Ssc Mains (Maths) Mock Test-6 (Solution)

8/13/2019 Ssc Mains (Maths) Mock Test-6 (Solution)

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SSC MAINS (MATHS) MOCK TEST6 (SOLUTIONS)

1.(A)

. .

. . .

0 13 0 07

0 25 0 075 0 2

243 243

7 49 343

=

5 0.13 5 0.07

0.25 2 0.075 3 0.2

(3 ) (3 )

(7) (7 ) (7 )

=

0.65 0.35

0.25 0.150 .06

3 3

7 7 7

=

(0.65 0.35)

(0.25 0.150 0.6)

3

7

1

1

3 3

77

2.(B) 3 3

2 216 16

3

2 23

2

1(4 )

16=

32

23

2 2

14

(4 )

33

14

4

164

64

4096 1

64

4097

64

3.(C) No. of digits required= [{(91)+1}1 + {(5010)+1}2]= 9 1 + 41 2 = 9+82 = 91

4.(D) Digit in the unit's place of(251)98+(21)59(106)100+ (705)35164 + 259= 1 + 1 6 + 5 4 + 9= 6

5.(D) The required number must also be divis-ible by (232+1) and among the options given,(296+1) is divisible by (232+1)

296+1 = 296+196

= (232)3 + (132)3 , which is divisible by 232+1[when nis odd, (an+bn) is always divisibleby (a+b) ]

6.(B) Given that,H.C.F. of the two numbers = 27So, Let the numbers are 27 xand 27 ywherexand yare co-prime nos. i.e. prime to eachother.

Now, A.T.Q27x + 27y = 216or, 27 (x + y) = 216 x + y= 8So, possible pairs of xand yare (1, 7) & (3, 5)So, The possible pairs of two numbers willbe (27, 189) & (81, 135) possible pairs ofThe possible no. of pairs is 2.

7.(B)221 428[(1931) ]

1932=

even0dd(1931)

1932

= even1931

;Remainder 11932

even(a 1)

;R 1a

8.(B) Term difference = 124 = 8

Value difference = 7014=56

Value difference per term (i.e. commondifference) = 56/8 = 7

So, first term = 4thterm 3 common dif-

ference = 14 3 7 = - 7

9.(B) Required height at the 1st bounce 3

324

Required height at the 2nd

bounce

23

324

Required height at the 3 rd

bounce

33

324

= 27 1

32 13 m64 2

10.(A) Remaining no. of total balls after 1st ball is

chosen = (12 + 6)-1 = 17 balls

Also,

Remaining no. of black balls after 1st ball

(which is black) is chosen = 121

= 11 black balls

So, The probability that the second ball is

also black = 11/17

11.(A) Let xbe the initial no. of people in the

company.

So, ATQ,

=x

x

35 5 32

345

or, 35x + 160 = 34x + 170x = 10

12.(*) Initial bowling average = 12.4

After improving bowling average by 0.2,

new bowling average = 12.4 0.2 = 12.2

Now, let xbe the number of wickets taken

before the last match

So, A.T.Q,

=. x

.x

12 4 26

12 24

or 12.4x + 26 = 12.2x + 48.8

0.2x = 22.8


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22.8

1140.2

x

No. of wickets taken before the last match= 114

13.(A) Average speed during the entire journey

=Total distance

Total time

3584km 3584km

2days 8hours 56hours

= 64 km/hour

Now, Average speed during the remaining

part (last 8 hr.) of journey

=3584 (1440 1608)

8km/hr.

=3584 30488

= 536/8

=67 km/hrSo, required difference = (67 64) km/hr

= 3 km/hr

3 km/hr. more

14.(A) Age Age height

9 yr. 9 3 4 ft.

(9+7) yr.=16 yr 16 4 4

4 ft3

=1

5 ft3

15.(A) Let the required ratio is x: y

So, A.T.Q,

=

192x 150 y 120194.40

x y 100

192 x 150y or, 162

x y

or, 192 x + 150 y = 162 x + 162 y

or, (192 162)x = (162 150)y

or, 30x = 12y

or, 5x = 2yx : y = 2 : 5

16.(B) weight of lead per kg in the new alloy

3 3 1kg

(5 4 2) 2 24 8

17.(B) Required average rate of interest per

annum

=

1 1 1 110 9 1 12 %

2 3 2 3

= (5 + 3 + 2) % = 10%

18.(C) Let the income of Sanjay two yrs. ago= Rs. x

Saving of Sanjay two yrs ago= 20% of Rs.x = Rs. x/5

Expenditure of Sanjay two yrs. ago.

