ssc mains (maths) mock test-13 (solution)

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

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SSC MAINS TEST - 13 (T-II) 2013 (SOLUTION)SSC MAINS TEST - 13 (T-II) 2013 (SOLUTION)SSC MAINS TEST - 13 (T-II) 2013 (SOLUTION)SSC MAINS TEST - 13 (T-II) 2013 (SOLUTION)

1. (*)7

5,

11

8,

3

2,

13

9

=4013

409

, 1803

1802

, 4511

458

, 727

725

=520

360,

540

360,

495

360,

504

360

In the above fractions, numerators are

equal, so the fraction with the greater

denominator will have the least value of

ratio and the fraction with the lowest value

of denominator will have the highest value

of ratio.

540

360<

520

360 <

504

360<

495

360

3

2<

13

9<

7

5 <

11

8

[The required ascending order of the

given fractions] Ans.

2. (D) 21

2

1

)347()31028(

1 1

2 2 2 22 25 ( 3) 2.5. 3 2 ( 3) 2.2. 3

5 3 (2 + 3 )1

5 3 32

1

5 3 34

32

5 3 (2 3 )

5 3 2 + 3 = 3 Ans

3. (B) Let x be a +ve integer.

Now, for x = 1

(A)x

x= 1,

(B)x

x 1=

1

2 = 2

(C)1x

x=

2

1= 0.5

and (D)3

2

x

x=

4

3 = 0.75

(B)x

x 1has the greatest value. Ans.

4. (D)Let the total number of swans be x.

The number of swans playing on the shore

of the pond = x2

7

Number of swans inside the pond = 2

x = 22

7x

2(x 2) = x7

4(x2 4x + 4) = 49x

4x2 16x + 16 49x = 0

4x2 65x + 16 = 0

On solving x = 16

Number of swans = 16 Ans.

5. (B)342

784

342

)7( 283

342

)343(28

According to remainder theroem,

342

)343(28

will have the same remainder

as 342

)1( 28

1 Ans.

6. (A) Let x represents number of students & yrepresents the number of rows.

Then,

No. of students in each row = y

x

Case : (I)

4

y

x (y 2) = x

2y2 4y = x ... (i)

Case : (II)

4

y

x (y + 4) = x

y2 + 4y = x ... (ii)

From eqn (i) & (ii)

2y2 4y = y2 + 4y


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2

y(y 8) = 0

y = 8

Total no. of students

x= 2(8)2 4 8

= 128 32

x = 96 Ans.

7. (C)Let the numbers be x and 4x.

Then, 84 21 = x 4x

4x2 = 1764

x2 = 441

x = 21Hence, the largest number is 4x = 84. Ans.

8. (A) Total one-digit number = 9, No. of digits =9

Total two-digit number = 90, No. of digits = 180

Total three-digit number = 900, No. of digits = 2700

Let there are x-4 digits numbers.

According to question,

90 + 180 + 2700 + 4 x = 3189

x = 4

28893189 = 75

Number of pages

= 9 + 90 + 900 + 75 = 1074 Ans.

9. (*) Sn

= 2

n[2a + (n 1)d]

S40

= 20[24 + 39 4]

= 20 164 = 3280 Ans.

10. (A) nth term of a GP = arn 1

8th term = 5 (2)8 1

= 5 (2)7

= 5 128

= 640 Ans.

11. (A)

Total students who are studying at leastone subject = 21 + 1 + 2 + 1 + 9 + 3 + 8

= 45

Students who are not studying any of thethree subjects. = 80 45 = 35 Ans.

12. (C) Let the average of 12 innings be x.

Also, 13

9612 x = x + 5

12x + 96 = 13x + 65

x = 31

His new average

= 13

963112

= 13

468= 36 Ans.

13. (A) No. of workers = x, No. of officers = y.

Case : (I)

x + y = 400 ... (i)

Case : (II)

2000 x + 10000 y = 400 3000

x + 5y = 2000

3000400

x + 5y = 600 ... (ii)Substracting (ii) from (i)

(x + 5y) (x + y) = 200

y = 50

No. of officers = y = 50

No. of workers = x = 400 50 = 350 Ans.

14. (A) Let their ages be x and y.

x

1 +

y

1= 5

yx

11

y + x = 5(y x) 6x = 4y

y

x=

3

2... (i)

Now, yx

xy

= 1

4.14

xy = 14.4(x + y) ... (ii)From Eqs. (i) and (ii),

x = 24 yrs. and y = 36 yrs.

