15(a + 7d) = 15 × 4 = 60

56. If ‘n’ arithmetic means are inserted between 7 & 71 and 5th arithmetic mean is 27, then ‘n’ is equal to: June-2013 a) 15 b) 16 c) 17 d) 18 Solution : Total terms = n + 2 27 a = 7 a = 5d = 27 ⇒ ⇒

71 ⇒ ⇒


8.16

⇒ ⇒ ⇒ ⇒

57. In a GP the sixth term is 729 and the common difference is 3, then the first term of GP is: June-2013 a) 2 b) 3 c) 4 d) 7 Solution :

729 729

3 729 [r = 23]

a = 3 71. If a, b, c are in Arithmetic Progression (A.P.), then the value of a-b+c is Dec-2015

a) 0 b) -b c) b d) c Solution : a, b, c are in Arithmetic Progression (A.P.) b - a = c - b -------- (1) So a - b + c ⇒ a + c - b = a + b - a (According to equation) = b.

72. Find the two numbers whose geometric mean is 5 and arithmetic mean in 7.5. Dec-2015 a) 10 and 5 b) 13.9 and 1.91 c) 12 and 3 d) None of the above Solution : (Hit and Trial Method)

73. The sum of n terms of the series log 2

32

loglogy

x

y

xx ............. is June-2016

a)

xy

y

xn

nloglog2

2 b)

y

xxyn

nloglog

2

c)

xy

y

xn

nloglog

2 d)

xy

y

xn

nloglog

2

Solution :

⋯………….

… . . ⋯ . 2 3 … . . 2 ⋯ . 1 log

1 2 3… . . 1 2 ⋯ . 1 1

21 22

2 1 1

2

2

2

75. If baaccb

1,

1,

1are in arithmetic progression then ,,, 222 cba are in progression June-2016

a) Arithmetic Progression b) Geometric Progression c) Both in arithmetic and geometric Progression d) None of these. Solution :


8.17

1 1 1 1

⇒1 1 1 1

⇒1 1 2

2

⇒2 2

⇒ 2 2 2 2 2 2 ⇒ 2 ⇒ 2 0 ⇒ 0 ⇒ (They are n A.P.)

76. The income of a person is `5,00,000 in the firm in the first year and he receives an increase of `15,000 per year for next 10 years. The total amount he receives in 10 years is: Dec-2016 (a) `56,75,000 (b) `72,50,000 (c) `15,67,5000 (d) None of these Solution : Sn after 10 years

22 1

102

2 500000 10 1 15000

= 5 [10,00,000 + 1,35,000] = ` 56,75,000 77. If the sum 50 + 45 + 40 + 35 + ________ is zero, then the number of terms is : Dec-2016

(a) 22 (b) 20 (c) 21 (d) 25 Solution :

0250 2 1 5

⇒ 0 100 5 5 ⇒ 0 105 5

∴ 21. 78. The number 2.353535 ________ in form is: Dec-2016

(a) (b) (c) (d)

Solution : 2.353535 in form

235 299

23399

80. The sum of first 20 terms of a GP is 1025 times the sum of first 10 terms of same GP then common ratio is : June-2017 a) √2 b) 2 c) 2 √2 d) 1/2 Solution : Let the common ratio be r. Sum of 1st 20 terms.

Sum of 1st no terms

A. T. Q.


8.18

11

10251

1

⇒ 1 1025 1 ⇒ 1 1025 1 Hit and Trial method -

r = 2. 81. The value C such that a, -3, b, 5, c are in A.P. is: June-2017

a) -7 b) 1 c) 13 d) 9 Solution : a, -3, b, 5, c d = - - 3 - a - 3 - a = + 3 or d = b + 3 ⇒- 3 - = a + b or d = c - 5 ∴a + b = - 6 ---- (1) b + 3 = c - 5 d = 5 - b ⇒ 5 3 ⇒ 8 ---- (2) a + b = - 6 c - b = 8 a + c = 2 ----- (3) ⇒ a = 2 - c are know c = a + (5 - 1)c ⇒ c = a + 4 (c - 5) ⇒ c = 2 - c + 4c - 20 ⇒ c = 2 + 3c - 20 ⇒ - 2c = - 18 ∴ c = 9.

Answer Key 1. a 2. c 3. a 4. a 5. b 6. a 7. a 8. c 9. a 10. d 11. b 12. b 13. a 14. c 15. b 16. b 17. b 18. a 19. a 20. a 21. b 22. a 23. c 24. c 25. c 26. b 27. b 28. b 29. a 30. d 31. b 32. c 33. c 34. b 35. d 36. c 37. c 38. b 39. c 40. a 41. b 42. a 43. d 44. a 45. a 46. b 47. a 48. c 49. a 50. a 51. c 52. c 53. c 54. a 55. a 56. a 57. b 58. d 59. a 60. a 61. b 62. c 63. b 64. b 65. d 66. c 67. b 68. c 69. d 70. a 71. c 72. b 73. d 74. c 75. a 76. a 77. c 78. d 79. a 80. b 81. d 82. a 83. 84. 85. c 86. a 87. a 88. a

Set Theory GOPAL BHOOT

9.1

CHAPTER SET THEORY

Past Exam Questions Sets

1. Out of 20 members in a family, 11 like to take tea and 14 like coffee Assume that each one likes at least one of the two drinks Find how many like both coffee and tea : Nov-2006 Ans:(d) (a) 2 (b) 3 (c) 4 (d) 5

Ans:-A ∪ B = 20 A = 11 B = 14

A + B – AB = 20 11 + 14 – AB = 20 AB = 5 2. In a group of 70 people, 45 speak Hindi, 33 speak English and 10 speak neither Hindi nor English

Find how many can speak both English as well as Hindi: Feb-2007 Ans.(c) (a) 13 (b) 19 (c) 18 ; (d) 28

Ans:- Hindu = 45 = A English = B = 33 ∩ = 10 Now A ∪ B = ∩ = S – ( ∩ ) = 70 - 10 = 60 A + B – AB = 60 45 + 33 – AB = 60 AB = 18

3. In a survey of 300 companies, the number of companies using different media - Newspapers (N), Radio (R) and Television (T) are as follows : n(N) = 200, n(R) =100, n(T) = 40, ∩ = 50,

,20)TR(n and25)TN(n 5)TRN(n May-2007 Ans:(d) Find the numbers of companies-using none of these media :

(a) 20 companies (b) 250 companies (c) 30 companies (d) 50 companies Ans:- N = 200 R = 100 T = 40 NT = 25 NR = 50 RT = 20 NRT = 5 N∪R∪T = N + R + T – NR – NT – RT + NRT = 200 + 100 + 40 – 50 – 20 – 25 + 5 = 250 None media = = ∪ ∪ = 300 – 250 = 50

4. In a town of 20,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C If 2% families buy all the three newspapers, then the number of families which buy A only is: Nov-2006 Ans:(a) (a) 6600 (b) 6300 (c) 5600 (d) 600 Ans:- A = 40% B = 20% C = 10% AC = 4% AB = 5% BC = 3% ABC = 2% Only A = = A – AB – AC + ABC = 40 – 5 – 4 + 2 = 33% No of families = 33% of 2000 = 6600.

5. Out of total 150 students, 45 passed in Accounts, 30 in Economics and 50 in Maths, 30 in both Accounts and Maths, 32 in both Maths and Economics, 35 in both Accounts and Economics, 25 students passed in all the three subjects Find the numbers who passed at least in any one of the subjects : Feb-2008 Ans:(b) (a) 63 (b) 53 (c) 73 (d) None Ans. : A = Account = 45 B = Economics is = 30 C = Maths = 50 AC = 30 BC = 32 AC = 35 ABC = 25 ∪ ∪ = A + B + C – AB – AC – BC + ABC = 53

6. If A = {p, q, r, s} B = {q, s ,t} C = {m, q, n} Find C – (A ∩ B) Dec-2008 Ans:(a) (a) {m, n} (b) {p, q} (c) {r, s} (d) {p, r} Ans:- A∩ B = {q,s} C – (A ∩ B) = m, n

7. If A = {x : x2 – 3x + 2 = 0}, B = {x : x2 + 4x – 12 = 0}, then B – A is Equal to June-2010 Ans:(a) (a) {-6} (b) {1} (c) {1,2} (b) {2, -6} Ans:-

(A) x2 – 3x + 2 = 0 (x – 2) (x -1) = 0

(B) x2 – 4x + 12 = 0 (x + 6) (x – 2) = 0

Set Theory GOPAL BHOOT

9.2

x = 2 , 1 A = {2 , 1}

x = – 6 , 2 B = {-6 , 2}

∴ B – A = {-6}

8. For any two sets A and B, )B'A(A = _______, where A’ represent the compliment of the set A Dec-2010 Ans:(a) (a) AB (b) AB (c) A'B (d) None of these Ans:-

∪ = shaded area A ∩ shaded = A ∩B

9. If AB, then which one of the following is true Dec-2010 Ans:(b) (a) AB =B (b) AB = B (c) AB =A1 (d) AB = Ans:- AB So, A ∪ B = B Option (b) is correct

10. There are 40 students, 30 of them passed in English, 25 of them passed in Maths and 15 of them passed in both Assuming that every Student has passed at least in one subject How many student’s passed in English only but not in Maths June-2011 Ans. (a) (a) 15 (b) 20 (c) 10 (d) 25 Ans:- E = 30 M = 25 EM = 15 E M = E – EM = 30 – 15 = 15

11. If A = (1, 2, 3, 4, 5), B = (2, 4) and C = (1, 3, 5) then (A – C) × B is Ans. (d) a) {(2, 2), (2, 4), (4, 2), (4, 4), (5, 2), (5, 4)} b) {(1, 2), (1, 4), (3, 2), (3, 4), (5, 2), (5, 4)} c) {(2, 2), (4, 2), (4, 4), (4, 5)} d) {(2, 2),(2, 4), (4, 2), (4, 4)} Ans:- A – C = {2 , 4} B = {2 , 4} (A – C) × B ={(2 , 2) (2 , 4) (4 , 2) (4, 4)}

12. For any two sets A and B the set (AUB’)’ is Equal to (where’ denotes compliment of the set) Dec-2011 Ans.(a) a) B – A b) A – B c) A’ – B’ d) B’ – A Ans:- (A ∪ B = A∩ B = B – A

13. The number of proper sub set of the set {3, 4, 5, 6, 7} is June-2011 Ans. (b) (a) 32 (b) 31 (c) 30 (d) 25 Ans:- 2n – 1 = 25 – 1 = 31

14. For a group of 200 persons, 100 are interested in music, 70 in photography and 40 in swimming, Further more 40 are interested in both music and photography, 30 in both music and swimming,20 in photography and swimming and 10 in all the three How many are interested in photography but not in music and swimming? Dec-2012 Ans.(d)

(a) 30 (b) 15 (c) 25 (d) 20 Ans:- M = 100 P = 70 S = 40 MP = 40 MS = 30 PS = 20 MPS = 10 MS = P – PM – PS + PMS = 70 – 40 – 20 + 10 = 20

15. Of the 200 candidates who were interviewed for a position at call centre, 100 has a two-wheeler, 70 has a credit card and 140 had a mobile phone, 40 of them had both a two-wheeler and a credit card, 30 had both a credit card and a mobile phone, 60 had both a two-wheeler and a mobile phone, and 10 had all three. How many candidates had none of the three? Dec-2013 Ans. (c) (a) 0 (b) 20 (c) 10 (d) 18 Ans. : Two wheeler = A = 100 Credit Card = B = 70 Mobile = C = 140 AB = 40 BC = 30 AC = 60 ABC = 10 ∪ ∪ = A + B + C – AB – BC – AC + ABC = 100 + 70 + 140 – 40 – 30 – 60 + 10 = 190 none = CBA = ( ∪ ∪ )C = S – ( ∪ ∪ ) = 200 – 190= 10

Set

16.

1. 11. 21.

Theory

In a clastudentsa) 13 Ans. : MathemCommeMaths ∩

MathsP(Math 50 P(M ∩

d 2. d 12. 22.

35

ass of 50 stus who opted

n = 50 matics = 35 erce = 37 ∩ Commerc

s Comm

hs ∪Comme C)

c 3. a 13. b 23.

37

udents, 35 opd for both M

b) 1

ce ?

merce erce )

d 4. b 14. a 24.

pted for MaMathematics15

= P(Maths)- P(Maths ∩ = 35 + 37 –= 72 – 50 = 22 (Ans.)

Aa 5.d 15.d 25.

9.3

athematics as and Comm

) + P(Comm∩ Commerc– P(M ∩ C)

)

Answer Keb 6.c 16.a

and 37 optemerce are :

c) 22

merce) ce) )

y a 7.c 17.

d for Comm

a 8.b 18.

GO

merce. The n Dec-d) 28

a 9. b 19.

OPAL BHO

number of s-2013 Ans.

b 10.a 20.

