ssc mains (maths) mock test-8 (solution)

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��MUKHERJEE NAGAR ��MUNIRKA ��UTTAM NAGAR�� DILSHAD GARDEN ��ROHINI��BADARPUR BORDER

SSC MAINS MOCK TEST - 8 (MATHS SOLUTION)SSC MAINS MOCK TEST - 8 (MATHS SOLUTION)SSC MAINS MOCK TEST - 8 (MATHS SOLUTION)SSC MAINS MOCK TEST - 8 (MATHS SOLUTION)

1.(B) Here first term; a = 4, common diffference;d = 3 & no. of terms; n = 20

So, sum up to 20 terms;

S20

= 20

2 [2 (4) + (20-1) × 3] = 10[ 8 + 57 ] = 650

2.(B) Given expression, 10x + y = 7 (x + y)or, 3x - 6y = 0

� � x = 2y Now, When each digit is increased by 3,

then, 10( x + 3) + (y + 3) = 6 (x +3 + y +3) +6� 4x - 5y = 9

or, 8y - 5y = 9 form eq. (i)y = 3 & x = 6

� The given number is 63.3.(D) Given experssion = 325 (1 + 3 + 32 + 33 )

= 325 × 40 = (324 × 3) (4 ×10) = (324 × 4) (3 ×10)

Which is divisible by 304.(A) LCM of 6, 9 and 12 = 36

� Number is the form of 36p + 4Since, the required number between 300 and

400So, the numbers will be 328 (when p = 9) and

364 (when p = 10)Required sum = 328 + 364 = 692

5.(C) Area of the courtyard = 3.78 × 5.25 sq m= 378 × 525 sq cm = 198450

� 198450 = 21 × 21 × 450450 sq marble stones shall be used of size21 cm × 21 cm

6. (C) 1st day = 4km, 2nd day = 4 × 1

2= 2km,

3rd day = 2 × 1

2= 1km

� Total distance S = 4 +2 +1 +1

2+

1

4....

Which is infinite GP with

a = 4, r = 1

2

Now, � r < 1

So, Sum; S = 1

a

r�=

4

11

2�

=4

1

2

= 8km

7.(C)

3 3 3

2 2 2

3a b c abca b c

a b c ab bc ca

� � ��

� � � � ��

So,3 3 3

2 2 2

(1.5) (4.7) (3.8) 3 1.5 4.7 3.8

(1.5) (4.7) (3.8) 1.5 4.7 4.7 38 3.8 1.5

� � � � � ��

= 1.5 + 4.7+ 3.8 =10

8.(A)7

8 7 4 52

x� � � ��

� ��

� 8 - 1

7 52

x� � � � � ��

� ��

�1

8 7 52

x� ��

�15

8 52

x� ��

� 8 - 15

2 + x = 5

�1

2+x = 5

9.(D) 23 + 43 + 63 + ..... + 203

= (2 ×1)3 + (2× 2)3 + (2 × 3)3 +....+ (2 ×10)3

= 8×13 + 8× 23 + 8× 33 +....+ 8 ×103

= 8× [13 + 23 +33 + 43 +....+103]= 8 × 3025 = 24200

10.(C) Required number of arrangements= 9 × 9× 8 ×7× 6× 5 × 4× 3× 2 ×1= 9 ×9!

11.(B) Investment and Income in 2000

= ` x and ` 1.20xInvestment and Income in 2000

= ` (x- 50000) and ` 1.20x

� Profit = 20 + 6 = 26%

Income in 2001 = ` (x - 50000) × (1.00+0.26)Thus, 1.26 (x - 50000) = 1.20x

x = ` 105000012.(B) Total population of town

= 15 × Number of males

Number of females =

7

8

� Number of males and females = 7x and 8xNumber of male children = 25% of 7x

= 25

100× 7x = 1.75x

Number of female children = 20% of 8x

=20

100× 8x = 1.6x

� Number of adult females = 8x � 1.6x = 6.4x� 6.4x = 235200

� x = 235200

6.4= 36750

� Total population of town = 15 × 36750 = 551250

13.(C) Given, x = 20%, y = 30%, z = 10%, A = 100.80

Required total money

= 100 100 100

(100 )(100 )(100 )