=

45 5xx x

Two years later now,

income of Sanjay = 120 6

Rs. x Rs. x100 5

and saving of Sanjay =x

Rs.5

Expenditure of Sanjay =

6 xRs. x x

5 5

So,% increase in the expenditure

4x x

5 100%4x

5

x

100% 25%4x

19.(B) Price is reduced by 20%

Consumption can be increased by

20100%

100 20= 25%

25% of initial consumption = 500 gm Initial consumption (ie. 100%) = 2000gm

= 2 kg

Original price of the Sugar per kg= Rs. 36/2kg= Rs. 18/kg

20.(B) Let the maximum marks = xCase (i) Pass marks = 32% of x + 16Case (ii) Pass marks = 36% of x 10from Case (i) & Case (ii), we get,

32% of x + 16 = 36% of x 10or, 4% of x = 26

or, 4

x 26100

x =

26 100

6504

So,

Pass% =

1632% 100 %

650

= 6 6

32% 2 % 34 %

13 13


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21.(D) Re. 1 50pcoins coins

Ratio of respective values = 13 : 11

Ratio of value of 1 coin of each = 2 : 1

So, Ratio of no. of coins = 132

: 111

= 13 : 22

total no. of coins = 210

So, No. of Re.1 coins =

13210

13 22

13210

35

= 78 coins

22.(A) By Alligation method

1

2

35 49

12

49

118

35

12

110

==

Required ratio of mixture =1 1

:18 10

= 10 : 18

= 5 : 9A.T.Q, Amount of the former mixture

= 3 litre

So, required of the later mixute

93

5 litre

25

5 litre

23.(D) Let the original number of boys and girlsbe 5x and 3x respectively and that of new

boys and girls be 5y and 7y respectively. 5x + 3x + 5y + 7y = 1200

2x + 3y = 300 ......... (i)

and5 5 7

3 7 5

x y

x y

25x + 25y = 21x + 49y4x = 24yx = 6y ......... (i)From equation (i),

4x + 6y = 6005x = 600x = 120Original no. of students= 8x = 960

24.(A) Ratio of first and second class fares= 3:1

and Ratio of no. of passengers = 1 : 50Ratio of total amount from 1st & 2nd classpassengers

= 3 1 : 1 50 = 3 : 50So, Amount collected from 2nd class

passengers =50

132552

=Rs. 1250

25.(C) ATQ,Ratio of money received by each

(Son : Daughter : Nephew)= 5x : 4x : xSo, Ratio of amount to

5 Sons : 4 daughters : 2 nephews= 25x : 16x : 2x

25x : 16x : 2x = Rs. 8600or, 43x = Rs. 8600x = Rs. 200Required money to each daughter= 4 200 = Rs. 800

26.(C)2 2

A 40 B 205 7

910

17C x (let)

5 7( 40), ( 20)

2 2A x B x

and,17

( 10)9

C x

5 7 17( 40) ( 20) ( 10)

2 2 9x x x

= 600x = 100

As share =5

. (100 40)2

Rs = Rs. 150

27.(D) Extra interest received in 4 years if the

rate of interest is increased by 1%

= (41)% of 1200= 4% of 1200= Rs. 48

Total amount received in 4 years if the rateof interest is increased by 1%

= 1632 + 48

= Rs. 1680

28.(C) Reparied gain =1

2 6 4 % 50004

of

12 2 % 5000

4of


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92 5000

4 100

= Rs. 225

29.(A) By method of alligation

-6% 14%

-4%

{14-(-4)}%= (14+4)%=18%

{-4-(-6)}%= (-4+6)%=2%

Ratio of Amount = 18:2 = 9:1 Quantity sold at 14% profit

=1

50 5

10

kg kg

Quantity sold at 6% loss

950 45

10kg kg

30.(C) n= 2 years, r = 10% C.I. = 525

. . 1 1100

nr

C I P

210

525 1 1

100

P 121100

100

P

525 100

21P

= Rs. 2500

Now, ATQ, n1= 4 years, r

1= 5%

So,1 1. .

