Ssc Tier II Maths Solution Paper 2

SSC (T-II) 2013 (SOLUTION)

1. (A) Unit digits in (7)4 = 1, therefore unit digitin (74)8 i.e. 732 will be 1.Hence, unit digits in (7)35

= 1 × 7× 7 × 7 = 3Again, unit digit in (3)4 = 1Therefore, unit digit in the expansion of (34)17

= (3)68 = 1

� Unit digit in the expansion of (3)71

= 1 × 3 × 3 × 3 = 7and unit digit in the expansion of (11)35

= 1

� Unit digit in the product of735 × 371 × 1155 = 1

2. (C) a = 100

1, b =

5

1, c =

10

1

or a = 0.01, b = 0.2, c = 0.1

� b > c > a3. (A) Product of the numbers

= HCF × LCM= 21 × 4641= 21 × 3 × 7 × 13 × 17= 3 × 7 × 3 × 7 × 13 × 17.

i.e. 273 and 357.4. (A) Let the number be x.

� x = 765k + 42� 17 × 45K + 17 × 2 + 8� 17(45K + 2) + 8� Remainder = 8

5. (B) Distance traversed by the extremity of the

minute-hand in one hour = 2 × 7

22× 10

Distance traversed by the extremity of theminute-hand in 3 days and 5 hour, i.e.in 77 hours

= 2 × 7

22× 10 × 77

= 22 × 220 = 4840 cmDistance traversed by the hour-hand in12 hour

= 2 × 7

22× 7 = 44 cm

Distance traversed by the hour-hand in77 hour

= 12

44 × 7 =

3

7711�=

3

847= 282.33 cm

Reqd. difference = 4840 – 282.33= 4557.67 cm

6. (D) The series obtained= 4 × 2 + 3, 4 × 5 + 3, 4 × 8 + 3, 4 × 11 + 3 ...= 11, 23, 35, ...

7. (A)

2

1

2

1

2

12

1

2

1

2

1

��

��

=

4

32

12

2

1��

= 2

1 ×

3

4 =

3

2

8. (C) Let a1, d

2 represent the Ist term and

common difference of AP1.

and a2, d

2 represent the Ist term and

common difference of AP2.

ATQ,

'n

n

S

S=

])1(2[2

])1(2[2

22

11

dnan

dnan

��

��

274

17

�

�

n

n=

22

11

)1(2

)1(2

dna

dna

��

��...... (i)

Now, 11

11

b

a=

22

11

10

100

da

da

�

�=

22

11

302

302

da

da

�

�

= 22

11

)121(2

)121(2

da

da

��

��

= '21

21

S

S

= 27214

1217

��

��

= 111

148 Ans.

9.(B) Sn

= 0.4 + 0.44 + 0.444 + ... + to n terms= 4[0.1 + 0.11 + 0.111 + ... + to n terms

= 9

4[0.9 + 0.99 + 0.999 + ... + to n terms

= 9

4��

�

�� termnto....

1000

999

100

99

10

9

=

=

=

��

�

�

�

��

�

��

��

��

��

�

10

11

10

11

10

1

n9

4

n

)r1(

)r1(aS

n

n�

��

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=

��

�

�

��

��

��

�

10

910

11

10

1

n9

4 n

= ��

�

��

��

��

n10

11

9

1n

9

4

= ��

�

��

n10

11n9

9

4 Ans.

10. (C) Number of women who have either nosestuds or ear right = (15 – 3) = 12

�Number of those who have both nose studsand ear rings = (8 + 7) – 12 = 3

11. (C) Ashok's present age = 26 – 6 = 20 yrs

�Pradeep's present age = 20 × 4

3= 15 yrs]

12. (A) Let the number of girls = x

� boys = x – 2According to question:-

x + x – 2 = 52

� x = 27

� No. of bags = 25Again:-

Let average no. of girls = y kg42 × 25 + 27xy = 52 × 52

� y = 27

25425252 ��

= 61.25~61 kg (Approx.)13. (D) Wrong calculated marks

= 35 × 75

= 35

86362520 �� =

35

2570= 73.42

14. (C) If a container contains y units ofliquid and x units of liquids is taken out.If this operation is repeated n times. Thefinal quantity of the liquid in thecontainer is

n

y

xy ��

�

��

��1

� 24 = 54

2

541 �

�

��

��

x

�

2

541 �

�

��

��

x=

54

24 =

9

4

� ��

��

��

541

x =

3

2

�54

x =

3

1

� x = 18 l

15. (C) Let the quantity of pure milk be x l.If 5 l of water is added to it,then,

Cost of (5 + x) l= Rs. (3x + 5)

� Profit = Rs. 15Given, 20% of 3x = 15

�5

3x= 15

� x = 25 l16. (D) Let the amount be Rs. x and rate is r%.

