Pinnacle SSC CGL Tier 2 Test Series Paper Code 707 Math Pinnacle SSC CGL Coaching Centre, Gurudwara Road, Model Town, Hisar, 9729327755 Page 1 Pinnacle SSC CGL Tier 2 Test Series Paper

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Pinnacle SSC CGL Coaching Centre, Gurudwara Road, Model Town, Hisar, 9729327755 www.ssccglpinnacle.com Page 1

Pinnacle SSC CGL Tier 2 Test Series Paper Code 707 Math

Answer Key

Q ans Q ans Q ans Q ans Q ans

1 d 21 a 41 b 61 b 81 d 2 a 22 a 42 a 62 a 82 a

3 b 23 b 43 c 63 d 83 c 4 b 24 a 44 c 64 a 84 c

5 d 25 a 45 b 65 c 85 c

6 d 26 a 46 b 66 c 86 c 7 c 27 d 47 a 67 a 87 d

8 a 28 b 48 b 68 c 88 c 9 d 29 b 49 b 69 b 89 d

10 b 30 c 50 b 70 c 90 b

11 d 31 a 51 a 71 c 91 a 12 a 32 b 52 d 72 d 92 b

13 b 33 d 53 c 73 a 93 b 14 c 34 a 54 a 74 b 94 c

15 d 35 d 55 c 75 c 95 b 16 b 36 a 56 a 76 a 96 a

17 c 37 b 57 d 77 a 97 d

18 b 38 c 58 c 78 c 98 b 19 c 39 c 59 b 79 c 99 d

20 a 40 c 60 a 80 c 100 a

Explanation:

1. d; Hint: See note.

Let the two-digit no. be 1Ox + y

According to question,

(10x+y)-(10y+x)=54

9x-9y=54; x-y = 6 but we cannot find the number.

2. (a) First find the difference among all and then HCF of the difference

19139 -12288 =6851 ; 28200-19139=9061 ; 28200-12288= 15912



HCF of 6851 , 9061, 15912 is 221

3. (b) A number is divisible by 18, if it is divisible by 2 and 9 both.

`A number is divisible by 2, if its unit’s digit is an even number.

Unit’s digit of the given number is an even number.

.*. The number to be added must be an even number so that sum of two numbers is even number.

Now, a number is divisible by 9, if sum of its digits is divisible by 9.

Sum of digits of 15762 is 21

If we add 6 then 21+6 =27 is divisible by 7

4. (b) ��

�� =

��

��= 12

5. (d) The least five digit number = 1000

Divide 10000 by 666 we get

Remainder = 10

Hence we add = 666 – 10 = 656

to the least five digit number

Required number = 10656.

6. (d) 10 is substracted

17 – 10 : 24 – 10 = 7 : 14 = 1 : 2

7. (c) Let the sum be Rs. x and the rate be R% p.a.

Then

xR

x4

3

2

25

100

.684

3

4

3

8

R

R

Rate = 6% p.a.



8. (a); 30×��

�= �� 345

20×��.��

��= 285

345 + 285 = 630

SP = 630×��

��

Per KG SP = ��

��×

��

�� =16.30

9. (d) 30 days

5 M + 7 B ( in 6 days ) can earn = 3825

5 M + 7 B ( in 1 day ) can earn = ��

�

2 M + 3 B ( in 4 days ) can earn = 1050

2 M + 3 B ( in 1 day ) can earn = ��

�

5 M + 7 B = 3825

6

2 M + 3 B = ��

�

THEN 1 M = 75 and 1 B = 37.5

Then 7 M + 6 B = 750

HENCE Rs. 750 Can be earn by ( 7 M + 6 B ) in = 1 day

Hence Rs. 22500 can be earn in = 30 days

10. (b) 2 = 17

13

13

41

1

4

13

11

1

xxx

x = 2 – 17

21

17

13

11. (d) In the second case, 3 pens are substituted by 3 books.

Increase in total cost = Rs. 241 - Rs. 232 = Rs. 9

Difference between cost of one book and one pen — — — Rs. 3

Cost ofI book = Cost of 1 pen + Rs. 3



Cost of 7 pens + cost of 4 books = Rs. 232

...-. Cost of 11 pens = Rs. 232- 4 x Rs. 3 = Rs. 220

.-. Cost of 1 pen = Rs. 222 = Rs. 20

.'. Cost of 1 book = Rs. 20 + Rs. 3 = Rs. 23

12. (a) Sum of five numbers = 42 × 5 = 210

Sum of eight numbers = 81 × 8 = 648

Average of all numbers = 6613

858

13

648210

13. (b) suppose the number is x.