15. (D) By the rule of alligation,

Quantity of rice sold at 10% gain

= 12

12 3 50 = 40 kg

Quantity of rice sold at 5% loss

= 312

3

50 = 10 kg Ans.

16. (A) Quantity of nitric acid = 10 10

1

= 1 litre Water = 10 1 = 9 litre

Suppose x litre water is added in thesolution.

Then, (10 + x) 100

4= 1

x = 15 litre Ans.


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17. (A) Let the average speed be x km/h.

Time taken by aircraft (t) = x

800

ATQ,

t 60

40=

40

800

x

x

800

40

800

x =

3

2

)40(

32000

xx = 3

2

x(x + 40) = 48000 x = 200 km/h Ans.

18. (C) Mohan can reach the middle in 12.5 min.

Puran can reach the middle in 25 min.

So, required time = 25 12.5

= 12.5 min Ans.

19. (D) Speed of X = 2

3Y

Distance covered by Y before catching = D m

2

3D = D + 300

2

1D = 300

D = 600 mTotal distance = 600 + 300 = 900 m Ans.

20. (C) Selling price of first shopkeeper

= 700 100

70

100

94 = ` 460.60

Selling price of second shopkeeper

= 700 80

100

84

100 = ` 470.40

Required difference

= 470.40 460.60 = ` 9.80

21. (D) Let the original value of fridge be ` x.

Then, cost price = 16

15x

Selling price = 100

110x

Gain percent = 100

16

1516

15

100

110

x

xx

= 17.33% Ans.

22. (D) Let the rate of flow of river be x km/h.

Then, downward speed = (10 + x) km/h.

ATQ,

)10(

91

x + )10(

91

x = 20

On solving, we have x = 3.

Hence, speed of flow of river is 3 km/h.

23. (A) Let the rate of filling tank be x m3 /min.

Then, the rate of empting the tank

= (x + 10) m3/min

ATQ,

x

2400

10

2400

x= 8

2400

)10(

10

xx= 8

x(x + 10) = 3000

x = 50 m3/min Ans.

24. (C) Let first pipe can fill the tank in x h.

Second pipe can fill the tank in (x 5) h

Third pipe can fill the tank in (x 9) h

ATQ,

x

1 +

5

1

x =

9

1

x

5

)5(

xx

xx = x 9

x2 5x = 2x2 23x + 45

(x 15)(x 3) = 0

x = 15 h as x = 3 h is not possible. Ans.

25. (B) Number of men = 5

2 25 = 10

Number of women= 5

3 25 = 15

Amount distributed among men and women

= 275 80%

= ` 220

Let the wages paid to a man be ` 5x and

to a woman be ` 4x, then

10 5x + 15 4x = 220

50x + 60x = 220

x = 2

Wages recieved by a woman= 2 4 = ` 8 Ans.

26. (C) 10% of profit = 100

10 800 = ` 80

Remaining Rs. 720 is divided in the ratio

= 1500 : 900 = 5 : 3

Share of Y =8

3 720 + 80 = ` 350 Ans.

27. (A)According to question,

%5%

2

18 of sum = 350


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sum = 5.3

350 100

= ` 10000 Ans.

28. (D) Cash price, CP = ` 39000

Cash down payment, DP = ` 17000

Balance due, after Ist instalment,

BD = ` 22000

P = value of instalment = ` 4800

n = no. of instalments = 5

R = rate of interest

ATQ,

12001

nRBD =

2400

)1(1

RnnP

1200

51

R22000 =

2400

41

R5 4800

1200

51

R11=

2400

41

R12

11 + 1200

55R = 12 +

1200

24R

1200

55R

1200

24R = 1

1200

31R = 1

R = 31

1200 = 38.71% Ans.

29. (B)

Profit received by Anu = ` 110 + Rs. 350

= ` 460

& Profit received by Bimla = ` 420

30. (A)Required % = 5 + 5 + 100

55

= 10.25% Ans.

31. (B)

Percentage of failed students

= 30% + 35% 27% = 38%

Percentage of passed students

= 100% 38% = 62%

Now, Let total number of students be x.

62% of x = 248

x = 248 62

100 = 400 Ans.