Then, Simple interest

100

rx �= 25

� x × r = 2500For true discount

100

)20( rx �� = 20

�100

20rrx ��= 20

� xr – 20r = 2000From Eqs. (i) and (ii), we get

2500 – 20r = 2000

� r = 25%From Eq. (i)

x × 25 = 2500

� x = 10017. (*) Let the CP price of pen = Rs. 1

CP of 40 pen = 36 × 1 = Rs. 36

�Selling price of 40 pen = 40 – 3% of 40

= 40 – 100

3× 40

= 10

12400 �

= Rs. 38.80

�% profit = ��

�

� �

36

3680.38× 100

= 36

280=

9

70= 7

9

7%

18.(C) Length of bridge = 1000 mLength of train = 500 mTotal length = 1000 + 500 = 1500 m

Speed of train = 1000

1500×

2

60= 45 km/h

19. (B) Distance (D) = Speed (S) × Time (T)

� D = 4 × ��

��

��

60

15T

� D = 4T + 1www.ssc-cgl2014.in



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and D = 6 ��

��

��

60

10T

D = 6T – 1Solving eqns. (i) and (ii),

T = 1 hD = 4 × 1 + 1 = 5 km

20. (D) It is clear from the question that when Acovers 500m, B covers 400 m i.e., A takesa lead of 100 m in every 500 m of distance.Therefore, a lead of 400 m will be takenin travelling a distance of 2000 m or inother words A passes B after every 2000m.Hence, total number of such pass

= 2000

5000=

2

5 = 2

2

1 times

21. (C) Let three numbers A, B and C are Rs. 12x,15x and 25x respectively.

� 12x + 15x + 25x = 312

� x = 52

312 = 6

� Required ratio = 615625

612615

��

��

= 610

63

�

�

= 10

3= 3 : 10

SHORTCUT METHOD:-

There is no need to calculate the value of x.

Required ratio = xx

xx

1525

1215

�

�

= x

x

10

3=

10

3= 3 : 10

22. (B) Ratio of investment of Sita, Gita and Rita is(5000 × 3 + 7000 × 9) : (4000 × 1 + 3000 × 11) : (7000 × 11)

= 78000 : 37000 : 77000= 78 : 37 : 77

�Share of Rita in profit

= 773778

77

�� × 1218 = Rs. 488.47

23. (A) Let the speed of man and current be x km/hand y km/h respectively.Then,

yx �

30 +

yx �

44 = 10 ... (i)

andyx �

40 +

yx �

55 = 13 ... (ii)

Solving eqns.(i) and (ii),y = 3 km/h

24. (C) Let the tap can fill the cistern in x h.

�8

8

�

�

x

x= 12

� 8x = 12x – 96� x = 24 h

�Capacity of cistern= 24 × 60 × 6 = 8640l

25. (A) Part of the cistern filled in 3 min

= 12

3 +

16

3 =

48

21 =

16

7

Let remaining 16

9 part was filled in x min

Then,12

x ×

8

7 +

16

x ×

6

5 =

16

9

� x ��

��

� �

96

57 =

16

9

� x = 16

9×

12

96 = 4.5 min

26. (A) Ratio of efficiencies of the three persons

= 6

24 :

8

24 :

3

24 – �

�

��

��

8

24

6

24

= 4 : 3 : 1

�Boy's share = )134(

1

�� × 600 = Rs. 75

27. (A) Let the cost price of geyser be Rs.x,then, x × 1.1 × 1.15 × 1.25 = 1265

x = 58125.1

1265

= Rs. 80028. (B) Let his increased income be x.

Then,

(x – 1200) × 100

80 ×

100

12= x ×

100

80×

100

10

� 12x – 14400 = 10%x = Rs. 7200

29. (D) Let the present value of what A owes to Bbe Rs. x.Then,

x + 1002

314

�

��x= 1573

� x + 100

21x = 1573

�100

121x= 1573

� x = Rs. 1300Let y be the present value of what B owes A.