A

x y z

� � ��

=100.80 100 100 100

80 70 90

� � ��

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=100800

8 7 9� �= ` 200

14. (A) Let Initial investments = 3x , 5x and 7xAfter one year(3x - 45600) : 5x : (7x + 337600)= 24 : 59:167

�3 45600 24

5 59

x

x

�� x = 47200

� initial investment of A = 47200 × 3 = Rs.141600

15.(B) Given, 14

15

a c

b d� � and

35

24

a c

b d� � or

35

24

a d

b c� �

(where a

b is greater fraction)

Now multiplying both the equations

14 35

15 24

ac ad

bd bc� � �

�2

2

49

36

a

b�

�7

6

a

b�

� the greater fraction is 7

6

16. (D) 15% of (A + B) = 25% of (A - B)

�15

100(A+B) =

25

100(A-B)

� 15 (A + B) = 25 (A - B)� 15 A+ 15B = 25A - 25B� 10A = 40B� A = 4B

Now let x % of B is equal to A

�100

x×B = A = 4B

� x = 400%17. (C) Let the third number be x,

A.T.Q.,

First number = 20

100× x =

5

x

& Second number = 50

100×x =

2

x

� Required percentage

= 100

5

2

x

x

�=

2100

5

x

x� � = 40%

18.(A) Let, Average age of 11 member hockey team= x years

� Total age of hockey team = 11x yreas.When captain aged 26 yrs and goal keeperaged 26 + 3 = 29 yrs. are excluded.Total age of remaining 9 players = 11x - (26+ 29) = (11x - 55) yrs.

Now, ATQ,

11 55

9

x �= x - 1

or, 11x - 55 = 9x - 9 = 2x = 44� x = 22 yrs

19.(B) Sum of temperature of first 3 days = 22ºC × 3

= 66ºC

Sum of temperature of last 3 days = 24 ºC × 3

= 72ºC

Sum of temperature of whole week = 23.5ºC × 7

= 164.5ºC

� The temperature of the last day

= (164.5 - 66- 72)ºC = 26.5ºC

20. (A) Let the present ages of mother and son be

x years and (45 -x ) years respectively.

Then, (x - 5) (45 -x -5) = 4 (x -5)

� x2 + 41x - 180 = 0

� x = 36

� The present ages of mother and son are

36 yrs & 9 yrs. respectively.

21.(A) Let x years ago the ratio of their ages

was 3 : 5.

So, A.T.Q.,

40 3

60 5

x

x

��

�

� 200 - 5x = 180 - 3x

� 2x = 20

� x = 10 yrs.

22.(A) Let the four parts into which 3150 is

divided are a, b, c and d.

�2 3 4 12

a b c d� � � =k

Then a = 2k, b = 3k , c = 4k and d = 12k

As a + b + c + d = 3150

� (2k + 3k +4k +12k) = 3150

� 21k = 3150

� k = 150

Hence the four parts are 300, 450, 600, 1800

So, the largest part is 1800

23.(A) Let the total monthly sales of companies

A and B be Rs. 2x and Rs.3x and their total

monthly expenditure be ` 3y & ` 4y.

Given that A's profit = 1/5 of sales = 2x /5

� 2x - 3y = 1

5(2x)

�4

5(2x) = 3y � y =

8

15x

Profit of company

B = 3x � 4y = 3x � 4× 8

15x =

13

15

x

Hence the ratio of the profits of the two

companies are

2

5x :

13

15

x= 6:13

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24.(B) Milk : water in first glass = 1 2

:3 3

= 4 : 8

and Milk: Water in second glass =1 3

:4 4

= 3 : 9

Milk in the vessel = 4 + 3 = 7

Water in the vessel = 8 + 9 = 17

Ratio of milk and water in the vessel = 7 :17

25.(D) Total quantity of milk = 2 × 0.9 + 5 × 0.8 +

9 × 0.7= 12.1 litre

Milk concentration in the resultant mixture

= 12.1

1002 5 9

��

= 75.625%

Water concentration in the resultant mixture

= 100- 75.625% = 24.735%

� Milk : Water = 75625

24735= 121: 39

26. (D) Let each day's salary = ` x

Given, 18x + 8× 2

x� 60 =1700

� 22x = 1760

22

� Monthly Salary = 1760

22× 30 = ` 2400

27.(B) As each works for 3 hrs.

