ssc mains (maths) mock test-12 (solution)

Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER

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MOCK TEST PAPER - 12 (T-II) 2013 (SOLUTION)

1. (B)21

1

+

32

1

+

43

1

+ ...

121120

1

=)12(

)12(22

+

)23(

)23(22

+

)34(

)34(22

..... + )120121(

)120121(22

= ( 2 1 ) + 3 2 ) + 4 3 )

+.... ( 120 119 ) + ( 121 120 )

= 1 + 121 = 11 1 = 10 Ans.

2. (A)2

5

5

10 + 125 =

2

5

5

10 +

1

55

=52

510205 =

22362

35

=

472.4

35

= 7.826 Ans

3. (D)

3/2

216

1

3/2

36

1

=

2

6

1

= (6)2 = 36 Ans.

4. (C) Required measurement

= HCF of 403 kg, 434 kg and 465 kg

= 31 kg Ans.

5. (C) Let R = Age of Randheer,

A = Age of Anup

So, ATQ,18

6R = A

Given, Mahesh = 5 yrs

Anup = (5 2) yrs = 3 yrs

R = 18 3 + 6 = 60 yrs Ans

6. (B) Let x be the number of buffaloes and

y be the number of ducks.

Then, total number of legs = 4x + 2y

and total number of heads = x + y

So, 4x + 2y= 2(x + y) + 24

2x = 24

x = 12 Ans.

7. (A) Cost of 9 lemons = 4 48 paise

= 144 paise

= cost of 4 mangoes

So, cost of 3 mangoes = 4

144 3

= 108 paise

= Cost of 5 apples

Cost of 9 oranges = 108 paise

Cost of1 orange =9

108

= 12 paise Ans.

8. (A) Sum of digits from numbers 1 to 10 = 46,

Sum of digits from numbers 11 to 20 = 56

&Sum of digits from numbers 21 to 29 = 63

Sum of digits of the given numbers

= 46 + 56 + 63 = 165

Again,

Sum of digits of the number 165 = 1 + 6 + 5 = 12

Now, when 12 is divided by 9, we get

remainder = 3 Ans.

9. (A) Let x be the number.

Then calculation done by the student

= 6

12x=112

x = 660

So, the correct answer = 6

660 + 12

= 122 Ans

10. (D) Let x be the number of students and

w be their average weight.

ATQ,

1

50

x

xw= w + 1

x + w = 49 .... (i)

Again,2

5050

x

xw= w + 1.5

1.5x + 2w = 97 .... (ii)

From equations (i) and (ii),

we get, w = 47 Ans

11. (A) Let x be the first number

x x x x x + ( + 1) + ( + 2) + ( + 3) + ( + 4)

5 = 41

or, 5x + 10 = 205

5x = 195

x = 39

So, Product of A and E = x (x + 4)

= 39 43

= 1677 Ans.


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12. (A) Let Sunil, Sumant and Surat get x,

y and z respectively.

So, 100

15x=

100

25y =

100

35z

x = 2y = 3z

x : y : z = 6 : 3 : 2

Required difference= 236

26

7700

= 2800 Ans

13. (C) Let the speed of train on level terrain

= x km/h

So, the speed of train through mountainous

terrain = (x 10)km/h

Mountainous Level299 km

111 km 188 km

ATQ,

x

188 +

10

111

x = 7

188 1880 111

( 10)

x x

x x

= 7

7x2 369x + 1880 = 0

x = 47 km/h Ans.

14. (D)TV Newspaper

65% 40%

Number of people who either watch TV

or read newspaper = 65% + 40% 25%

= 80%

Number of people who neither watch

TV nor read newspaper.

= (100 80)%

= 20% Ans.

15. (D) Let required amount of second solution

to be added = x L

So,

x

x

20

52015= 10

300 + 5x = 20 + 10x

x = 5

100 = 20L Ans.

16. (A) Milk in mixture

= 6 100

25 + 4

100

30 =

100

270

Required percentage = 100

)46(270

= 27% Ans.

17. (B) Let x be the total score in the innings.

