CAT 2009 - ebook

CAT 2009 50 Practice Questions

1. The total number of integer pairs (x, y) satisfying the equation x + y = xy is

0

1

2

3

None of these

2. A pond is inhabited by only four types of fishes, namely A, B, C and D. There are 25

percent more fishes of type A than type C, 10 percent fewer fishes of type B than type A,

and 20 percent of the fishes are of type D. If there are 144 fishes of type B, then how

many fishes are there in the pond?

407

540

532

601

600

3. If log32, log3(2x – 5), log3(2

x – 7/2) are in arithmetic progression, then the value of x is

equal to

5

4

2

3

6

4. For how many values of a natural number ‘n’, the LCM of 66, 88 and ‘n’ is 1212?

1

6

24

25

36

5. Consider the following two curves in the x-y plane y = x2 – 6x + 8 and y = –x2 + bx + c. If

the maxima of one curve is the minima of the other curve, then what is the value of ‘b’?

6

5

3

8

7

6. A hoarding is fitted above a building. The height of the building is 12 m. When I look at the

lights fitted on top of the hoarding from a nearby bus stop, the angle of elevation is 50° and

when I look at the top of the building (terrace) from the same bus stop, the angle of elevation

is 45°. If the height of a flat on each floor is equal to the height of the hoarding, what is the

maximum number of floors in the building? (tan 50° = 1.1917).

2

4

5

7

None of these

7. The remainder, when (1523+ 2323) is divided by 19, is

4

15

0

18

19

8. Two cyclists are travelling at a speed of 3 km/hr and 10 km/hr around a circular track.

What should be the ratio of number of distinct points that they would meet on that track if

they travel in the same direction first and in the opposite direction next?

7 : 13

3 : 10

3 : 7

30 : 13

Cannot be determined

9. Given that w, x, y and z are natural numbers such that 4w = xyz, 4x = wyz and 4z = wxy.

Which of the following is a possible value of (w + x + y – z)2?

25

9

36

1

49

10. At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he

added p% of the goats at the beginning of the year and sold q% of the goats at the end of the

year where p>0 and q>0. If Shepard had nine dozen goats at the end of year 2002, after

making the sales for that year, which of the following is true?

p = q

p < q

p > q

p = q/2

None of the above

11. N persons stand on the circumference of a circle at distinct points. Each possible pair of

persons, not standing next to each other, sings a two-minute song one pair after the other. If

the total time taken for singing is 28 minutes, what is N?

5

6

7

8

9

12. An equilateral triangle BP is drawn inside a square ABCD. What is the value of the angle

APD in degrees?

75

90

120

135

150

13. What is the ratio of the length of PQ to that of QO?

1:4

1:3

3:8

3:4

None of these

14. For which of the following values of ‘p’, (x + p)(x + 10) + q = (x + m)(x + n) + r for all

values of x, where m, n are integers and q and r consecutive integers with q > r?

30

12

24

8

Both (2) and (4)

15. Which one of the following conditions must p, q and r satisfy so that the following system

of linear simultaneous equations has at least one solution, such that

p + q + r = 0?

x + 2y – 3z = p

2x + 6y – 11z = q

x – 2y + 7z = r

5p – 2q – r = 0

5p + 2q + r = 0

5p + 2q – r = 0

5p – 2q + r = 0

None of the above

16. Let g(x) = max(5 – x, x + 2). The smallest possible value of g(x) is

4.0

4.5

1.5

2.0

None of the above

17. Find the number of positive integral solutions of the equation (xy)z = 64.

14

15

16

17

18

18. How many three-digit numbers are there such that no two adjacent digits of the number

are consecutive?

592

516

552

600

596

19. If the length of the largest straight rod that can be put inside a cuboid is 10 m, then the

surface area of the cuboid cannot be more than

100 m2

200 m2

400 m2

600 m2


20. Consider 4 (dimensionless) flies, 2 males and 2 females. They are situated at the four

corners of a square of side 10 meters, the two males at diagonally opposite ends. At any time,

each fly tries to reach the male-female fly in front of her-him using the shortest possible

route. Since the flies are flying towards another, they will meet each other at a certain time at

the center of the square. What is the length of the path that each has traveled at the moment

they reach each other?

