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SSC MAINS (MATHS) MOCK TEST–6 (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 .063 3

7 7 7

=

(0.65 0.35)

(0.25 0.150 0.6)3

7

1

13 3

77

2.(B)

3 32 216 16

3

2 232

1(4 )

16 =

322

32 2

14

(4 )

33

144

16464

4096 164

409764

3.(C) No. of digits required= [{(9–1)+1}×1 + {(50–10)+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)35–164 + 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 n is 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 x and 27 y wherex and y are co-prime nos. i.e. prime to eachother.Now, A.T.Q27x + 27y = 216or, 27 (x + y) = 216 x + y = 8So, possible pairs of x and y are (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 ;Re mainder 1

1932

even(a – 1) ;R 1a

8.(B) Term difference = 12–4 = 8Value difference = 70–14=56Value difference per term (i.e. commondifference) = 56/8 = 7So, first term = 4th term– 3 × common dif-ference = 14 – 3 × 7 = - 7

9.(B) Required height at the 1st bounce 3324

Required height at the 2nd

bounce

23324

Required height at the 3 rd

bounce

33324

= 27 132 13 m64 2

10.(A) Remaining no. of total balls after 1st ball ischosen = (12 + 6)-1 = 17 ballsAlso,Remaining no. of black balls after 1st ball(which is black) is chosen = 12–1

= 11 black ballsSo, The probability that the second ball isalso black = 11/17

11.(A) Let x be the initial no. of people in thecompany.So, ATQ,

= xx

35 5 32 345

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

12.(*) Initial bowling average = 12.4After improving bowling average by 0.2,new bowling average = 12.4 – 0.2 = 12.2Now, let x be the number of wickets takenbefore the last matchSo, A.T.Q,

= . x .x

12 4 26 12 2

4or 12.4x + 26 = 12.2x + 48.80.2x = 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 3584km2days 8hours 56hours

= 64 km/hourNow, Average speed during the remainingpart (last 8 hr.) of journey

=3584 – (1440 1608)

8 km/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

= 15 ft3

15.(A) Let the required ratio is x : ySo, A.T.Q,

=

192 x 150 y 120 194.40x y 100

192 x 150yor, 162

x y

or, 192 x + 150 y = 162 x + 162 yor, (192 – 162)x = (162 – 150)yor, 30x = 12yor, 5x = 2y x : y = 2 : 5

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

3 3 1 kg(5 4 2) 2 24 8

17.(B) Required average rate of interest perannum

=

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.

=

4–5 5xx x

Two years later now,

income of Sanjay = 120 6Rs. x Rs. x100 5

and saving of Sanjay = xRs.5

Expenditure of Sanjay =

6 xRs. x – x5 5

So,% increase in the expenditure

4x – x5 100%

4 x5

x 100% 25%

4x19.(B) Price is reduced by 20%

Consumption can be increased by

20 100%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 26

100

x =

26 100 650

4So,

Pass% =

1632% 100 %650

= 6 632% 2 % 34 %

13 13


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21.(D ) Re. 1 50pcoins coins

Ratio of respective values = 13 : 11Ratio 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 =

13 21013 22

13 21035

= 78 coins22.(A) By Alligation method

12

35

49

12

49

118

35

12

110==

Required ratio of mixture = 1 1:

18 10= 10 : 18= 5 : 9

A.T.Q, Amount of the former mixture= 3 litre

So, required of the later mixute

935

litre

255

litre

23.(D) Let the original number of boys and girlsbe 5x and 3x respectively and that of newboys and girls be 5y and 7y respectively.5x + 3x + 5y + 7y = 1200 2x + 3y = 300 ......... (i)

and 5 5 73 7 5

x yx y

25x + 25y = 21x + 49y 4x = 24y x = 6y ......... (i)From equation (i),4x + 6y = 600 5x = 600 x = 120 Original no. of students= 8x = 960

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

and Ratio of no. of passengers = 1 : 50 Ratio 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 to5 Sons : 4 daughters : 2 nephews= 25x : 16x : 2x25x : 16x : 2x = Rs. 8600or, 43x = Rs. 8600x = Rs. 200Required money to each daughter= 4 × 200 = Rs. 800

26.(C)2 2A 40 B 205 7

9 1017

C x (let)

5 7( – 40), ( – 20)2 2

A x B x

and, 17 ( –10)9

C x

5 7 17( – 40) ( – 20) ( –10)2 2 9

x x x

= 600x = 100

A’s share = 5. (100 – 40)2

Rs = Rs. 150

27.(D) Extra interest received in 4 years if therate of interest is increased by 1%