OOT

such (c)

a a

Functions - Assignment GOPAL BHOOT

10.1

CHAPTER FUNCTIONS - ASSIGNMENT

Past Exam Questions Functions & Relations

1. Let R is the set of real numbers, such that the function f: RR and g : R R are defined by f(x) = x2 + 3x+1 and g(x) = 2x – 3 Find (fog): Feb-2007

(a) 4x2 + 6x + 1 (b) x2 + 6x + 1 (c) 4x2 – 6x + 1 (d) x2 – 6x+l

Ans:- f(x) = x2 + 3x + 1 g (x) = 2x - 3 fog = f{g(x)} = f(2x – 3)

= (2x – 3)2 + 3(2x – 3) + 1 = 4x2 – 6x + 1

2. If R is the set of real numbers such that function RRf : is defined by f(x) = (x + l)2, then find (fof) : May-2007

(a) 11x 2 (b) 1x2 (c) 22 11x (d) None Ans:- f (x) = (x + 1)2 f of = f {f(x)} = f(x + 1)2 = {(x + 1)2 + 1}2

4. Let ƒ: R R be such that ƒ(x) = 2x then ƒ(x + y) equals : Nov-2006 (a) ƒ(x)+ƒ(y) (b)ƒ(x)ƒ(y) (c)ƒ(x)÷ƒ(y) (d) None of these

Ans:- f(x) = 2x f(x + y) = 2x+y f(x) . f(y) = 2x . 2y = 2x+y (b) is correct

7. If f (x) = x2 + x-1 and 4f (x) = f (2x) then find 'x' Dec-2008 (a) 4/3 (b) 3/2 (c) –3/4 (d) None of these Ans:- f(x) = x2 + x – 1 4f(x) = 4(x2 + x – 1) = 4x2 + 4x – 4 Now, f(2x) = (2x)2 + 2x – 1 = 4x2 + 2x – 1 Again 4f(x) = f(2x) 4x2 + 4x – 4 = 4x2 + 2x – 1 2x = 3 x = 3/2

11. If F: A → R is a real valued function defined by ,x

1)x(f then A = _____ June-2010

(a) R (b) R – {1} (c) R–{0} (d) R – N

Ans:- f(x) = x can take all real nos value except R – {0}

13. If f:R→R, f(x) = x+l, g: R→R g(x) = x2+l then fog(–2) equals to Dec-2010 (a) 6 (b) 5 (c) –2 (d) None Ans:- f(x) = x + 1 g(x) = x2 + 1 fog(-2) first g(-2) = (-2)2 + 1 = 5 f{g(-2)} = f(5) = 5 + 1 = 6

16. If 2x1

xx)

f( and

2x1

x)x(g

Find fog? June-2011

(a) x (b) x

1 (c)

2x1

x

(d) 2x1x

Ans:- f(x) = √ g(x) = √

fog = f{g(x)} = f√

= = x

17. f(x) = 3 + x, for – 3 < x < 0 and 3 – 2x for 0 < x < 3, then value of f(2) will be Dec-2011 a) – 1 b) 1 c) 3 d) 5 Ans:- Given f(x) = 3 + x - 3 < x < 0 = 3 – 2x 0 < x < 3 Now, f(2) = 3 – 2 × 2 = - 1

23. If xxgxxf 7,2 , then g of (x) = ______________ June-2013

a) xx x 7.2.7 b) 27 x c) x749 d) None of these Ans:- f(x) = x + 2 g(x) = 7x

Functions - Assignment GOPAL BHOOT

10.2

g of (x) g{f(x)} g(x + 2) = 7x + 2 = 7x × 49

24. If

x

xxf

1

1log , then

21

2

x

xf is equal to : June-2013

a) f(x) b) 2f(x) c) 3f(x) d) –f(x)

Ans:- f(x) = log f = log

= log

= log

= = 2 log = 2f(x)

26. If ,5

25)(

2

x

xxf then )5(f is Dec-2013

(a) 0 (b) 1 (c) 10 (d) not defined Ans. : If x = 5 then denominator is 0 So, f(x) is not defined. [Remember, this is not a question of limits]

Answer Key 1. c 2. c 3. a 4. b 5. a 6. a 7. b 8. b 9. d 10. a 11. c 12. d 13. a 14. c 15. c 16. a 17. a 18. b 19. b 20. c 21. c 22. c 23. c 24. b 25. a 26. d 27. a 28. b 29. b 30. b 31. c 32. c 33. a 34. c 35. a 36. d 37. d 38. b 39. a 40. c 41. c 42. b 43. a 44. b 45. b 46. b 47. b

Calculus GOPAL BHOOT

11.1

CHAPTER

CALCULUS Past Exam Questions Basic Concepts of Differential And Integral Calculus 1. The slope of the tangent at the point (2, - 2 ) to the curve x2 + xy + y2 – 4 = 0 is ;given by: Nov-2006

Ans:(b) (a) 0 (b) 1 (c) -1 (d) None

Ans: So, 4 0 (1)

( Slope of the tangent at any point to the curve is equal to at that point)

∴ Differentiating (1) both side we get

Or 2 2 0 0Or 2 2

Or Or 2, 2 1

2. The derivative of x2 log x is : Nov-2006 Ans:(c)

(a)1 + 2 log x (b) 2 1og x (c) x (l + 2 1og x) (d) None of these Ans: So, Let log x Differentiating both sides we get

= x2 (log x) + log x (x2) = x2 × + log x × 2

= x + x log x = x(1+ log x)

3. )(1

0 xx ee dx is: Nov-2006 Ans:a

(a) 1ee (b) ee 1 (c) 1ee (d) None Ans: So, =

∴ = = 1 1=

4. 33

2

)2(

8

x

x dx is equal to: Nov-2006 Ans:(b)

(a)3

4 (x3 + 2)2 + C (b) C2x

3

4 23

(c) 3

4(x3 + 2)2 + C (d) None of these

Ans: So,

Let 2 Now, differentiating both side the above eq. we get 3

dx =

= ∵ 2

=

=

= ∵

= 2

5. If x = y log (xy), then dx

dy is equal to: Feb-2007Ans:(b)


11.2

(a) )log1( xyx

yx

(b) )log1( xyx

yx

(c) )log(log yxx

yx

(d) )log(log yxx

yx

Ans: So, x = y log (xy)

Differentiating both sides we get Or,1 = log + log

Or, 1 = log Or, 1 = log

Or, 1 = [ log(xy) + 1] Or,

6. If y = 2x+ x

4, then y

dx

dyx

dx

ydx

2

22 yields Feb-2007Ans:(c)

(a) 3 (b) 1 (c) 0 (d) 4

Ans: So, y = 2x +

Differentiating both sides we get 2

Again differentiating we get

A T P Putting the value of and and y

= x2 2 2 = 2 2

= 2 2 ∴ = 0

7. Evaluate: 22 ax

dx Feb-2007Ans:(b)

(a) C)axxlog(2

1 22 (b) C)axxlog( 22

(c) Caxx(log 22 (d) C)axx(log2

1 22

Ans: log(x + √ ) + c [ by formula]

8. The value of

2

0 2 xx

x dx is : Feb-2007Ans:(d)

(a) 0 (b) 3 (c) 2 (d) 1

Ans: Let I = √

√ …..(1)

= √

√ ∴

= √

√ √ …...(2)

Adding (1) and (2) we get

2I = √ √

√ √2I = 2I = [x]2

02I = 2-0 I = 1

9. If f(x) = xk and f ' (1) = 10, then the value of k is : May-2007Ans:a

(a) 10 (b) -10 (c) 1/10 (d) None

Ans: f(x) = xk f1(x) = k. xk-1 f1(1) = 10 Or, 10 = k(1)k-1 Or, 10 = k

Or. k = 10 [ ∴1 to the power any thing is 1]

10. Given x = 2t + 5 ; y = t2 - 2, then dx

dy is calculated as : May-2007Ans:a

(a) t (b) 1/t (c) -1/t (d) None


11.3

Ans: x = 2t +5 ……(1) = 2

y = t2 – 2 ……(2) = 2t

= /

/

11. The integral of (e3x + e-3x) / ex is: May-2007Ans:(b)

(a) Cee xx

42

42

(b) Cee xx

42

42

(c) e2x – e-4x + C (d) None of these

Ans:

= = =

12. dxex x32 is: May-2007Ans:(c)

(a) x2e3x – 2xe3x + 2e3x + C (b) Ceexe x

xx

333

29

.

3

(c) Ceexex x

xx

3332

27

2

9

.2

3

. (d) None of these

Ans: Applying by parts we get

= 2 . [ Taking as first function] = 2

= c

Again applying by parts we get

= e 1.

= . . . ..

=

13. dxx

x

2

121

2 : May-2007Ans:a

(a) 2

5log e (b) 12log5log ee (c)

5

2log e (d) None of these

Ans:

dxx

x

2

121

2 Let 1 + x2 = z

2x dx = dz = =

= loge 5 – loge 2 = loge

[ Dierentiating we get] When X Z 2 5 1 2

14. If xy = yx, then gives : Aug-2007Ans:(c)

(a) )log(

)log(

xxyy

yyxx

(b) )log(

)log(

yyxy

xxyx

(c) )log(

)log(

xxyx

yyxy

(d) None of these

Ans: xy = yx Talking log on both sides we get

Or, Or,

Or, Or, Or,

15. If x3 – 2x2y2 + 5x + y = 5, then dx

dy at x = 1 and y = 1 is: Aug-2007Ans:a


11.4

(a) 4/3 (b) -5/4 (c) 4/5 (d) -4/3 Ans: x3 – 2x2y2 + 5x + y = 5 Differentiating both sides we get

Or, 3 4 4 5 0Or, 1 4 4 3 5

Or, Or, ,

– Or, ,

= =

16. The value of e

x

x

1

)log1( dx is: Aug-2007Ans:(b)

[Given Loge = 1] (a) 1/2 (b) 3/2 (c) 1 (d) 5/2

Ans:

Let 1 + log x = z [differentiating we get] dx = dz

when x = e , z = 2 when x = 1 , z = 1 = = 21= 4 1

17. Find 32

3

)1(x

xdx: Aug-2007Ans:(b)

(a)

])1(

12

4

122

2

x

x (b)

])1(

12

4

122

2

x

x (c)

])1(

12

2

122

2

x

x (d)

])1(

12

2

122

2

x

x

Ans:

.

Let, + 1 = z [ z – 1]

2x dx = dz x dx = dz

= . = = =

= =

18. If y = nmxx )( 22 then dx

dy: Nov-2007Ans:a

(a) 22 mx

ny

(b) ny (c)

22 mx

ny

(d) None

Ans: y = √ Differentiating both sides we get

√ 1 √

= √

√

√

√ =

√

19. If xy (x – y) = 0, find : Nov-2007Ans:a

(a) )2(

)2(

xyx

yxy

(b) )2(

)2(

xyy

yxx

(c) )2(

)2(

yxx

xyy

(d) None

Ans: xy(x – y) = 0 x2y – xy2 = 0 differentiating both sides we get

or, 2 2 0 or, 2 2

or,

or, = = or, =

20. If y = .....x

x then dx

dyis: Nov-2007Ans:(c)


11.5

(a) x

y

log

2

(b)xy

y

log2

2

(c)

)log2(

2

xyx

y

(d) None

Ans: √√ …..∞

Taking log on both sides we get

xy

xxy log2

log.....log

Or, √

√ √ xgtoyy log

Or, √ Or, √

Or, √ Or,

21. dxax 22

1is: Nov-2007Ans:(c)

(a) log (x – a) – log (x + a) + C (b) Cax

ax

log

(c) Cax

axlog

a2

1

(d) None of these

Ans: (c) log [by formula]

22. The value of

1

0 )2)(1( xx

dx is: Nov-2007Ans:(b)

(a) 4

3log (b)

3

4log (c) 12log (d) None

Ans:

Or, 3 2 2 2 =

= = log

= log – log = log =

23. If .........!

...................!3!2

132

n

xxxxy

n

then ydx

dy is equal to: Nov-2007Ans:(c)

(a) 1 (b) -1 (c) 0 (d) None

Ans: y = ex ∵ 1!

!

⋯!