100

P r nS I

2500 5 4

100

= Rs. 500

31.(D) Let Rs. xbe the sum borrowedSo, Rs. xafter two years will become

Rs. x

25 441x

1 Rs.100 400

Present value of Rs.441x

882 Rs.800x

Present value of Rs. 882 = Rs. 800

Now, Amount of Rs. x after one year

5 211 .

100 20

xx Rs

Present value of = Rs.21

20

xafter one year

= Rs. x

Present value of Rs. 882 due after oneyear

20 882.840

21

xRs

x

Required sum= Rs. (800 + 840) = Rs. 164032.(A) Let Rs. x be the marked price of the shirt.

ATQ, Difference of discounts = 2%

2% of x = 15

2

15100

x

15 100

.7502

x Rs

33.(C) Let the first CP of the commodity be Rs.100

First SP = Rs. 110Second CP = Rs. 90.

Gain =2

6 %3

=50

%3

Second SP

50100 % .90

3of Rs

35090

3 100

= Rs.105

Difference of SPs = Rs. (110 105) = Rs. 5 If the difference is Rs.5, then CP = Rs. 100So, If the difference be Rs. 2, then

100 2 .405CP Rs

34.(C) For the first trader,

Let the CP of the article = Rs. 100

SP = Rs. 120Now, For the second trader,SP of the article = Rs. 120

& Gain = 20%Let the CP be Rs. x.

120

100 20120

x

6120 20 245x

120 24 .96x Rs Gain = Rs. 24Now when difference of gains = Rs. 4,then SP = Rs. 120

So, When the difference = Rs. 85,

then SP =120

854

= Rs. 2550

35.(A) Let the C.P. = Rs. 100,When sold at 3/4 th of S.P. the loss is 4%.

S.P. in this case = Rs. 96

= 3/4 times Actual selling Price.


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Actual selling price = (96)4/3 = Rs. 128If Gaurav sells at the actual S.P. then hemakes a profit of Rs. 28 on a cost price of

Rs. 100 i.e. profit = 28%.36.(A) Let the marked price of the article= Rs. x

Single discount for successive discounts of30% and 20%

30 2030 20 %

100

(50 6)% 44% discountNow, A.T.Q.

(100 44)% 2240of x

56

2240100

x

2240 100

.400056x Rs

37.(D) Let the printed price of the book = Rs. x.

So, Selling price = 90% of x

9

.10

x

Rs

Now, if the CP of the book = Rs. y. (let)Then, A.T.Q,

112 9

100 10

xy

or,

9 100 45

10 112 56

y

x

Required ratio = 45 : 5638.(A)Let x= number of months (from starting)

after which C joined the business.So, Ratio of shares of Profit

= 30,00012 : 40,0008: 50,000 x= 32 : 36 : 5x

Cs share5

36 32 5

x

x

5

68 5

x

x

given,5

68 5

x

x=

15000

49000 x= 6

C joined the business (i.e. 6-4) =2 monthsafter joining of B

39.(A) Let 1 childs 1 days work =x& 1 adults 1 days work = y

Then,1

1216

x 1

192x

and1

812

y 1

96y

Work done in 3 days by 16 adults

1 116 3

96 2

part

Remaining work1

2 part

Now, (6 adults + 4 children)s 1 days work

6 4 1

96 192 12

i.e. 1

12work is done by them is 1 day

So,1

2work will be done by them in

=1

122

days

= 6 days

40.(C)Time taken by A to complete the work

4 3

2

= 6 days

& Time taken by B to complete the work

=6 5

103

days

A and B together will complete the work

in6 10

6 10

days =3

34

days.

41.(C) Completed road in 80 days by 280 workers

=7

2km= 3.5 km

Remaining road to be completed in 20days = 1.5 kmLet, total x workers are needed to constructthe 1.5 km road in 20 days.