Then, y + y × 2

1 ×

100

14 = Rs. 1444.50

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� y + 100

7y = Rs. 1444.50

� y = 107

10050.1444 �= Rs. 1350

Hence, B pay Rs. 50 to A.30. (A) Let the amount given at 4% per annum

be Rs. x.�Amount given at 5% per annum

= (1200 – x)

100

24��x +

100

25)1200( �� x= 110

�100

120002 �� x= 110

� x = Rs. 500Also, (1200 – x) = 1200 – 500

= Rs. 70031.(A) Let money invested at 5% be Rs. x.

�100

51��x +

100

81)10000( �� x = 688.25

� 5x – 8x + 80000 = 68825� 3x = 11175� x = Rs. 3725

32. (C) Let principal amount be Rs. 100.

Then, SI = 100

320100 �� = Rs. 60

and CI = 100

3

100

201 �

�

��

�� – 100

= 100 ×

3

5

6��

��

� – 100 =

5

364

�CI – SI = 5

364 – 60 =

5

64

If difference is Rs. 5

64, Principal = Rs. 100

If difference is Rs. 48,

Principal = 64

5100�× 48 = Rs. 375

33. (C)��

��

��

��

��

��

100

Pr

1001

2

P

rP

= 5

6

� ��

��

��

1001

r=

5

6

� r = 20%34. (D) Train with a speed of 54km/h passes the

man in 20s.

�Length of the train = 54 × 18

5 × 20

= 300 mLet the length of platform be x m.

Then, (300 + x) = 54 × 18

5 × 36

x = 540 – 300 = 240 m35. (A) Length of train = 12 × 15 = 180 m

Time = 18 s

Speed = 18

180 = 10 m/s

New distance = 15 × 10 = 150 m

�Required time = 10

150 = 15s

36. (B) Let the CP of the pen and book be Rs. xand Rs. y respectively.

� 0.95x + 1.15y = (x + y) + 7� 0.15y – 0.05x= 7 ... (i)and 1.05x + 1.1y = (x + y) + 13� 0.05x + 0.1y = 13 ... (ii)Solving Eqs. (i) and (ii), we get

y = Rs. 8037. (D) Let the cost price of articles be Rs.x.

Then, selling price of article = 0.88x.Marked price of article

= 80

88.0 × 100 × x = 1.1 x

New selling price of article = 1.045x

�Profit per cent = x

xx �045.1 × 100 = 4.5%

38. (A) Let the time of meet = t h

Now,

��

��

��

60

2015 t + 20t = 450

� t = 13h

Distance from A = 15 ��

��

��

3

113 = 190 km

39. (A) Speed of Ramesh = 3x km/hrSpeed of Suresh = 4x km/hrLet the distance = DATQ,

x

D

3 –

x

D

4=

2

1

� ��

��

�

12

1

x

D=

2

1

� D = 6x

Time of Ramesh = x

D

3=

x

x

3

6 = 2 h

Time of Suresh = x

D

4=

x

x

4

6= 1.5 h

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40.(*) Work done by A & B in 16 days = 30

16 =

15

8

Remaining work = 1 – 15

8 =

15

7

�15

7 work is done by B alone in 44 days.

� 1 work is done by B alone in

= 157

44

= 7

1544� = 94.29 days

41. (C) SP = 700

70 MP

SP For Tarun = Rs. 8750

Labelled price = 100

125 ×

100

70MP = 8750

� MP = Rs. 1000042. (A) Let the CP of the article be Rs. x.

Then,SP = x × 1.12 × 1.1

Given, x × 1.12 × 1.1 = 616

� x = 232.1

616= Rs. 500

43. (B) Let CP = Rs. x and SP = Rs. y.� y × 7% = x × 8%and y × 9% = x × 10% + 1

and 100

9× y =

100

10 × x + 1

� 7y = 8x

and 9y = 10x + 100

� 9 × 7

8x= 10x + 100

� x = Rs. 35044. (B) Let the numbers be 3x and 4x.

Then,16x2 = 8 × (9x)2 – 224

� 16x2 = 72x2 – 224� 56x2 = 224� x2 = 4� x = 2Hence, numbers are 6, 8.