⋯⋯∞ = ex

A T P ex – ex = 0

24. The slope of the tangent to the curve 24 xy at the point, where the ordinate and the abscissa are equal, is: Nov-2007Ans:a

(a) -1 (b) 1 (c) 0 (d) None Ans: y = √4

Differentiating both side we get = √

When x =

1242

222,2

2,2

dx

dyy


11.6

y = x 24 xy 24 yy 2,2 x

22 42 y 22 y 2y

25. The value of 3

2

3

2

)()5( dxxfdxxf is : Nov-2007Ans:(b)

(a) 1 (b) 0 (c) -1 (d) None

Ans. : :

a

b

a

bdxxbafdxxfformulaby )()(0

26. dxx

exe

log

is; Nov-2007Ans:(b)

(a) x-1 + C (b) x + C (c) x2 + C (d) None

Ans:

Let = z Or, = z Or, log = z [differentiating we get]

Or, dx = z = = = x + c

27. Differentiate )( xxe : June-2008Ans:(c)

(a) (1 + log x) (b) xx (1 + log x) (c) xxe (1 + log x)xx (d)

xxe (1 + log x) Ans: Talking log on both sides we get log y = log Or, log y = Again taking log we get Or, log (log y) = x log x [ Differenting both side we get ]

Or, Or, = y log y ( 1 + log x) Or, 1

28. If xm yn = (x + y)m+n , then find dx

dy : June-2008Ans:(b)

(a) x/y (b) y/x (c) xy (d) None Ans: Talking log on both sides we get Or, log log Or, log log

Or, 1 1

Or,

Or,

Or, Or,

29. Evaluate )2)(1(

1

xx dx: June-2008Ans:a

(a) Cx

x

1

2log (b) Cxx 12log

(c) Cx

x

2

1log (d) None


11.7

Ans:

=

x2 – 3x + 2 = 2 2 = =

= log =

30. dxx 4

1

52 and the value is: June-2008Ans:(c)

(a) 10 (b) 3 (c) 30 (d) None

Ans: 2 5 = 2 5 = 5

= 16 + 5(4) – 1 – 5 = 36 – 6 = 30 31. If f(x) = ax xn then find ƒ’(x) Dec-2008Ans:(c)

(a) f(x) [a + log a] (b) f(x)

a

x

alog (c) f(x)

a

x

alog (d) f(x)[a + x log a]

Ans: So, f(x) = f 1(x) = log =

32. dxxx 1

15

Dec-2008Ans:(b)

(a) Cx

x

1

log5

5

(b) Cx

x

1

log5

15

5

(c) Cx

x

1

log3

15

5

(d) Cx

x

5

5 1log

3

1

Ans:

Let 1 5

= =

=

= log

= log

33. Find the value of dxxx

3

3

28 June-2009Ans:(c)

(a) 1 (b) – 1 (c) 0 (d) None of these

Ans: √8 Let 8 - 2x dx = 2z dz x dx = - z dz

= = 33= 3 3 27 27 = 0

34. If x3 y2 = (x – y)5 Find dx

dyat (1,2) June-2009Ans:a

(a) –7/9 (b) 7/9 (c) 9/7 (d) –9/7 Ans: Differentiating both sides we get

Or, 3 2 5 1

Or, 3 2 5 5

Or, 2 5 5 3

Or,


11.8

Or, ,

=

35. Evaluate dxcx x. June-2009Ans:(b)

(a) ex(x + 1) + c (b) ex(x – 1) + c (c) ex + c (d) x – ex + c Ans: Integrating by by parts we get = 1. [ talking x as first function] = 1

36. Find

dxx

x

32

3

1 June-2009Ans:(c)

(a) Cxx 1222 12/114/1 (b) Cxx

12/114/1 212

(c) Cxx 1222 12/114/1 (d) None of these

Ans: 1/4(x2 + 1)-2 – 1/2 (x2 + 1)-1 + c [Same as Q – 17]

37. dxx

x

1 Dec-2009 Ans:(b)

(a)

1

3

12 2

1xx (b)

1

3

12 2

1xx (c)

2

1

3

12 xx (d) None of these

Ans: √ √

= √ √

= / / = /

/

/

/ = 2 / 1

38. dxx

x

1

0 1

1 Dec-2009Ans:a

(a) 2 log 2 – 1 (b) 4 log 2 – 1 (c) 2 log 2 (d) None of these

Ans:

= =

=

1

0

1

0 1

11201

12

0

1dx

x

xdx

x

xx

1

0

1

0 1221

x

dxdx

0

11log2

0

121 xx

1log2log20121 2log221 12log2 Ans. a

39. x = 2 t + 5 and y = t2 – 5, then dx

dy =? Dec-2009Ans:a

(a) t (b) –1/t (c) 1/t (d) 0

Ans: x = 2t + 5 y = t2 – 5 = 2 = 2t /

/

40. x = at2 y = 2 at, dx

dy =? Dec-2009Ans:a

(a) 1/t (b) –1/t (c) t (d) None of above

Ans: x = at2 y = 2at = 2at 2a /

/ 1/t

41. Find the second derivative of 1 xy Dec-2009Ans:(b) (a) 1/2 (x + 1) –1/2 (b) –1/4(x + 1) –3/2 (c) 1/4 (x + 1) –1/2 (d) None of these


11.9

42. 1343 xx

dx Equal to June-2010 Ans:(b)

(a) cxx 23

23

134327

2 (b) cxx 2

32

31343

27

2

(c) cxx 23

23

13433

2 (d) None of these

Ans : √ √

= √ √

√ √ √

= √ √ = √3 4 √3 1

= 3 4 / 3 4 / = /

/ +

/

/ + c

= 3 4 / 3 1 /

Ans: Y = √ 1 Differentiating both sides we get √

Again differentiating we get √

= - /

= - ¼(x +1 )-3/2

43.

2

1 2 2x

xdx____________ June-2010Ans:a

(a) 2log (b) 3log (c) 2

1log (d)

3

1log

Ans:

Let x2 + 2 = z when x = 2 , z = 6 2xdx = dz when x = 1 , z = 3

Xdx = dz/2 = = log 63

= 1/2 [log 6 – log3] =1/2 log2 = log√2

44. If x2 + y2 = 4 then June-2010Ans:(b)

(a) 0122

2

2

dx

dy

dx

ydy (b) 01

2

2

2

dx

dy

dx

ydy

(c) 012

2

2

dx

dy

dx

ydy (d) 012

2

2

2

dx

dy

dx

ydy

Ans: x2 + y2 = 4 differentiating both sides we get 2x + 2 = 0

or, 2 = -2x or, = -x again differentiating we get

or, = -1 or, 1 0

45. The cost function for the production of x units of a commodity is given by C(x) = 2x3 – 15x2 + 36x +

15 The cost will be minimum when “x” is equal to Dec-2010 Ans:a (a) 3 (b) 2 (c) 1 (d) 4 Ans: c(x) = 2x3 – 15x2 + 36x +15 differentiating we get

c1(x) = 6x2 – 30x + 36 again differentiating we get c11(x) = 12x – 30 for minimum value c1(x) = 0 or, 6x2 – 30x + 36 = 0 or, x2 – 5x + 6 = 0 or, x2 –3x –2x + 6 =0 or, x(x–3) – 2(x-3) = 0 x = 2 or 2 putting x = 3 and x = 2 in c11(x) we get c11(3) = 36 – 30 = 6>0 c11(2) = 24 – 30 = –6<0 x is minimum at x = 3


11.10

46. dx3x2x

4x6

is equal to Dec-2010Ans:(c)

(a) 22 log(x – 3) – 16(x – 2) (b) 11 log(x – 3) – 8(x – 2) (c) 22 log(x – 3) – 16log(x – 2) (d) 22 log(x – 3) + 16 log(x – 2)

Ans:

=

= 3

Let x2 + 5x + 6 = z 2x + 5dx = dz = 3 11

= 3 log| 3 2 | 11

= 3 log | 3 2 |/

/

/

= 3 log | 3 2 | 11 log

= 3 log 3 2 11 log 2 11 log 3 = 3 3 3 log 2 11 log 2 11 log 3

= 14 log (x + 3) – 8 log (x + 2) + c

47.

dxxx 2log1

1is equal to Dec-2010Ans:(c)

(a)

cx

2log12

1 (b) c

x

log1

1 (c) c

x

log1

1 (d) None of these

Ans:

1 + log x = z

= = =

48. Solve : dx)ee(1

1

xx

June-2011 Ans. a

(a) 0 (b) 1 (c) 12 (d) None of the above

Ans: = = = [ ex + e-x] 11

= e1 + e–1 – e–1 – e1 = 0

49. Solve :

dx.x

xlog3

2x

June-2011 Ans. (b)

(a) C)x(log2

3 3 (b) C)x(log3

1 3 (c) C)x(log6

1 3 (d) C)x(log7

3 3

Ans: =

= Let log x = z

= =

50. If )1('then;C)x( 3

x ff ? June-2011. (b)

(a) 1/6 (b) -1/6 (c) 5/6 (d) -5/6

Ans: f(s) = = !

! ! =

!

= 2 = 3 3 2

y1(x) = 3 6 2 y1(1) = 3 1 6 1 2 = –1/6


11.11

51. Given, y = dxee alogxxloga ; then dx

dy June-2011Ans. (b)

(a) xa ax (b) xa ax (c) alogaax x1x (d) None of the above

Ans: y = = = Differentiating both side we get

= Caaax xa log1

52. If ______)x(and0)1(;x

2x3)x(

32 fff' June-2011Ans. (c)

(a) 2x3

x 23

(b) 2xx 23 (c) 2xx 23 (d) None of these

Ans: )(xf = 3x2 -

Integrating both side we get

f(x) = 3 2 = 3 = x3 + + c f(1) = 1 + +c

0 = 2 + c c = –2 f(x) = x3 + x–2 – 2

53.

1

1

dxx

x = _________ Dec-2011 Ans. (b)

a) -1 b) 0 c) 1 d) 2

Ans.

1

0

0

1

1

1

dxx

xdx

x

xdx

x

x

1

0

0

1

dxdx = 1 + 1 = 2

54.

2

log

2x

dx

d = _________ Dec-2011Ans. a

a) 1 b) 0 c) 1/2 d) xx 2log.2

Ans. 2

log

2x

dx

d

Formula

xa a

x

log

So, xx

2log

2 1)( xdx

d

55.

dx

x

edx

x

e xx

.12.

1 23 ___________ Dec-2011Ans. (c)

a) 0 b)

Cx

ex

212

c)

Cx

ex

212

d)

Cx

ex

212

Ans.

dxx

edx

x

e xx

23 12.

1 )()(4 xfxfe

)(xfe x (So Ans. c)

56. If 2

2

dx

ydthenxY x ______________ Dec-2011Ans. a


11.12

a) )log1()log1( xdx

dYx

dx

dY b) )log1()log1( x

dx

dx

dx

dY

c) )log1()log1( xdx

dYx

dx

dY d) )log1()log1( x

dx

dx

dx

dY

Ans. Y = xx Putting log both the side we get,

xxy loglog xxy loglog xdx

dxx

dx

dxy

dx

dloglog.log

xx

x

dx

dy

ylog.

1 xy

dx

dylog1 )log1[.)log1[

2

2

xdx

dyx

dx

dy

dx

yd

= xdx

dyx

dx

dylog1log1 (Ans.)

57. If ,25)( 2xxg then 1

)1()(1

x

gxgLimx

is equal to _______ June-2012 Ans. (b)

(a) 0 (b) 24/1 (c) 24 (d) None of these

Ans. 225)( xxg then, 01

)1()(1

x

gxgLimx

1

12525 2

1

x

xLimx

)1(

2425 2

1 x

xLimx 2425)1(

12

2

1

xx

xLimx 24

1

2425

12

x

x(Ans.)

58. If x = c t, y = c/t, then dx

dyis equal to: June-2012Ans. (c)

(a) 1/t (b) tt.e (c) 21/t (d) None of these

Ans. x = ct t

cy c

dt

dx 2 tc

dt

dy

c

tc

dtdx

dtdy

dx

dy 2)[

/

/

21

t

59.

1

0 2)]1([ xbax

dx ___________ June-2012Ans. (d)

(a) a/b (b) b/a (c) ab (d) 1/ab

Ans.

1

021 xbax

dx zxbax )1( dzdxba

ba

dzdx

a

b z

dz

ba 2

1

abbabab

a

ba

1111

2

11

60. If dx

dyeey axxa then,loglog June-2012Ans. (b)

(a) xa ax (b) aaxa xa log1 (c) 11 xa axxa (d) nx ax

Ans. axxa eey loglog xa ax ee loglog xa ax

.log1 aaaxdx

dy xa

61. 2 3 5 ___________________ Dec-2012 Ans. (b)

(a)

(b)

(c)

(d)

62. For the functions y = x3 -3x, the value of at which is zero, is Dec-2012Ans. (c)

(a) ±1 (b) ±3 (c) ±6 (d) None of these


11.13

Ans. xxy 33 033 2 xdx

dy 12 x 1x

xdx

yd6

2

2

= 6

63. The equation of tangent to the curve, f = x2 – 3x + 2, at the point (2,7) is ________ Dec-2012Ans.

(c) (a) y = 2x – 3 (b) y = 10x (c) 10x – 13 (d) y = 10 Ans. 232 xxf 32)( xxf 134)( 7,2 xf

1)(2

xfx

zy 27 xy 5 xy

Formula

)(1

1 xfxx

yy

64. If y =

, then = ___________ Dec-2012Ans. (c)

(a) (b) (4x2 – 5) – (3 + 5x2) (c) (d) 8x – 10

Ans.

2

2

53

45log

x

xy

22 53log5log xx 22 53

10

45

8

x

x

x

x

dx

dy

22 53

10

54

8

x

x

x

x

(Ans.)

65. If y = ,log xy then dx

dyis equal to: June-2013 Ans. (b)

a) yx log

1

b)

yxx log

1

c)

yx log1

1

d)

xy log

1

Ans. y

xy log

y

xy

log

log xyy loglog

xdx

dy

dx

dyy

dx

dy log.loglog.

xdx

dyy

dx

dy

yy

1.log.

1.

x

ydx

dy 1log1 yxdx

dy

log1

1

=

yxx log

1

(Ans.)

67. If x = log t, y = te , then dx

dy June-2013Ans. (b)

a) 1/t b) tet. c) 2/1 t d) None of these

Ans. teytx ,log tedt

dy

tdt

dx ,

1 te

dtdx

dtdy

dx

dy t /

/ (Ans.)

68. dxxxx 5.3.2 23 ___________ June-2013Ans. (b)

a) Cxxx

270log

5.3.2 23

b) Cxxx

360log

5.3.2 23

c) Cxxx

180log

5.3.2 23

d) Cxxx

90log

5.3.2 23

Ans. dxxxx 532 23 Let yxxx 532 23 Putting log both the side

yxxx log5log3log2log 23


11.14

69. The points on the curve y = ,123 xxx where the tangent is parallel to x – axis are Dec-2013 Ans. a

a)

27

32,

3

1 and (1, 0) b (0, 0) and (1, 0) (c) (1, 0) and (1, 1) (d) (0, 1) and (1, 1)

Ans. 123 xxxy 0123 2 xxdx

dy 00tan dx

dy 0133 2 xxx

0133 2 xxx 0113 xxx 0113 xx 3

11

xorx

0,10,1 yx

27

32,

3

1

27

32,

3

1yx

70. _______________ Dec-2013Ans. a

(a) a

a x

log2

2

(b) a

a x

log

.2 2

(c) 2

log.2 aa x

(d) None of these

Ans. Formula, ca

adxa

xx log. So c

a

adxa

xx log2

22 (Ans.)