So, the highest score = 9

2x

And, the next highest score

= 9

2 of the remaining runs

= 9

2

xx

9

2

So, ATQ, 9

2x

9

2

xx

9

2= 8

x x + 9

2x =

2

98

x = 22

998

= 162 Ans.

18. (B) (13 + 22 + 33 + ... 153) (1 + 2 + 3 + ... + 15)

= 15 2

)15313(

2

1615

= 1245 120

= 1125 Ans.

19. (B) Let total marks be 100.

So, the minimum marks required to be

pass = 40% of 100 = 40 marks

ATQ, Marks obtained by A = 40 40 100

10

= 36 marks

&, Marks obtained by B = 36 9

100

100

36

= 36 4

= 32 marks

So, Marks obtained by C

= (36 + 32) (36 + 32) 10017

700

= 68 28 = 40 marks Ans.

20. (B) Let x be the monthly salary of Mohan.

ATQ,

10

4x +

10

6x

100

50+

10

3x

100

30 + saving = x

10

4x+

10

3x +

100

9x + 630 = x

630 = 100 70 9

100

x x x


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630 = 100

21x

x = 3000 Ans

21. (B) the total number of men employed =50

So, total number of wommen employed = 100

& total number of children employed = 150

Now, let the men, women and children

wages are 6x, 3x and 2x respectively.

50 6x + 100 3x + 150 2x = 4500

or, 900x = 4500

x = 5

Per day wages paid to a man, a woman

and a child are 30, 15 and 10.

Weekly wages paid to a man, a woman

anda child are 210, 105 and 70 Ans.

22. (B) Let the present and last year salaries of

Mahesh and Suresh are x, x' and y, y'

respectively.

'

'

y

x=

5

3,

x

x'=

3

2

andy

y' =

5

4

y

yx

x

'

'

=

5

43

2

'

'

y

x

x

y=

12

10

5

3

x

y=

12

10

x

y=

36

50

Also, x + y = 43000

x + 36

50x = 43000

x = 86

3643000= 18000 Ans.

23. (D) Actual amount to be paid

= 20000

3

100

101

= 26620

Amount that Rohit have to pay now

= 20000

2

100

101

100

151

= 27830

So, required extra amount

= 27830 26620

= 1210 Ans.

24. (A) Let the value of second rate of interest

be x% and equal amounts be P each.

So, ATQ,

P

6

100

51

= P

3

1001

x

2

100

51

=

1001

x

100100

105105

=

100

100 x

110.25 = 100 + x

x = 104

1% Ans

25. (B) Upstream speed = 2 km/h

Downstream speed = 10

1 60 = 6km/h

Speed in stationary water

= 2

62 = 4 km/h

Required time = 4

5 = 1 hr 15 min Ans.

26. (D) Let the man went up stream for x km.

He turned back downstream for (x 2) km.

So,)5.15.4(

x +

2

(4.5 1.5)

x

= 2 h 10 min

6

22 xx =

6

12 hour

3x 2 = 13

x = 5 km Ans.

27. (A) Let x% be the other discount.

So, 65 100

90

100

)100( x = 56.16

100 x = 56.16 100 100

65 90

100 x = 96

x = 4% Ans.

28. (A) Let cost price of the article = x.

So,

x 100

5.117 + 11.55 = x

100

92

100

130


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x = )5.11713092.0(

55.11

100

= 550 Ans.

29. (B) 12.5% of profit= 100

5.12 880 = 110

Remaining 770 is divided in the ratio= 5000 : 6000 = 5 : 6

Profit to Anu = 11

5 770 + 110

= 460

Profit to Bimla = 6

11 770

= 420

Required profits are 460 and 420 Ans.

30. (B) Ratio to capitals= 1

3 :

3

2 = 1 : 2

Ratio of profits = 5

3 :

5

2 = 3 : 2

Let Y's money was used for x months.

(1 9) : (2 x) = 3 : 2

x = 3 months Ans.

31. (B) A's 1 day work = 40

1

So, A's 5 day work = 40

1 5 =

8

1

So, remaining work= 1 8

1 =

8

7

B completed the remaining work in 21 days

So, 1 day's work of B = 218

7

=

24

1

Both (A + B)'s 1 day work = 40

1 +

24

1

= 120

53 =

120

8=

15

1

Hence, both complete the work in 15 days. Ans.