Sangeeta's work for 3 h = 3× 1

6=

1

2 part

So, Sangeeta gets Rs. 1

3202

� ��

= ` 160

Manisha's works for 3h = 3 × 1

8=

3

8 part

So, Manisha gets `3

3208

� ��

= ` 120

Rekha gets = 320 � (160 + 120)

= 320 � 280 = ` 40

28.(D) Let the person invest amount x and y into

two different rates of interest.

�12 1

100

x � �+

10 1

100

y � �= 130

� 12x + 10y = 13000 (i)

and 12 1

100

y � �+

10 1

100

x � �= 134

� 12y + 10x = 13400 (ii)

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

x = ` 500 and y = ` 700

29.(B) Let sum = P,

As amount = 2P � SI = P

and time = 8 yr.

� Rate = 100 SI

Sum Time

��

=100

8

P

P

��

% = 1

12 %2

30. (D) Cost price of article = ` x

and selling price of article = ` y

y × 7

100= x ×

8

100� y =

8

7

x

ATQ,

y × 9

100� x ×

10

100=1

�8

7

x×

9

100�

10

x=1

�18

175

x

10

x� = 1

�36 35

350

x x�=1

� x = 350

31.(C) Cost price of an aticle = `10

11

Selling price of an article = `11

10

� Profit =11

10�

10

11=

121 100

110

�= `

21

100

� Profit percent =

21100

11010

11

�=

2100

110×

11

10=21%

32.(B)Let x km/h be the speed of the boat in stillwater .speed of the boat downstream=(x+2)km/h

and speed of the boat upstream= (x - 2) km/hSo, A.T.Q.,

20 20 110

2 2 60x x� �

� �

20( 2 2) 11

( 2)( 2) 6

x x

x x

� � ��

� �

� 11x2 - 44 = 240x� (x - 22)(11x + 2) = 0

or, x - 22 = 0 � x = 22km/h

33. (B) Distance = Difference × Sum of speed

Difference in speed

= 165 × 155

15= 1705 km

34. (C) Speed of bus = 20 50 18

60 5

�� = 60km/h

35. (A) Since the car runs at 7

11th of its own

speed, the time it takes will be 11

7th of its

usual speed.Let the the usual time taken by t h.

Then we can write, 11

7t = 22

� t = 22 7

11

�= 14 h

� Time saved = 22 � 14 = 8 h36.(A) P = ` 9000, r = � 10%

n = 3years

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So, A = 9000

310

1100

� ��

= 9000× 9

10×

9

100×

9

10 = ` 6561

37.(D) Here, R1 = 8%, R

2 = 10%, R

3= 12%

P = ` 6000T = T

1 + T

2 + T

3 = 3yr.

T1 = T

2 = T

3 = 1 yr.

� A = 6000 8

1100

� ��

101

100

� ��

121

100

� ��

= `108 110 112

6000100 100 100

� ��

= `27 11 28

600025 10 25

� �� = ` 7983.366

� Compound Interest = 7983.366 – 6000= Rs. 1983.366

38. (C)The speeds of the trains be a m/s and bm/s.

Now, when they are moving in the samedirection.� Relative speed = (a - b) m/s

� a - b = 100 80

18

�=10

Similarly a + b = 100 80

9

�= 20

By solving Eq. (i) and Eq. (ii), we geta = 15m/s and b = 5m/s

� Speed of faster train = 15 m/sec.

39.(B) In �BCD,

BD2 = BC2 + CD2 = x2 + x2

� BD = 2x

� BD = AE = 2x

� OE = OA + AE = 2 .x + 2 .x = 2 2x

� BA = DE = x

� In �ODE,OD2 = OE2 + DE2

� Minimum distance

� OD = 2

22 2.x x� = 2 28x x� = 3x km

40. (A) � Water that leaks in 5.5 min = 2.25 tones

� Water that leaks in 60 min = 2.25

605.5

�

= 1350

55tones

After pumping water that is left in boat in

60min.

= 1350

55- 12 =

690

55tones

� 92 tones water that remains in boat in

55

690× 92 =

22

3 hr

� Required speed = 77

22

3

= 231

22= 10.5km/h

41.(D) Let the capactiy of the tank = x litres

A.T.Q.,

Quantity of water emptied by the leak in

1 hour = 20

x litres

Quantity of water filled by the tap in 1 hour

= 120 litres

According to the question,

20 30

x x� = 120 � 240

10 15

x x� �

�3 2

30

x x�= 240

�30

x= 240

� x = 240 × 30 = 7200 liters

42.(B) A's investments for 3 yrs.