12.5 mts

10 mts

7.5 mts

5 mts

None of these

21. ‘S’ denotes the sum to infinity and ‘Sn’ denotes the sum to ‘n’ terms of the series 1 + 4/5

+ 16/25 + ...+ If S - Sn < 5/2.

then the least value of ‘n’ is

2

3

4

5

6

22. If x + r = 1, p + 1 = n, r + n = k, r = 8 and x + p + k = 30, the value of k is

17

22

23

15

11

23. A picnic invites two kinds of charges: bus fare, which is independent of the number of

people attending the picnic and buffet lunch, which increases directly with an increase in the

number of people. The charges are calculated to be Rs. 165 per head when there are 200

invitees and Rs. 170 per head when there are 150 invitees. What would be the charges per

head when there are 100 invitees?

Rs.175

Rs.180

Rs.185

Rs.190

Rs.195

24. A combo pack having a bulb and a tubelight costs Rs. 52. If the cost of the bulb drops by

20% and the cost of the tubelight escalates by 50%, the pack would cost Rs. 50. Find the

price of a tubelight?

Rs. 8

Rs. 9

Rs. 10

Rs. 12

Rs. 15

25. A test has 50 questions. A student scores 1 mark for a correct answer, -1/3 for a wrong

answer, and - 1/6 for not attempting a question. If the net score of a student is 32, the

number of questions answered wrongly by that student cannot be less than:

6

12

3.3

9

None of these

26. If f(x) = 2x2 + (ab2 + ac2 – 2abc)x + abc and the minimum value of f(x) is at x = –54,

then the what is the minimum value of (a + 2b – 2c)? (Given that a > b > c)

18

10

12

20

– 20

27. The function f (x) = | x - 2 | + | 2.5 - x| + |3.6 -x|, where x is a real number, attains a

minimum at:

x = 2.3

x = 2.5

x = 2.7

x = 4.1

None of these

28. Three positive real numbers ‘x’, ‘y’ and ‘z’ exist such that they are in an arithmetic

progression and the product of x, y and z is 25. If the common difference of the arithmetic

progression is 2 sqrt(5), then find the value of (x + y + z).

10 + 2sqrt(5)

15

8 + 4sqrt(5)

20

10

29. Given that the cost price of 10 oranges is equal to the cost price of 1 kg of apples and the

cost price of 12 apples is equal to the cost price of 1 kg of oranges. If the selling price of 15

oranges is equal to the selling price of 1 kg of apples, then the selling price of 1 kg of oranges

is equal to selling price of (Assume that all the apples are identical and this holds true for the

oranges as well.)

8 apples

9 apples

10 apples

12 apples

9 apples

30. On Madagascar Island, there are x dodos in a particular year. 30 dodos in a thousand of

the original population die every year and 25 dodos in a thousand of the original population

are born every year. In how many years will the population of dodos halve itself ?

200

125

100

50

None of these

31. A cube is divided into 8 equal cubes. Each of these cubes is further sub-divided into 8

equal cubes. If the original cubes sides are painted blue, then what is the probability that

exactly 2 sides of a small cube is painted blue ?

3/8

1/16

1/4

3/4

None of these

32. A man buys shares at a discount of Rs.x. Later he sells all but 10 of the shares he

purchased at a premium of Rs.x. If his investment was Rs.4500 and proceeds from the sale

were Rs.6250, how many shares did he buy originally ? [Assume face value of shares as

Rs.100.]

50

40

60

90

None of these

33. The diameter of the smaller circle is equal to the side of the square and the diagonal of the

square is equal to the diameter of the bigger circle. If the circles are concentric, then their

areas are in the ratio.

1:2

2:3

1:sqrt(2)

1:4

None of these

34. A positive whole number M less than 100 is represented in base 2 notation, base 3

notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in

exactly two out of the three cases the leading digit is 1. Then M equals

31

63

75

91

87

35. How many even integers n, where 100 < n < 200, are divisible neither by seven nor by

nine?