32. (C) Let the maximum marks be x.

Then,

296 259 = 5% of x

100

5x = 37

x = 740 Ans.

33. (C) Let Ram's monthly income be ` x.

Total savings = x 100

80

100

85

100

70

x = 9520 80

100

85

100

70

100

= ` 20000

34. (B) Total candidates = 2000

No. of boys = 900

No. of girls = 1100

No. of students who passed

= 100

90032 +

100

110038

= 288 + 418 = 706

No. of students who failed

= 2000 706 = 1294

Required percentage

= 2000

1294 100 = 64.7% Ans.

35. (D) Only the option (d) gives the difference of

votes between two candidates as 308. Ans.

36. (B) Let the cost price of colour printer and

computer system be ` x and ` y respectively.

According to question,

x 100

120 + y

100

90= x + y

0.2x = 0.1y ..... (i)

x 100

85 + y

100

105= x + y 800

0.05y= 0.15x 800

From Eqs. (i) and (ii),

x = ` 16000 Ans.

37. (D) Let the cost price of book be ` x.

Then, (1.2x 18) 0.8x = 0.25 0.08x


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0.4x 18 = 0.20x

x = 20.0

18= ` 90 Ans.

38. (A)Cost price of transistor = ` 320

Selling price of transistor

= 320 1.15 = ` 368

Marked price of transistor

= 368 + 32 = ` 400

Required percentage of profit

= 320

320400 100 = 25% Ans.

39. (A) Let the cost price of 12 apples be ` 100.

Then, selling price of 100 apples will be ` 800.

CP of 1 apple = 12

100= `

3

25

and SP of 1 apple = 100

800= ` 8

Loss =

8

3

25 = `

3

1

Loss percent = CP

Loss 100

=

3

253

1

100

= 3

1

25

3 100 = 4% Ans

40. (A) Let the distance covered by the first train

be x km/h.

As both trains have travelled for same

time.

50

x=

60

120x

60x = 50x + 6000

x = 600

Total distance = x + (x + 120)

= 1320 km Ans.

41. (D) Let the speed of train be x km/h.

As both the persons are walking in the

same direction of train.

So, (x 4.5) 8.4 = (x 5.4) 8.5

0.1x = 8.1

x = 81 km/h Ans.

42. (A) Let Ashokan can finish the work in x days.

Then,

Nitin can finish the work in 3x days.

3x x = 40

x = 20 days

and 3x = 60 days

So, together they can finish the work in

6020

6020 days = 15 days Ans.

43. (D) Men : Women : Boys = 15 : 24 : 36

= 5 : 8 : 12

Convert women and boys in terms of men

8 women= 5 men

12 women = 8

512 =

2

15men

12 boys = 5 men

6 boys =12

5 6 =

2

5men

Total women and boys in terms of men

=12

5 +

2

5 =

2

20 = 10 men

Let the number of men required = x

Then, (x + 10) =630

25.281215

= 18

x + 10 = 18

x = 8 men Ans.

44. (B) Let they make x pieces per day.

Then,x

360

4

360

x= 1

360

)4(4

xx= 1

x(x + 4) = 1440 = 36 40 x = 36

Required number of days

= 36

360= 10 days Ans.

45. (C) LCM of (3 + 2), (7 + 3)

and (11 + 4) is 30.

Let the capacity of each container be 30 l

Quantity of milk after mixing

=

15

11

10

7

5

3 30 = 61 l

Quantity of water after mixing

=

15

4

10

3

5

2 30

= 29 l

Required ratio = 61 : 29 Ans.


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46. (D) As the sum of money that are to be divided

among A, B and C and between E and F

are not given. So, the amount that B

receive cannot be determined. Ans.

47. (D) New ratio = 3 100

150 : 5

100

160 : 7

100

150

= 2

9 : 8 :

2

21 = 9 : 16 : 21 Ans.

48. (B)18 carat gold = 4

3 pure gold =

4

3 24 = 18

20 carat gold = 6

5 pur gold =

6

5 24 = 20

Required ratio = 18 : 20 = 9 : 10 Ans.49. (D) Let the income of two persons be ` 4x and

` 5x and their expenses be ` 7y and ` 9y

respectively.

Then,

4x 7y = 50 .... (i)

and 5x 9y = 50 .... (ii)

On solving Eqs. (i) and (ii), we get

x = 100 and y = 50

The income of persons are ` 400 and` 500. Ans.