So,280 8 0 3.5

x 20 1.5

x =80 1.5

28020 3.5

x = 480 workers No. of more workers needed

= (480280) people

= 200 workers42. (B) Ratio of wages of A, B, C

= (65): (46): (94) = 30 : 24: 36

= 5 : 4 : 6

As Share =5

180015

= Rs. 600

43.(D)Part of tank filled in one hour by inlet pipe

1 1 1

12 15 60 part

So, the inlet pipe can fill the tank in 60 hrs.

Inlet pipe fills water at the rate of 5 litres

per minute


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Capacity of tank= (60 60 5) litres = 18000 litres

44.(A) Part filled by 1stpipe from 8 a.m. to 11 a.m.

=3 1

15 5 part

Part filled by 2nd pipe from 9 a.m. to 11 a.m.

=2 1

12 6 part

Part filled till 11 a.m.

1 1 6 5 11

5 6 30 30

part

At 11 a.m. 3rd pipe is also opened to empty

it.

Now, time taken by 3 pipes together tocompletely empty the full cistern =

1 1 10hrs.

1 1 1 1

4 15 12 10

So, Time required to empty11

30filled part

=11 11

1030 3

hr =2

33

hrs.

i.e. 3 hours 40 minutesi.e. at 11. 40 a.m.

45.(D)Let the speed of the bus = xkm/hr

Then to take a lead of 60m,he will have to

cover a distance of (60+40)m= 108m, withthe speed of (30x) km/hr in 20 sec.100m = (30x) km/hr 20 sec.

100 =(30 x) 1000

203600

1 (30 )

10 180

x

or , 180= 30010x

10x = 120 x= 12 km/hr

46. (C)

r

Circumference = 2 r

Time taken for one round=40

8= 5 min.

Now, new radius = 10 r

So, New circumference = 2 10r = 20 r

So Required time =20 r

52 r

minute

= 50 minutes

47.(B) In such type of questions, required ratio of

the speeds of the two trains =9 3

24 = 3 : 2

48.(A) Let the speed of goods train = xkm/hr

Distance travelled at the speed of 80 km/

hr in 4 hours.

= 4 80 = 320 km

320

10x = 32 km/hr

49.(B) Let the length of train or platform = xmetre.

Speed = 90km/hr

=90 5

18

metre/sec.

= 25 metre/sec.

Distance covered in 60 sec.= 25 60 = 1500 metres

Now, according to question,

2x = 1500

x = 750 metre50.(A) Speed in still water = 12km/hr

speed against the current

412

3 = 9km/hr (80 min=

4

3hr)

Speed of current = 12 9 = 3km/hr

Speed with the current = 12 + 3 = 15 km/hr

So, required time =45 45

15 9 = 8 hr.

51.(A) Let the cost of one saree = Rs. x

and the cost one shirt = Rs. y

According to question

20x + 4y = 1600

x + 2 y = 800 ........... (i)

and

x + 6 y = 1600 ........... (i)

on solving equations (i) and (ii), we get

x = 400; y = 200;

cost of 12 shirts= 12 200 = Rs. 2400


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52.(D)

4 21 1 1

1 (1 )

x xA

x x x x

)1(

1

11

)1)(1(

2

22

xxx

x

x

xx

= A

)1(

1

1)1(

)1)(1)(1(2

2

xxx

x

x

xxx

= A

A = 1

53.(*) 1/3( 2 1)x

3 2 1x

Now, 31 1

2 1

x

1 2 12 1 2 1

= 2 12 1

2 1

Now,3

3

1 ( 2 1) ( 2 1)x

x

= 2 1 2 1 2 54.(C) Let p(x) = (x+1)7 + (2x + k)3

(x+2) is a factor of p(x)

p(2) = 0 [by factor Theorem]

(2 + 1)7 + (2 2 + k)3 = 0 (1)7 + (k4)3 = 0

(k 4)3

=1 k 4 = 3 1 1

5k

55.(C) (y z) + (z x) + (x y) = 0

[a3 + b3 + c3 = 3abc if a + b + c =0]

(y z)3+ (z x)3+ (x y)3= 3(y z) (z x)(x y)