40

37

39

38

41

36. Let g(x) = max(5 – x, x + 2). The smallest possible value of g(x) is

4.0

4.5

1.5

2.0

None of the above

37. How many three digit positive integers, with digits x, y and z in the hundred’s, ten’s and

unit’s place respectively, exist such that x < y, z < y and x > 0?

245

285

240

320

280

38. The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, f, f, f, f, f, f…. is

u

v

w

x

z

39. The number of non-negative real roots of 2x – x – 1 = 0 equals

0

1

2

3

4

40. In a certain examination paper, there are n questions. For j = 1, 2 …n, there are 2n-

j students who answered j or more questions wrongly. If the total number of wrong answers is

4095, then the value of n is

12

11

10

9

8

41. Let T be the set of integers 3, 11, 19, 27,…451, 459, 467 and S be a subset of T such that

the sum of no two elements of S is 470. The maximum possible number of elements in S is

32

28

29

30

27

42. There are 6 boxes numbered 1, 2, …6. Each box is to be filled up either with a red or a

green ball in such a way that at least 1 box contains a green ball and the boxes containing

green balls are consecutively numbered. The total number of ways in which this can be done is

5

21

33

60

55

43. Consider the set P = –5, –3, –1, 1, 3, 5…. consisting of 1998 numbers. If ‘a’ be the

average of the elements in P and ‘b’ be twice the average of the first 1998 natural numbers,

then which of the following is equal to (a – b)?

–6

5

–8

–7

–9

44. If a0 = 2, a1 = 1 and a n + 1 × a n – 1 = a n – 1 – a n + 1, then find the value of a1000.

1/1001

1/2002

1/2

2 /1001

3/1003

45. If the length of the largest straight rod that can be put inside a cuboid is 10 m, then the

surface area of the cuboid cannot be more than

100 m2

200 m2

400 m2

600 m2


46. A function f(n) is defined for all positive real values of ‘n’ such that f(n + 2) = f(n + 1) +

f(n).

If f(1) = f(2) = 1, then find the highest common factor of f (12) and f (30)

f(11)

f(4)

f(6)

f(3)

f(5

47. There were ten friends four of whom are Sanjay, Salim, Reena and Teena. A team of six is

to be formed such that Reena & Teena are never together but Sanjay and Salim are always

together. How many teams can be made?

110

68

77

76

108

48. A point P is randomly chosen at a distance of 5 cm from the center of a circle. A chord AB

is drawn passing through P. C is a point on the circumference of the circle such that AC = BC.

If the radius of the circle is 13 cm, then which of the following cannot be a value of

area(∆ABC)?

96 cm2

216 cm2

169 cm2

84 cm2

Both (2) and (3)

49. Raju went to a shop to buy a certain number of pens and pencils. Raju calculated the

amount payable to the shopkeeper and offered that amount to him. Raju was surprised when

the shopkeeper returned him Rs. 24 as balance. When he came back home, he realized that

the shopkeeper had actually transposed the number of pens with the number of pencils. Which

of the following is certainly an invalid statement?

The number of pencils that Raju wanted to buy was 8 more than the number of pens

The number of pens that Raju wanted to buy was 6 less than the number of pencils.

A pen cost Rs.4 more than a pencil.

A pencil cost Rs.12 less than a pen.

None of the above

50. ABCD is a square of side 1 unit. P is a point on AB such that AP:PB = 1:3. If PC and BD

intersect at a point X inside the square, then find the length of line segment PX.

2/3 units

15/28 units

11/21 units

4/7 units

None of these

Solutions 1. Answer: 3

Feedback:

Given equation is x+y=xy, so only two pairs satisfies this and those are (0,0) and (2,2).

Hence answer option is (3).