72. If nxnx beaey , then 2

2

dx

yd is equal to ________. June-2014Ans. a

a) n2 y b) – n2y c) ny d) None of these

Ans. nxnx beaey nxnx bneanedx

dy nxnx ebneandx

yd 22

2

2

yn2

73. The value of definite integral dxx2

0

1 __________ Dec-2014Ans. (d)

a) 0 b) 1/2 c) 3/2 d) 1

Ans. dxx 2

0

1 dxxdxx 2

0

1

0

11 0

2

21

2

2

22

xx

xx

0

2

11

2

1122

.)(1

2

1

2

1Ans

75. The value of

2/1

0 23 x

dxis June-2015 Ans. (c)

a) 1 b) 2/31 c) 23 d) 32

Ans.

2/1

0 23 x

dx 3 – 2x = z - 2dx = dz

2

32

1

z

dz

2

3

12

12

11

2

1

z

)23(2

3 z

76. The value of dxxe x2

0

2

is June-2015Ans. (d)

a) 1 b) e – 1 c) (e/2) – 1 d) 12

1 4 e


11.15

Ans. dxex x2

0

2

zx 2 2x dx = dz dzez

2

02

1

0

4

2

1 ze 12

1 4 e (Ans.)

77. If ,qpqp yxyx then dx

dyis equal to _______ June-2015Ans. (c)

a) p

q b)

y

x c)

x

y d)

q

p

Ans. qpqp yxyx

Putting log both the side we get, qpqp yxyx loglog

)log(loglog yxqpyx qp )log()(loglog yxqpyqxp

yxdx

dqpy

dx

dqx

dx

pd log)(loglog

dx

dy

yx

qp

dx

dy

y

q

x

p1.

dx

dy

yx

qp

yx

qp

x

p

dx

dy

y

q..

x

p

yx

qp

dx

dy

yx

qp

dx

dy

y

q

.. )(

)()(.

yxx

yxpqpx

yx

qp

y

q

dx

dy

)()(

)()(

yxx

pypxqxxp

yxy

yqpyxq

dx

dy

qypyqyqx

yxy

yxx

pypxqxxp

dx

dy

)(

)(

x

y

qypyqxqx

pypxqxxp

x

y (Ans.)

78. If 44 xyexy then dx

dy ________ June-2015Ans. (b)

a) x

y b)

x

y c)

y

x d)

y

x

Ans. 44 xyexy 0.4. 4

y

dx

dyxy

dx

dyxexy

ydx

dyxye

dx

dyxe xyxy 4.4 yeyxxe

dx

dy xyxy 44.

xxe

yey

dx

dyxy

xy

4

4

x

y

e

e

x

yxy

xy

4

4.

Answer Key

1 b 2 c 3 a 4 b 5 b 6 c 7 b 8 d 9 a 10 c 11 b 12 c 13 a 14 c 15 a 16 b 17 b 18 a 19 a 20 c 21 c 22 b 23 c 24 a 25 b 26 b 27 c 28 b 29 a 30 c 31 c 32 b 33 c 34 a 35 b 36 c 37 b 38 a 39 a 40 a 41 b 42 b 43 a 44 b 45 a 46 c 47 c 48 a 49 b 50 b 51 b 52 c 53 b 54 a 55 c 56 a 57 b 58 c 59 d 60 b 61 b 62 c 63 c 64 c 65 b 66 a 67 b 68 b 69 a 70 a 71 b 72 a 73 d 74 b 75 c 76 d 77 c 78 b 79 a 80 81 s 82 c 83 b 84 a 85 a 86 c 87 b 88 a 89 b 90 a 91 b 92 a 93 a 94 c 95 b 96 c 97 c 98 a 99 a 100 c 101 102 103 104 105 106 107 108 109 d 110 b 111 A 112 a 113 a 114 b 115 a 116 c 117 b

Probability GOPAL BHOOT

15.1

CHAPTER PROBABILITY

Past Exam Questions Probability

1. There are six slips in a box and numbers 1, 1, 2, 2, 3, 3 are written on these slips. Two slips are taken

at random from the box. The expected values of the sum of numbers on the two slips is : Nov-2006Ans:(c) (a) 5 (b) 3 (c) 4 (d) 7 Ans. Cases 1, 1 2, 3 1, 2 2, 4 1, 3 3, 1 Total Cases = 9 2, 1 3, 2 2, 2 3, 3

X Sum Cases Favour Prob XP 2 (1, 1) 1/9 3 (1, 2), (2, 1) 2/9 4 (1, 3), (2, 2), (3, 1) 3/9 5 (2, 3), (3, 2) 2/9 6 (3, 3) 1/9 1 36/9 = 4

2. A letter is taken out at random from the word RANGE and another is taken out from the word PAGE. The probability that they are the same letters is : Nov-2006Ans:(b) (a) 1/20 (b) 3/20 (c) 3/5 (d) ¾ Ans. A, A or G, G or E, E

4

1

5

1 +

4

1

5

1 +

4

1

5

1 =

20

3

3. An urn contains 9 balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of same colour is: Nov-2006Ans:(c) (a) 3/27 (b) 20/31 (c) 5/84 (d) None Ans. P (all blue) + P (all black)

= 3

93

3

C

C +

39

34

C

C =

84

5

2 Red 3 Blue 4 Black

4. A card is drawn from a well shuffled pack of 52 cards. Let E1, "a king or a queen is drawn" & E2 : "a queen or a jack is drawn", then : Nov-2006Ans:(a) (a) E1 and E2 are not independent (b) E1 and E2 are mutually exclusive

(c) E1 and E2 are independent (d) None of these Ans. 1EP = King or queen (8 cards out of 52) = 8/52

2EP = queen or jack = 8/52 21 EEP = P (Common is queen) = 4/52

Here 2121 . EPEPEEP So, not independent

5. In a non - leap year, the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays is : Nov-2006 Ans:(c) (a) 4/7 (b) 2/7 (c) 3/7 (d) 1/7 Ans. Non Leap year 365 days (52 weeks + 1 extra day) it may be Monday, Tuesday or Wednesday or Thursday or Friday or Saturday or Sunday

P (extra is 53rd Sunday or Tuesday or Thursday) = 7

3

7

1

7

1

7

1


15.2

6. If A and B are two events and P(A) = , P(B) = , ∩ then the value of ∪ is :

Ans. BAP = BA1 De Morgan law = 1 - 434

1

7. The probability that there is at least one error in an account statement prepared by A is 0.3 and for B and C, they are 0.4 and 0.45 respectively. A, B and C prepared 20, 10 and 40 statements respectively. The expected number of correct statements in all is : Nov-2006 Ans:(c) (a) 32 (b) 45 (c) 42 (d) 25

Ans. error A = 0.30 B = 0.40 C = 0.45

No error 70.0A 60.0B 55.0C N = 20 Statements 10 Statements 40 Statements Expected Frequencies = N × probability. = 20 × 0.70 + 10 × 0.60 + 40 × 0.55 = 42.

8. From a pack of cards, two are drawn, the first being replaced before the second is drawn. The chance that the first is a diamond and the second is king is : May-2007 Ans:(a)

(a) 1/52 (b) 3/2704 (c) 4/13 (d) 3/52 Ans. P (Diamond in first, King m 2nd) independent events as With Replacement

= P(A) × P(B) = 52

1

52

4

52

13

9. The theory of compound probability states that for any two events A and B : May-2007Ans:(b)

(a) BPAPBAP (b) A/BPAPBAP

(c) A/BPAPBAP (d) BAPBPAPBAP Ans. ABPAPBAP )()( 10. The probability of getting qualified in IIT- JEE and AIEEE by a the student are respectively 1/5 and

3/5. The probability that the student gets qualified for one of the these tests is : May-2007Ans:(a) (a) 17/25 (b) 22/25 (c) 8/25 (d) 3/25

Ans. IIT AIEEE

Pass A = 5

1 Pass B =

5

3

P(Select atleast one test) = P(atleast 1) = 1-P(none)

= 1 - BAP = 1 - 25

17

5

2

5

4 (independent)

11. Amitabh plays a game of tossing a dice. If the number less than 3 appears, he is getting ` a, otherwise he has to pay ` 10. If the game is fair, find a : May-2007Ans:(b) (a) 25 (b) 20 (c) 22 (d) 18

Ans. Game is Fair Expected gain is 0 Case Gain (X) Probability (P) XP Number less than 3 (1, 2)

a

6

2

6

2a

Otherwise (3, 4, 5, 6)

- 10

6

4

6

40

6

402 a

Expected gain .06

402

a Solving a = 20.

12. Suppose E and F are two events of a random experiment. If the probability of occurrence of E is 1/5 and or probability of occurrence of F given E is 1/10, then the probability of nonoccurrence of at least one of the events E and F is : Aug-2007Ans:(d) (a) 1/50 (b) 1/25 (c) 13/50 (d) 49/50

Ans. 5

1EP

10

1

F

EP


15.3

Find PFEP (non occurrence of at least one)

10

1

E

FP

10

1

EP

FEP

10

1

51

FEP

So, P(FE) = ,50

1

Now FEPFEP 1 (By De Morgan’s Law) = 1 - 50

49

50

1

13. A bag contains 8 red and 5 white balls. Two successive draws of 3 balls are made without replacement. The probability that the first draw will produce 3 white balls and second 3 red balls is :

Aug-2007Ans:(c) (a) 6/255 (b) 5/548 (c) 7/429 (d) 3/233 Ans. 8 R (3 Balls) 3 balls and 3 ball

5 W (3 Ball) Without replacement = dependent events P(3 White 1st Draw 3 Red in 2nd draw)

=

BA

RWP 21 33 =

alreadywhite3assumingred33

33

1

21

W

RPWP

(3 White) 8 Red 8 Red (3 Red) 5 White 2 White

= 3

3

3

3

10

8

13

5

C

C

C

C = 429

7 (Ans.)

14. A box contains 12 electric lamps of which 5 are defectives. A man selects three lamps at random. What is the expected number of defective lamps in his selection ? Aug-2007Ans:(a) (a) 1.25 (b) 2.50 (c) 1.05 (d) 2.03 Ans. 5 defectives Expected no. of defectives

7 good E (x) No. of defectives (X) Probability

0 (0 defectives 3 good) 220

35

12

7

3

3 C

C


105

12

75

3

21

C

CC


70

12

75

3

12

C

CC

3 (3 defectives) 220

10

12

5

3

3 C

C

X P XP 0 35/220 1 105/220 2 70/220 3 10/220 xp = 1.25 Ans.

15. Three identical dice are rolled. The probability that the same number will appear on each of them is: Nov-2007Ans:(c)

(a) 1/6 (b) 1/12 (c) 1/36 (d) 1 Ans. P (1 , 1, 1) + P(2, 2, 2) + …… P(6, 6, 6)

3 throws are independent

=

6

1

6

1

6

1 +

6

1

6

1

6

1......

6

1

6

1

6

1


15.4

= 216

1

216

1

216

1

216

1

216

1

216

1

= 36

1

216

6

(a) 0.15 (b) 0.25 (c) 0.254 (d) 0.55 16. Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and

in at least one of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only : Nov-2007Ans:(a)

(a) 0.15 (b) 0.25 (c) 0.254 (d) 0.55 Ans. Fail

Stats 30% Math 35% atleast 1 = 45%

A = 0.30 B = 0.35 45.0) BAP

45.0)()()( ABPBPAP 0.30 + 0.35 – P(AB) = 0.45 P(AB) = 0.20 P (fail in Math only) = BAP i.e. he pass in stats = P(B) – P(AB) = 0.35 – 0.20 = 0.15 (Correct Answer) Alternatively

17. An article consists of two parts A and B. The manufacturing process of each part is such that

probability of defect in A is 0.08 and that B is 0.05. What is the probability that the assembled product will not have any defect? Nov-2007Ans:(d) (a) 0.934 (b) 0.864 (c) 0.85 (d) 0.874

Ans. (defective) A = 0.08 (defectives) B = 0.05 (good) A = 0.92 (good) B = 0.95 P (assembled part have no defect) = P (Both Parts good) = BAP = AP . BP (assume independent events) = 0.92 × 0.95 = 0.874

18. Daily demand for calculators is having the following probability distribution : Nov-2007Ans:(c) Demand : 1 2 3 4 5 6 Probability : 0.10 0.15 0.20 0.25 0.18 0.12

Determine the variance of the demand. (a) 2.54 (b) 2.93 (c) 2.22 (d) 2.19

Ans. X Probability XP PX 2

1 0.10 2 0.15 3 0.20 4 0.25 5 0.18 6 0.12 _____ ______ _____ ______


15.5

Variance = 2xE - 2xE

= px2 - 2xp = 2.22

19. If 10 men, among whom are A and B, : stand in a row, what is the probability that there will be exactly 3 men between A and B ? Feb-2008Ans:(d) (a) 11/15 (b) 4/15 (c) 1/15 (d) 2/15 Ans. Probability =

— — — — — — — — — — Total no of arrangements = n10 Cases favourable (m) three men between A & B

A — — — B — — — — — In 28 — A — — — B — — — — ” — — A — — — B — — — ” — — — A — — — B — — ” — — — — A — — — B — ” — — — — — A — — — B ” m = 286

Probability = 8910

286

10

286

=

15

2 Ans.

20. The probability of an event can assume any value between: Feb-2008Ans:(a) (a) 0 and 1 (b) -1 and 0 (c) -1 and 1 (d) None of these Ans. Probability lies between 0 and 1 Ans. (a)

21. The odds are 9 : 5 against a person who is 50 years living till he is 70 and 8 : 6 against a person who is 60 living till he is 80. Find the probability that at least one of them will be alive after 20 years: Feb-2008 Ans:(c) (a) 11/14 (b) 22/49 (c) 31/49 (d) 35/49

Ans. 50 year man live 20 more years odd against 9 : 5

14

5)( AP

60 year man live 20 more years odds against 8 : 6

14

6)( BP P(atleast one) = )( BAP

= P(A) + P(B) – P(AB) = 14

6

14

5

14

6

14

5 = .