= ( 250000 × 3 + 100000 × 2 + 100000 ×1)

= ` 1005000

B's investment for 3 yrs

= 350000 × 2 = ` 700000

C's investment for 3 yrs. = 350000 × 1

= ` 350000

� Ratio of investment of A : B: C

= 1005000 : 700000 : 350000

= 3: 2 :1

B's profit = 2

3 2 1� �× 1500000

= ` 500000

43.(A) Work done/hour by a female, a male and a

child are x, y and z, unit respectively.

So, 8x = 6y = 12z

� x = 3

4y and z =

2

Y

9 males can complete a work in 6 days

working 6 h/ day.

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� Work done =9 ×6 ×6y = 324y

Work done by 12 males, 12 females and 12

children in 1 day = 8h/day

= (12y + 12x + 12z)× 8

=3

12 12 124 2

Yy y

� �� × 8 = 216y

Days required to finish work = 324

216

y

y = 11

2 days

44. (*) 1 man's one day's work = 1

100

1 man's six day's work =3

50

10 men's six day's work = 3

5

Remaining2

5of the work is done by 15

women in 6 days.

� Whole work is done by 15 women in 6 5

2

�

= 15 days

� One women alone can finish the work in

225 days

45.(B) In such type of question

CP =profit)percent(100loss)percent(100

profit)percent(100costTotal

��

�

=720 119

85 119

��

= 720 119

204

�= ` 420

46.(C) Cost price of each table watch = 250 + 2500

150

=800

3

Labelled pirce = ` 320

SP = 320 - 5% of 320 = 304

Profit percent =

800304

3

800

3

� �� ×100

=112

3×

3

800×100 = 14%

47.(C)5

16

� ��

of time taken by Sneha

= 1 hour 15 minutes

� Time taken by Sneha

= 1 hours 15 minutes × 6

= 7 hours 30 minutes

48.(A) � At 45g per man per day the provision is

for 16 weeks for 220 men

� 1 g per man per day provision is for 1 week

= 220 × 45 ×16

� 32 g per man per day the provision is for

24 weeks = 220 45 16

33 24

� ��

= 200

Number of men to go out = 220 -200 = 20

49.(C) The part of field cultivated by A in 1 day

=2

5 6�=

1

15

The part of field cultivated by B in 1 day

B = 1 1

3 10 30�

�

� The part of field cultivated by A and B

together in 1 day = 1

15+

1

30=

3

30=

1

10

�4

5part of field is cultivated by A and B

together in 1 day = 4

5

1

10

days = 4 10

5

� = 8 days

50.(D) X : Y : Z = ` (16000 × 3 + 11000 × 9 : 12000

× 3 + 17000 × 9 : 21000 × 6)

= 7 : 9 : 6

� (Y's share - Z's share)

= `9 6

26400 2640022 22

� ��

= ` (10800- 7200) = ` 3600

51.(D) Speed of boat = x m/sec

Speed of stream = Y m/sec

Then,

=1

3

� x = 2y

� x : y = 2 : 1

52.(B) Gain = 2 min + 4 min 48s = 6min 48s

= 408 seconds

Hour = (7× 24 + 2 ) = 170 hrs.

� Clock gains = 408

170= 2.4 s/h

� It will gain 2 min or 120s in 120

2.4h. = 50h

� Clock will show correct time 2pm to

Tuesday

53. (*) � �� ,, are the roots of polynomial

3x3 – 5x2 – 11x – 3

� �� = 3

)5(��=

3

5

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�� = 3

11�

�� = 3

)3(�� =

3

3 = 1

Now,

�� 3333 = ( �� )

[ �� 222 ]

� �� 3333 = ( �� )

[( �� )2 –

�� 222

– �� ]