2. Answer: 2

Feedback:

Since there are 10 percent fewer B than A,

A :B = 10 : 9

Number of fishes of type 144*10/9 = 160

Since there are 25 percent more A than C,

A : C = 5 : 4

or

Number of fishes of type C=160×4/5 =128

Since there are (160 + 144 + 128) = 432 fishes which are A,

B, or C type, and these account for 80 percent of fishes in the

pond, the pond must have 432 × 100 = 540 fishes.

80

Hence option (2) is the correct choice

3. Answer: 4

Feedback:

Using log a – log b = log a/b, 2 / (y–5) = (y–5)/(y –3.5) where y = 2x

Solving we get y = 4 or 8 i.e. x = 2 or 3. It cannot be 2 as log of negative number is not

defined (see the second expression).

4. Answer: 4

Feedback:

Given LCM of 66, 88 and ‘n’ is 1212

⇒ LCM of (26 × 36, 224 and ‘n’) is 224 × 312

So, ‘n’ can take the following values

20 × 312, 21 × 312, 22 × 312, …, 224 × 312.

So, the number of values that ‘n’ can take is 25.

5. Answer: 1

Feedback:

6. Answer: 3

Feedback:

7. Answer: 3

Feedback:

We can rewrite the numerator as (-4)23 + 423. Hence we get remainder = 0.

Hence answer option is (3).

8. Answer: 1

Feedback:

If they travel in the same direction they will met at 10 – 3 = 7 distinct points.

If they travel in opposite direction they will meet at 10 + 3 = 13 distinct points.

Hence the ratio = 7 : 13

9. Answer: 1

Feedback:

10. Answer: 3

Feedback:

The number of goats remain the same.

If the percentage that is added every time is equal to the percentage that is sold,

then there should be a net decrease. The same will be the case if the percentage added is less

than the percentage sold.

The only way, the number of goats will remain the same is if p > q.

11. Answer: 3

Feedback:

Total songs are 14. Two people can be selected in nC2 ways but there will be n pairs which are

adjacent .So answer will be => nC2 – n =14, on solving we will get n=7. Hence answer option

is 3.

12. Answer: 5

Feedback:

13. Answer: 2

Feedback:

Ratio of the length of PQ and OQ will be 1:3. Hence answer option (2).

14. Answer: 5

Feedback:

If (x + p) (x + 10) + q and (x + m) (x + n) + r are equal, then we

can write

(x + p) (x + 10) + q = (x + m)(x + n) + r

Þ (x + p)(x + 10) + 1 = (x + m)(x + n) .....(Since q – r = 1)

Þ x2 + (10 + p) x + 1 + 10 p = x2 + (m + n) x + mn

Þ m + n = (10 + p) and 1 + 10p = mn ...(ii)

Solving the above two equations, we get

Þ (m – 10)(n – 10) = 1 ...(i)

If m and n are both integers, then solution for (m, n) can only

be (11, 11) and (9, 9)

From (ii), we get the corresponding values of p as 12 and 8

Hence, option (5) is the correct choice

15. Answer: 1

Feedback:

It is given that p+ q+ r = 0, if we consider the first option, and multiply the first equation by

5,

second by –2 and third by –1, we see that the coefficients of x, y and z all add up-to zero.

Thus, 5p – 2q – r = 0

No other option satisfies this

16. Answer: 5

Feedback:

We can see that x + 2 is an increasing function and 5 – x is a decreasing function. This system

of equation will have smallest value at the point of intersection of the two. i.e. 5 – x = x + 2

or x = 1.5. Thus smallest value of g(x) = 3.5

17. Answer: 3

Feedback:

(xy)z = 26 = 43 = 82 = 641

When z = 1, xy = 64. The total number of factors of 64 is 7.

z = 2, xy = 8. The total number of factors of 8 is 4.

z = 3, xy = 4. The total number of factors of 4 is 3.

z = 6, xy = 2. the total number of factors of 2 is 2

Hence, total number of positive integral solutions is 16.