Acc. the question ,

3��-4x=x+50

3��-5x-50=0

3��-15x+10x-50=0

(3x+5)(x-5)=0

Hense x=5

14. (c) Total weight of 120 students

= (56 × 120) kg = 6720 kg.

Let the number of boys be x.

Then, the number of girls = (120 – x).

60x + (120 – x) × 50 = 6720

60x + 6000 – 50x = 6720

10x = (6720 – 6000) = 720 x = 72.

Number of boys = 72.

Number of girls = (120 – 72) = 48



15. (d) Ratio of Speeds = 2 : 3 : 4

Ratio of times taken = 3:4:64

1:

3

1:

2

1

16. (b) Sum of the ages of 40 students

= (18 × 40) years = 720 years.

Sum of the ages of 60 students

=

60

2

118 years =

60

2

37years

= 1110 years.

Sum of the ages of 20 new students

= (1110 – 720) years

= 390 years.

Average age of 20 new students

=20

390years =

2

119 years.

17. (c) Let the shares of W, X, Y and Z be Rs. 3x, Rs. 7x, Rs. 9x and Rs. 13x respectively.

W + Y = 11172 Þ (3x + 9x) = 11172

12x = 11172 x = 931.

Required difference = Rs. (13x – 7x) = Rs. 6x

= Rs. (6 × 931) = Rs. 5586

18. (b) Cost price of the table : cost price of chair = : = 3 • 2

Let cost price of a table = Rs. 300

Then cost price of a chair = Rs. 200

Total profit = 3 x 15- 2 ×10 = 45- 20 = Rs. 25



But actual profit = Rs. 150 =Rs.25 × 6

Cost of a table = 300 × 6 = Rs. 1800

19. (c)

By short cut

200

1SI

CI R

200

1600

618 R

R = 6%

20. (a) B’s income = Average of A & B + Average of B & C - Average of A & C

= Rs. (5500 + 4500 - 4000) = Rs. 6000

21. (A) Rs. 10000

By short cut,

D = P × ��

��

P = �� × ��

�� =Rs. 10000

22. (a) 13 : 14

�� - �� : �� + �� = 27 – 1 : 27 +1 =26 : 28 = 13 : 14

23. (B) 5%

A = P ( 1 + �

�� )�

441 = 400 ( 1 + �

�� )�

��

�� = ( 1 +

�

�� )�

��

�� = ( 1 +

�

�� )

r = 5%



24. (a) Number of women = (296000 – 166000)

= 130000.

Total number of literates

= .148000100

50296000

Literate men = .116200166000100

70

Literate women = (148000 – 116200) = 31800

25. (a) Rs. 625

Same as que. 21

26 (a) 10% decrease

Let us assume the price is RS.100

Decrease = 25% = Rs. 25

Now the price = Rs. 75

Increase = 20% = 75 × ��

�� = Rs. 15

Now the price = 75 + 15 = Rs. 90

Hence price decrease by 10%

27. (d) Minimum passing marks=120+80=200 = 40%ofthe maximum marks

Maximum marks = 500

28. (b) 11 days

Let us assume the work = 100 units

A B

Time taken (days) 25 20

1 day work (units) 4 5



1 day work by both A and B = 9 units

5 days work by both = 45 units

Remaining works = 100 – 45 = 55 unit

5 unit is done by B in = 1 day

55 units is done by B in = 11 days

29. (b) If B’s salary is Rs. 100, then A’s salary = Rs. 110% of Rs. 100 = Rs. 110.

But salary of A =Rs. 440 = Rs. 110 × 4

Hence salary of B = Rs. 100 × 4 = Rs. 400

If C’s salary is Rs. 100, then B’s salary = 80% of Rs. 100 = Rs. 80.

But salary of B = Rs. 400 = Rs. 80 × 5

Hence salary of C = Rs. 100 × 5 = Rs. 500

30. (c) Peter completes in 1 day =12

1

piece of work Mohan is 50% more efficient than Peter

Mohan completes in 1 day =8

1

100

150

12

1 piece of work.

Mohan takes 8 days to complete the work.