49

31

22. An urn contains 6 white and 4 black balls. 3 balls are drawn without replacement. What is the expected number of black balls that will be obtained ? Feb-2008 Ans:(a) (a) 6/5 (b) 1/5 (c) 7/5 (d) 4/5 Ans. 6 White Expected (no of black balls)

4 Black (3 Ball drawn) = E(x) no of Black Balls Probability 0 Black (& 3 White)

120

20

10

6

3

3 C

C

1 Black (& 2 White)

120

60

10

64

3

21

C

CC

2 Black (& 1 White)

120

36

10

64

3

12

C

cC

3 Black (and no white)

120

4

10

4

3

3 C

C


15.6

No of Black (x) Probability XP 0

120

20

1

120

60

2

120

36

3

120

4

E(x) =

5

6 Ans.

23. If P(A) = p and P (B) = q, then : June-2008Ans:(c) (a) P (A / B) q / p (b) P (A / B) p / q (c) P ( A / B) p / q (d) P (A / B) q / p Ans. qBPpAP )(,)( We known, P(AB) P(A) Divide both sides by P(B)

)(

)(

BP

ABP

)(

)(

BP

AP

B

AP

q

p

24. The probability that a trainee will remain with a company is 0.8. The probability that an employee earns more than ` 20,000 per month is 0.4. The probability that an employee, who was a trainee and remained with the company or who earns more than ` 20,000 per month is 0.9.What is the probability that an employee earns more than ` 20,000 per month given that he is a trainee, who stayed with the company? June-2008Ans:(b) (a) 5/8 (b) 3/8 (c) 1/8 (d) 7/8 Ans .Given, P(Trainee remains in Co) = 0.80 = A P(earns more than 20000) = 0.40 = B.

Also given P(trainee remains in Co. or earns more than 20000) = 0.90 = BP Now, )()()()( ABPBPAPBAP 0.90 = 0.80 + 0.40 – P(AB) P(AB) = 0.30 Now, we find P (employee earns more than 2000 given he was trainee in the Co.)

= 8

3

80.0

30.0

)(

)(/

AP

ABPABP

25. A random variable X has the following probability distribution : A June-2008ns:(a) X : -2 3 1 P(X = x) : 1/3 1/2 1/6

Find E and E (2X + 5) (a) 6 and 7 respectively (b) 5 and 7 respectively (c) 7 and 5 respectively (d) 7 and 6 respectively


- 2

3

1

3

2

3

4

3

2

1

2

3

2

9

1

6

1

6

1

6

1

E(x) = 1 6 = 2xE


15.7

Now .62 xE Also E(2x + 5) = 2E(x) + E(5) [E (Constant) = Constant] = 2 × 1 + 5 = 7

26. The limiting relative frequency of probability is : Dec-2008Ans:(c) (a) Axiomatic (b) classical (c) statistical (d) Mathematical

Ans. Statistical Definition of probability says P(A) = n

fn lim

27. If a probability density function is 1 0 10

Dec-2008Ans:(c)

(a) ∞ (b) 0 (c) (d) - ∞

Ans. In Probability Density function

E(x) =

dxyprobabilitx )( =

dxxxf )(

When x lies between 0 & 1 otherwise f(x) = 1 f(x) = 0.

= 1

0

)( dxxfx + dxxfx )(

= 1

0

1 dxx + dxx 0

= 1

0

xdx + 0

= 12

2

x =

2

1

2

0

2

1 22

Ans. (Correct Answer)

28. If: x : -2 3 1 P (x) : 1/3 1/2 1/6 Then find E (2x + 5) Dec-2008Ans:(a) (a) 7 (b) 6 (c) 9 (d) 4


- 2

3

1

3

2

3

4

3

2

1

2

3

2

9

1

6

1

6

1

6

1

E(x) = 1 6 = 2xE

Now .62 xE Also E(2x + 5) = 2E(x) + E(5) [E (Constant0 = Constant] = 2 × 1 + 5 = 7

29. If A and B are two independent evens and P(AUB) = 2/5; P(B) = 1/3. Find P(A). June-2009Ans:(d) (a) 2/9 (b) -1/3 (c) 2/10 (d) 1/10

Ans. 3

1)(

5

2)( BPBAP

P(A) + P(B) – P(AB) = 5

2 P(A) + P(B) – P(A) . P(B) =

5

2

P(A) + 5

2

3

1)(

3

1 AP Solving P(A) =

10

1

30. A bag contains 12 balls of which 3 are red 5 balls are drawn at random. Find the probability that in 5 balls 3 are red. June-2009Ans:(d)


15.8

(a) 3/132 (b) 5/396 (c) 1/36 (d) 1/22 Ans. 3 Red 5 drawn

9 other colours

P(3 red and 2 other) = 22

1

512

29

33

C

CC

31. A random variable X has the following probability distribution. June-2009Ans:(d) X 0 1 2 3

P(x) 0 2K 3K K Then, P (x < 3) would be: (a) 1/6 (b) 1/3 (c) 2/3 (d) 5/6

Ans. X Probability 0 0 We know Total Probability = 1 1 2 K 0 + 2K + 3K + K = 1

2 3 K K = 6

1

3 K P(X )3 = P(X = 0, 1, 2)

1 = 0 + 2K + 3K = 5K = 6

5.

32. P(A) = 2/3; P(B) = 3/5; P(AB) = 5/6. Find P (B/A) Dec-2009Ans:(b) (a) 11/20 (b) 13/20 (c) 13/18 (d) None

Ans. 3

2)( AP P(B) =

5

3

P )( BA = 6

5 P(A) + P(B) – P(AB) =

6

5

6

5)(

5

3

3

2 ABP

30

13)( ABPSolving Now )/( ABP =

20

13

32

3013

)(

)(

AP

BAPAns.

33. If P (AB) = P (A) × P (B), then the events are: Dec-2009Ans:(a) (a) Independent events (b) Mutually exclusive events (c) Exhaustive events (d) Mutually inclusive events

Ans. If P(A )B = P(A) × P(B) the events are independent 34. E (XY) is also known as: Dec-2009Ans:(b)

(a) E (X) + E (Y) (b) E (X) E (Y) (c) E (X) – E (Y) (d) E (X) E (Y) Ans. If x and y are independent E(xy) = E(x) . E(y)

35. In a pack of playing cards with two jokers probability of getting king of spade is June-2010Ans:(d) (a) 4/13 (b) 4/52 (c) 1/52 (d) 1/54

Ans.52 cards + 2 jokers = 54 cards P (King of Spade) = (Hukum ka Badshah) = 54

1

36. Consider two events A and B not mutually exclusive, such that P(A) = 1/4, P(B) = 2/5, ∪, is June-2010Ans:(c)

(a) 3/7 (b) 2/10 (c) 1/10 (d) None of the above

Ans. 2

1)(

5

2)(

4

1)( BAPBPAP

P(A) + P(B) – P(AB) = 2

1

2

1)(

5

2

4

1 ABP Solving, P(AB) =

20

3.

Now )( BAP = P(A) - P(AB) = 10

1

20

3

4

1 Ans.

37. If x be the sum of two numbers obtained when two die are thrown simultaneously then P(x 7) is June-2010 Ans:(b) (a) 5/12 (b) 7/12 (c) 11/15 (d) 3/8

Ans. Refer two Dice Rule.


15.9

Sum 7 8 9 10 11 12

Probability 636

+ 536

+ 436

+ 336

+ 236

+ 136

=

38. E (13x +9) = ___________. June-2010 Ans:(c) (a) 13x (b) 13E(x) (c) 13E(x) + 9 (d) 9 Ans. E(13x + 9) = 13E(x) + 9.

39. A dice is thrown once. What is the mathematical expectation of the number on the dice ? Dec-2010Ans:(c) (a) 16/6 (b) 13/2 (c) 3.5 (d) 4.5

Ans. Points Probability (P) XP 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6

1 3.5 Ans. 40. If P(A/B) = P(A), then A and B are Dec-2010Ans:(c)

(a) Mutually exclusive events (b) Dependent events (c) Independent events (d) Composite events

Ans. P(A/B) = P(A) )()(

)(AP

BP

ABP P(AB) = P(A) – P(B) independent events.

41. A bag contains 3 white and 5 black balls and second bag contains 4 white and 2 black balls. If one ball is taken from each bag, the probability that both the balls are white is________. Dec-2010Ans:(b) (a) 1/3 (b) 1/4 (c) 1/2 (d) None of these

Ans. 3 White 4 White 5 Black 2 Black independent events

P(white in 1st White in 2nd) = 41

6

4

8

3 Ans.

42. The odds in favour of A solving a problem is 5;7and Odds against B solving the same problem is 9:6. What is the probability that if both of them try, the problem will be solved ? Dec-2010Ans:(a) (a)117/180 (b) 181/200 (c) 147/180 (d) 119/180

Ans. odds in favour 5 : 7 A = 12

7,

12

5A

odds against 9 : 6 B = 15

9,

15

6B

P(Problem solved) = P(atleast one solved)

= 1 – P(none) = 1 - )( BAP = 1 - 15

9

12

7 (independent) =

180

117

43. Consider Urn 1 : 2 white balls, 3 black balls. Urn II: 4 white balls, 6 black balls .One ball is randomly transferred from first to second Urn, then one ball is drawn from II Urn. The probability that drawn ball is white is Dec-2010Ans:(c) (a) 22/65 (b) 22/46 (c) 22/55 (d) 21/45 Ans. Urn = Box 2 white 4 White 3 Black 6 Black White required Bhoot formula (2nd event depends on 1st) Let A = White transfer B = Black transfer


15.10

A = 5

2 B =

5

3

Let E means white from second bag AEP / P(White in 2nd assuming white in 1st) Bag 5 White

6 Black = .11

5

BEP / = P(White in 2nd assuming black in 1st) 4 White

7 Black = .11

4

Bhoot Formula P(E) = )/()()/()( BEPBPAEPAP

= .55

22

11

4

5

3

11

5

5

2

44. If ∪ , ∩ June-2011Ans. (d)

(a) P(A).P(B) (b) P(A) + P(B) (c) 0 (d) P(B) Ans. )(APBAP P(A) + P(B) – P(AB) = P(A)

P(B) – P(AB) = 0 So P(AB) = P(B) 45. In how many ways a team of 5 can be made out of 7 Boys and 8 Girls, if 2 Girls are compulsory to

form a Team. June-2011Ans. (c) (a) 2,646 (b) 1,722 (c) 2,702 (d) 980

Ans. Team of 5(2 girls compulsory) This is a question of combination 7 Boys 8 Girls

3 2 girls 2

83

7 CC

2 3 girls 3

82

7 CC

1 4 girls 4

81

7 CC 0 5 girls

58

07 CC

Total 2702 Ans.

46. A bag contains 5 Red balls, 4 Blue balls and ‘m’ Green Balls. If the random probability of picking two green balls is 1/7. What is the no. of green Balls (m). June-2011Ans. (c) (a) 5 (b) 7 (c) 6 (d) None of the above Ans.

5 red P(2 green) = 4 blue

m green

Total m + 9 1

29 82

17

19 8

17

Solving, m = 6. (by trial & error or quadratic equation) 47. The probability of Girl getting scholarship is 0.6 and the same probability for Boy is 0.8. Find the

probability that at least one of the categories getting scholarship. June-2011Ans. (c) (a) 0.32 (b) 0.44 (c) 0.92 (d) None of the above. Ans. Girl = 0.6 , Boy = 0.8

7

1


15.11

A = 0.6 B = 0.8

P(atleast 1) = 1 – P(none) = 1 – = 1 – 0.4 × 0.2 = 0.92 Ans. 48. If 15 persons are to be seated around 2 round tables, one occupying 8 persons and another 7 which

they can be seated. June-2011Ans. (d)

(a) !

! (b)

!

! (c) 7!. 8! (d) 2. 6! 7!

