Ssc Mains (Maths) Mock Test-4 (Solution)

8/13/2019 Ssc Mains (Maths) Mock Test-4 (Solution)

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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINIBADARPUR BORDER

SSC MAINS (MATH) MOCK TEST - 4 (SOLUTION)

1. (D) Total no. of marbles kept in 50th box

= sum of factors fo 50= 1 + 2 + 5 + 10 + 25 + 50= 93

2. (C) Let x be the required number

and abe its first partso, (xa) will be its second part

Now, A.T.Q,0.8a= 0.6 (x a) + 3

or, 0.8a+ 0.6a= 0.6x+ 3or, 1.4a = 0.6x + 3

4.1

36.0

xa (i)

also 0.9a+ 6 = 0.8 (x a)or, 0.9a+ 0.8a = 0.8 x- 6

or, 1.7a = 0.8x 6

7.1

68.0

xa -------------- (ii)

from (i) & (ii) ,

7.1

68.0

4.1

36.0

xx

1.02x + 5.1 = 1.12x 8.4

0.1x = 13.5

1.0

5.13x = 135

A l t er n a t i v e me t hod

3)(5

3

5

4 axa

35

3

5

3

5

4

aa --------(i)

also )(5

46

10

9axa

65

4

5

4

10

9

xa --------(ii)

from (i) & (ii), we get,

5

346

5

4

10

893

5

3xx

or,5

42

25

28

10

51

50

51 xx

or,5

42

10

51

50

51

50

56

x

550

10135 x

x = 1353. (A) Let the two digit number be 10y+ xwhere

x> y

10x+ y 10y x = 63 9x 9y= 63

x y = 7 x= 7, 8, 9 andy= 0, 1 , 2

possible values of x are 7,8,9

4. (A) For every 4n ;

n! will be divisible by 8

remainder will be zero[ because for 4n , 8 will be a factor of n!]

So, remainder of 1! + 2! + 3! + 4! + ....... + 100!

wil be equal to the remainder of 1! + 2! + 3!only

1! + 2! + 3! = 1 + 2 + 3 = 9

and8

9; R = 1

5. (D) H.C.F. = x and L.C.M. =y

A B = x ySo, A + B = (A + B) - 3AB(A + B)

= (x+ y) 3xy(x+ y)

= x + y + 3xy (x+ y) 3xy(x+ y) = x + y

6. (A) 2/11

3/1

1

3

1

2

1

yxyxyx

7. (C) Required average age of 2 persons

= 442

)38(3430

yrs.

8. (*) Required average age just before the birthof the youngest member

=110

)1010()2010(

=9

100= 11.11 yrs.

9.(C) Weight of first member = xkgWeight of second member = (x +2)--------------

Weight of fifth member = (x + 8) kg

10. (A) Average speed =takentimeTotal

covereddistanceTotal

=min)151515(

40)km8(6


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=min45

km54=

hr.60

45

km54

= 72 km/hr.11. (C) Age of the captain

= (11 30 ) {(5 29) + (5 27)}= (330 280) yrs.

= 50 yrs.12.(B) (2m+ 4b) 10 = (4m+ 5b) 6

20m+ 40b = 24m + 30b

4m= 10b 2m= 5b So,5b = 2 40

b=2 40

5

= 16

Required ratio = 40: 16 = 5 : 213.(B) In case I.

Let the number of shirts of brand B be x.Let the cost of a shirt of brand B be ` 1

Original cost = 4 2 + x = ` (8 +x )

In case II.

4x + 2x= (8 + x)140

100= (8 + x)

7

5

20 + 10x = 56 + 7x

10x 7x = 56 20 = 36

3x= 36 x= 12 Required ratio = 4 : 12 = 1 : 314. (C) Half the sum = 18

total sum = 36

Nos. will be 6, 12 and 18So, ratio of squares = 6 : 12 : 18

= 36 : 144 : 324

15. (B) A.T.Q,

RJA

5

2

46

A : J : R

= 6 : 4 :2

5

= 12 : 8 : 5

Apoorva's share

= 2250)5812(

12

= Rs. 1080

16. (A) S.P. at 20% profit = Rs. 9/litre

Cost of mixture = 100120

9 = Rs. 7.5/litre

Now, let the ratio of milk and water in themixture = x:y

yx

yx

)0()10(= 7.5

10x= 7.5 (x+ y)

2.5x= 7.5 y

x= 3y

x :y = 3 : 117.(D) Ratio of equivalent capitals of A, B and

C for 1 month= (40500 6 + 45000 6 ): (45000 12)

: (600006+450006)= (405+450): (450 2 ): (600 + 450)= 855 : 900 : 1050

= 171 : 180 : 210= 57 : 60 : 70

Sum the ratios = 57 + 60 + 70 = 187

Required difference =70 57

187

= 56100

=13

187 56100 = ` 3900

18. (A) Let the cost price of 1 orange = Re. 1.