56.(D)1 1 1 1

1 1 1 11 2 3x x x x

1 2 3 4

1 2 3

x x x x

x x x x

4x

x

57.(B) 5 12 13x x x

The above statement is true only for 2x

x = 22 = 4

58.(C) 3( 3 2) ,a

3( 3 2)b

a.b =3

( 3 2)( 3 2)

3 33 2 (1) 1

=

1 1

( 1) ( 1)a b

=1 1

1 1a b

1 1

1

b a

ab b a

=

2

1 1

a b

a b

2

2

a b

a b

[ ab =1]

= 1

59.(B) x y za b c k (Say)

zyx kckbka111

,,

2b ac

yk

2

= zxk

11

)(

2 1 1y x zk k

2 z x

y zx

2zxyz x

60.(C) let p(x) = ax3 + 3x2 8x + b

(x + 2) is a factor of p(x)

P(2) = 0 a (2)3 + 3(2)2 8(2) + b =0 8a + 12 + 16 + b = 0

8a + b + 28 = 0 .............. (i)Again, (x2) is factor of p(x)

P(2) = 0 a(2)3 + 3(2)2 8.2 + b = 0

8a + b 4 = 0 .............. (ii)On adding (1) & (2), we have

2b + 24 = 0

b = 12

On substituting b = -12 in (2)

8a 12 4 = 0

a = 2

61.(C) sinB =2

1= sin30

030B

Now, 3 cos B - 4cos3B

= 3 cos 300 - 4cos3300


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3 3 33 4

2 8

3 3 2 3 0

2 2

Another method

33cos 4cosB B= cos3B

= -cos 3 300

= - cos900

= 0

62.(A) tanA =1 A = 450

tan B = 3 = B = 600

Now, cosA. cosB + sinA.sinB

cos450.cos600 +sin450.sin600

1 1 1 3

2 22 2

=22

13

63.(B)

P C

B300 m

Pole

tan =12

5

5

12

AB

BP

5

300 12

AB

BC

------------- (1)

tan B =4

3

3

4

AB

BC ------------- (2)

On dividing (1) by (2), We have

5 4 5

300 12 3 9

BC

BC

9BC = 5BC + 1500

BC =4

1500= 375m

Height of the pole = AB =3

4BC

=3

3754

=1125

281.254

m

64.(A) (sinA+cosecA)2 + (cosA + secA)2

=sin2A + cosec2A + 2sinA.cosecA+ cos2A+ sec2A + 2cosA.secA

= sin2A +cos2A + cosec2A + sec2A + 2.1 + 2.1= 1 + 1 + cot2A +1 + tan2A + 4

= 7 + cot2A + tan2A65.(D)

B300 m

Tower

b m

In ABQ

AB

QAtan

AB

btan

tanb

AB

In ABP

AB

PBtan

tan

tanb

PB

PQ =

tan.tan

b

= cottanb


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66.(C)

h

2

In CBD,

tan =h

BD

BD = cot

tanhh

In CBA,

tan =BA

CB=

DABD

CB

=

2cot

hh

h

(BD= hcot )

tan =

2

coth

h

h

1 1cot

2 tan

cot +1

2= cot

cot cot =1

2

67.(B)

Let the height of the tower be xunit

In CBA

tan CB x

BA BA

BA =tan

x=xcot ....................... (i)

In DBA

tan =DB

BA=

cot

h x

x

( BA = x cot )

xcot = tan

h x= ( )h x cot

x(cot cot ) = hcot

cot

cot cot

hx

tan1 1

tan tan

h

=

tantantantan

tanh

tan

tan tan

h

68.(B) 6 6 4 42(sin cos ) 3(sin cos ) 1q q q q

= 3 3 2 2

2 2 2 42 sin cos 3 sin cos 1

q q q q

32 2 2 2 2 22 sin cos 3sin cos (sin cos )q q q q q q

2

2 2 2 23 sin cos 2sin .cos 1q q q q

3 2 22 1 3sin .cos .(1)q q -3 [(1)2 sin2q.cos2q] + 1

= 2 6sin2q cos2q 3 + 6sin2q cos2q + 1

2 3 1 0

69.(C ) 2 3sin sin sin 1 3 2sin sin 1 sin

On squaring both sides,

422 cos)]sin1)([(sin

2 2 2 4(1 cos )(2 cos ) cos

]coscos44)[cos1( 422 = 4cos

2 4 2 44 4 cos cos 4 cos 4cos 6 4 cos cos

6 4 2 cos 4 cos 8cos 4 0 6 4 2cos 4cos 8cos 4


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BC2= BO2+ CO2 pythograms Theorms )CD2= CO2+ DO2