� 13333 �� = ��

�

��

��

�

��

�(3

3

5

3

52

= ��

��

� ��

3

113

9

25

3

5

= ��

��

��11

9

25

3

5

= ��

��

� �9

9925

3

5

333 �� = 34

124

3

5��

= 327

620�

= 27

81620 � =

27

701

54. (C) � x = 2 and x = – 2 are zeroes of the

polynomial 2x4 – 3x3 – 3x2 + 6x – 2

� )2( �x and )2( �x are factors of

2x4 – 3x3 – 3x2 + 6x – 2

� (x2 – 2) is a factor of 2x4 – 3x3 – 3x2 + 6x – 2

Now,

Other factor = 2x2 – 3x + 1for other zeroes

2x2 – 3x + 1 = 0� 2x2 – 2x – x + 1 = 0

� 2x(x – 1) –1(x – 1) = 0

� (2x – 1) (x – 1) = 0

� x = 2

1, 1

55. (B) � a – b, a, a + b are zeroes of x3 – 3x2 + x + 1

� sum of zeroes = 1

)3(�� = 3

� a – b + a + a + b = 33a = 3

a = 1

Product of zeroes = 1

1�

(a – b) (a) (a + b) = – 1

� (1 – b) (1) (1 + b) = –1

� 1 – b2= – 1

� 2 = b2

b = 2

Hence, a = 1, b = 2

56. (B) � (x – 1) is a factor of 4x3 + 3x2 – 4x + k

� Remainder P(1) = 0

� 4(1)3 + 3(1)2 – 4 × 1 + k = 04 + 3 – 4 + k = 0

k = –357. (C) Volume of the cuboid

= 12ky2 + 8ky – 20k= 4k[3y2 + 2y – 5]= 4k[3y2 + 5y – 3y – 5]= 4k[y(3y + 5) –1(3y + 5)]= (4k)(y – 1)(3y + 5)

3rd dimension = 3y + 558. (B) Let the present age of Jacob = x yrs.

the present age of his son = y yrsTheir age 5 years ago

Jacob = x – 5 yrshis son = y – 5 yrs

Case I:

x – 5 = 7(y – 5)x – 7y + 30 = 0 .... (1)

Five years later their ageJacob = x + 5 yrshis son = y + 5 yrs

Case II:

x + 5 = 3(y + 5) .... (2)On solving eqns. (1) & (2)

x = 40, y = 10Hence their present age = 40 yrs & 10 yrs

59. (A) � The given linear equations have nosolution.

�2

1

a

a =

2

1

b

b�

2

1

c

c

�12

3

�k =

1

1

�k�

12

1

�k

� 3k – 3 = 2k – 1k = 2

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60. (B) Let A & B are two friends.Case I:

A + 100 = 2(B – 100)A – 2B = –300 .... (1)

Case II:

B + 10 = 6(A – 10)6A – B = 70 .... (2)

On solving eqns. (1) & (2), we haveA = Rs. 40 & B = Rs. 170

61. (A) Let 2x = 3y = 6–z = k

� 2 = xk1)( , 3 =

yk1

)( , 6 = zk1

)( �

Now, 2 × 3 = 6

� xk1)( .

yk1

)( = zk1

)( �

�yx

11� =

z

1�

Hence x

1+

y

1 +

z

1 = 0

62.(B) 4 cot �= 3 � cot �= 3

4

sin cos

sin cos

� � ��

=

cos1

sincos

1sin

��

=1 cot

1 cot

� ��

=

31

4

31

4

� ��

=

1

47

4

=1

7

63.(C) As, 3sin � - 4sin3 �= sin3 � So, 3 sin 15º - 4 sin3 15º = sin3 (15º)= sin 45º

=1

2

64.(C) The maximum value of a sin � + b cos � is

2 2a b� &, here a = 1 , b = 1

So, Maximum value = 1 1� = 2

65.(A) sin (�+� ) = 21 cos ( )� �� =16

125

� =19

25=

3

5

cos(� -� )= 21 sin ( )� � �� =25

1169

� =144

169 =

12

13

� tan ( � + � )=sin( )

cos( )

� � �� =

3 5

5 4

� ��

=3

4

tan ( � -� )=sin( )

cos( )

� � �� =

5 13

13 12� =

5

12

� tan (2� )= tan [( � +� ) +( � -� )]

=tan( ) tan( )

1 tan( )tan( )

� � � � � � �� =

3 5564 12

3 5 3314 12

��

� �

66. (A) sin 3A = cos(A – 30)

� cos(90 – 3A) = cos(A – 30)

� (90 – 3A) = A – 30

� 120 = 4A

� A = 30º

Now tan2A = tan 2 × 30º

= tan 60º = 3

67. (C) Let the balloon subtends an angle � at

the eye of the observer at P.