= (4×1)% 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 = 12 6 – 4 % 50004

of

12 2 % 50004

of


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92 50004 100

= Rs. 22529.(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

10kg kg

Quantity sold at 6% loss

9 50 4510

kg kg

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

. . 1 –1100

nrC I P

210525 1 –1100

P 121 –100100

P

525 10021

P = Rs. 2500

Now, ATQ, n1= 4 years, r1 = 5%

So, 1 1. .100

P r nS I 2500 5 4

100

= Rs. 500

31.(D) Let Rs. x be the sum borrowedSo, Rs. x after two years will become

Rs. x 25 441x1 Rs.

100 400

Present value of Rs. 441x 882 Rs.800

x

Present value of Rs. 882 = Rs. 800Now, Amount of Rs. x after one year

5 211 .100 20

xx Rs

Present value of = Rs. 2120

x after one year

= Rs. x

Present value of Rs. 882 due after oneyear

20 882 .84021

x Rsx

Required sum= Rs. (800 + 840) = Rs. 1640

32.(A) Let Rs. x be the marked price of the shirt.ATQ, Difference of discounts = 2% 2% of x = 15

2 15

100x

15 100 .750

2x Rs

33.(C) Let the first CP of the commodity be Rs.100 First SP = Rs. 110Second CP = Rs. 90.

Gain = 26 %3

=50 %3

Second SP

50100 % .903

of Rs

350 903 100

= 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 .405

CP 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 20

120x

6120 – 20 245

x

120 – 24 .96x Rs Gain = Rs. 24Now when difference of gains = Rs. 4,then SP = Rs. 120So, When the difference = Rs. 85,

then SP = 120 85

4 = 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 ofRs. 100 i.e. profit = 28%.

36.(A) Let the marked price of the article= Rs. xSingle discount for successive discounts of30% and 20%

30 2030 20 – %100

(50 – 6)% 44% discountNow, A.T.Q.(100 – 44)% 2240of x

56 2240

100x

2240 100 .4000

56x Rs

37.(D) Let the printed price of the book = Rs. x.So, Selling price = 90% of x

9.10

xRs

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

112 9100 10

xy

or, 9 100 45

10 112 56

yx

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

after which C joined the business.So, Ratio of shares of Profit= 30,000×12 : 40,000×8: 50,000× x= 32 : 36 : 5x

C’s share 5

36 32 5x

x

5

68 5x

x

given, 5

68 5x

x=

1500049000 x = 6

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

39.(A) Let 1 child’s 1 day’s work =x& 1 adult’s 1 day’s work = y

Then, 112

16x

1192

x

and 18

12y

196

y

Work done in 3 days by 16 adults

1 116 396 2

part

Remaining work 12

part

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

6 4 196 192 12

i.e. 112

work is done by them is 1 day

So, 12

work will be done by them in

= 1122

days

= 6 days40.(C)Time taken by A to complete the work

4 32

= 6 days

& Time taken by B to complete the work

= 6 5 10

3

days

A and B together will complete the work

in 6 106 10 days =

334

days.

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

= 72

km= 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 80 3.5x 20 1.5

x = 80 1.528020 3.5

x = 480 workers No. of more workers needed

= (480–280) people= 200 workers

42. (B) Ratio of wages of A, B, C= (6×5): (4×6): (9×4) = 30 : 24: 36

= 5 : 4 : 6

A’s Share = 5 1800

15 = 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 litresper minute


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

44.(A) Part filled by 1st pipe 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 5 115 6 30 30

part

At 11 a.m. 3rd pipe is also opened to emptyit.Now, time taken by 3 pipes together tocompletely empty the full cistern =

1 1 10 hrs.

1 1 1 1– –4 15 12 10

So, Time required to empty 1130 filled part

= 11 111030 3

hr =233 hrs.

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

45.(D)Let the speed of the bus = x km/hrThen to take a lead of 60m, he will have tocover a distance of (60+40)m= 108m, withthe speed of (30–x) km/hr in 20 sec.100m = (30–x) km/hr × 20 sec.

100 = (30 – x) 1000 20

3600

1 (30 – )10 180

x

or , 180= 300–10x 10x = 120 x = 12 km/hr

46. (C)

r

Circumference = 2 r

Time taken for one round= 408 = 5 min.