32. (D) Let (A + B) together complete the work inx hours.

A completes the work in (x + 8) days.

& B completes the work in

2

9x days.

Using formula,

2

252

2

9)8(

x

xx

= x

x2 + 8x + 2

9x + 36 = 2x2 +

2

25x

x2 = 36 x = 6 days Ans.

[* Read days as hours]

33. (C) Let the present population of male and

female be x and y respectively.

ATQ, x + y = 5500

and100

111x +

100

120y = 6330

111x + 120y = 633000

From equations (i) and (ii), we get

y = 2500 Ans.

34. (C) Let the ouput be x and percentage be a.

Then,

x

2

1001

a= 2x

1 + 100

a = 2

100

a = 2 1

a = ( 2 1) 100%

= 100( 2 1)% Ans.

35. (A) Let the cost of the product = x.

So, ATQ,

x 100

110

100

115

100

125 = 1265

x = 1265 100

110

100

115

125

100

= 800 Ans.

36. (C) Total work = 124 120 = 14880 men-days

Work completed in 64 days

= 3

2 14880 men-days

= 9920 men-days

Remaining work for remaining 60 days

= (14880 9920) men-days

= 4960 men-days

Now, 1

11

W

DM=

2

22

W

DM

9920

64120=

4960

602 M

M2

= 64

Number of workmen who can be

reduced =120 64 = 56 Ans.

37. (C) Till 3 pm, part of tank filled

= 8

2 +

12

1 =

4

1 +

12

1 =

12

4 =

3

1part


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Remaining part = 1 3

1 =

3

2

Now, let x be the time taken by all three

pipes work to fill the remaining part of tank

1 1 1

8 12 6

x = 3

2

48

)846( x=

3

2

48

2x=

3

2

x = 16 hour

Hence, the tank will be filled at 1 pm

+ 16 hour = 5 am. Ans.

38. (D) Let leakage alone can empty the full

cistern in x h.

Then,

2

92

9

x

x

= 5

2

9x = 5x

2

45

2

1x =

2

45

45 h Ans.

39. (D) ATQ, original (prime) cost of article = 6.

Original raw material cost = 2

So, New cost of raw material = 2 5

12

= 4.80

Also, original manufacturing expenses

= (6 2) = 4

So, new manufacturing expenses

= 4 4

5 = 5

New cost of article = 4.80 + 5

= 9.80 Ans

40. (A) Let the shares of Anita, Bindu and

Champa are 11x, 18x and 24x respectively.

So, 1105 = 11x + 10 + 18x + 20 + 24x + 15

1105 = 53x + 45

x = 20

Amount received by Champa

= 24x + 15

= 24 20 + 15

= 495

41. (A) Total marks got by the students in

8 subjects = 8 57 = 696

Total marks got by the student in

6 subjects = 6 85 = 510

Remaining marks in 2 subjects

= 696 510 = 186

Let the highest marks be x, then the

second highest marks will be x 2.

x + x 2 = 186

2x = 188

x = 94

Highest marks = 94 Ans.

42. (C) Total score of first three friends

= 15 3 = 45

and total score of last three friends

= 16 3 = 48

Total score of four friends

= 45 + 19 = 64

Score of first friend

= 64 48 = 16

So, required percentage

= 48

16 100% = 33

3

1% Ans.

43. (B) When a value is first increased and then

decreased by the same percentage, then

the initial value is always decreased by

100

2x%. (irrespective of initial value)

So, loss percent = 100

)15( 2= 2.25% Ans.

44. (B) Let rate of popualtion increase = R% per annum

So, 4800 = 3600

3

1001

R

3

4=

3

1001

R .... (i)

Now, the population after 3 years

= 4800

3

1001

R

= 4800 3

4[from (i)]

= 6400 Ans.

45. (D) Let the two numbers are x and y.