In �OMP,

PO

MO= 2

sin�

PO

r = 2

sin�

PO = r cosec 2�

Now,

In �ONP

sin = PO

ON =

2eccosr

ON

�

� ON = r cosec 2� sin

� The height of the ballon

= sin cos2

� r ec

68. (C) Let 'A' the point h m above the lake

where the angle of elevation of the cloud

is ' � ' and the angle of depression in the

lake is ' '

In �ABN

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tan = AB

BN [Now BN = BC + NC

= h + x + h = (x + 2h)m

tan = AB

hx 2�

� AB = (x + 2h) cot

[� In a plane mirror, image distance = object

distance]

� MC = NC

x + h = NC

From (1) & (2), we have

x cot � (x + 2h)cot

x(cot � – cot ) = 2hcot

� x = ��

cotcot

cot2h

Height of the cloud above the lake = x + h

= 2 cot

cot cot

!�

� � !h

h

= 2 cot cot cot

cot cot

!� � � !� � !

h h h

= ��

cotcot

cotcot hh

cot cot

cot cot

� � � ��

h = tan tan

tan tanh� � � ��

69. (B) Let T1 & T

2 represents the two towers.

In � ABT2

tan30º = AB

AT2

3

1 =

310

AT2

� T2A =

3

310= 10m

70. (B) Let W1 & W

2 are two window of a house

which are at the height of 6m & 2m above

the ground

Let AC = x cm� W

1P = W

2Q = AC = x m

� QP = 4 m

In �BPW1

tan 60º = PW

BP

1

3 = PW

4BQ

1

�

BQ + 4 = 3 × W1P = 3 × x m

� BQ = 3 x – 4 m

In �BQW2

tan 30º = QW

BQ

2

3

1=

x

x 43 �

� x = 3x – 34

� –2x = – 34

� x = 32

Height of the ballon = BQ

= 43 �x

= 4323 �� = 6 – 4 = 2mHeight of the balloon above the ground

= 2 + 4 + 2 = 8 m71.(C) Let sides of a square be a.

Then, AC = a 2 and AO = OC = a

2

Here, AM = a

2

� LM = a

2-

a

2and OM =

a

2

In � OML

Centres at:

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

9


tan2

�=

a a

22a

2

�

=

2 1

21

2

�

= 2 -1

� tan �= 2

2tan2

1 tan2

�

��

=2 2 1

1 (2 1 2 2)

�

� � �

= 2 2 1

1 3 2 2

�

� �=

2 2 1

2 2 2

�

�

� tan �=1

72.(B) From O draw OL "AB and OM "CD & joinOA and OC

AL = 1

2, AB = 5 cm, OA = 13cm

OL2 = OA2 - AL2 = (13)2 - (5)2

= (169 - 25) = 144

� OL = 144 = 12 cm

Now, CM = 1

2×CD = 12 cm and OC = 13cm

� OM2 = OC2 - CM2 = (13)2 - (12)2

= (169- 144) = 25

� OL = 25 = 5cm

� ML = OM + OL = (5 + 12)cm = 17cm

73.(A) In �ABC

A = 1

2×base × altitude =

1

2×b × AC

AC = 2A

b

Using Pythagorus theorem,

AC2 + AB2 = BC2

� BC = 2

2

2

4Ab

b�

Again in �ABC, A = 1

2× BC × AD

� AD = 2 4

2

2A

4A b

b

�= 2 4

2Ab

4A b�

74. (C) � XY || AC,

� �BXY & �BAC are similar

(by AA similarity.)

� XY divides �BAC into two parts of equal

area.