45. (A) P = 49

362

2

�

�

x

x =

)7)(7(

)6)(6(

��

��

xx

xx

Q = 7

6

�

�

x

x

� Q

P=

)7(

)6()7)(7(

)6)(6(

�

��

��

x

x

xx

xx

=7

6

�

�

x

x

46. (B) Sn

= pn + qn2

Sn – 1

= p(n – 1) + q(n – 1)2

= pn – p + qn2 – 2qn + q

an

= Sn – S

n – 1

= pn + qn2 – (pn – p + qn2 – 2qn + q)= p + 2qn – q

Common difference= a

2 – a

1

= (p + 4q – q) – (p + 2q – q)= 2q Ans.

47.(C) a, b and c are in GP and zyx cba

111

�� .

Let, kcba zyx ��

111

� a = kx

b = ky

c = kz

Now:-b2 = ac [a, b and c are in G.P.]

or, (ky )2 = kx + z

� 2y = x + z

48. (C)

= 3

4

12

2

1

34

32

2

1

��

��=

7

13

49. (C)zx

y

�=

z

xy �

� yz = xy + x2 – yz – xz ... (i)

Also,y

x=

zx

y

�

� x2 – xz = y2 ... (ii)Using Eqs. (i) and (ii), we get

yz = xy – yz + y2

�2yz = xy + y2

� 2z = x + y ... (iii)Only option (C) satisfies the Eq. (iii).

50. (A) CP = x (say)34% of CP = 26% of SP

x26

34= x

13

17= SP

% profit = 10013

1317

�

�

x

xx

= 30.77%www.ssc-cgl2014.in



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51. (B) Given roots are real and equal.

� B2 – 4AC = 0

� [–2(a2 – bc)]2 – 4(c2 – ab)(b2 – ac) = 0

� 4(a4 + b2c2 – 2a2bc – c2b2 + ac3 + ab3 – a2bc = 0

� 4a(a3 + b3 + c3 – 3abc) = 0

� a3 + b3 + c3 = 3abc

52. (B)a

a

a

aa

�

��

�

� ��

1

)1(

1

2/12/12/1

�)1(

1

)1)(1(

2 2/1

a

a

aa �

��

��

�

�a

aa

�

��

1

)1)(1(2 2/1

�a

aa

�

��

1

112 2/12/1

�a

aa

�

��

1

)(22 2/12/1

�a�

��

1

222 =

a�1

2

53. (A) Both the expressions are divided by (x – 2).Hence f

1(2) = f

2 (2).

(� Both remainders are same)

� p × (2)4 – 3(2)3 + 20 = 4(2)2 + 7(2) – p

� 16p – 24 + 20 = 16 + 14 – p

� 17p = 34

� p = 2

Hence, the value of 'p' is 2.

54. (A) )(333 baabbax �� = 25

)2( 5

�3)( bax � = 2125

�35x =

352

� x = 2

55.(A) Joining point O to three vertices A, B and C.

Now,

Area of ( �OBC + �OCA + �OAB) = area of

�ABC.

�2

1(a × 8 + a × 7 + a × 6) =

4

3a2 =

2

21a

a = 3

3

3

42�

� a = 314

� Hence, area of triangle ABC.

= 4

3a2

= 4

3( 314 )2 m2

= 254.6 m2

56. (D) Let A1B

1 be the ladder placed against the

wall.A

1B

1= length of ladder = 25 m

B1C = 7 m

Now, ladder foot is drawn out from B1 to B

2

such that B1B

2 = x m, then the top of the

ladder comes down from A1 to A

2 and A

1A

2

= 2

xm (as per question),

So, A1B

1 = A

2B

2 = 25 m and A

2C = Y (say)

In �A1B

1C

A1C = 2

1211 )()( CBBA �

� A1A

2 + A

2C = 22 725 �

�2

x + Y = 24 ... (i)

In �A2B

2C

(A2C)2 + (B

2C)2 = A

2B

22

� Y2 + (x + 7)2 = 252

�

2

224 �

�

��

��

x+ (x + 7)2 = 252 [from (i)]

� 576 + 2

2x – 24x + x2 + 49 + 14x = 625

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�4

5 2x = 10x

� x = 8 m57. (C) Let the circumference of outer circle

C1 = 62.832 m

circumference of inner circleC

2 = 37.6992 m

The area between two circles = A1 – A

2 = ?

Here, using the correlation formula forcircumference and area.