C.P. of 1 banana = Rs.4

3

and C.P. of 1 apple = Rs.2

3

New prices:

1 orange = Re. 1.1

1 banana100

110

4

3 = Rs. 0.825

1 apple100

110

2

3 = Rs. 1.65

Original price of (4 bananas + 2 apples +3 oranges)= Rs. (3 + 3 + 3) = Rs. 9New price of (4 banana + 2 apples + 3

oranges)= Rs. (4 0.825 + 2 1.65 + 3 1.1)

= Rs. (3.3 + 3.3 + 3.3) = 9.9

Percentage increase

= 1009

99.9

= 10%

19. (A) Let the no. be x

% change

= %100resultoriginal

resultchanged-resultoriginal

= %1005

55

x

xx

= %10025

25

x

xx

= %100

25

24

x

x= 96%


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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINIBADARPUR BORDER20. (C) 20% decrease in price

25% increase in consumption(when expenditure is constant)

increased amount of sugar = 25% of 20kg = 5 kg

Total amount now of sugar = (20 + 5) kg = 25 kg

21. (C) Votes got by Rahul Gandhi

= (100-10)% of5

4of total voters

= 90% of5

4of total voters

=5

4

10

9 of total voters

= 25

18

of total voters

= 216 voters -------- (i)Now, Votes got by Varun Gandhi

= (100-20)% of th

5

41 of the total voters

= 80% of th5

1of total voters

=5

1

5

4 of total voters

=254 of total voters

= 418

216 = 48 voters

So, total no. of votes polled = (216 + 48) votes = 264 votes

22. (C) 10% reduction in price

11.11% increase in quantity(when expenditure is constant)

A.T.Q,

11.11% of original amount of wheat = 50 gm

Original amount (ie.100%) of wheat= (50 9) gm= 450 gm

23. (A) C.P. of each book sold by publisher

=4002000

000,70

= ` 43.75S.P. of each book sold by publisher

= (100-30)% of `75= `52.5

So, % gain = %100

75.43

75.435.52

( S.P. > C.P.) = 20% gain

24.(C) If the C.P. of wrist watch be ` xthen

C.P. of wall clock = ` (390 x )

So,

x x10 (390 ) 15

100 100

= 51.50

10x+ 5850 15x= 5150

5x= 5850 5150 = 700

x=5

700= ` 140

C.P. of wall clock = 390 140= ` 250

Required difference = 250 140 = ` 110

25.(A)

(40 20)% = Re1

So, 120% =1

20120= ` 6

26. (B) Required Marked price

=)5.12100(

100

100

120210

=5.87

120210

= `28827. (D) Let the C.P. = x

So, S.P. in 1st case = 1.05xnow, C.P. in 2nd case = 0.95xand S.P. in 2nd case = 1.05x 2

Now, A.T.Q. 0.95x 1.1 = 1.05 x- 2or, 1.045x = 1.05x 2

1.05x 1.045x= 2

0.005x= 2

400005.0

2x

28.(C) Discount on ` 36000 =3600 7

100

= ` 2520

Discount on first ` 20,000 =20000 8

100

= ` 1600

Discount on next ` 10,000 =10,000 5

100

= ` 500

Discount on remaining ` 6,000

= 2520 (1600 + 500) ` 420

Required percent =420 100

6000

= 7%

29. (B) Net discount given by A

= %100

255255

= 28.75%

& Net discount given by B


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= %100

12161216

= 26.08%

A is giving more discount

it is more profitable to purchase the fan

from A.