Now, new radius = 10 r

So, New circumference = 2 ×10r = 20 r

So Required time = 20 r 52 r

minute

= 50 minutes47.(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 = x km/hrDistance travelled at the speed of 80 km/hr in 4 hours.

= 4 × 80 = 320 km

32010

x = 32 km/hr

49.(B) Let the length of train or platform = x metre.Speed = 90km/hr

= 90 5

18

metre/sec.

= 25 metre/sec. Distance covered in 60 sec.= 25 × 60 = 1500 metresNow, according to question,2x = 1500 x = 750 metre

50.(A) Speed in still water = 12km/hrspeed against the current

4 123

= 9km/hr (80 min= 43 hr)

Speed of current = 12 – 9 = 3km/hrSpeed with the current = 12 + 3 = 15 km/hr

So, required time =45 4515 9

= 8 hr.

51.(A) Let the cost of one saree = Rs. xand the cost one shirt = Rs. yAccording to question

20x + 4y = 1600 x + 2 y = 800 ........... (i)andx + 6 y = 1600 ........... (i)on solving equations (i) and (ii), we getx = 400; y = 200; cost of 12 shirts= 12 × 200 = Rs. 2400


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52.(D)4 21– 1 1

1 (1– )x x Ax x x x

)1(1

11)1)(1(

2

22

xxxx

xxx

= A

)1(1

1)1()1)(1)(1(

2

2

xxxx

xxxx

= 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 1

2 –1

2 –1

Now, 33

1– ( 2 1) – ( 2 –1)xx

= 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 + (k–4)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 11 1 1 1

1 2 3x x x x

1 2 3 41 2 3

x x x xx x x x

4xx

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 11

b aab b a

= 2

1 1a b

a b

22

a ba b

[ ab =1]

= 159.(B) x y za b c k (Say)

zyx kckbka111

,,

2b ac

yk2

= zxk11

)(

2 1 1y x zk k

2 z xy 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, (x–2) 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 have2b + 24 = 0b = –12On substituting b = -12 in (2)8a –12 – 4 = 0 a = 2

61.(C) sinB = 21

= sin30º

030B Now, 3 cos B - 4cos3B= 3 cos 300 - 4cos3300


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3 3 33 – 42 8

3 3 2 3– 02 2

Another method33cos – 4cosB B

= – cos3B= -cos 3× 300

= - cos900

= 062.(A) tanA =1 A = 450

tan B = 3 = B = 600

Now, cosA. cosB + sinA.sinBcos450.cos600 + sin450.sin600

1 1 1 32 22 2

=

2213

63.(B)

A

P C B300 m

Pole

tan = 125

5

12ABBP

5

300 12AB

BC

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

tan B = 43

34

ABBC

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

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

5 4 5300 12 3 9

BCBC

9BC = 5BC + 1500

BC = 41500

= 375m

Height of the pole = AB = 34

BC

= 3 3754

= 1125 281.25

4m

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 + tan2A

65.(D)

A B300 m

Tower

b m

In ABQ

ABQA

tan

ABb

tan

tanbAB

In ABP

ABPB

tan

tan

tan bPB

PQ =

tan.tan b

= cottanb


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66.(C)

h2

In CBD,

tan = h

BD

BD =

cottan

hh

In CBA,

tan = BACB

= DABDCB

= 2

cot hh

h

(BD= hcot )

tan =

2cot hh

h

1 1cot2 tan

cot +12

= cot

cot – cot = 12

67.(B)

Let the height of the tower be x unitIn CBA

tan CB xBA BA

BA = tan

x = x cot ....................... (i)

In DBA

tan =DBBA

= coth x

x ( BA = x cot )

xcot = tan

h x= ( )h x cot

x(cot –cot ) = hcot

cot

cot cothx

tan

1 1–tan tan

h

=

tantan

tantantan

h

tantan – tan

h

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

= 3 3 2 22 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

22 2 2 2–3 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 + 12 – 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 = 4cos2 4 2 44 – 4cos cos – 4cos 4cos

6 4– cos cos 6 4 2– cos 4cos – 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 AB2 + CD2 = AD2 + BC2

73.(A)

A

B D 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)= 12

ar ( ACD)

On adding (1) & (2)

ar ( BEC) = 12

ar ( ABD)+12

ar ( ACD)

=12

ar ( ABC)

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

other adjacent angle = 23 x0,

Now, 0 0 02 1803

x x

[Adjacent angles of a gm aresupplementary]

021 1803

x

x = 0 03180 1085

Smallest angle = 0 02 2 108 723 3

x

75.(B) BCE CBD BDC (Exterior angle of a is equal to the sum ofopp. interior angles. )650 = 280 + BDC

65 – 28 = BDC

370 = BDCAlso, ABDC & BD works as a transversal.