AD2= AO2+ OD2

Now, AB2+ CD2= AO2+ BO2+ CO2+ DO2

AD2+ BC2= AO2+ OD2+ BO2+ CO2

Hence AB

2

+ CD

2

= AD

2

+ BC

2

73.(A)

A

BD C

E

E is the mid point of AD

BE is the median.

ar ( BED) = ar ( ABE)

=

1

ar( ABD)2 .............. (1)[A medians divides each into two parts ofequal areas]Similarly, we can write

ar ( CED) = ar (AEC)=1

2ar ( ACD)

On adding (1) & (2)

ar ( BEC) =1

2ar ( ABD)+

1

2ar ( ACD)

=1

2

ar ( ABC)

74.(C) Let one of the two adjacent angles be of x0,

other adjacent angle =2

3x0,

Now, 0 0 02 180

3x x

[Adjacent angles of a gm aresupplementary]

021 1803

x

x=0 03180 108

5

Smallest angle =0 02 2 108 72

3 3x

75.(B) BCE CBD BDC

(Exterior angle of a is equal to the sum ofopp. interior angles. )

650= 280+ BDC

65 28 = BDC

370 = BDC

Also, ABDC & BD works as a transversal.

70.(C) tan a

qb

sin

cos

q a

q b

2

2

sin

cos

a q a

b q b

Applying C & D, we have

2 2

2 2

sin cos

sin cos

a q b q a b

a q b q a b

71.(B) D C

A B

O

P

Q

ABCD is a gm whose diagonal BD=18cm.Let both the diagonals bisect at 'O'

DO = OB = 9 cm.

DO and BO are medians of ADC & ABC

Also P & Q are centroids of ADC & ABC

1PO BO3

& 1QO DO3

[centroid of a divides each median in theratio of 2 : 1]

1PO 9

3 &

1QO 9

3

= 3 cm & = 3 cm

PQ = PO + QO = 3 + 3 = 6cm

72.(B)

ABCD is a quad. where diagonals AC & BD

intersect each other at O. Also AC BDAB2 = AO2+ BO2 (By using


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BDC = ABD (Alternate interior angles)370 = ABD

76.(D) Let the side of the square be xcmLenght of the rectangle = (x + 5) cm

its breadh = (x 3) cm

ATQx2 = (x + 5) (x 3)

x2 = x2 3x + 5x 152x = 15

x =15

7.52

cm

Perimeter of the rectangle = 2 (l + b)

= 2 [(7.5 + 5)+ (7.5 3)]= 2 [12.5 + 4.5]

= 2 17 = 34 cm.77.(D) EF || DC (Given)

EGF CGD (by AA Similarity)

EG EF

GC DC

5

10 18

EF

EF =10

518 = 9 cm

78.(C) ||C F A B

BCF ABC (alternate interior angles) = 850 (Given)

BCE = BCF + ECF= 850 + 200

= 1050

BAD BCF

(Angles in the alternate segment)= 1050

79.(*) coordinates of the mid-point of (4,2) & (6,4)

4 6 2 4,

2 2

= (5, 3)

Equation of the required straight line is

3 3 3 ( 5)

5 5y x

6 3 ( 5)

10y x

6x 30 = 10y 30 6x 10y = 0 3x 5y = 0

80.(A) The equation the circle is

(x +1) (x + 2) + (y1) (y + 3) = 0

x2 + 3x + 2 + y2+ 2y 3 = 0

x2 + y2+ 3x + 2y 1=0

On comparing with the standard eqn. ofcircle.

x2 + y2+2gx + 2fy + c = 0

g =3

2,f= 1, c = 1

rad. of the circle = 2 2 g f c

=

223 1 (1)

2

=9 17

24 2

Area of the circle = 2r

=

217

2

17

4 sq. unit.