A = �4

2C

� A1 – A

2=

�4

21C

– �4

22C

� A1 – A

2=

�

�

4

22

21 CC

� A1 – A

2=

�

��

4

))(( 2121 CCCC

� A1 – A

2=

1416.34

)6992.37832.62)(6992.37832.62(

�

��

� A1 – A

2= 201 m2

58. (A)

Let AOB be one quadrant of a circle radius.OA = OB = 10 m

Two circles OKB and OKA are made on theabove radiii as diameter. So, if E and D arecentres.

OD = BD = OE = EA = 5 mSince two circles are equal, k will bemidpoint of arc AKO and OKB.

� �KDO = 90º

� OK = 25

Area common to both circle = 2 × area ofsegment OXK (or OYK)Using the formula, for central angle = 90ºArea of segment OXK = 0.285 × r2

= 0.285 × 52

= 7.13 m2

Area common to both circle= 2 × 7.13= 14.26 m2

59. (B) Let A = 2 pens are red.B = 2 pens are blue.

than P(A) = 9

5 ×

8

4 =

72

20 =

18

5

P(B) = 4

9 ×

3

8 =

12

72 =

6

1

Now, A and B are mutually exclusive events.Hence,P(selecting either 2 red pens or blue pens)= P(A + B) = P(A) + P(B)

= 18

5 +

6

1=

4

9

Hence, the required probability is 4

9.

60. (B) sec2 � cosec2 � – [tan2 � + cot2 � ]

= (1 + tan2 � ) (1 + cot2 � ) – [tan2 � + cot2 � ]

[since sec2 � = 1 + tan2 � ]

= 1 + tan2 � + cot2 � + tan2 � cot2 � – tan2 � –

cot2 �

= 1 + 1= 2Hence, the required value is 2.

61.(B)Let AB be the man standing out side a housewhich has a window CD. C (top point) and D(bottom point) of window are viewed frompoint A.

In right �DAE,

tan 45º = AE

DE=

3

DE

� DE = 3tan 45º = 3 m

In right �CAE,

tan 60º = AE

DE=

AE

DECD � =

AE

CD 3�

� 3 = ��

��

��1

3

CD

Therefore, the length of the window is 2.2 m.

62. (C) (1 – sin2 A) = Asec

8.0

� cos2 A . sec A = 0.8

� cos A = 0.8 = 5

4

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� sin A = 5

3

� tan A = 4

3

So, tan A + Acos

1=

4

3 +

4

5 =

4

8 = 2

63. (A) tan � = 3 cot �

� tan � = �tan

3

� tan2 � = 3

� tan � = + 3

Since � is an acute angle � � = 60º

Now, sin2 � + cosec2 � – 2

1cot2 �

= sin2 60º + cosec2 60º – 2

1cot2 60º

=

2

2

3��

�

�

��

�

� +

2

3

2��

��

� –

2

1×

2

3

1��

��

�

= 4

3 +

3

4 –

6

1 =

12

23

64. (C)

Let ABCD be a rectangular grass plot whichhas a gravel path of width = W = 2.5 m(shaded portion)length of plot = l = 112 mbreadth of plot = b = 78 mHere, the path is inside the rectangularplot.Using the formula:-

= 2W(l + b – 2W)= 2 × 2.5 (112 + 78 – 2 × 2.5)= 925 m2

Now, cost of construction the path= 925 × 3.40 = Rs. 3145

65. (A)

Since the triangle is an equilateral, so

each angle = 60º = � , r = 15

� area of sector = 2

º360r��

�

= 360

º60×

7

22× 225

= 117.85 m2.

66. (D)

Let ABCO be the right cone whose slantheight = l = AB = ACheight = h = 200 m

In �ABO, �ABO = 30º

sin 30º = l

200

� l = 200 × 2 = 400 m

��

�

��

2

1º30sin

cos 30º= l

r

� r = l cos 30 º = 400 × 2

3 = 3200 m

��

�

��

2

3º30cos

Now,Area of curved surface = � rl

= 7

22× 200 × 3 × 400

= 435312 m2.67. (A) Let, the side of equilateral triangle is 'a' m.

Area (A) = 4

3a2 m2

� Cost of paving = 10 × 4

3a2 rupees

Similarly, perimeter (P) = 3a m� Cost of fencing = 25 × a rupeesAccording to the question,

10 × 4

3a2 = 25 × 3a

� a = 310www.ssc-cgl2014.in



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� a = 17.32 mHence, the side of the equilateral triangleis 17.32 m.

68. (C)

Let ABCD is a rectangle whosediagonal = d = 25 m

area = A = 168 m2 length = l = ? breadth = b = ?