Then,

their AM =2

yx = 5

x + y = 10 ... (i)

and their GM= xy = 4

xy = 16

Now, (x y)2 = (x + y)2 4xy

= 100 64 = 36

or, x y = 6 ... (ii)


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From equations (i) and (ii),

x = 8 and y = 2

So, the required numbers are 2 & 8 Ans.

46. (B) Let the first term and common term of

the AP be a and d respectively.

ATQ, (a + 5d) + (a + 14d) = (a + 6d) +

(a + 9d) + (a + 11d)

2a + 19d = 3a + 26d

a + 7d = 0

8th term is 0. Ans.

47. (A) Let M stands for Mohan and R stands for

Ram.

Case : I (M + 30) = 2(R 30)

M 2R= 90 ... (i)

Case : II (R + 10) = 3(M 10)

R 3M = 40 ... (ii)

Solving Eqs. (i) and (ii), we get

M = 34 and 62 Ans.

48. (B) Given,

zyx

x

2 =

zyx

y

2 =

zyx

z

2 = a

x = a(2x + y + z), y = a(x + 2y + z)

z = a(x + y + 2z)

x + y + z = a(4x + 4y + 4z)

4a = 1 a = 4

1 Ans.

49. (C) Roots of a quadratic equation

ax2 + bx + c = 0 are real if b2 4ac > 0

Option (a): 3x2 4x + 5 = 0

b2 4ac = (4)2 4(3)(5)

= 44 < 0

Hence, roots are not real.

Option (b): x2 + x + 4 = 0

b2 4ac = (1)2 4(1)(4)

= 1 16 = 15 < 0

Hence, roots are not real.

Option (c): (x 1) (2x 5) = 0

x = 1 and x = 2

5> 0

Hence, roots are real.

Option (d): 2x2 3x + 4 = 0

b2 4ac = (3)2 4(2)(4)

= 9 32 = 23 < 0.

Hence, roots are not real. Ans.

50. (C) 2x2 7xy + 3y2 = 0

2

2

y

x7

y

x + 3 = 0

y

x=

a

acbb

2

42

= 7 49 24

2 2

= 4

57 = 3,

2

1

y

x=

1

3and

y

x=

2

1Ans.

51. (A) Let f(x) = 2x3 ax2 (2a 3)x + 2

(x + 1) is a factor of the above expression,

Remainder f (1) = 0

f (1) = 2(1)3 a(1)2 (2a 3) 1+ 2 = 0

2 a + 2a 3 + 2 = 0

a 3 = 0

a = 3 Ans.

52. (B) Let the common base be x m.

Then, area of triangle = area of parallelogram.

2

1 x Altitude = x 100

Altitude = 200 m Ans.

53. (D) For two similar triangles ABC and PQR.

A

B C

P

Q R

10 cm

PQ

AB=

24

36

AB =24

36 10 = 15 cm Ans

54. (C) R

S

N M

OQ SQ = 23 cm

17

cm

17cm

P

8

In PQO,

(17)2 = (8)2 + (OQ)2

(OQ)2 = 289 64 = 225

OQ = 15

OS = 23 15 = 8

Now in ORS,

(RS)2 = (17)2 (8)2

= 289 64 = 225


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Ph: 011-27607854, 08860-333-333 7

RS = 15 cm

Hence, length of other chord

= 15 2 = 30 cm Ans.

55. (D) Let the three points be A

3

8,0 , B(1, 3)

and C(82, 30)

Then, AB =

22

3

83)01(

=

3

10

BC = 22 )330()182(

= 7296561

= 7290 = 1027

CA =

22

3

830)082(

=9

)82(10 2 = 10

3

82

Now, AB + BC = 3

10 + 1027 =

3

1082

Since, AB + BC = AC, it means points A,

B and C are collinear. Ans.

56. (A) Given, AD2 = AB2 + BC2 + CD2

A

BC

D

?90

In ABC,

AC2 = AB2 + BC2

AD2 = AC2 + CD2

C = 90 Ans.57. (A) Let the length and breadth of a cuboid be

8x cm and 7x cm respectively.

Volume of cuboid = l b h

1120 = 8x 7x 5

x2 = 578

1120

= 4

x = 2

Length and breadth of a cuboid are 16 cm

and 14 cm by 2 cm.