70.(C ) tan aqb

sincos

q aq b

2

2sincos

a q ab q b

Applying C & D, we have2 2

2 2sin cossin – cos –

a q b q a ba 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 & ABCAlso P & Q are centroids of ADC & ABC

1PO BO3

& 1QO DO3

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

1PO 93

& 1QO 93

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

72.(B)

ABCD is a quad. where diagonals AC & BDintersect 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 x cmLenght of the rectangle = (x + 5) cmits breadh = (x – 3) cmATQx2 = (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 EFGC DC

510 18

EF

EF = 10518

= 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 – 5

y x

6– 3 ( – 5)10

y x

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

80.(A) The equation the circle is(x +1) (x + 2) + (y –1) (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 = 32

, f = 1, c = –1

rad. of the circle = 2 2 –g f c

= 2

23 1 – (–1)2

= 9 1724 2

Area of the circle = 2r

=

2172

174

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 : 482.(A) Let the parallel sides of the trapezium be

5x cm & 7x cm

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 = 5×10×20 cm3

Volume of the required solid wooden cube= 5 × 10 × 20 × x3 cm3

(where x3 is 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= 6x2 sq. unit.

Side of the new cube = 3x unitits area = 6×(3x)2 = 6 × 9x2= 54x 2 sq. unit.


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

% increase in area = 2 2

254 – 6 100

6x x

x

=2

248 100 800%6

xx

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 – BD2

2 a 3a – a4 2

AI : ID : 2 : 1

AI=23 AD=

23 ×

3 a2

=a3

unit.

ID = 1 1 3 1AD a a3 3 2 2 3

unit.

Now, Area of circumcircle – area of incircle= 44

2 2a a– 443 2 3

222 1 1– a 44

7 3 12

1214

722

a2 = 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 21

22 9 – r 147

21

748 7 81 – r22 14

17 = 81–r1

2

r12 = 81–17 = 64

1r 64 8cm thickness = 9–8 = 1 cm

87.(B) Area of the tank = 2(lb+bh+lh)–lb=2[25×12+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 coneC= rl

V = 13 r

2h

3V = r2hNow, qv2–c2h2+3Vh3

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

= 2r4h2 + 2r2l2h2+2r2h4

= 2r2h2(r2–l2) + 2r2h4

= 2r2h2×–h2+ 2r2h4

= 2r2h4+ 2r2h4

= 0

89.(B) Area of quad. ABCD = 12

× diagonal ×

(Sum of offsets onthe given diagonal)

= 1 16 (9 7)2

= 8 × 16 = 128 cm2

90.(*) r = 10 cmh = 48 cm

48 cm

10 cm

20 cm

h1

Volume of the water in the conical vessel= Volume of the water in the cylindricalvessel.

13 r

2h =r12h1

13 ×(10)2×48= (20)2×h1

110 10 48h20 20 3

= 4 cm


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91.(A) r = 72

cm

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

= 10000001663200

m3

= 1.6632 m3

= 1663.2 l

92.(D) ar ( GBD) = 16 ar( ABC)

6 = 16 ar ( ABC)

ar ( ABC) = 36cm2

93.(B) Area allotted for residential purpose

144 5 2acres.360

Area allotted for road purpose =

=36 15 acres.360 2

Required ratio = 2 : 12

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

zone.

= (108 18) 5

360 1005

= 1.75 100

5

= 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.

3 acres4

96.(C) Total land allotted for residencial &

commercial purpose = 144 54 5

360

198 5 990360 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 distributionof loan in 2008 is nearest to the averagedistribution.

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

120 –113 100113

= 700 226 %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 laonsby all banks in 2010.


Ph: 011-27607854, 08860-333-333 14

SSC MAINS (MATHS) MOCK TEST–6 (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

3x 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

10100 10 × 100% = 11.11%

i.e. by 19 th of the intial price

also, 20% profit is required So, Finally the rate of rice must be

150 × (1 9)

9

×120100 = 150×

10 69 5

= ` 200

51. Correct answer is option (3) 1255 2 7x = 3 2x - 7 = 35

or, 2x = 243 + 7

x = 2502 = 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)

= 213

r r : 323

r : 2r r

= 13 :

23 : 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. = ` xThen, 2 (x - 9400) = (10600 - x )

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

ratio of C:D should be 56 :

34

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

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