81.(B) BDE BCA (both s are equilateral equi angular) ar ( BDE) : ar ( ABC) = BD2 : BC2

221 BC :BC

2

2 21 BC : BC

4

= 1 : 4

82.(A) Let the parallel sides of the trapezium be5x cm & 7xcm

its area =1

[5 7 ] 142

x x

336 = 12x 7

336

7 12

x

x= 4smaller of the parallel sides = 5x cm

= 5 4 =20 cm.

83.(B) Volume of the wooden block = 51020 cm3

Volume of the required solid wooden cube= 5 10 20 x3cm3

(where x3is an unknown no.) only 8 is the smallest perfect cube

x3= 8 No. of wooden block = 8

84.(C) Let the original area of the cube

= 6x2sq. unit.Side of the new cube = 3xunit

its area = 6(3x)2 = 6 9x2= 54x2sq. unit.


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% increase in area =2 2

2

54 6100

6

x x

x

=2

2

48100 800%

6

x

x

85. (C) A

B C

I

D

Incentre & circumcentre of an equilateral

is same.let a unit be the side of ABC

Then AD = 2 2AB BD

22 a 3a a

4 2

AI : ID : 2 : 1

AI=2

3AD=

2

3

3a

2=

a

3unit.

ID =1 1 3 1

AD a a3 3 2 2 3

unit.

Now, Area of circumcircle area of incircle= 44

2 2a a 44

3 2 3

222 1 1 a 44

7 3 12

12

14

7

22a2= 44

2 44 12 7a 5622 3

Area of the = 2 23 3a 56 14 3cm4 4

86.(A) Let r1: internal radius

r2: external radius

h : height of the pipe

Volume of the metal = 2 22 1r h r h

748 =2 2

1

229 r 14

7

2

1

748 781 r

22 14

17 = 81r

1

2

r12 = 8117 = 64

1r 64 8cm thickness = 98 = 1 cm

87.(B) Area of the tank = 2(lb+bh+lh)lb

=2[2512+12 6 + 25 6] 25 12= 2[300 + 72 + 150] 300= 2 [522] 300

= 1044 300= 744 m2

Cost of plastering = Rs. 744 75Rs. 55800

88.(A) h : heightc : curved surface areaV = Volume of the cone

C= rl

V =1

3r2h

3V = r2hNow, qv2c2h2+3Vh3

= (r2h)2-(rl)2h2 + (r2h)h3

= 2r4h2 + 2r2l2h2+2r2h4

=2r2h2(r2l2) + 2r2h4

= 2r2h2h2+ 2r2h4

= 2r2h4+ 2r2h4

= 0

89.(B) Area of quad. ABCD =1

2 diagonal

(Sum of offsets on

the given diagonal)

=1

16 (9 7)2

= 8 16 = 128 cm2

90.(*) r = 10 cm

h = 48 cm

48 cm

10 cm

20 cm

h1

Volume of the water in the conical vessel

= Volume of the water in the cylindricalvessel.

1

3r2h =r

12h

1

1

3(10)248= (20)2h

1

110 10 48

h

20 20 3

= 4 cm


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91.(A) r =7

2cm

h = 12 cmVolume of the water in pipe in 1 sec. =r2h

=27

27

722 12

= 66 7 cm3

Volume of water stored in (3600 seconds)1 hr.

= 66 7 3600 cm3

= 1663200 cm3

=1000000

1663200m3

= 1.6632 m3

= 1663.2 l

92.(D) ar ( GBD) =1

6ar( ABC)

6 =1

6ar ( ABC)

ar ( ABC) = 36cm2

93.(B) Area allotted for residential purpose

1445 2acres.

360

Area allotted for road purpose =

=36 1

5 acres.360 2

Required ratio = 2 : 12

= 4 : 194.(A) % of area allotted for water area and green

zone.

=

(108 18)5

360 1005

=1.75

1005

= 35%

95.(D) Land allotted for green zone

=108

5 1.5acres360

Land allotted for commercial purpose

= 54 5 0.75acres360

difference = 1.5 0.75 = 0.75 acres.