Hence, length exceeds its breadth by 2 cm. Ans.

58. (C) Let length and breadth of blackboard

be x m and (x 8) m respectively.

ATQ, x (x 8) = (x + 7) (x 12)

x2 8x = x2 5x 84

x = 3

84= 28 cm

Length = 28 cm

Breadth = x 8 = 20 m Ans.

59. (C) Let the outer radius of pipe be r.

Then, volume of the metal

= 28 (r2 82)

1496 = 28 (r2 82)

r2 64 = 2228

1496

7

r2 = 17 + 64 = 81

r = 9 cm Ans.

60. (D) Let dimensions of the rectangular stone

be 3x, 2x, x respectively.

Volume of stone= 10368

3x 2x x = 10368

6x3 = 10368

x3 = 1728

x = 12

Dimensions are 36 dm, 24 dm, 12 dm.

Entire surface area of a stone.

= 2(lb + bh + hl)

= 2(36 24 + 24 12 + 12 36)

= 2(864 + 288 + 432)

= 3168 dm2

Total cost of polishing

= 3168 0.02

= 63.36 Ans.

61. (A) Total area of two parks

= (82 + 62) = 100

Area of bigger park

= 100

r2 = 100

r2 = 100

r = 10 m Ans

62. (D) Total area of playground

= 750 2 rh

= 750 2 7

22

2

70 150

= 2475 104 cm2

= 2475 m2

Total cost of levelling = 2475 2

= 4950 Ans.

63. (C) Since AD bisects A.

DC

BD

AC

AB

x

x

89

6

9x = 48 6x

x = 15

48


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x = 3.2 cmA

B CD

9 cm40

406 cm

8 cm

x cm

DC = 8 3.2 = 4.8 cm Ans.

64. (A)

90 m

A

B

C

60

Let AB be the string.

AB

BC = sin 60

BC = 90 2

3= 45 3 m Ans.

65. (D) cos6 x + sin6 x + 3sin2 x cos2 x

=(cos2 x)3 + (sin2 x)3 + 3sin2 x cos2 x(sin2x

+ cos2 x)

= (cos2 x + sin2 x)3

= 1 Ans.

66. (A) Given, a sec + b tan =1 ... (i)

and, a2 sec2 b2 tan2 = 5 ... (ii)

From Eqs. (i) and (ii),

a sec b tan = 5 ... (iii)

From Eqs. (i) and (iii),

a sec = 3 ... (iv)

b tan = 2 ... (v)

From Eq. (iv),

a2 sec2 = 9

a2(1 + tan2 ) = 9

a2

2

41

b= 9

a2b2 + 4a2 = 9b2 Ans

67. (A)

A B

C

D E

60 m

30

x m

150

m

Let the height of the shorter tower be x m.

Then, from CDE,

tan 30 = 60

)150( x

3

1=

60

150 x

(150 x) = 3

60= 2 3

x = 150 20 3

= 116 m (approx) Ans.

68. (D) x = 2 + 22/3 + 21/3

The value of x is greater than 2, let us try

it for x = 2 and 3.

For x = 2, x3 6x2 + 6x = 4

For x = 3, x3 6x2 + 6x = 9

x3 6x2 + 6x < 0 for x = 2 and 3

Hence, option (d) is the answer. Ans.

69. (A) t 2 = 3 4 + 3 2

(t 2)3 = 4 + 2+ 3 3 4 3 2 ( 3 4 + 3 2 )

t3 23 3 2 t(t 2) = 6 + 6t 12

t3 6t2 + 6t 2 = 0 Ans.

70. (B) Sum of roots = a + b = 6

Product of roots = a b = 6

Now, a2 + b2 = (a + b)2 2ab

= 36 12 = 24 Ans.

71. (A) Using AM > GM

2

33 xx > xx 3.3

3x + 3x > 2 Ans.