50. (C)

Hence, the next number of the series will

be 27. Ans.

51. (B)8

x=

23

32

On applying componendo and dividendo:-

8

8

x

x=

23

233

... (i)

Again,12

x=

23

22

12

12

x

x=

32

323

... (ii)

From Eqs. (i) + Eqn (ii)

8

8

x

x +

12

12

x

x=

23

233

23

323

= 23

323233

= 23

)23(2

= 2 Ans.

52. (A) a = 3 + 22

a

1=

223

1

+

223

223

= 89

223

= 3 22

aa

1 = 6

3

246 1

a

aaa = 3

3 11

aaaa

=

3

3 1

aa +

aa

1

=

31

aa

aa

13 +

aa

1

63 3 6 + 6 = 216 18 + 6 = 204 Ans.53. (A) x = 997, y = 998, z = 999

Now,

zxyzxyzyx 222 =

})()(){(2

1 222 xzzyyx

= 2

1{(1)2 + (1)2 + (2)2}

= 2

1(1 + 1 + 4) = 3 Ans.

54. (D)a

a3

15 = 5

Multiplying both sides by 5

3

aa

5

13 = 3

Squaring both sides

aa

aa

5

132

25

19

2

2 = 9

22

25

19

aa = 9

5

6 =

5

39Ans.

55. (C) 22 1

xx + 2 = 3

22 1

xx = 1

x4 + 1 = x2

x4 x2 + 1 = 0


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Now, x206 + x200 + x90 + x84 + x18 + x12 + x6 + 1

= x200(x6 + 1) + x84(x6 + 1) + x12(x6 + 1) + 1(x6 + 1)

= (x6 + 1)(x200 + x84 + x12 + 1)

= (x2 + 1) (x4 x2 + 1) (x200 + x84 + x12 + 1)

0 Ans.

56. (A) a3 + b + c3 3abc

=

=

=2

1(32 + 52 + 12)

=2

1 35 = 17.5 Ans.

57. (A) If 4

2

2

2 ytx

y

x is a perfect square

It must be equal to

22

2

22y

y

xy

y

x

= xy

y

x

22

2

On comparing, we have

tx = x

t = 1 Ans.58. (D) 8x3 ax2 + 54x + b

(2x 3)3 = 8x3 3 2x 3(2x 3) 27

= 8x3 36x2 + 54x 27

On comparing with the given expression

a = 36, b = 27 Ans.

59. (B) a = 3b 7 .... (i)

a 3b = 7

a + 3 = 3(b + k) 7

a + 3 = 3b 3k 7

a 3b = 3k 10

7 = 3k 10

k = 1 Ans.

60. (C) y = |x| 5

|x| y 5 = 0

Area of two traingles formed by

|x| y 5 = 0

=

ab2

c2

2

= ab

c 2=

11

)5( 2

= 25 Sq. unit Ans.

61. (D)

In APY and BPX

X = Y = 90A tangent is always perpendicular to the

radius through the point of contact.

APY ~ BPX [By AA similarity]

BX

AY=

PB

AP

2

5=

PB

AP

P divides AB externally in the ratio of 7 : 2.

62. (D) DE || BC (given)

ADE and ABC are similar.

)(

)(

ABCar

ADEar =

2

AB

AD

2

1=

AB

AD

AB

AD=

2

1

BD

AD=

12

1

Ans.

63. (C) Length of the common tangent

= 22 )36( d

8 = 22 )36( d

or d2 = 64 + 81 = 145

d = 145 cm Ans.

Distance between their centres = 145 cm

64. (B) Let the number of sides in polygon = n

The sum of interior angles

= (n 2) 180

ATQ,

5 172 + (n 5) 160= 180n 360


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860 + 160n 800 = 180n 360

60 + 360= 180n 160n

20n = 420

n = 21 Ans.

65. (C) The sum of any two sides of a triangle

is greater than the 3rd side.

2 + 3 > 5, which is wrong.

2 + 3 > 6, which is wrong.

(2, 3, 5) or (2, 3, 6) will not form a triangle.Triplets (3, 5, 6) & (2, 5, 6) are true for the

sides of a triangle.

= 2 triangles Ans.