Using the correlation formula

(l + b)2 = d2 + 2A [Refer 12.4]

and (l + b)2 = d2 – 2A

� (l + b)2 = (25)2 – 2 + 168

l + b = 31 (i)

Similarly,

� (l + b)2 = (25)2 – 2 × 168

l – b = 17 (ii)

From (i) and (ii)

l = 24 m = length of the rectangle

b = 7 m = breadth of the rectangle

69. (A) The given points A, B and C are collinear.

So, these area of the triangle formedby these points will be zero.

� (m + 1)(3 – 2m) + (2m + 1)(2m – 1) +

(2m + 2)(1 – 3) = 0

� 2m2 – 3m – 2 = 0

� 2m2 – 4m + m – 2 = 0

� 2m(m – 2) + 1(m – 2) = 0

� (2m + 1) (m – 2) = 0

m = –2

1 or m = 2

70. (B) Let P(x, y) be the required point dividingthe line AB externally in the ratio 5 : 3.

�BP

AB =

3

5 =

m

1 (say)

Using the formula (section)

x = ml

mxlx

�

� 12

where x = 2, x = –3

�35

2335

�

��x = –

2

21 = –10.5

and y = ml

myly

�

� 12

where, y1 = –3, y

2 = 7

y = 35

)3(375

�

��= 22

Hence, the required point P is (– 10.5, 22).71. (D) Using :-

22

21 dd � = 2(a2 + b2)

� a2 + b2 = 2

1[(10)2 + (17.78)2] = 208... (i)

and 2(a + b) = 40 (given)� a + b = 20 ..... (ii)From (i) and (ii), we have

a = 12, b = 872. (B) Let ABCD be a rhombus whose

perimeter = p = 36 marea = A = 72 m2

perimeter p = 4 × a� 36 = 4a� a = 9 m

Area, A = a × h� 72 = 9 × h

� h = 9

72m

h = 8 m

73. (C)

Given that:-

DA

BD=

DC

DA.... (i)

= DA2 = BD × DC .... (ii)

In right triangle AOB and AOC:-

AB2 = AD2 + BD2 .... (i)

and, AC2 = AD2 + DC2 .... (ii)

Adding (A) and (B):-

AB2 + AC2 = 2AD2 + DC2

= 2 × BD × DC + BD2 + DC2

[from (ii)]

Thus in �ABC;

AB2 + AC2 = BC2

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74. (D) Use length of the rope

= )12(2

��

ndn

Where d = 2 cm

n = 140

length = 2

]281[140214.3 ��

= 1236.40 m

75. (C) Let ABCDEF be a regular hexagonalpyramid whose base

ABCDEF is a regular hexagon.

If a = each side of regular hexagon,

then, 6a = 30

� a = 5 m

height of pyramid = OX = h = 20 m

Volume (V) = 3

1Ah

= 204

36

3

1 2 ��

�

�

��

�

�� a

= 2310 a

= 25310 �

= 250 × 1.732 m3

= 433 m3

76. (B) It is given that

S = 2.5 k, where k stands for curvedsurface area.

� 2 � r (h + r) = 2.5(2 � rh)

�h

rh �= 2.5

� 7 + r = 2.5 × 7 [since h = 7 cm]

� r = 10.5 cm

Now,

Volume V = � r2h

= 7

22× (10.5)2 × 7

= 2425.5 cm3

77.(A)

Let ABCD be a rectangular grass plotwhose length = l = 80 m

breadth = b = 60 mTwo roads of width W = 10 m (shaded part)are crossing each other at the middle ofplot.Area of roads = W(l + b – W)

= 10(80 + 60 – 10)m2

= 1300 m2

Cost of gravelling the roads= rate of gravelling / m2 × area of roads= Rs. 2 × 1300= Rs. 2600

78. (C) Let r = radius of hemisphere bowl� 2 � r = 176� r = 28 cmVolume of the quantity in hemispherical

punch bowl = 3

3

2

2

1��

= 328

3

1�� cm3

Volume of the bowl in which food is to be

served = 32

3

2�� cm3.