18. Answer: 4

Feedback:

Case : I

Digit at ten’s place is (2, 3, 4, 5, 6, 7 or 8)

If we keep 2 at ten’s place _2_

⇒ Hundred’s place can be filled in 7 ways (2, 4, 5, 6, 7, 7, 9)

⇒ Unit digit can be filled in 8 ways (0, 2, 4, 5, 6, 7, 8, 9)

Number of ways = 7 × 7 × 8 = 392 ways

Case: II

Digit at ten’s place is (1)

_1_

⇒ Hundred’s place can be filled in 8 ways (1, 3, 4, 5, 6, 7,

8, 9)

units place can be filled in 8 ways (1, 3, 4, 5, 6, 7, 8, 9)


Case: III

Digit at ten’s place is (0 or 9)

If we keep 9 at ten’s place _9_

⇒ Hundred’s place can be filled in 8 ways (1, 2, 3, 4, 5, 6,

7, 9)

⇒ unit’s digit can filled in 9 ways (0, 1, 2, 3, 4, 5, 6, 7, 9)


⇒ Required total number of ways = 392 + 64 + 144 = 600

19. Answer: 2

Feedback:

Let the length, breadth and the height of the cuboid be L, B

and H. We are given that L2 + B2 + H2 = 102 = 100.

Using A.M. ≥ G.M. property, we can write

⇒L2 + B2 + H2 ≥ (LB + BH + LH)

⇒ 2 × (L2 + B2 + H2 ) ≥ 2 × (LB + BH + LH)

⇒ Total surface Area ≤ 200.

20. Answer: 2

Feedback: Because all flies constantly fly perpendicular to another fly, they all travel the

shortest distance to each other, which is 10 meter. All flies make a kind of spiral flight to the

center of the square, and during this flight, the flies constantly form a square until they meet

in the center. The flies all travel 10 meter

21. Answer: 3

Feedback:

22. Answer: 3

Feedback:

Adding all the equations, we get

2(x + p) + 2r = 30.

x + p = 7

Also x + p + k = 30 ⇒ k = 23 Answer: (3)

23. Answer: 2

Feedback:

Let the bus fare = Rs. x

and buffet lunch per head = Rs. y

x + 200y = 165 × 200 …(i)

x + 150y = 170 × 150 …(ii)

Then, x + 100y = 2 × (ii) – (i) = 51000 – 33000 = 18000

∴Cost per head = Rs. 180

24. Answer: 4

Feedback:

Cost of a tubelight = Rs. t and cost of a bulb = Rs. b

t + b = 52

Cost of bulb drops by 20% and cost of tubelight increases by

50%

⇒ 1.5t + 0.8b = 50

Solving the two equations in ‘t’ and ‘b’,

t + b = 52

1.5t + 0.8b = 50

Multiplying first equation by 4 and second equation by 5,

4t + 4b = 208

7.5t + 4b = 250

Subtracting,

3.5t = 42

⇒ t = 12

Thus, cost of a tubelight = Rs. 12

25. Answer: 3

Feedback:

Let x be the number of questions answered correctly, y be wrong attempts and z be number of

questions not attempted. Then, net score 32 = x – y/3 – z/6 =32; 6x -2y-z = 192. Also, x + y

+ z = 50. Adding, we get 7x - y = 242, or y = 7x - 242. By hit and trial, we get y = 3 for x =

35. y cannot be less than 3 for integral values of x.

26. Answer: 1

Feedback:

f(x) = 2x2 + a(b – c)2 + abc.

As we know that for an equation ax2 + bx + c = 0,

the minimum or maximum value of f(x) lies at x = –b/2a

Therefore,–a(b – c) 2/4 = – 54 or a(b – c)2 = 216 ...(i)

Now, the sum a + (b – c) + (b – c) will be minimum when

a = b – c = 6

Hence, the minimum value of a + 2(b – c) = 6 + 6 + 6 = 18

Hence, option (1) is the correct choice.

27. Answer: 2

Feedback:

f(x) = |x - 2| + |2.5 - x| + |3.6 - x|. This can attain minimum value when either of the terms

= 0.