72. (B) a + b + c = 0

a2 = (b + c)2 or a = b c

Given expression

bca

a

2

2

+ 2

2

b

b ca +

abc

c

2

2

= bccb

cb

2

2

)(

)( +

2

2 ( )

b

b c b c +

2

2 ( )

c

c b b c

=bccb

cb

22

2)(+

bccb

b

22

2

+ bcbc

c

22

2

= bccb

cbbccb

22

2222 2=

bccb

bccb

22

22 )(2

= 2 Ans.


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73. (C) Sum of all the exterior angles of the

polygon is 360.

x

75

105

65

180 x

90

90 115

x + 75 + 90 + 115 = 360

x = 80 Ans.

74. (B)

D

C

A

B

Given A = 2C

Also, A + C = 180

( ABCD is cyclic)

C = 60

and A = 120

Now, B D = 3

1A = 40

B = 110

D = 70

Minimum difference between any two

angles is 10. Ans.

75. (C) Given, OBC = OCB = 37

[ Angles opposite to equal

sides of a triangle are equal]

COB = 180 (37 + 37)

= 106

BAC = 2

1COB =

2

106

= 53 Ans.

76. (C) In OC1C

2,

(OC1)2 = (OC)2 + (CC

1)2

(r + 1)2 = (PC OP)2 + 1

(r + 1)2 = (2 r)2 + 1

r2 + 1+ 2r = 4 + r2 4r + 1

6r = 4

r =3

2Ans.

77. (B)AB

BD=

AC

DC

A

B CD

DC

BD =

5

3

BD : DC = 3 : 5

Now, BD = )53(

3

6

= 8

18 =

4

9 = 2.25 cm Ans.

78. (C) Area of four walls = 2h(l + b)

l = 2b and h = 4 m

120= 8 3b

b = 24

120 = 5 and l = 10

Area of floor = l b = 10 5 = 50 m2Ans.

79. (C) Initial length of rectangle = L

New length of rectangle = 2

3L

Initial breadth of rectangle = B

New breadth of rectangle=3

1B

Initial Area = L B

New Area = 2

3L

3

1B =

2

1LB

Hence, change in area is 2

1. Ans.

80. (D) Let water in the tank is h deep.

3 2

3 h = 8 2.5

h = 33

25.28

= 4.44 ft Ans.

81. (A) Let the water level rises by x m.

90 40 x = 150 8

x = 4090

8150

= 0.3333 m

= 33.33 cm Ans.


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82. (B) Curved Surface Area of 50 pillars

= 50 2 3.14 2

50 400

= 314 104 sq. cm. = 314 sq. cm.

Labour charges= 314 0.50

= 157 Ans.

83. (A) Total surface area of remaining solid

= curved surface area of the cylinder

+ area of base + curved surface area

of cone

= 2 rh + r2 + rl

= 2 8 15 + (8)2 + 8 17

= 240 + 64 + 136 = 440 cm2 Ans.

84 . (C)Let the breadth and height of b and h

respectively.

Then, 70 2.2 = 14 b

b = 11 m

Also, 14 b h = 70 11

h =1114

1170

= 5m Ans.

85. (B) Total cost of plastering inner surface

= 2 7

22

2

7 22.5 3

= 1485 Ans.

86. (C)

A B

C

D

2h

d

Let BC = h be the height of the pillar, then

CD = 2h.

Also, let BAC = CAD = and AB = d

In ABC,

tan = d

h

and in ABD,

tan2 =d

h3

2tan1

tan2=

d

h3

21

2

d

h

d

h

= d

h3

2

3= 1

2

d

h

d

h=

3

1Ans.

87. (D) Given,

A

A

cos

sin=

7

4

2

cos

sin7

3cos

sin7

A

AA

A

=

47 3

74

7 27

=

24

34

=

6

1

88. (C)

30 45

m

DC A

B

h

x

)13(50 -

Distance travelled in 2h, CD = )13(50 m

Let AB be the building and its height be h

and AD be x.

x

h = tan 45 h = x .... (i)

50( 3 1)

h

x = tan 30

h = 3

x+

50( 3 1)

3

.... (ii)

From eqs. (i) and (ii),

h 3

h =

50( 3 1)

3

h( 13 ) = 50( 3 1)

h = 50 m Ans.