� ar( �BXY) = ar(quad. XYCA) = 2

1ar( �BAC)

�)BAC(ar

)BXY(ar

�

� = 2

2

BA

BX

�)BXY(ar2

)BXY(ar

�

�= 2

2

BA

BX

2

1 = 2

2

BA

BX

BA

BX =

2

1

� 1 – BA

BX = 1 –

2

1

�BA

BXA �# =

2

12 �

�AB

AX=

2

12 � =

2 2

2

�

75. (B)

�)COD(ar

)AOB(ar

�

�= 2

2

CD

AB = 2

2

CD

)CD2(

[� AB = 2CD]

� ar( �AOB) : ar( �COD) = 4 : 1

76. (D)

4BL2 = 4BA2 + AC2 (1)

Again, CM2 = CA2 +

2

AB2

1��

��

�

Centres at:

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

10


= CA2 + 4

1AB2

4CM2 = 4CA2 + AB2 (2)

Adding (1) & (2)(4BL2 + CM2) = 4(BA2 + CA2) + AC2 + AB2

4(BL2 + CM2) = 5AB2 + 5AC2

= 5[AB2 + AC2]= 5 × BC2

[by Pythagorus theorem]

77. (B) In �ACB & �DCE

$A = $CEB (given)

$ACB = $DCE (common)

� By AA similarity,

� ACB ~ �DCE

�DE

AB =

DC

CB =

AC

CE

�x

9=

10

78 �

� x = 15

109� = 6

78. (B) � QT & RT are bisectors of $ PQR

& $ PRS respectively.

$ TRS = 2

1$ PQR + $QTR (1)

(Ext. angle property)Also,

$ PRS = $ PQR + $QPR

2

1$ PRS =

2

1$PQR +

2

1$QPR (2)

$ TRS = 2

1$ PQR +

2

1$QPR

From (1) & (2)

2

1$ PQR + $QTR =

2

1$PQR +

2

1$QPR

$QTR = 2

1$QPR

79. (C) � (llgm ABCD) & (llgm ABMN) are on the same base & between the same

parallels.

� ar(llgm ABCD) = ar(llgm ABMN)

� ar(llgm ABCD) = 80 sq. unit

Again, � APN & llgm (ABMN) are on thesame base & between the same parallels.

� ar( � APN) = 2

1ar(llgm ABMN)

= 2

1 × 80 sq. unit

= 40 sq unit.

80. (*)

In �BME

$ 1 = 180º – x

$ 2 = 180º – y

� $CEB = 180º – ($1 + $2)

$CEB = 180º – [180º – x + 180 – y] = x + y - 180º = x + y – %

81. (B) Suppose (–4, 6) divides AB in the ratioof K : 1

By section formula

– 4 = 1

6–13

�&�&

K

K

–4K – 4 = 3K – 6–7K = – 2

K = 7

2

� Required ratio = 2 : 7

82. (B)

� Coordinates of mid point of AC= Coordinates of mid pp. of BD.

��

��

� ��

2

62,

2

1 x= �

�

��

� ��

2

5,

2

43 y

�2

1 x� =

2

7&

2

62 �=

2

5 y�

� x = 6 y = 3

83.(B) Curved surface of tomb = % rl = 22

7×14× 50

= 2200m2

� Cost of white washing = 2200× 0.80

= ` 176084.(C) Let the sides of the two cubes are x and y So,

A.T.Q,3

3

27

64

x

y� =

3

3

(3)

(4)

�x

y = 3

4

Now,

� surface area of the cube = 6× (side)2

Centres at:

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

11


� Ratio of their surface areas =

2

2

6

6

x

y

= 2

2

6 3

6 4

��

=9

16= 9: 16

85.(C) Let ED = x

Now, AC = 2 28 6� =10

In � AED,

AE2 = AD2 - x2 = 36 -x2 (i)

And in � CFD,

CF2 = (8)2 - (10 -x )2 (ii)From Eqs. (i) and (ii), we get

36 -x2 = 64 - (10 -x )2 (�AE = FC)� 36 -x2 = 64 - (100 +x2 - 20x )

� 20x = 72

� x = 18

5

� From Eq. (i)AE2 = 36 -

218

5

� ��

AE2 = 36 - 324

25=

900 324

25

�

�

Area of rec tangle ABCD

Area of recetangle AEFC=

8 6

2410

5

�

�=1

86.(C)

15

Capacity of bucket= Volume of frustrum of cone

= h

3

%[R2 +r2 + Rr]

= 22

7×

24

3[(15)2 + 52 + 15 × 5]cm3

= 22

7× 8 [225 + 25 + 75]cm3 =

176

7(325)cm3

= 8171.43cm3

87. (D) Side of the square field = 31684

= 178 mPerimeter of the square field = 4 × 178

= 712 mLength of the wire required to cover thefield once = 105% of 712 m

= 1.05 × 712

= 747.6 m

Total length of the wire = 4 × 747.6

= 2990.4 m

88. (B) Total surface area of the closed cylindrical

petrol tank = 2 % r (h + r)

= 22

2 2.1(4.5 2.1)7

� � �

= 2 × 22 × 0.3 × 6.6

= 87.12 m2

�12

1 of the total steel wasted away

�12

11 of the total steel was used to make the

tank.