31. (a) MP =90

10016560= Rs. 18,400

and CP =90

10016560= Rs. 14,400

If no discount is allowed, profit%

= 10014400

1440018400

= %.9

727

32. (b) 3 days



A can do a work in = 10 days

B can do = 15 days

Hence C will do the work in = �� ×��

�� = 6 days

They can do the work together = �� × �� ×�

�� ×�� ×�� ×�� =3 days

Or

Let us assume total work = 30 units

A B C

Time taken (days) 10 15 6

1 day work 3 2 5

Hence C will complete the work in 6 days

Now let us assume the total work = 30 units

A B C

Time taken (days) 10 15 6

1 day work (units) 3 2 5

Total work in one days by all = 10 units

Total number of the days taken by them = 3 days

33. (d) 17.25% gain

Total cost price of house for A = Rs. 10000

Selling price for A = 10000 × ��

�� = Rs. 11500

COST price for B = Rs. 11500

Selling price for B RS= 11500 × ��

�� = Rs. 9775

Hence cost price for A = Rs. 9775

Total profit of A = 11000 - 9775 =Rs. 1725



Profit percentage = ��

�� × 100 = 17.25%

34. (a); 48510 cu cm, 5610 sq cm

Hint: r=7 cm, R = 28 cm and h=45 cm

Capacity =Volume of the frustum

= �

��ℎ(�� + �� + Rr )

= �

�×

��

�× 45 ( 28� + 7� + 28 × 7 ) = 48510 cu cm

Surface area of the bucket = �(� + �)� + �(�� + �� ) = 5610 ��

35. (d) 20%

Total CP= Rs(1200 + 200)= Rs 1400 and SP =Rs 1680

SP> CP, Hence profit and profit % = ��

�� × 100=20%

36. (a) 3750cub m, 12.5m

Total depth = 15m

If 500 cubic meters of water be drawn off the tank, the level of the water in the tank goes down by 2 meters.

Then capacity of tank = ��

� × 15 = 3750 ��

Let us assume that breath is b. then lbh = 3750 cu m.

Hence b = ��

�� =

��

��×�� = 12.5m

37. (b) Part filled by both in 1 min =

20

1

30

1

= .12

1

60

5

60

)32(

Both will fill the tank in 12 minutes

38. (c) RS. 1440

discount = 20 + 10 - �� ×��

�� = 28%

total discount = 2000 × ��

�� = Rs. 560



hence selling price = 2000 – 560 = Rs. 1440

39. (c) Let the slower pipe fill the tank in x minutes.

Then, the faster pipe will fill it in 3

xminutes.

Part filed by both in 1 min

= .431

xxx

But, part filled by both in 1 min. = .36

1

36

14

x x = (36 × 4) = 144.

Hence, the required time is 144 minutes

40. (c) Two equal sides are base and height of the triangle.

.'. Area of the triangle = Base x Height =-�

�x8x8= 32cm2

41. (b) 14% gain

= 100×��

��×

��

�� =114 ; so gain of 14%

42. (a) 5:6

Ratio of curved surface area = 5:8 ; Ratio of height = 3:4 ; Ratio of radius = 5:8 \ 3:4 = 5:6

43. (c) 35cm

Ratio between areas of two circles =1:25

Ratio between radii of two circles= 1:5

.'. Radius of second circle=5 x 7 cm = 35 cm

44. (c) Rs900, Rs 1008

Cost price = ��

�� ×100 = Rs. 900

Initial profit = 12%



Initial selling price = 900 + 900 × ��

�� = Rs. 1008

45. (b) 1% decrease

Rate of change of area = l +b + lb/100

= 10 -10 -�� ×��

�� = -1

Hence area will decrease by 1%

46. (b) x = 180 – 40 = 140º

47. (a) -�

�

sum of the roots = ��

�

48. (b) ABC = 90

y + 90 + 50 = 180

y = 40º

49. (b) 204

��

�� = �� +

�

�3 + a +

�

�

=204

50. (b) 40cm

Perimeters of two squares are 24 cm and 32 cm.

Hence sides of two squares are 6cm and 8cm.

Hence areas of two squares are 36�� and 64��.

Hence area of third square is 100��

Hence side of third square is 10cm and perimeter of third square is 40cm.