89. (B)Let AB = h m and DC = EB = x m

A

B

CD

E30

6010 m

x m

(10 + )mh


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Ph: 011-27607854, 08860-333-333 11

Here, pole DE = 10 mAC = tower = 10 + h

In AEB, we have

tan 30 = EB

AB=

3

1 =

x

h... (i)

x = h3

In ACD, we have,

tan 60 = CD

AC =

x

h10

3 = x

h10

x = 3

10 h ... (ii)

From eqs. (i) and (ii), we get

3 h = 3

10 h

3h = 10 + h

h = 5 cm

AC = 10 + h = 15 m Ans.[Remember the result in the above : heightof tower = 1.5 height of the pole]

90. (C) When x = 0sin x + cosec x = 0 + =

When x = 2

sin x + cosec x = 1 + 1 = 2

When2 < sin x + cosec x <

Hence, sin x + cosec x > 2 Ans.

91. (C)

19861987198819891990

44.5046.0057.0082.00

138.00

44.5090.50

147.50229.50367.50

YearTotal Sales ofall company

Cumulative Sales

Required ratio = 50.367

50.90 =

735

181 =

4

1Ans

92. (A)Average sales of Celia

= 5

505.145.161550.18

= 5

50.114 = 22.90 Ans.

93 (A) Growth rate for

Alpha = 20

2025 100 = 25%

Baron = 50.11

50.1100.15 100 =30.43%

Celia = 5.14

5.1450 100 = 244.8%

Dowby = 36

3648 100 = 33.33%

Growth is least for Alpha. Ans.94. (B) For the year 1986, total sales = 44.5

For the year 1987, total sales = 46For two years show the closest sales.

95. (A)Required ratio = 2

50 = 25 : 1 Ans.

96. (A)

A

Candidate who joinedMBA = 8550

Candidate who completedMBA = 5700

B

C

D

E

F

G

8550 = 1881 5700 = 1026

8550 = 1282 5700 = 969

8550 = 855 5700 = 741

8550 = 1454 5700 = 912

8550 = 684 5700 = 513

8550 = 1026 5700 = 855

8550 = 1368 5700 = 684

22

100

18

100

15

100

17

100

10

100

13

10017

100

16

1008

100

9

10012

100

15

10016

100

12

100

Required percentage of

A = 1881

1026 100 = 54.55%

B = 1282

969 100 = 75.59%

C = 855

741 100 = 86.67%

D = 912

1454 100 = 62.72%

E = 684

513 100 = 75%

F = 1026

855 100 = 83.33%

G= 1368

684 100 = 50%

97. (C) Required percentage = 75% Ans.(See solution of last question)

98. (B) Required ratio = 1368

684= 1 : 2 Ans.

99. (D) Required percentage

=8551282

741969

100

= 2137

1710 100 = 80% Ans.

100. (C) Required difference= (A + D) (C + E)

[Completed joined]= (1026 + 912) (855 + 684)= 1938 1539= 399 Ans.


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Ph: 011-27607854, 08860-333-333 12

1. B

2. A

3. D

4. C

5. C

6. B

7. A

8. A

9. A

10. D

11. A

12. A

13. C

14. D

15. D

16. A

17. B

18. B

19. B

20. B

21. B

22. B

23. D

24. A

25. B

26. D

27. A

28. A

29. B

30. B

31. B

32. D

33. C

34. C

35. A

36. C

37. C

38. D

39. D

40. A

41. A

42. C

43. B

44. B

45. D

46. B

47. A

48. B

49. C

50. C

51. A

52. B

53. D

54. C

55. D

56. A

57. A

58. C

59. C

60. D

61. A

62. D

63. C

64. A

65. D

66. A

67. A

68. D

69. A

70. B

71. A

72. B

73. C

74. B

75. C

76. C

77. B

78. C

79. C

80. D

SSC MAINS (MATHS) MOCK TEST -12 (ANSWER SHEET)

81. A

82. B

83. A

84. C

85. B

86. C

87. D

88. C

89. B

90. C

91. C

92. A

93. A

94. B

95. A

96. A

97. C

98. B

99. D

100. C

ssc mains (maths) mock test-12 (solution)

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