3acres

4

96.(C) Total land allotted for residencial &

commercial purpose =144 54

5360

198 5 990

360 360

acres

= 2. 75 acres =432 acres

97.(B) Average distribution of loan

87 104 113 120 424

106crores.4 4

It is clear from the table that the distribution

of loan in 2008 is nearest to the averagedistribution.

98.(B) % increse of disbursement of loans by allbanks from 2009 to 2010

120 113100

113

=700 22

6 %

113 113

99.(D) Total disbursement of loans by (in cores)

A & B

C & D

45 56 63 71

42 48 50 49

2007 2008 2009 2010

disbursement of loans by A & B is neverequal to the disbursement of loans by C & Din any year.

100.(B) It is clear from the table that the bank Bdistributes more than 30% of the total laons

by all banks in 2010.


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SSC MAINS (MATHS) MOCK TEST6 (ANSWER KEY)

1. (A)

2. (B)

3. (C)

4. (D)

5. (D)

6. (B)

7. (B)

8. (B)

9. (B)

10. (A)11. (A)

12. (*)

13. (A)

14. (A)

15. (A)

16. (B)

17. (B)

18. (C)

19. (B)

20. (B)

21. (D)

22. (A)

23. (D)

24. (A)

25. (C)

26. (C)

27. (D)

28. (C)

29. (A)

30. (C)

31. (D)

32. (A)

33. (C)

34. (C)

35. (A)36. (A)

37. (D)

38. (A)

39. (A)

40. (C)

41. (C)

42. (B)

43. (D)

44. (A)

45. (D)

46. (C)

47. (B)

48. (A)

49. (B)

50. (A)

51. (A)

52. (D)

53. (*)

54. (C)

55. (C)

56. (D)

57. (B)

58. (C)

59. (B)

60. (C)61. (C)

62. (A)

63. (B)

64. (A)

65. (D)

66. (C)

67. (B)

68. (B)

69. (C)

70. (D)

71. (B)

72. (B)

73. (A)

74. (C)

75. (B)

76. (D)

77. (D)

78. (C)

79. (*)

80. (A)

81. (B)

82. (A)

83. (B)

84. (C)

85. (C)86. (A)

87. (B)

88. (A)

89. (B)

90. (*)

91. (A)

92. (D)

93. (B)

94. (A)

95. (D)

96. (C)

97. (B)

98. (B)

99. (D)

100. (B)

7. Correct answer is option (3) 12 yrs.

Let the ages of the boys are 4x, 5x& 7x A.T.Q.

4 5 7

3

x x x = 16

or, 16x= 48 x = 3 Age of the youngest boy = 4x

=12 yrs.23. Correct answer is option (1)

10% of rice was spoiled rate of rice must be increased by

10

100 10 100% = 11.11%

i.e. by1

9 th of the intial price

also, 20% profit is required

So, Finally the rate of rice must be

150 (1 9)

9

120

100= 150

10 6

9 5 = ` 200

51. Correct answer is option (3) 125

5 2 7x = 3 2x- 7 = 35

or, 2x= 243 + 7

x=250

2= 125

52. Correct question should be

MOCK TEST PAPER - 3 CORRECTIONS


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32x +1 - 3x= 3x +3- 32

& according to the above correct question,correct answer is option (4) -1 and 2

87. Correct answer is option (2) 1 : 2 : 3 Vol.(cone) : Vol. (hemisphere) : Vol. (cylinder)

=21

3r r :

32

3r : 2r r

=1

3 :

2

3 : 1

= 1 : 2 : 3

MOCK TEST PAPER - 5 ( CORRECTIONS)

14. Accroding to question printed in Englishlanguage answer is option (D) 20yrs.

but, Accroding to question printed in Hindilanguage answer is option (A) 27yrs.

20. Correct option is (A) 9800correct solution

Suppose the cost price of T.V. = ` x

Then, 2 (x- 9400) = (10600 - x )

2x- 18800 = 10600 - x 3x = 29400 x = 980033. In the question,

ratio of C:D should be5

6:3

4

and with this correction,answer should be option (C)

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Ssc Mains (Maths) Mock Test-6 (Solution)

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