66. (C)

In ADE,AE = AB AE = AD

E = D = (say)

A = + = 2

Similarly, In BCF,BF = AB BF = BC

C = F = (say)

B = 2

In OEF,

A + B = 1802 + 2 = 180

+ = 90

EOF = 90

ED CF Ans.67. (B) BC = AD

Distance= BC

In AOB,

tan 30= AO

350

3

1=

AO

350

AO = 150 m

AD = 300 m

Speed = 2

300= 50 m/min

= 2

5 m/s

= 9 km/h Ans.

68. (B)Let the shortest side be x m.

The hypotenuse = (2x + 3)m

Let third side = y m

x + y + 2x + 3 = 6x y = (3x 3) mNow, (x)2 + (3x 3)2 = (2x + 3)2

x2 + 9x2 + 9 18x = 4x2 + 9 + 12x 6x2 30x = 0 x = 5 mas x 0 Three sides are 5 m, 12 m and 13 m.

69. (B)In AQC,

AQ2 = AC2 + CQ2

In PCB, BP2 = PC2 + CB2

AQ2 + BP2 = AC2 + CQ2 + PC2 + CB2

= (AC2 + CB2) + (CQ2 + PC2)

= AB2 + PQ2

= AB2 +

21

2AB

AQ2 + BP2 = 4

ABAB4 22

4(AQ2 + BP2) = 5AB2 Ans.70. (B) OABC is a rhombus with centre O.

Let diagonal of the rhombus be

OB = 2x and AC = 2y

Radius of the circle = OB = OA = OC = 2x


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9

In POC,

OC2 = OP2 + PC2

(2x)2 = x2 + y2

4x2 = x2 + y2

3x2 = y2 x = 3

y

Also, area of rhombus

2

1 (2x)(2y) = 332

x = 4

Radius of circle = 2 4 = 8 m Ans.

71. (C) a sin + b cos = cOn squaring,

a2 sin2 + b2 cos2 + 2ab sin cos = c2

...(i)

Let a cos b sin = k

a2 cos2 + b2 sin2 2ab sin cos= k2

... (ii)

From equation (i) & (ii),

a2(sin2 + cos2 ) + b2(sin2+ cos2 ) = c2 + k2

k2 = a2 + b2 c2

k = 222 cba Ans.

72. (D) sec =x

x

4

14 2

b = 4x, h = 4x2 + 1

Then,

P = 222 )4()14( xx

= 224 161816 xxx

= 1816 24 xx

= 4x2 1

sec + tan = x

x

4

14 2 +

x

x

4

14 2

= x

xx

4

1414 22

= x

x

4

8 2= 2x Ans.

73. (B) cos

sin1

1

sin1

1= 4

cos

2sin1

sin1sin1 = 4

cos

2cos

2= 4

cos =2

1

cos = cos 60

= 60 Ans.

74. (D) un

= cosn + sinn

Now,

2u6 3u

4 + 1

= 2(cos6 + sin6 ) 3(sin4 + cos4 ) + 1

= 2(cos2 + cos2 )3 3sin2 cos2

(sin2 + cos2 ) 3{(sin2 + cos2 )2

2sin2 cos2 )} + 1

= 2 6 sin2 .cos2 + 6sin2 cos2 3 + 1

= 2 3 + 1 = 0 Ans.

75. (B) (a2 b2) sin + 2ab cos = a2 + b2

22

22

ba

ba

sin + 22

2

ba

ab

cos = 1

Let 22

22

ba

ba

= sin & 22

2

ba

ab

= cos

Then, above equation becomes

sin2 + cos2 = 1

sin = 22

22

ba

ba

cos = 222

ba

ab

tan = ab

ba

2

22 =

1

2ab(a2 b2) Ans.

76. (D)

2

4

cos

cos +

2

4

sin

sin = 1

By taking

= , it satisfies the above equation

2

4

cos

cos +

2

4

sin

sin = 1 Ans.


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10

77. (B) l cos2 + m sin2 =

2

2

cot

cos

2

2

sin

sin1

l cos2 + m m cos2 = 1 + 1 cos2

l cos2 + cos2 m cos2 = 2 m

cos2 = 1

2

ml

m

or, sec2 = m

ml

2

1

or, tan2 +1 = m

ml

2

1

or, tan2 =m

mml

2

21

or, tan2 =m

l

2

1

tan = m

l

2

1 Ans.

78. (C) x sin a y cos a = 0

x sin a = y cos a

cos

sin=

x

y

xy3 + yx3 = xy

xy(y2 + x2) = xy

x2 + y2 = 1 Ans.