Case I: When |x - 2| = 0 then x = 2, the value of f(x) = 0 + 0.5 + 1.6 = 2.1

Case II: When |2.5-x| = 0 then x = 2.5; value of f(x) = 0.5 + 0 + 1.1= 1 .6.

Case III: When |3.6-x| = 0 then x=3.6; value of f(x) = 1.6 + 1.1 +0 = 2.7. Hence the

minimum value of f(x) is 1.6 at x = 2.5.

28. Answer: 2

Feedback:

Let, the values of ‘x’ and ‘z’ be denoted by (y - 2 sqrt(5)) and

(y + 2 sqrt(5))

25 = (y – 2 sqrt(5) ) (y) (y + 2 sqrt(5) )

⇒ y3 = 25 + 20y

y 5 ⇒ =

Therefore, x + y + z = 3y = 15

29. Answer: 1

Feedback:

30. Answer: 3

Feedback: Every year 5 in a thousand, i.e. 0.5% of the dodos are eliminated from the

population. In 100 years, 50% of the population will be eliminated.

31. Answer: 1

Feedback: Out of 64 cubes, 24 are painted 2 on the exactly two sides. Hence ratio = 3/8.

32. Answer: 3

Feedback: Let the man buy z shares. z(100 – x) = 4500 and (z – 10)(100 + x) = 6250.

Solving we get z = 60.

33. Answer: 1

Feedback: Area of smaller circle =pie a^2. Area of bigger circle = pie(asqrt(2))^2 =

2piea^2. Ratio of their areas = 1 : 2.

34. Answer: 4

Feedback:

Since the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by

either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of

numbers are 61, 91.

Among these only 31 and 91 are a part of the answer choices.

Among these, (31)10 ? (11111)2?(1011)3?(111)5

Thus, all three forms have leading digit 1.

Hence the answer is 91.

35. Answer: 2

Feedback:

There are 99 integers in all, of which 49 are even.

From 101 to 199, there are 14 multiples of 7, of which 7 are even.

There are 11 multiples of 9, of which 6 are even.

But there is one integer (i.e. 126) that is a multiple of both 7 and 9 and also even.

Hence the answer is (49 – 7 – 6 + 1) = 37

36. Answer: 5

Feedback:

We can see that x + 2 is an increasing function and 5 – x is a decreasing function. This system

of equation will have smallest value at the point of intersection of the two. i.e. 5 – x = x + 2

or x = 1.5. Thus smallest value of g(x) = 3.5

37. Answer: 3

Feedback:

If y = 2 (it cannot be 0 or 1), then x can take 1 value and z can take 2 values.

Thus with y = 2, a total of 1 X 2 = 2 numbers can be formed. With y = 3, 2 X 3 = 6 numbers

can be formed. Similarly checking for all values of y from 2 to 9 and adding up we get the

answer as 240

38. Answer: 4

Feedback:

The number of terms of the series forms the sum of first n natural numbers i.e.

n(n + 1)/2.

Thus the first 23 letters will account for the first (23 x 24)/2 = 276 terms of the series.

The 288th term will be the 24th letter viz. x.

39. Answer: 3

Feedback:

2x – x – 1 = 0

2x – 1 = x

If we put x = 0, then this is satisfied and we put x = 1, then this is also satisfied.

Now we put x = 2, then this is not valid.

40. Answer: 1

Feedback:

Let us say there are only 3 questions. Thus there are 23–1 = 4 students who have done 1 or

more

questions wrongly, 23–2 = 2 students who have done 2 or more questions wrongly and 23–3 = 1

student who

must have done all 3 wrongly. Thus total number of wrong answers = 4 + 2 + 1 = 7 = 23 – 1

= 2n – 1.

In our question, the total number of wrong answers = 4095 = 212 – 1. Thus n = 12.

41. Answer: 4

Feedback:

Tn = a + (n – 1)d

467 = 3 + (n – 1)8

n = 59

Half of n = 29 terms

29th term is 227 and 30th term is 243 and when these two terms are added the sum is more

than 470.

Hence the maximum possible values the set S can have are 30.