51. (a) Marked price of the article = 21 × ��

�� = Rs. 30



52. (d) Ratio of volumes of two cylinders = 2� × 5 ∶ 3� × 4 = 5 : 9

53. (c) 22.86%

CP = Rs. 1404

Profit = Rs. 196

SP = Rs.(1404 + 196 ) =Rs. 1500

if the article is sold at 15% more on selling price

then SP = Rs.( 1500 + 1500 × ��

��) = Rs 1725

hence profit is = Rs. 321

PROFIT % = 22.86%

54. (a) Case I : If 1 = 2, 1 = 2 = 3 = 4

Number of distinct angles is one.

55. c) Rs9375

D = P × ��

�� hence P = D ×

��

�� = 15 ×

��

�� = 15 × 625 = Rs. 9375

56. (a) 60° E

C

D

A B

O

Given COA = 36º and DOA =

3

1COA then DOA = 12°

Then ∠�� = 24°

BOC = 180° - COA = 180° – 36° = °

∠�� = �

� × 1 44° = 36°

DOE =∠�� + ∠�� = 24° + 36° = 60°



57. (d) 2

Roots of the equation x2 – 7x + 10 = 0 are 2 and 5.

Roots of the equation x2 – 10x + 16 = 0 are 2 and 8.

Hence common root is 2.

58. (c) 13.5s

Length of train = 75m

Speed of train = 20 kmph = ��

� mps

Time taken by train to cross a man standing at platform = 75 / ��

� = 13.5s

59. (b) 4

We know that 5� + 12� = 13�

Hence √� = 2 then x = 4

60. (a) 32.5%

M P = Rs. 480

Discount = Rs. 120

S P = Rs. (480 – 120 ) = Rs. 360

Another discount = 10% = Rs. 36

Total discount = Rs. 156 = 32.5%

61. (b) FDG = KCD (corresponding angles)

= ECA (vertically opp. angles)

ECA = 55º

EAC = 40º

E = 180º – (55º + 40º) = 85º

xº = 85º (corresponding angles)



62. (a) Area of field = 90 x 50= 4500 m2

Area of field dug out = 25 x 20 = 500 m2

.'. Area of remaining field = 4500m2-500 m2=4000m2

Volume of the earth dug out = 25 x 20 x 4 = 2000 m3

2000

Field will rise by = 0.5 metre

63. (d) 1

��65 = �� 25

��25 + ��25 = 1

64. (a) Let the numbers are x and y.

, Then x +y = 17 and �� - �� = 85

X - y = �2 − �

2

x + y =

��

�� = 5

65. (c) sec2 + cos2 =

2cos2cos

1

=

2

cos

1cos

+ 2 2.

66. (c) rl = × (12)(13) = 156 cm2

67. (a) 1

tan20tan40tan50tan70 = tan20tan40cot 40cot 20 = 1



68. (c) (1- x6)

(x+ 1) x (1-x) x (x2 – x + 1) x (x2 + x +1) = (x3 + 1) x (1- x3) = (1- x6)

69. (b) a = b

70. (c) The required ratio

=)111111(2).248(

)282448(2

= 7 : 24

71. (c) Relative speed = (72 – 60) km/h = 12 km/h

= 12 × 18

5m/s =

3

10m/s

Total distance covered = (240 + 240) m

= 480 m

Required time =

10

3480 = 144 sec

= 2 min 24 sec

72. (d) 30(√3 + 1 )

A

AB = 60m



Let us assume that PM = h

∠PBM = 45

�ℎ�� PM = BM = H ℎ�� AM = h + 60

��

�� = tan30

�

�� =

�

√�

h (√3 – 1) =60 h = 30(√3 + 1 )

73. (a) 3

4

if x: y = 7: 3 then putting x = 7 and y = 3 in ��

�� , we find the answer is

�

� .

74. (b) 5 )32( cm

Volume of shell = 4

1× original cylinder

Volume of empty space

= 4

3× original cylinder

r2 × (1000) = 4

3× (10)2 (1000)

r2 = 4

3×100 r = 35 cm

Hence thickness is = 10 - 35 cm = 5 )32( cm

75. (c) 4 km/h

By given condition, 4 =1

6

1

10

xx

x = 4 km/h.

76. (a) 1 : 1 : 1

Given that a + b + c = 4√3 . Here taking a = b = c we get a = b = c = �√�

� then a2 + b2 + c2 = 16.



Hence a : b : c = 1 : 1 : 1

77. (a) Total distance from temple = 22 )240(x

Where x =3º60tan

hh

So distance =2

2

22

)240(3

h

hbut ,)240(

3

h

=3

1

3

1

)240(3

22

2

h

h

After solving h = 60 6 metre.