79. (C) 1 +

2= 90

2

= 90 1

In ABC, tan 1=

BD

AB

tan 1 =

x

AB... (i)

In ABC, tan 2 =

y

AB

tan(90 1) =

y

AB

cot 1 =

y

AB

tan 1 =

AB

y... (ii)

from equation (i) and (ii),

x

AB=

AB

y

(AB)2 = xy

AB = xy Ans.

80. (A)3x + 4y = 12

x = 0, y = 3 A(0, 3)

y = 0, x = 4 B(4, 0)

6y + 8y = 60

x = 0, y = 7.5 C(0, 7.5)

y = 0, x = 10 D(10, 0)

Area of Trapezium

= Area of COD Area of AOB

= 2

1 10 7.5

2

1 4 3

= 37.5 6 = 31.5 Ans.

81. (A) Curved surface area of a frustum

= ! (r + R)L

Here, R = 2

8= 4m, r =

2

4 = 2 m

and l = 22 )( rRh

l = 22 26 = 40

Curved surafce area

= 7

22 (2 + 4) 40

= 119.26 = 118.4 m2 (approx.) Ans.

82. (C) Required difference = (l b h) ! r2h

= 10 10 21 7

222

2

10

21

= 2100 1650 = 450 cm3 Ans.

83. (A)Total surface area of prism

= lateral surface area + 2 (area

of base)


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11

Here, s = 2

cba =

2

132021 = 27

Required area = (21 + 20 + 13) 30 + 2

)1327)(2027)(2127(27

= 54 30 + 2 147627

= 1620 + 2 126

= 1872 sq. m Ans.

84. (C) Curved surface area of cone = Area of

sector of circle

! rl = !R2 360

120

Here, l = R

r = 15 360

120 = 5 cm

h = 25225 = 210 cm

Volume of cone = 3

1! r2h

= 3

1 ! 25 210

= 2250 ! /3 cm3 Ans.

85. (B) The error percent in area

= + 5 6 100

65 = 1.3% Ans.

86. (D) Volume of the ball

= Volume of raised water

3

3

4r! = 2(12) (6.75)

r3 = 729r = 9 cm Ans.

87. (C) Let the diameter of third ball be 2r.

3

4!

3

2

3

=

3

4!

3 3 31.5 2 2

2 2 2

r

8

27 =

64

27 + 1 + r3

r3 = 8

27

64

27 1

r3 = 64

6427216 =

64

125

r = 4

5 cm

2r = 4

10= 2.5 cm

88. (D) Total number of smaller cubes

= 111

444

= 64

Increase in surface area

= 64 6 (1)2 6 (4)2

=384 96 = 288 cm2

Required percentage

= 96

288 100 = 300% Ans.

89. (B) Radius of the circle = 2

14 = 7 cm

Area of circle = 22

7 7 7 = 154 cm2 Ans.

90. (B) Volume of cube = 2

1 25 20 4

= 1000 cm3

Side of cube = 10 cm

Total surface area = 6 (10)2

= 600 cm2

91. (B) Ratio of processing cost of water for

industrial, energy and domestic usage

= 3 : 5 : 2

In 2006, water usage for industrial,

energy and domestic = 25, 26 and 35

trillion litres

In 2009, water usage for industrial,

energy and domestic

= 49, 35. 30 trillion litres.

Ratio of processing cost for abovementioned usage in 2006 to that in 2009

= 230535349

235526325

= 60175137

7013075

= 382

275= 0.72 Ans.

92. (A) Usage in energy related sector in 2006

= 26 trillion litres

Usage in energy related sector in 2009

= 35 trillion litres

Required percentage increase

= 35 26

26

100 = 34.6% Ans.


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12

93. (C) Required percentage

= 25303549490

490

100

= 629

490 100 = 77.90% Ans.

94. (D) Required difference

=

= 629

30 100

531

35 100

= 4.77% 6.59%

= 1.82% (Ignore ve) Ans.

95. (B) Percentage increase in usage from 2006

to 2009.