30.(C) Single equivalent discount = %100

pqqp

31. (A) Required ratio = (15 22) : (11 25)= 330 : 275= 6 : 5

32. (D) Let Rs. x= quaterly paymentgiven, r= 16% per annum

So, rate of interest per quarter

=4

16= 4% per quarter

Also, No. of quarters in 2 years = 4 2

= 8 quartersSo,

8x+ (7 + 6 + 4 + 3 + 2 + 1) 4% of x= 2280

or, 8x+

2

87 4% of x= 2280

2

)1(...321

nnn

or, 8x+ 112% of x= 2280or, 800% of x+ 112% of x= 2280

912% of x= 2280

So, x=912

1002280 = 250

Direct Method:-Quarterly Payment

=

2

7848100

2280100

=2241600

22280100

= 250

33. (D) Let xbe the required annual payment.So, also, r = 10% p.a. and t = 3 years

1.11%10

So, (1 + 1.1 + 1.21)x= 3310 (1.1)3

or, 3.31x = 3310 1.331

x=31.3

331.13310

= Rs. 1331

Direct Method:-

Annual payment =

1331

1000

121

100

11

10

3310

= )100011001210(

13313310

= 1331

34.(C) A = P

TR

1100

3 = 1R

3

1100

On squaring both sides,

9 = 1R

6

1100

35. (A) Cash price of refrigerator

= 1500 +

121

1001003

11

101020

1331

1000990

=

1331

990000)11100300()12110200(1500

=

1331

990000110330012342001500

= 1500 +1331

3327500

= 1500 + 2500= 4000

Alternative Method:-Cash price of refrigerator

= 32 )1.1(

990

)1.1(

1003

1.1

10201500

=331.1

990

21.1

1003

1.1

10201500

= 1500 + 2500

= 400036. (D) When B works normally then days

taken by B to complete the work

=1220

1220

days

= 30 days

Now, If B does the work only half a day daily

B will take twice the total days to

momplete the whole work alone

Now No. of days taken by B= (30 2) days

= 60 days

So,


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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINIBADARPUR BORDERNow days taken by (A + B) together to do the

whole work =6020

6020

= 15 days

37.(A) Let time taken by B in completing the work= xday

Time taken by A (x 10) days

x x

1 1 1

10 12

x x

x x

10 1

10 12

24x 120 = x2 10x

x2 34x+ 120 = 0

x2 30x 4x+ 120 = 0

x (x 30) 4(x 30) = 0

(x 4) (x 30) = 0

x 30 because x438. (C) Let days taken by A to do the whole work

= xdays So, days taken by B to do the whole work

= (x 5) daysand, days taken by C to do the whole work

= (x 9) daysNow, ATQ,

Days taken by (A + B) together to do thewhole work= Days taken by C alone to do the whole work

)5(

)5(

xx

xx= x 9

or, x2 5x= (x 9)(2x 5)or, x2 5x= 2x2 5x 18x+ 45

or, x2 18x+ 45 = 0or, x2 15x 3x+ 45 = 0or, x(x 15) 3(x 15) = 0

or, (x 3) (x 15) = 0 x= 3 or 15but x= 3 is not possible as x 5 or x 9 will

become negative.

So, x= 15 days

39.(B) Time1

cross sectional area of the pipe

Time2

1

d4

Time 21

d

1

2

t

t=

21

2 2

(d )

(d )

So, t2 = t

1

2

1

2

d

d

t1= 40 minutes, d

1= d, d

2= 2d

t2= 40

2d

2d

t2 = 40

21

2

t2= 10 minutes

time taken by a pipe of diameter 2d

for doing the same job = 10 minutes40.(C) Let the capacity of the tank be xgallons.

Quantity of water filled in the tank in 1minute when all the pipes A, B and C are

opened simultaneously =x x

320 24

According to the question,x x x

320 24 15

x x x

320 24 15

x x x6 5 8

3120

3x= 3 120

x=3 120

1203

gallons

41. (B) Let speed of car = xkm/hr.Here, Distance covered by the car in 27 minutes

= Distance covered by the sound in (28 minutes 30 seconds 27 minutes)

xkm/hr

60

27

hr.

=

hr/km

5

18330

hr

60

5.1

x= 330 27

60

60

5.

5

18

= 66

Speed of car = 66 km/hr.

42. (D) In the race between Sonu and Monu.Distance travelled by Sonu and Monu

in same time = 600 mtr. & (600

60)mtr= 600 mtr. & 540 mtr.