Ans. This is a question of combination & permutation First 8 person in 1st Round table, balance 7 in 2nd table

(a) or Vice Versa means First 7 in 1st Round table, balance 8 in 2nd table

(b) Required ways = a + b =

=

= Ans. 49. A coin is tossed 5 times, what is the probability that exactly 3 heads will occur. June-2011Ans. (a)

(a) 5/16 (b) 1/32 (c) 5/36 (d) 3/32 Ans. Binomial Problem

n = 5 throws prob of head = prob of tail =

P(3 heads = f(3) = =

50. Two unbiased dice are thrown. The Excepted value of the sum of numbers on the upper side is; Dec-2011Ans. (b) a) 3.5 b) 7 c) 12 d) 6

Ans. For one dice, expected points 3.5 For 2 dice, expected points = 7. 51. One card is drawn from a pack of 52, what is the probability it is a king or queen Dec-2011Ans.(b)

a) 11/13 b) 2/13 c) 1/13 d) None of these

Ans. P(King or queen) = P(K) + P(Q) – P(K∩ 0

52. In a packet of 500 pens, 50 are found to be defective. A pen is selected at random. Find the probability that it is non defective. Dec-2011Ans. (c) a) 8/9 b) 7/8 c) 9/10 d) 2/3

Ans.

defect 50 Prob =

non 450

53. Four married couples have gathered in a room. Two persons are selected at random amongst them, find the probability that selected persons are a gentleman and a lady but not a couple. Dec-2011Ans.(b) a) 1/7 b) 3/7 c) 1/8 d) 3/8 Ans. 4 gents P(gents ∩ lady) + P(lady ∩ gents) 4 ladies = P(gents) P(lady/gent)

+ P(lady). P(gents/lady)

2.0B

4.0A

)( BAP

177181578

CC

188)17(1587

CC

716156171578

CC

6715671515157778

CCCC equaland

671527

C

2

1p

2

1q

2323

2

1

2

155

33

CC qp

16

5

4

1

8

110


15.12

54. Let A and B two events in a sample space S such that P(A) = ; , ∪ Find

∪ June-2012Ans. (b) (a) 3/4 (b) 1/4 (c) 3/16 (d) None of these.

Ans. ∩ 1 ∪ 1

55. A card is drawn out of a standard pack of 52 cards. What is the probability of drawing a king or red colour ? June-2012Ans.(c) (a) 1/4 (b) 4/13 (c) 7/13 (d) 1/2

Ans. P(King or red) = P(King) + P(red) – P(King ∩red)

56. A player tosses two fair coins, he wins ` 5 if 2 heads appear, ` 2 if one head appears and ` 1 if no head occurs. Find his expected amount of winning. June-2012Ans. (a) (a) 2.5 (b) 3.5 (c) 4.5 (d) 5.5 Ans. HH, HT, TH, TT Case gain = x Prob. xp 2 heads 5 1/4 5/4 1 heads 2 2/4 4/4 no head 1 1/4 1/4 2.5

57. Aruan & Taruan appear for an interview for two vacancies. The probability of Aruan’s selection is 1/3 and that of Tarun’s selection is 1/5 Find the probability that only one of them will be selected. June-2012Ans. (a) (a) 2/5 (b) 4/5 (c) 6/5 (d) 8/5

Ans. ( independent)

= P(A). =

58. A company employed 7 CA’s, 6 MBA’s and 3 Engineer’s. In how many ways the company can form a committee, if the committee has two members of each type. June-2012Ans. (d) (a) 900 (b) 1,000 (c) 787 (d) 945

Ans. This is a combination question

59. Two dice are thrown together. Find the probability of getting a multiple of 2 on one dice and multiple of 3 on the other. Dec-2012Ans. (b)

(a) 2/3 (b) 1/6 (c) 1/3 (d) None of the above

Ans. P(2, 4, 6) × P(3, 6) (independent)

60. The odds against A solving a certain problem are 4 to 3 and the odds in favour of B solving the same

problem are 7 to 5. Dec-2012Ans. (b) What is the probability that the problem will be solved if they both try? (a) 15/21 (b) 16/21 (c) 17/21 (d) 13/21

Ans. A odds against 4:3 P(A) = B odds against 7:3 P(B) =

P(problem solved) = P(atleast 1 solved = 1- P(none) = 1 –

= 1- 1 (independent events) = 64/84 = 16/21

61. Find the expected value of the following probability distribution Dec-2012Ans. (b)

x: -20 -10 30 75 80 p(x): 3/20 1/5 1/2 1/10 1/20

(a) 20.5 (b) 21.5 (c) 22.5 (d) 24.5 Ans. x p xp

)( BAPBAP

)().()( BPAPBP 5/25

1

3

2

5

4

3

1

9453152123

26

27 CCC

BAP


15.13

-20 3/20 -3 -10 1/2 -2 30 1/2 15 75 1/10 7.5 80 1/20 4 21.5

62. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball form the bag

is twice that of a red ball, find the number of blue balls in the bag Dec-2012Ans. (b) (a) 10 (b) 12 (c) 14 (d) 16

Ans. Red 6 Prob (Blue) =

Blue x Prob (Red) =

Prob (Blue) = 2 × Prob (Red)

63. The probability of selecting a sample of size ‘n’ out of a population of size N by simple random

sampling with replacement is: June-2013Ans. (b)

a) 1/N b) 1/Nn c) n

N C

1 d)

!

1

nCnN

Ans. With Replacement no of samples Nn Probability = nN

1

64. A box contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue? June-2013Ans.(a) a) 10/21 b) 11/21 c) 2/7 d) 5/7

Ans. 2 red 3 green 2 balls 2 blue

P(none ball blue) = 21

10

27

25

C

C

65. The odds that a book will be favourable received by 3 independent reviewers are 5 to 2, 4 to 3 and 3

to 4 respectively. What is the probability that out of 3 reviewers a majority will be favourable? June-2013Ans. (a) a) 209/343 b) 209/434 c) 209/443 d) 209/350

Ans. Odds in favour A 5 : 2 B 4 : 3 C 3 : 4 P(A) = 5/7 P(B) = 4/7 P(C) = 3/7

P(majority favours) = P(2 favours) or P(all 3 favours)

=

66. A player tosses 3 fair coins. He wins ` 5 if three heads appear, ` 3 if two heads appear, ` 1 if one head occurs. On the other hand, he losses ` 15 if 3 tails occur. Find expected gain of the player: June-2013Ans.(b) a) 0.15 b) 0.25 c) 0.35 d) 0.45

7/2)( AP 7/3)( BP 7/4)( CP

ABCPBCAPCBAPCABP

343

209

7

3

7

4

7

5

7

3

7

4

7

2

7

3

7

3

7

5

7

4

7

4

7

5


15.14

Ans. Binomial n = 3, p = (head), q = (tail)

Cases Prob x(gain) xp 3 head f(3) = 1/8 5 5/8 2 head f(2) = 3/8 3 9/8 1 head f(1) = 3/8 1 3/8 0 head f (0) = 1/8 -15 -15/8 2/8 = 0.25

67. Find the probability of drawing a spade on each of two consecutive draws from a well shuffled pack of cards, without replacement June-2013Ans.(b) a) 2/51 b) 3/51 c) 4/51 d) 5/51

Ans. Without Replacement P(spade in 1st ∩ spade in 2nd) = .

68. If P(A) = 0.45, P(B) = 0.35 and P(A&B) = 0.25, then P(A/B) =? Dec-2013Ans. (c) (a)1.4 (b)1.8 (c) 0.714 (d) 0.556

Ans. P(A) = 0.45 P(B) = 0.35 P(A∩B) = 0.25 P / ∩ = .

.0.714 (Ans.)

69. The probability of a cricket team winning match at Kanpur is 2/5 and losing match at Delhi is 1/7 what is the Probability of the term winning atleast one match? Dec-2013Ans.(b) (a) 3/35 (b) 32/35 (c) 18/35 (d) 17/35

Ans. P (Cricket team winning at Kanpur) = 5

2 P(Cricket team losing at Delhi) =

7

1

Let P(A) = 5

2 P(B) =

7

6

5

3AP

7

1)( BP

P(winning atleast 1 match) = 1 – BA = 1 - 7

1

5

3 =

35

31 =

35

32

70. Find the expected value of the following probability distribution. Dec-2013Ans. (b)

X: -20 -10 30 75 80 P(x): 3/20 1/5 ½ 1/10 ½0

(a) 20.5 (b) 21.5 (c) 22.5 (d) 24.5 Ans. X P(X) XP -20 3/20 -3 -10 1/5 -2 30 ½ 15 75 1/10 15/2 80 1/20 4_________ ∑ 21.5

71. Two coins are tossed simultaneously. Find the probability of getting exactly one head Dec-2013Ans. (d) (a) 3/4 (b) 2/3 (c) 1/4 (d) 1/2 Ans. HH, TH, TT, HT

P(getting exactly 1 head) = 2

1

4

2

72. If a pair of dice is thrown then the probability that the sum of the digit is neither 7 nor 11 is _______.

June-2014 Ans. (d) a) 1/6 b) 1/18 c) 2/9 d) 7/9

Ans.P(that is either 7 or 11) = P(7) + P(11) – P(7∩11) 036

2

36

6

36

8


15.15

P(neither 7 or 11) = 1 36

8

9

7

18

14

36

28(Ans.)

73. An urn contains 2 red and 1 green balls. Another urn contains 2 red and 2 green balls. An urn was

selected at random and then a ball was drawn from it. If it was found to be red then the probability that it has been drawn from urn one is ______. June-2014Ans. (c) a) 4/7 b) 3/7 c) 2/3 d) 7/12

Ans. 2 Red 2 Red

1 Green 2 Green

2

1)( AP

2

1)( BP P(that selected ball is red from urn 1) =

3

2(Ans.)

74. For any two events A1 ,A2 let , , ∩ , then A1, A2 are June-2014

Ans. (c) a) Mutually Exclusive but not independent events b) Mutually Exclusive and independent events c) Independent but not Mutually Exclusive d) None of these

Ans. 8

3

3

221 APAP

4

121 AAP Two Event A & B are independent when,

BPAPBAP 2121 APAPAAP 8

3

3

2

4

1

4

1

So, it is Independent but Not Mutually Exclusive. (Ans.)

75. An unbiased die is thrown twice. The probability of the sum of numbers obtained on the sum of numbers obtained on the two faces being divisible by 4 is: Dec - 2014Ans. (b) a) 7/36 b) 1/3 c) 11/36 d) 1/4

Ans. Sample Space :-

D(divisible by 4) = 36

9 =

4

1(Ans.)

77. A discrete random variable X takes three values -1, 2 and 3 with probabilities Dec - 2014Ans. (c)

,3

1)3(,

3

1)2(,

3

11 ppp then XE is :

a) 3/2 b) 5/2 c) 2 d) 9/2 Ans. | | Prob XP

1 13 1

3

2 13 2

3

3 13 1

1

3

2

3

1XP

1 2 33

2 .

78. The sum of numbers obtained in throw of a dice twice is S. Probability of S will be maximum is S is

June - 2015 Ans.(b) a) 5 b) 7 c) 6 d) 8

Ans. The probability of S will be maximum in throw of a dice twice = .36

6 When the sum is 7

So, S = 7


15.16

79. An unbiased coin is tossed three times. The expected value of the number of heads is June-2015 Ans. (c) a) 2.5 b) 1.0 c) 1.5 d) 2.0 Ans. HHH, TTT, HTT, THH, HTH THT, HHT, TTH

X [Getting head] PROB xp 3 heads 1/ 3

8 2 heads 3

8 68

1 heads 38 3

8

128

1.5 .

80. For two events A1 and A2, let P(A1) = 2/3 and P(A2) = 3/8 and 21 AAP = 1/4, then June-2015

Ans. (c) a) Mutually exclusive but not independent b) Mutually exclusive and independent c) Independent but not mutually exclusive d) None of these

Ans. 8

3

3

221 APAP

4

121 AAP Two Events A & B are independent when,

)()( BPAPBAP )( 2121 APAPAAP

4

1

8

3

3

2

4

1 So, it is Independent but not Mutually exclusive

83. A bag contains 15 one rupee coins, 25 two rupees coins and 10 five rupees coins, if a coin is selected at random than probability for not selecting a one rupee coin is : Dec - 2015Ans. (d) a) 0.30 b) 0.20 c) 0.25 d) 0.70 Solution : One Rupee Coins Two Rupee Coins Five Rupee Coins

15 25 10 Probability of selecting all coins = 1 Probability of selecting one rupee

Coins 1 0.30

None = 1 - All = 1-.30 = 0.70 84. Three coins are together, the probability of getting exactly two head is : Dec - 2015Ans. (b)

a) b) c) d) None

Solution : No. of exactly two heads in three coins = 3 HHT HTH THH Total cases = 2 × 2 × 2 = 8

∴ 2

85. If two letters are taken at random from the word “HOME”, what is the probability that none of the letters would be vowels? Dec - 2015 Ans. (a)

a) b) c) d)

Solution :

Total Cases = 6

No. of cases in which none would be vowel = 1.


15.17

∴ .

86. In a game, cards are thoroughly shuffled and distributed equally among four players. What is the probability that a specific player gets all the four kings ? June - 2016Ans. (b)

a) 13

53

1348

413

C

CC b)

1352

948

44

C

CC c)

1352

454

413

C

CC d)

132

939

44

C

CC

Solution : Total distributive of cards = 13

52C

If a specific person gets all four kings then the probability = 44C

Remaining = 948C

∴ Ans. 13

524

49

48

C

CC

87. A bag contains 4 Red and 5 Black balls. Another bag contains 5 Red and 3 Black balls. If one ball is drawn at random from each bag. Then the probability that one Red and one Black drawn is - :June - 2016Ans. (c)

a) b) c) d)

Solution : In one bag ______ 4R 5B

In other bag _____ 5R 3B

If one red is drawn from one bag, and one blue is drawn from other bag, then

If one blue is drawn from one bag, and one red is drawn from other bag, then

∴ 12 2572

3772

88. If

1B

APthen

6

5B)P(Aand

5

3P(B),

3

2P(A) is June - 2016Ans. (a)

a) b) c) d)

Solution :

We know ∩

∴′

∩ ′

91. If two events A, b P(A) = 2

1, P(B) =

3

1 and

3

2)( BA then )( BAP is : Dec-2016Ans. (b)

a) 1/4 b) 1/6 c) 2/3 d) 1/2 Solution :

∪ ∩

⇒23

12

13

∩

⇒23

3 26

∩

⇒23

56

∩


15.18

⇒ ∩56

23

∴ ∩ .