Domestic = 629

30 100

531

35 100

= 4.77 6.59 = 1.82%

[ignore ve]

Industrial = 629

49 100

531

25 100

= 7.79 4.71 = 3.08%

Others = 629

25 100

531

35 100

= 3.97 6.59 = 2.62%

[ignore ve]

Energy =629

25 100

531

26 100

= 5.56 4.89 = 0.66% Ans.

96. (C) One lakh Ans.

97. (A) Required factor = 400

320= 0.8 Ans.

98. (D) In 1990 it shows the least increasing

over the preceding year. Ans.

99. (A) Average sales

= 6

400440420400320340

= 6

2320 = 386.66

Sales are above average in 1988, 1989,

1990 and1991 and below average in 1986

and 1987

Required ratio = 4 : 2 = 2 : 1 Ans.

100. (A) Average sales from 1988 to 1991

= 4

400440420400

= 415 Ans.


=============================================================

13

1. *

2. D

3. B

4. D

5. B

6. A

7. C

8. A

9. *

10. A

11. A

12. C

13. A

14. A

15. D

16. A

17. A

18. C

19. D

20. C

21. D

22. D

23. A

24. C

25. B

26. C

27. A

28. D

29. B

30. A

31. B

32. C

33. C

34. B

35. D

36. B

37. D

38. A

39. A

40. A

41. D

42. A

43. D

44. B

45. C

46. D

47. D

48. B

49. D

50. C

51. B

52. A

53. A

54. D

55. C

56. A

57. A

58. D

59. B

60. C

61. D

62. D

63. C

64. B

65. C

66. C

67. B

68. B

69. B

70. B

71. C

72. D

73. B

74. D

75. B

76. D

77. B

78. C

79. C

80. A

SSC MAINS (MATHS) MOCK TEST -13 (ANSWER SHEET)

81. A

82. C

83. A

84. C

85. B

86. D

87. C

88. D

89. B

90. B

91. B

92. A

93. C

94. D

95. B

96. C

97. A

98. D

99. A

100. A


=============================================================

14

54. (C) In the given question please readpolynomial 2x4 3x3 2x2 + 6x + 2

as polynomial 2x4 3x3 3x2 + 6x 2 and according to the correct polynomial, the correct answer will be option (C).89. (*) The given solution is correct,

according to which the answer is 4000m3,which is present as both the options(A) & (B)

11. (C) Let the five consecutive odd nos. bex, x + 2, x + 4, x + 6 & x + 8.

So, ATQ,5x + 2 + 4 + 6 + 8 = 41 5

or, 5x = 205 20

x = 5

185= 37

So, product of 1st & last terms (A & E)= 37 (37 + 8) = 1665 Ans.

37. (B) In the solution given,Duration of time = 16 hours

(which is correct)and point of time = 3 pm + 16 hours

= 7 am Ans.42. (D) A + B + C = 15 3 = 45

& B + C + D = 16 3 = 48 D A = 3or, 19 A = 3 A = 16

CORRECTIONS FOR MOCK TEST - 12

So, The required % = %10016

16 = 100% Ans.

68. (D)or, x = 31

3

2

222

or, x 2 = 31

3

2

22

So,

" 3

1

3

2

3

1

3

23

3

12

3

2

3 222.2.322)2(x

or, (x 2)3 = 4 + 2 + 3.2(x2)or, (x 2)3 = 6 + 6(x 2)or, x3 6x2 + 12x 8 6 6x 12 = 0or, x3 6x2 + 6x 26 = 0

x3 6x2 + 6x = 26 Ans.

69.(*) (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)or, (1)2 = x2 + y2 + z2 + 2(1) x2 + y2 + z2 = 3 Ans.

76.(C) The given solution is totally correct. but the correct option is C (12 cm). Ans.


88.(A) Correct question should be "cos2 + 2cos " & according to this question the correct

option is A. Ans.


21.(B) Quantity of dry fruits (without water) in 100kg fresh fruit = 32% of 100 kg = 32 kg.So, required quantity of dry fruit (dry fruitcontains 80% fruit and 20% water)

=80

10032kg = 40 kg Ans.

46(B).

Now, C takes 1 hour 54 minutes = 114 minSo, time taken by B to cover the same

Distance = 114

3min = 38 min. Ans.


97. (D) Can not be determind as the populationis not known.

100.(D)Can not be determind as the populationis not known.

ssc mains (maths) mock test-13 (solution)

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