78. (c) He covers 10 km in 1 hour (i.e. in 60 minutes)

He will take 6 minutes in covering 1 km.

He rests for 4 minutes after every km.



Time taken =(6+4)minutes= 10 minutes for every km

Time taken (for first 9 km)= 9 * 10 = 90 minutes

Time taken to cover last one km = 6 minutes

.". Total time taken = 90 + 6 = 96 minutes

Hint: Rest time after 10th km is not added as he has reached his destination.

Alternative Method:

He covers 10 km in 1 hour (i.e. in 60 minutes) and rests for 4 minutes after every km.

He will take rest9 times.

Total time taken = 60 + 9 ×4 = 96 minutes

79. (c) 30°

∠�� = ∠�� = 90°

THEN ∠�� + ∠�� = 180°

LET ∠�� = � �ℎ�� ∠�� = 5�

Hence x + 5x = 180°

X = 30°

80. (c) 10 21 meters

A

B 45 C

Here BC = 10m

∠�� = 45 THEN AB = BC HENCE AB = 10m

Now ��

�� = sin 45 HENCE

��

�� =

�

√� HENCE AC = 10√2m

Hence length of tree = 10 + 10√2 = 10 21 meters

81. (d) Distance covered by the first train in 3 hours = 3 * 80 = 240 km.

Second train gains = 110-80 = 30 km/h



Second train will meet the first train after ��

�� =8 hours.

Distance covered by the faster train in 8 hours = 8 x 110 = 880 km

82. (a)

10� − 5� = √75 = 5√3

Chord length is double = 10√3

83. (c) -1 +4+3 =6

84. (c)

∠�� = 140°

∠�� = 70 ° ; opposite angle = 180-70 =110°

85. (c) �� + �� + �� + 3�� = (� + � + �)(�� + �� + �� + �� − �� − ��)

A +b +c= �

�+

�

�+

�

� =��

��

86. (c)

87. (d)

88. (c) 15% of the total in 2009 are Engineers which is 60 Total in 2009 = 400 MBA's are 30% US are 15%

Difference = 100

15× 400 = 60

89. (d) In 2010 : Total = 800

maximum difference = 45 – 10 = 35%.

10 5

40°

140° A

B

70° D

c E



100

35× 400 = 60.

90. (b)

91. (a) Ratio of circum radius and in-radius for equilateral triangle is always 2:1 , so 5cm is in-radius

92. (b) �� = 180° − 160° = 20° ;��.�� = ��°

�� =

��°

��° =18°

93. (b)

√�

� � = ℎ ;

√�

� × 6 = ℎ ; ℎ =

3√3 ; ��

�� ℎ��ℎ� �� √3.�� ℎ� ℎ��ℎ� �� 2: 1.

94. (c)

�� ∠�� ℎ�� 60° �ℎ��ℎ �� = 30° ; ∠�� 40° ; ∠�� ℎ�� .�.20° ,∠�� 180− 20− 30

= 130°

95. (b) 3� + �

� �= 5 multiply both sides by

�

�

2� + �

�� =

��

� �� ℎ ��

8�� +�

��+ 3 × 2� ×

�

�� (��

�) =

��

��

8�� +�

��=

��

��= 30

��

��

96. (a)

Length of transverse circle = �� (��) � = 8 = ��

� ; � = √145

97. (d) add 1+1+1 both sides

�

��+ 1 +

�

��+ 1

�

��+ 1 = 1 + 3 = 4



We get ��

��+

��

��+

��

��= 1 + 3 = 4

�

�� +

�

��+

�

��= 4

98. (b) In these triangle ∠� = ∠� , �� ∠� = ∠� ; �ℎ�� ∠� = ∠� �ℎ��ℎ �� 180− 69− 37 = 74°

99. (d) Suppose ∠� = 90° �ℎ�� 90°, �� 90°��

�� 180°. So value will be 0+0+0+ =0 as �� 90° = 0

100 A.

AD = 27 cm

Centroid divides AD in to 2:1 .

DO will be �

�× 27 =9 ; DN = 12 ; ON=12-9 =3 cm

A

B C

N O

D

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Pinnacle SSC CGL Tier 2 Test Series Paper Code 707 Math Pinnacle SSC CGL Coaching Centre, Gurudwara Road, Model Town, Hisar, 9729327755 Page 1 Pinnacle SSC CGL Tier 2 Test Series Paper