42. Answer: 2

Feedback:

GRRRRR, RGRRRR, RRGRRR, RRRGRR, RRRRGR, RRRRRG

GGRRRR, RGGRRR, RRGGRR, RRRGGR, RRRRGG

GGGRRR, RGGGRR, RRGGGR, RRRGGG

GGGGRR, RGGGGR, RRGGGG

GGGGGR, RGGGGG

GGGGGG

Hence 21 ways.

43. Answer: 4

Feedback:

S = (–5) + (–3) + (–1) + 1 + 3 + .......

S = (2 × 1 – 7) + (2 × 2 – 7) + (2 × 3 – 7) + (2 × 4 – 7) + .......

Tn = (2n – 7)

∴ a – b = –7

44. Answer: 4

Feedback:

Given that a0 = 2, a1 = 1 and

For n = 1:

a2 a0 = a0 – a2.

⇒ 2a2 = 2 – a2

⇒ a2 =2/3

Similarly, a3 =1/2=2/4, a4=2/5, etc.

Observing the terms a2, a3, a4.... etc, one can easily figure

out that an is 2/(n+1)

So a1000 = 2 /1001

45. Answer: 2

Feedback:

Let the length, breadth and the height of the cuboid be L, B

and H. We are given that L2 + B2 + H2 = 102 = 100.

Using A.M. ≥ G.M. property, we can write

⇒L2 + B2 + H2 ≥ (LB + BH + LH)

⇒ 2 × (L2 + B2 + H2 ) ≥ 2 × (LB + BH + LH)

⇒ Total surface Area ≤ 200.

46. Answer: 3

Feedback:

The function f(n) is nothing but the Fibonacci Series 1, 1, 2, 3,

5, 8, 13, 21, 34, 55, 89…

Here if we find the HCF of any two numbers in the series, we

get an interesting pattern.

Lets find the HCF of f(6) and f(9), i.e. 8 and 34 respectively.

The HCF of 8 and 34 is 2, which is also equal to f(3).

Similarly lets find the HCF of f(10) and f(5), i.e. 55 and 5

respectively. The HCF of 55 and 5 is 5, which is also equal to

f(5). After checking for other numbers as well we can conclude

that the HCF of f(a) and f(b) is f(h), where ‘h’ is the HCF of ‘a’

and ‘b’.

Therefore, the HCF of f(12) and f(30) is f(HCF of 12 and 30),

i.e. f(6).

Hence option (3) is the correct choice.

47. Answer: 2

Feedback:

Case I: When Sanjay and Salim are selected and Reena and

Teena are not selected.

Total number of ways in which the team can be selected

= 6C4 = 15.

Case II: When Reena or Teena is selected and both Sanjay

and Salim are not selected.


= 2 × 6C5 = 12.

Case III: When Reena or Teena is selected and both Sanjay

and Salim are selected.


= 2 × 6C3 = 40.

Case IV: When none of Reena, Teena, Sanjay and Salim are

selected in the team.

Total number of ways in which the team can be selected = 1.

Therefore, the total number of ways in which the team can be

selected = 15 + 12 + 40 + 1 = 68.

48. Answer: 4

Feedback:

Chord AB passes through point P but its length is not fixed.

After choosing the point C, area of ∆ABC can be calculated.

Let CM, be the altitude on side AB. As AC = BC,

Area of ∆ABC = 1/2×AB*CM

The minimum value of area(∆ABC) is obtained when AB has

the shortest possible length, as shown in the figure below. In

this case, point M and P are same.

Hence, (4) is the correct option.

49. Answer: 5

Feedback:

x = Price of a Pen and

y = Price of a Pencil

a = Number of pens Raju wanted to purchase

b = Number of pencils Raju wanted to purchase

s = bill amount calculated by Raju

We have two equations:

ax + by = s ...(i)

bx + ay = s + 24 ...(ii)

Subtracting (i) from (ii), we get

⇒(b − a )(x − y) = 24 = 23 × 3.

Clearly, all the statements can be valid.

Hence option (5) is the correct choice

50. Answer: 2

Feedback:

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