92. A bag contains 6 white and 5 red balls. One ball is drawn. The probability that is drawn. The probability that it is red is : Dec-2016Ans. (a) a) 5/11 b) 6/11 c) 1/11 d) None of these Solution : The probability that a red ball is drawn = 5/11.

93. For two events, A, B let P(A) = 3

2, P(B) =

8

3 and

4

1)( BAp then A and B are : Dec-2016Ans.

(b) a) Mutually exclusive but not independent b) Independent but not mutually exclusive c) Mutually exclusive and independent d) None of these Solution : It is not mutually exclusive because ∩ 0. It is independent because.

∩

94. Let A and B are two events P(A) = , P (B) = and ∩ , then P(B/A) will be: June-2017

(Ans: c) a) 7/8 b) 1/3 c) 1/8 d) 8/7 Solution :

∩ /

⇒112

23 /

∴ / .

95. What is the probability of having at least one 'SIX' from 3 throws of an unbiased die? June-2017 (Ans: d)

a) b) c) 1 d) 1

Solution : xnx

xn qPCXP ..)0(

0300

3 .. qPC

3

6

5.1.1

3

6

5

Probability of having at least

3

6

511

Answer Key

1 c 2 b 3 c 4 a 5 c 6 b 7 c 8 a 9 b 10 a 11 b 12 d 13 c 14 a 15 c 16 a 17 d 18 c 19 d 20 a 21 c 22 a 23 c 24 b 25 a 26 c 27 c 28 a 29 d 30 d 31 d 32 b 33 a 34 b 35 d 36 c 37 b 38 c 39 c 40 c 41 b 42 a 43 c 44 d 45 c 46 c 47 c 48 d 49 a 50 b 51 b 52 c 53 b 54 b 55 c 56 a 57 a 58 d 59 b 60 b 61 b 62 b 63 b 64 a 65 a 66 b 67 b 68 c 69 b 70 b 71 d 72 d 73 c 74 c 75 b 76 d 77 c 78 b 79 c 80 c 81 c 82 a 83 d 84 b 85 a 86 b 87 c 88 a 89 d 90 a 91 b 92 a 93 b 94 c 95 d 96 b 97 d 98 a 99 a 100 a 101 102 103 104 105 106 107 108 a 109 a 110 b 111 c 112 a 113 c 114 a 115 b 116 d 117 c

Probability (Theoretical) Distribution GOPAL BHOOT

16.1

CHAPTER PROBABILITY (THEORETICAL) DISTRIBUTION

Past Exam Questions Theoretical Distributions 1. Parameter is a characteristic of: Nov - 2006 Ans:(a)

(a) Population (b) Sample (c) Probability distribution (d) Both (a) & (b)

Ans:- Population 2. What is the probability of making 3 correct guesses in 5 True - False answer type questions? Nov-

2006 Ans:(c) (a) 0.4156 (b) 0.32 (c) 0.3125 (d) 0.5235

Ans: All answers are independent prob of success is constant Apply Binomial

n = 5 P = prob of success (correct Answer) =

q = P(3 correct) = f(3) = ( = 0.3125

3. The I.Q.'s of army volunteers in a given year are normally distributed with Mean = 110 and Standard Deviation = 10. The army wants to give advance training to 20% of those recruits with the highest scores. What is the lowest I.Q score acceptable for the advanced training? The value of Z for the area 0.3 = 0.84. Nov - 2006Ans:(b) (a) 0.84 (b) 118.4 (c) 138.4 (d) 115.4

Ans: Normal distribution

Mean =110 SD = 10 z =

Top 20% students will be given advanced training

For top 20% students minimum z = 0.84 = 0.84

For top 20% students minimum x = 118.4 So, minimum score for top 20% students is 118.4

4. The number of calls arriving at an internal switch board of an office is 96 per hour. Find the probability that there will be : Nov - 2006Ans:(d)

(i) not more than 3 calls on the board, (ii) at least three calls in a minute on the board. [Given: = 0.2019] (a) 0.08 and 0.92 respectively (b) 0.19 and 0.92 respectively

(c) 0.92 and 0.13 respectively (d) 0.92 & 0.22 respectively

Ans: Average 96 calls per hour = 1.6 calls per minute

m = 1.6 passion f(x) = .

! =

. . .

!

P (not more than 3 calls)

= f(0) + f(1) + f(2) + f(3) = . . .

! +

. . .

! +

. . .

! +

. . .

!

= e-1.6 [ 1 + 1.60 + 1.28 + 0.683] = 0.92 P(atleast 3 calls) = 1 + [f(0) + f(1) + f(2)] = 1. e-1.6 [1 + 1.60 +1.28] = 0.22

5. For a normal distribution with mean 150 and S.D. 45; find Q1 and Q3: Nov - 2006Ans:(b) (a) 119.35 and 190.65 respective (b) 119.65 and 180.35 respective (c) 180.35 and 119.65 respective (d) 123.45 and 183.65 respectively

Ans: Mean 150 SD 45 Q1 = mean – 0.675 SD 150 – 0.675 × 45 = 119.625 Q3 = mean + 0.675 SD 150 + 0.675 × 45 = 180.375

6. The probability density function of a normal variable x is given by : Nov - 2006Ans:(c)

6.1e


16.2

(a) (b)

(c) (d) None of these.

Ans: Option (c) is correct

7. The Interval covers : May - 2007Ans:(d) (a) 95% area of normal distribution (b) 96% area of normal distribution (c) 99% area of normal distribution (d) All but 0.27% area of a normal distribution

Ans:(d) Interval mean - 3 SD and mean + 3 SD covers all except 0.27% area of normal distribution 8. The overall percentage of failure in a certain examination is 0.30. What is the probability that out of

a group of 6 candidates at least 4 passed the examination ? May - 2007Ans:(a) (a) 0.74 (b) 0.71 (c) 0.59 (d) 0.67

Ans: Fail = 0.30 Pass = 0.70 n = 6 p = 0.70 q = 0.30 f(x) = px qn-x = (0.70)x(0.30)6-x

P(atleast 4 passed) = f(4) +f(5) + f(6) = (0.70)4 (0.30)2 + (0.70)5 (0.30)1 + (0.70)6 (0.30)0 = 0.74

9. A manufacturer, who produces medicine bottles, finds that 0.1% of the bottles are defective. The bottles are packed in boxes containing 500 bottles. A drug manufacturer buys 100 boxes from the producer of bottles. Using Poisson distribution, find how many boxes will contains at least two defectives: May - 2007Ans:(c)

[Given: e-0.5 = 0.6065] (a) 7 (b) 13 (c) 9 (d) 11

Ans: n = 500 p = probability of defective = 0.1% = 0.001 m = np = 500 0.001 = 0.5

f(x) = .

! =

. . .

!

p(atleast 2 defective)

= 1 – [ f(0) + f(1)] = 1 –[ . . .

! +

. . .

! ] = 1 - . (1 + 0.5) = 0.09025

Expected no of boxes = N prob = 100 0.09025 = 9 boxes 10. The number of methods of fitting the normal curve is : Aug - 2007Ans:(c)

(a) 4 (b) 3 (c) 2 (d) 1 Ans:(c) Learn 2 method of fitting normal curve 11. Suppose that weather records show that on an average 5 out of 31 days in October are rainy days.

Assuming a binomial distribution with each day of October as an independent trial, then the probability that the next October will have at most three rainy days is : Aug - 2007Ans:(b)

(a) 0.4403 (b) 0.2403 (c) 0.3403 (d) None

Ans: Prob of raining October = p Prob of no rain

n = 31 f(x) = (

p(at most 3 rainy day) = f(0) + f(1) + f(2) + f(3)

= ( + ( + ( + (

= [ + + .

+ . .

. .]

12. If 5% of the families in Kolkata do not use gas as a fuel, what will be the probability of selecting 10 families in a random sample of 100 families who do not use gas as fuel ? Aug - 2007Ans:(d)

[Given : e -5 = 0.0067] (a) 0.038 (b) 0.028 (c) 0.048 (d) 0.018

Ans: n is very large and p is very small n = 100 p = 5% = 0.05 m = np = 100 0.05 = 5 conversion from Binomial to poisson

x0fore.2

1)x(f

22

2)x(

xfore.2

1)x(f

22

2)x(

xfore.

2

1)x(f

2x

2

1

)3,3(

331C


16.3

f(x) = .

! f(10) =

.

! =

.

. . . . . . . . . = 0.018

13. If the 1st quartile and Mean Deviation about median of a normal distribution are 13.25 and 8 respectively, then the mode of the distribution is : Aug - 2007Ans:(a)

(a) 20 (b) 10 (c) 15 (d) 23 Ans: 1st Quartile = 13.25

MD about median = 8 = 0.80 SD SD = 10 1st Quartile = mean – 0.6755D = 13.25 Mean – 0.675 10 = 13.25 Mean = 20 Mean = median = mode = 20

14. If 15 dates are selected at random, then the probability of getting two Sundays is: Nov - 2007Ans:(a) (a) 0.29 (b) 0.99 (c) 0.49 (d) 0.39

Ans: Sunday = = p Not a Sunday = = q n = 15

p(2 Sunday) = f(2) = 215C p2 q13 = 0.29

15. If X is a Poisson variate with P(X = 0) = P(X = 1), then P(X = 2) =: Nov - 2007Ans:(c) (a) 1/6e (b) e/6 (c) 1/2e (d) e/3 Ans: Poisson P(x = 0) = P(x = 1) f(0) = f(1)

. =

.

m = 1 now, f(2)

= .

= .

=

16. A sample of 100 dry battery cells tested to find the length of life produced the following results:

What percentage of battery cells are expected to have life less than 6 hours? Nov - 2007Ans:(a)

[Area under the normal curve from z = 0 to z = 2 is 0.4772] (a) 2.28% (b) 2.56% (c) 4.56% (d) 1.93%

Ans: Mean = 12 SD = 3 z =

p(x <6) = p( ) = p( z < -2)

= 0.5 – 0.4772 =0.0228 =2.28% 17. The method usually applied for fitting a binomial distribution is known as : Nov - 2007Ans:(c)

(a) Method of probability distribution (b) Method of deviations (c) Method of moments (d) Method of least squares.

Ans: Method of moments used for fitting Binomial to a Frequency Distribution 18. If X follows a normal distribution with 50 and 10 What is the value of 60/ 50 :

[Area under the normal curve from z = 0 to z = 1 is 0.3413]. Nov - 2007Ans:(a)

(a) 0.6826 (b) 0.7354 (c) 0.1983 (d) 0.5492

Ans: Mean = 50 SD = 10 and z =

P(x 60/ 50 = ∩

P(A/B) = =

=

= .

. = 0.6826

19. In a certain manufacturing process, 5% of the tools produced turn out to be defective. Find the probability that in a sample of 40 tools, at most 2 will be defective : Nov - 2007Ans:(d)

[Given : e-2 = 0.135] (a) 0.555 (b) 0.932 (c) 0.785 (d) 0.675

Ans: n = 40 p = 0.05 f(x) =.

n = np = 40 × 0.05 = 2 use Poisson p (at must 2) = p (×≤ 2)= .

! .

! .

!

= 0.675 20. Examine the validity of the following :

.3,12 hourshoursx


16.4

Mean and standard Deviation of a binomial distribution are 10 and 4 respectively.Nov - 2007Ans:(a) (a) Not valid (b) Valid (c) Both (a) & (b) (d) Neither (a) nor (b)

Ans: Mean = 10 SD = 4 Variance = 16 But in Binomial, variance is less than mean. So, the given statement was incorrect.

21. An experiment succeeds twice as often as it fails. What is the probability that in next five trials there will be at least three successes ? June - 2008Ans:(c)

(a) 33/81 (b) 46/81 (c) 64/81 (d) 25/81

Ans: [p:q =2:1]

Prob of success p = Prob of fail q =

N = 5 Binomial f(x) = . . = . .

P(atleast 3 success) = f(3) + f(4) + f(5) = 5 5 5

=

22. The probability than a man aged 45 years will die within a year is 0.012. What is the probability that of 10 men, at least 9 will reach their 46th birthday? June - 2008Ans:(b)

[Given : e-0.12 = 0.88692] (a) 0.0935 (b) 0.9934 (c) 0.9335 (d) 0.9555 Ans: Prob of death (p) = 0.012 Probability is very small

n = 10 m= np = 10 × 0.012 = 0.12 at least 9 men are alive atmost 1 die = f(0) + f(1)

= . .

!

. .

!= . 1 0.12 = 0.88692 × 1.12 = 0.9934

23. For a certain normal variate X, the mean is 12 and S.D. is 4. Find P (X > 20): [Area under the normal curve from z = 0 to z = 2 is 0.4772] June - 2008Ans:(d)

(a) 0.5238 (b) 0.0472 (c) 0.7272 (d) 0.0228 Ans: Mean = 12 SD = 4

P( × ≥20) = P = P(z ≥ 2) = 0.5 – 0.4772 =0.0228

24. In Poisson Distribution, probability of success is very close to : June - 2008Ans:(b) (a) -1 (b) 0 (c) 1 (d) None

Ans: p = Probability of success is very small (close to 0.) 25. If x and y are two independent standard normal variables, then the distribution of x/y is:Dec-2008 Ans:(a)

(a) Normal Distribution (b) Exponential Distribution (c) Couchy's Distribution (d) Binomial Distribution Ans: x and y are standard normal variables and x/y follows normal distribution 26. If X and Y are two independent random variables such that X~λ2

m and Y~ λ2n, then the distribution

of (x + y) is Dec - 2008 Ans:(d) (a) normal (b) standard normal (c) T (d) Chi-square

Ans: X~ λ2n Y~ λ2

n

X follows chi-square Y follows chi-square Independent So, (X+Y) also follows chi-square.