No. of persons served

= 3

3

23

2

283

��

��

= 1372

79. (D)22

3 yxyx �� = 81 = 34

� x2 – xy + y2 = 4 ... (i)

and x y3 3

2 � = 256 = 28

� x3 + y3 = 8 ... (ii)Dividing (ii) by (i) x + y = 2

80. (B) The given equation is

c + y

yd �= e – 1 +

y

f

� c + y

d –

y

y= e – 1 +

y

f

� c + y

d – 1 = e – 1 +

y

f

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�y

d –

y

f= e – c

�y

fd �= e – c

� y = ce

fd

�

�

81. (B) The given equation is

bx2 – ax + log2 my = 0

Now,

Sum of the roots = x1 + x

2

= b

a�

= b

a... (i)

The given relation is

22

21 xx � = a2

� (x1 + x

2)(x

1 – x

2) = a2

�b

a(x

1 – x

2) = a2

[From equation (i)]

� x1 – x

2 = ab ... (ii)

From equation (i) and equation (ii)

x1

= ��

�

��ab

b

a

2

1 =

b

ba

2

)1( 2 �

x2

= ��

�

��ab

b

a

2

1 =

b

ba

2

)1( 2�

Hence, the roots are

= b

a

2(b2 + 1),

b

a

2(1 – b2)

82.(*) The given expression is

= 2)( lk � + 2)( m + 2)( n + m2 lk � –

n2 m – n2 lk �

= 2nmlk ��

So, the square root of the given expressions

= + nmlk ��

83. (C))º75º90º360sin(2)º75º90cot(2

)º75º360tan()º75º180cos(

��

��

= º75cos2º75tan2

º75tanº75cos

��

�� =

2

1

84. (D) Let OT = height of tower = h metres

PQ = width of the river

Where P = point of the near shore to tower.

Q = point of the far shore to the tower.

� ZTA = Aº (angle of depression)

�ZTQ = Bº (angle of drepression)

Then, � ZTA = � TPO = Aº

� ZTQ = � TQO = Bº

Now,

In � TOP , tan A = OP

h

� OP = h cot A .... (i)

In � TQO, tan B = OQ

h

= OQOP

h

�

� OP + PQ = h cot B .... (ii)

From (i) and (ii),

PQ = h(cot B – cot A)

85. (C)

h2 = a2 + b2

since, a and h are consecutive integers,

h = a + 1

� (a + 1)2 = a2 + b2

� b2 = 2a + 1

� a = 2

12 �b

� h =2

12 �b

So, sin � = h

a=

b

b

2

2

1

1

�

�

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86. (A)

In 11DAP� ,

tan 60º= 1

11

AD

DP=

1

32000

AD

3 = 1

32000

AD

� AD1

= 2000 m

In 22DAP� ,

tan 30º= 2

22

AD

DP=

2

32000

AD

3

1=

2

32000

AD

� AD2

= 6000 m�D

1D

2= 6000 – 2000 = 4000 m

Speed of the Jet = 18

4000×

5

18

= 800 km/h87. (C)

Let K1 be the reflection of kite K.

Then AK = h (say) = AK1

D is a point above the surface AC of thelake such that CD = d = AE

Then, In �KED,

tan 30º = ED

KE=

ED

dh �... (i)

In �K1ED,

tan 60º = ED

EK1 =ED

dh �... (ii)

Dividing (i) by (ii), (ED get cancelled)

�3

1=

dh

dh

�

�� h = 2d

88. (B) Let the no. of sides of the polygon = nSo, there are n interior angles which arein A.P.Now,Sum of n interior angles

= Sn =

2

n[2a + (n – 1)d]

Where,a = 120ºd = 5

� Sn

= 2

n[2 × 120º + (n – 1)5º] ... (i)

But, also, using the formula,Sum of interior angles for n – sidedploygon = (2n – 4) × 90ºFrom (1) and (2)

(2n – 4) × 90º =2

n[2 × 120º + (n – 1)5º]

� n2 – 25n + 144 = 0� (n – 16)(n – 9) = 0

n = 16 or 916th interior angle

= a + (n – 1)d, where a = 120º, n = 16= 120 + (16 – 1) × 5= 195º, which is greater than 180º.

Since no interior angle of a regularpolygon can exceed 180º.

So, n = 16 is not valid.

Hence, no.of sides of the polygon is 9.

89. (A) Volume of earth dug out

= � r2h = 7

22×

2

2

5.3��

��

� × 12 = 115.5m3

The earth dug out is exactly spread is forma platform of height h m.