�12

11 of total steel = 87.12

Total steel= 11

1212.87 �= 95.04m2

89. (B) Speed of the flowing water = 2 km/h

= 60

10002�m/min

Length of the water stored in 1 min in the

river = 6

200m

Volume of the water = lbh

= 3406

200�� = 4000 m3

90. (B)

� � AOB ~ �OMN

�OM

OA =

MN

AB

�24

12=

7

AB

� AB = 7

2 cm

Volume of the upper part = hr3

1 2%

Centres at:

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

12


= 122

7

2

7

7

22

3

1�� = 154 cm2

91. (B) For the Frustum For the cylinder

r1 = 9 cm r = 4 cm

r2 = 4 cm h = 10 cm

h = 12 cm

Area of the sheet required

= area of frustum + area of cylinder

= % (r1 + r

2)l + 2 % rh

= 7

22[(9 + 4) × 13 + 2 × 4 × 10]

= 7

22[169 + 80]

= 7

22 × 249

= 782.57 cm2

92. (A)

= 3

8% +

3

16% =

3

24% = %8 cm3

Volume of the cylinder = hr2%= % × 2 × 2 × 4

= 16 % cm3

Required difference = 16 % – 8 %= 8 % cm3

= 25.12 cm3

93.(C) Slum population of A in 1991

= 35% of 91.9 lakh

= 35

100× 91.9 lakh

= 32.165 lakh = 32 lakh

94.(C) Difference = 21% of 25.5 lakh -10% of 29.2

lakh

=22

100×25.5 lakh -

10

100×29.2 lakh

= 5.355 - 2.920 = 2.435 lakh

95.(A) Highest slum population is 38%. It is

persent in B.

96.(A) Compostion of Bengalis is 12%.

So the % of Tamilians with respect to

Bengalis will be nearly 8

12×100 = 67% .

97.(C) Hindi speaking population

=18

100×413 = 7.43 = 8 million (aporox.)

98.(A) It is clear from the pie diagram that the

answer is Punjabi and Hindi speaking =

17+18 = 35%

99.(B) Other 4 %

� 2% of it =2

100×4 = 0.08

� Percentage of increase in Punjabis =

0.08

17×100 = 0.5 nearly

100.(C) Percentage of Punjabis = 0.5%

Increase in millions = 0.5

100×413 = 0.20

Centres at:

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

13


1. (B)

2. (B)

3. (D)

4. (A)

5. (C)

6. (C)

7. (C)

8. (A)

9. (D)

10. (C)

11. (B)

12. (B)

13. (C)

14. (A)

15. (B)

16. (D)

17. (C)

18. (A)

19. (B)

20. (A)

21. (A)

22. (A)

23. (A)

24. (B)

25. (D)

26. (D)

27. (B)

28. (D)

29. (B)

30. (D)

31. (C)

32. (B)

33. (B)

34. (C)

35. (A)

36. (A)

37. (D)

38. (C)

39. (B)

40. (A)

41. (D)

42. (B)

43. (A)

44. (*)

45. (B)

46. (C)

47. (C)

48. (A)

49. (C)

50. (D)

51. (D)

52. (B)

53. (*)

54. (C)

55. (B)

56. (B)

57. (C)

58. (B)

59. (A)

60. (B)

61. (A)

62. (B)

63. (C)

64. (C)

65. (A)

66. (A)

67. (C)

68. (C)

69. (B)

70. (B)

71. (C)

72. (B)

73. (A)

74. (C)

75. (B)

76. (D)

77. (B)

78. (B)

79. (C)

80. (*)

81. (B)

82. (B)

83. (B)

84. (C)

85. (C)

86. (C)

87. (D)

88. (B)

89. (B)

90. (B)

91. (B)

92. (A)

93. (C)

94. (C)

95. (A)

96. (A)

97. (C)

98. (A)

99. (B)

100. (C)

SSC MAINS MOCK TEST -8 ANSWER KEY

ssc mains (maths) mock test-8 (solution)

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