27. If the mean of a Poisson variable X is 1, what is P (x = at least one)? Dec - 2008 Ans:(c) (a) 0.456 (b) 0.821 (c) 0.632 (d) 0.254

Ans: Poisson mean=1

f(x) = .

! .

!

!

P(at least 1) = 1 – f(0) = 1 ! = 1 – 0.368 = 0.632

28. What is the probability of getting 3 heads if 6 unbiased coins are tossed simultaneously? Dec-2008Ans:(a) (a) 0.3125 (b) 0.25 (c) 0.6875 (d) 0.50

Ans: Prob of head = p = q = n = 6


16.5

f(x) = f(x) = = 0.3125

29. In a Poisson distribution P (x = 0) = P (X = 2). Find E (x). June - 2009Ans:(a) (a) (b) 2 (c) -1 (d) 0

Ans: f(0) = f(2) . .

1 = m = √ = Expectation of x

30. Shape of Normal Distribution Curve: Dec-2009 Ans:(b) (a) Depends on its parameters (b) Does not depend on its parameters

(c) Either (a) or (b) (d) Neither (a) nor (b) Ans. (b) Shape of normal distribution curve does not depend on its parameters. It is always bell

shaped. 31. For binomial distribution E(x) = 2, V (x) = 4/3. Find the value of n. Dec-2009Ans:(d)

(a) 3 (b) 4 (c) 5 (d) 6

Ans: Mean = 2 Variance =

np = 2 npq =

n 2q =

n = 6 q = p =

32. What are the parameters of binomial distribution? Dec-2009Ans:(c) (a) n (b) p (c) Both n and p (d) None of these

Ans : (c) n and p are parameters 33. The Variance of standard normal distribution is June-2010 Ans:(a)

(a) 1 (b) µ (c) (d) 0 Ans: (a) Variance of normal dist = Variance of standard normal = 1 34. For a Poisson distribution P (x = 3) = 5 P(x = 5), then S.D. is June-2010Ans:(d)

(a) 4 (b) 2 (c) 16 (d) Ans: P(× = 3) = 5.P(× = 5) f(3) = 5 f(5)

.

!. .

!

m2 = 4 m = 2

SD =√ √ 35. For a Binomial distribution B (6, p), P(x = 2) = 9p(x = 4), then P is June-2010Ans:(d)

(a) 1/2 (b)1/3 (c) 10/13 (d) ¼ Ans: B(6,p) n = 6 prob of success = p

P(× = 2) = 9P(× = 4) f(2) = 9. f(4) . . . . . p q 9. p 9 Taking square rout q =3p

p:q= 1:3 p = q =

36. In Binomial distribution n = 9 and P = 1/3, what is the value of variance: June-2010Ans:(c) (a) 8 (b) 4 (c) 2 (d) 16

Ans: Variance = npq = 9 = 2

37. If standard deviation of a poisson distribution is 2, then its Dec-2010Ans:(c) (a) Mode is 2 (b) Mode is 4 (c) Modes are 3 and 4 (d) Modes are 4 and 5

Ans: SD = 2 Variance = 22= 4 = m Modes are m and m-1 = m is an integer = Modes 4 and 4 – 1 = 4 & 3

38. The area under the Normal curve is Dec-2010Ans:(a) (a) 1 (b) 0 (c) 0.5 (d) -1

Ans:(a) Area under normal curve is 1 39. For a normal distribution is equal to Dec-2010Ans:(a)

(a) 0.9973 (b) 0.9546 (c) 0.9899 (d) 0.9788 Ans:- P( 3 ∪ 3 Mean-3SD & Mean+3SD COVERS 99.73% AREA (0.9973) 40. If for a Binomial distribution B (n. p,)the mean = 6 and Variance = 2 then "p" is Dec-2010Ans:(a)

2

2

2

3x3P,,N 2


16.6

(a) 2/3 (b) 1/3 (c) 3/5 (d) ¼ Ans: Mean 6 variance is 2

np = 6 npq = 2

6q = 2 q = p =

41. If the inflexion points of a Normal Distribution are 6 and 14. Find its Standard Deviation ? June - 201Ans. (a) (a) 4 (b) 6 (c) 10 (d) 12.

Ans: Mean - SD = 6 Mean + SD = 14

-2SD = - 8 SD = 4 42. In a Binomial Distribution, if mean is k-times the variance, then the value of ‘k’ will be _____. June

- 2011 Ans.(d)

(a) p (b) 1/p (c) 1 – p (d)

Ans:Mean = k. Variance

np = k.npq 1 = k.q k =

43. If x~N (3,36) and y~N (5,64) are two independent Normal variate with their standard parameters of distribution, then if (x + y) ~ N (8,A) also follows normal distribution. The value of A will be ______. June - 2011Ans. (a) (a) 100 (b) 10 (c) 64 (d) 36

Ans: x~N( ) x~N( ) x~N(3, 36) x~N(5, 64) Mean = 3 Mean 5 Variance = 36 (independent) Variance 64 (x + y) follows normal with Mean = 3+5 = 8 Variance of (x +y) = Variance of x + variance of y + 2 cov (x, y) = 36 + 64 + 2×0 = 100 [ For independent variable covariance = 0]

44. The mean of Binominal distribution is 20 and Standard deviation is 4 then; Dec-2011 Ans. (a) a) n = 100, p = 1/5, q = 4/5 b) n = 50, p = 2/5, q = 2/5 c) n = 100, p = 2/5, q = 4/5 d) n = 100, p = 1/5, q = 3/5

Ans: Mean = 20 SD= 4 np=20 npq =4

n× 20 npq = 16

n =100 20q =16 q = p =

45. A Company has two cars which it hires out during the day. The number of Cars demanded in a day has poisson distribution with mean 1.5. Then percentage of days on which only one car was in demand is equal to Dec-2011Ans. (b) a) 23.26 b) 33.47 c) 44.62 d) 46.40

Ans: Mean = 1.5 = m

Poisson f(x) = .

! .

.

= . . .

= /

P(one car in demand) = /

= f(1) = . . .

= √

= 0.2232

= . = 0.2232 46. The binominal distribution with mean 3 & variance 2 is : Dec-2011Ans. (c)

a) b) c) d)

Ans: In binomial all probabilities are found by expansion of (q + p)n ( ineed q, p, n)

9

7

1

7

2

n 9

6

1

6

2

n 9

3

1

3

2

n 9

5

1

5

2

n


16.7

Mean = 3 variance = 2

np = 3 npq = 2

n = 3 3q = 2

n = 9 q = p =

n = 9

expansion (q + p)n = ( 9 option is correct

48. If x is a Poisson variate and E(x) = 1, then P(x > 1) is June-2012Ans. (b)

(a) (b) (c) (d)

Ans: E(x) = 1 = m P(× >1) = 1 – f(0)

=1 - . 1

49. The mean and the variance of a random variable X having the probability density function

is. June-2012Ans. (a)

(a) 4, (b) 4,√

(c) 2, 2 (d) 2,

Ans: f(x) =

√ f(x) =

√

√

= √

Squaring . 2 =

= ½ 4

Mean = 4 variance = Parameter 4, ½

50. In a Normal Distribution Dec-2012 Ans. (c) (a) The first and second quartile are equidistant from median (b) The second and third quartiles are equidistant from the median (c) The first and third quartiles are equidistant from the median (d) None of the above Ans. : (c) as normal distribution is symmetric 51. If parameters of a binomial distribution are n and p then, this distribution tends to a Poisson

distribution when Dec-2012Ans. (d) (a) n→ ∞, p → 0 (b) p →0,np = (c) n → ∞, np = (d) n → ∞, p → 0, np = Ans. : (d) n → ∞, p → 0, np = 52. If a random variable x follows Poisson distribution such that E(x) = 30, the variance of the

distribution is Dec-2012Ans. (c) (a) 7 (b) 5 (c) 30 (d) 20 Ans. :Poisson E(x) = 30 = n Variance = m = 30 53. In a normal distribution quartile deviation is 6, the standard deviation will be Dec-2012Ans. (b) (a) 4 (b) 9 (c) 7.5 (d) 6 Ans. : Quartile Deviation = 0.675 SD = 6 SD = 8.89 (approx 9) 54. The mode of the Binomial Distribution for which the mean is 4 and variance 3 is equal to? June-

2013 Ans. (a) a) 4 b) 4.25 c) 4.5 d) 4.1

Ans. :Mean = 4 variance = 3 np = 4 npq = 3 4q = 3

Q = ¾ P =

4 n = 16

Mode = m = (n+1) p = (16+1) = 4.25 Not an integer Mode is largest integer = 4

21

1

e 11 e 121 e e

2

51

xxxXP ,/})4(exp{)( 2


16.8

55. For Poisson Distribution: June-2013Ans. (b) a) Mean and Standard Deviations are equal b) Mean and variance are equal c) Standard Deviation and variance are equal d) Both (a) and (b) are correct

Ans. :(b)In Poisson, Mean = Variance 56. Which of the following is not a characteristic of a normal probability distribution?June-2013Ans. (b)

a) Mean of the normally distributed population lies at the centre of its normal curve. b) It is multi-modal c) The mean, median and mode are equal d) It is a symmetric curve

Ans. : (b) because only 1 mode = mean 57. An approximate relation between quartile deviation (QD) and standard deviation (S.D.) of normal

distribution is: June-2013 Ans. (d) a) 5 QD = 4 SD b) 4 QD = 5 SD c) 2 QD = 3 SD d) 3 QD = 2 SD

Ans. : QD = 0.675 SD = approximately SD 3QD = 2SD

58. In a binomial Distribution with 5 independent trials, probability of 2 and 3 successes are 0.4362 and 0.2181 respectively. Parameter ‘p’ of the binomial distribution is: June-2013Ans. (b) a) 3/4 b) 1/3 c) 2/3 d) ¼

Ans. : n = 5 f(2) = 0.4362 f(3) = 0.2181 5C2 p

2 q3 =0.4362 5C3 p3 q2 =0.2181

2181.0

4362.0

)3(

)2(23

35

322

5

qpC

qpC

f

f

1

2

p

q 1:2: pq

3

1

3

2 pq

59. In a certain Poisson frequency distribution, the probability corresponding to two successes is half the probability corresponding to three successes. The mean of the distribution is Dec-2013 Ans. (a) (a) 6 (b) 12 (c) 3 (d) 2.45

Ans. f(2) = )3(2

1F

.23

)(

2

1

1)2(

)( 32

meme mm

2

3

6m

m M = 6 (Ans.)

60. Mean & Variance of a binomial variance are 4 and respectively then 1 will be ____.

June-2014 Ans. (a)

a) b) c) d) None of the above.

Ans. 4,3

4 npnpq

4 × q = 3

4 n =

2

34

q = 3

1 p = 6,

3

1,

3

2 nq

0600 qPCXP on =

60

06

3

1

3

2

C =

6

3

1

729

728

729

1729

3

110

6

XP

61. 5,000 students were appeared in an examination. The mean of marks was 39.5 with Standard Deviation 12.5 marks. Assuming the distribution to be normal, find the number of students recorded more than 60% marks June-2014Ans. (c)

[Given: When Z = 1.6, Area of normal curve = 0.4494]


16.9

a) 1,000 b) 505 c) 253 d) 2,227 Ans. Mean = 39.5 Standard deviation = 12.5

64.15.12

5.396060

zpZPXP = 0.5 – 0.4494 = 0.0509

Expected frequency = 5000 × 0.0506 = 253 63. If six coins are tossed simultaneously. The probability of obtaining exactly two heads are : Dec-2014

Ans. (c)

a) b) c) d) None of these

Ans. n = 6 P = 2

1

2

1Q

P(X = 2) P(X = 2) xnxx

n QPC

42

26

2

1

2

1

C

44

1

4

1

!4!2

!6

44

1

4

1

!4!2

!4563

.)(64

15

416

15Ans

Answer Key

1 a 2 c 3 b 4 d 5 b 6 c 7 d 8 a 9 c 10 c 11 b 12 d 13 a 14 a 15 c 16 a 17 c 18 a 19 d 20 a 21 c 22 b 23 d 24 b 25 a 26 d 27 c 28 a 29 a 30 b 31 d 32 c 33 a 34 d 35 d 36 c 37 c 38 a 39 a 40 a 41 a 42 d 43 a 44 a 45 b 46 c 47 a 48 b 49 a 50 c 51 d 52 c 53 b 54 a 55 b 56 b 57 d 58 b 59 a 60 a 61 c 62 a 63 c 64 a 65 c 66 c 67 c 68 c 69 c 70 c 71 a 72 c 73 c 74 c 75 c 76 c 77 c 78 a 79 b 80 c 81 c 82 c 83 a 84 b 85 a 86 c 87 88 89 90 91 92 93 94 a 95 d 96 b 97 c 98 d 99 c 100 c

Ratio & Proportion 1.1 GOPAL BHOOT CHAPTER RATIO AND … · 2019-09-28 · Ratio & Proportion 1.1...

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Transcript of Ratio & Proportion 1.1 GOPAL BHOOT CHAPTER RATIO AND … · 2019-09-28 · Ratio & Proportion 1.1...