� 115.5 = 10.5 × 8 × h

� h = 1.375 m

90. (C)

Let OABCD be the right pyramid whoseAB = DC = 24 cm

BC = AD = 18 cm

If Z is the mid point of side BC, then OZ =slant height = l = 17 cm

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In �OXZ, �X = 90º

� OX = height = 22 XZOZ �

� h =

22

217 �

�

��

��

DC

� h = 22 1217 �

� h = 12.04 cmNow,

Volume = 3

1Ah

= 3

1× 24 × 18 × 12.04

= 1733.76 cm3

= 1733.5 cm3 (approx.)

91. (A) Required difference

��

��

� ��

10012000

CD – �

�

��

� ��

10015000

EA

��

��

� ��

100

222512000 – �

�

��

� ��

100

10515000

= 5640 – 2250 = 3390

92. (B) There is only one absolute decrease i.e. E.

93. (D) A + B = 360 + 1560 = 1920 i.e.approximately equal to F type of tyres in

October i.e. 1950.

94. (B) Required difference= (750 + 2250 + 3750 + 1950 + 1800)– (360 + 1560 + 2640 + 960 + 1440)= 10500 – 6960 = 3540

[Note : D and E decreasing in sales]95. (D) Number of tyres D and E sold in September

= 3000 + 2040 = 5040Number of tyres D and E sold in October

= 3000 + 1500 = 4500Required percentage

= 4500

5040 × 100 = 112%

96. (C) Leather goods turnover in Tanzania

= 40 × 100

25 = $ 10 million

Leather goods turnover in Africa= 30 million

Rest of Africa turnover for leather goods= $ 30 million – 10 million= $ 20 million

Required percentage = 30

20× 100

= 66.67% = 67%97. (C) Jewellery items turnover in Tanzania

= 40 × 100

20 = $ 8 million

Jewellery items turnover in Africa= $ 10 million

� Jewellery items turnover in rest ofAfrica

= $ 10 million – $ 8 million= $ 2 million

Garments items turnover in Tanzania

= 40 × 100

30= $ 12 million

Garments items turnover in Africa= $ 40 million

� Garments items turnover in rest of Africa= 40 – 12 = $ 28 million

Total of Jewellery items and Garmentsitems turnover in rest of Africa

= 2 + 28 = $ 30 millionTurnover from Tanzania from Electricaland leather goods

= 40 × 100

10 + 40 ×

100

25

= 4 + 10 = $ 14 million

Ratio = 14

30 = 2.14 : 1

98. (B) Turnover from Tanzania for Electricalgoods and Handicraft together

= 40 × 100

10 × 40 ×

100

15

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= 4 + 6 = $ 10 millionTurnover of Garment in Tanzania

= 40 × 100

30 = $ 12 million

Turnover of Garment in Africa= $ 40 million

Turnover from rest of Africa for Garments= $ 40 million – $ 12 million= $ 28 million

According to question,28 × x = 10

� x = 28

10= 0.36 times

99. (D) Turnover from Jewellery and Garmentstogether from Tanzania

= 40 × 100

20 + 40 ×

100

30

= $ 20 millionTurnover from the rest of Africa forElectrical and Leather goods together

= (15 + 30) – ��

��

� ��

100

251040

= 45 – 14 = $ 31 millionRequired percentage

= 31

20× 100 = 65%

100. (C) Turnover from rest of Africa for Electrical

goods = 15 – 40 × 100

10

= $ 11 million

Now, 11 × 100

120 = $ 13.2 million

Turnover from Tanzania for handicrafts

items = 40 × 100

15 = $ 6 million

� Required ratio = 2.13

6

= 6 : 13

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1. A2. C3. A4. A5. B6. D7. A8. C9. B10. C11. C12. A13. D14. C15. C16. D17. *18. C19. B20. D

21. C22. B23. A24. C25. A26. A27. A28. B29. D30. A31. A32. C33. C34. D35. A36. B37. D38. A39. A40. *

41. C42. A43. B44. B45. A46. B47. C48. C49. C50. A51. B52. B53. A54. A55. A56. D57. C58. A59. B60. B

61. B62. C63. A64. C65. A66. D67. A68. C69. A70. B71. D72. B73. C74. D75. C76. B77. A78. C79. D80. B

SSC MAINS (MATHS) MOCK TEST (ANSWER SHEET)

81. B82. *83. C84. D85. C86. A87. C88. B89. A90. C91. A92. B93. D94. B95. D96. C97. C98. B99. D100. C

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Ssc Tier II Maths Solution Paper 2

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Transcript of Ssc Tier II Maths Solution Paper 2