In the same time,Ratio of distance travelled by Sonu &

Monu = 10 : 9Similarly,

In the same time,

Ratio of distance travelled by Monu &Bablu = 500 : (500 25)

= 500 : 475= 20 : 19

So,In the same time,Ratio of distance travelled by Sonu,

Monu & Bablu


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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINIBADARPUR BORDER= 10 20 : 9 20 : 9 19= 200 : 180 : 171

When Sonu travels 200 m, Bablu willtravel 171 m

So, When Sonu travels 400 m, Bablu will

travel 342 m In 400 m race between Sonu & Bablu,

Required No. of metres by which Sonuwill win the race

= 400 m 342 m= 58 m

43. (A) Let x= length of the faster train (in mtr.)

So, 36 seconds =kmph

x

)2040(

10

x = 36 second 20 18

5m/sec

= 200 mtr.

44. (A) Speed of boat in still water =2

SS updown

=2

yx= 0.5 (x+ y)

45.(C) Expression= 4 + 44 + 444 +........... to nterms

= 4 (1 + 11 + 111 +.... to nterms)

=4

9(9 + 99 + 999 + ...... to nterms)

=4

9

[(10 1)+(100 1) + (1000 1)+........ to n

terms]

=4

9[(10 +102 + 103 +......... to n terms) n]

=4

9[10(1 + 10 +102+ .......... to n terms) n]

=40

9

n10 1

9

4

9n

[1 +10 +102+ ........ to n terms =n10 1

9

]

= 4081

(10n 1) 49

n

46. (B) 180 = 2 2 3 3 5a3b= abc

a2 = bc

a3b= abc= 180 = 12180 1

13180

c = 147. (C) a + b + c = 0

b + c = aOn squaring both sides, we get

(b+ c)2= a2

b2+ c2= 2bc= a2

a2+ b2+ c2 +2bc= 2a2

a2+ b2+ c2 = 2a2 2bc= 2 (a2 bc)

a bca b c

a bc a bc

22 2 2

2 2

2

= 2

48. (D) (a 1) b a2 3 2

a = 3 : a 1 = b

3 1 : b b = 2

a+ b= 3 + 2 = 549.(B) OP = 2

OQ =3

2

B

32

X

Q

PO

X

0,

(2.0)

PQ = OP OQ 2 2

=

2

2 322

=4

49

=16 9

4

=

25

4=

5

2= 2.5cm

50.(A)

A

B

XO

X

Y

Y

(12.0)

(0-,9)

Putting x= 0 in 9x- 12y= 108,we get, y = 9

Putting y= 0 in 9x 12y= 108,we get, x= 12

OA = 12, OB = 9

AB = OA OB 2 2

= 2 212 9

= 144 81

= 225

= 15 units


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51.(A) xx

21

= 3

xx

1 = 3

On cubing both sides.

x3+x31

3 xx

1 = 33

x3+x31

= 33 33 = 0

x6+1 = 0

x206+ x200+ x90 + x84+ x18 + x12 + x6 + 1= x200(x6+1) + x84 (x6+1) + x12(x6 +1)+(x6 +1) =0

52. (A)76167616

7

=7327973279

7

=732)7()3(732)7()3(

7

2222

= 22 )73()73(

7

=)73()73(

7

=

72

7=

2

1

53. (B) 22

22

23

23

yxyx

yxyx

=

xyxyyx

xyxyyx

)2(

)2(22

22

=xyyx

xyyx

2

2

)(2

)(2=

1)24(2

1)6(22

2

=

1322

1362

=65

71

12

12

-

+=x

12

121

+

-==

x

y

12

12

12

12

+

-+

-

+

12

12

12

12

+

--

-

+

and = 1xy

x + y=

and =x y 24=

=6

54. (D)a

x

2=

ba

b

2

Apply C & D,

ax

ax

2

2

=

ab

ba

3..... (1)

Again,

b

x

2=

ba

a

2

Apply C & D,

bx

bx

2

2

=

ba

ba

3..... (2)

Now,

bx

bx

bx

ax

2

2

2

2

=

ba

ba

ba

ba

3

)(

3

=ba

baba

33

=ba

ba

22= 2

55. (C)

cot

1

cos

1

cot

1

cos

1

=

cos

sin

cos

1

cos

sin

cos

1

=

cos

sin1

cos

sin1

=

2

2

cos

sin1=

2

2

cos

cos= 1

56. (C)

tan1

5cos4

cosec

9

= 2

22sec

5cos4sin9

= 2222 cos5cos4sin4sin5

= )cos(sin4)cos(sin5 2222

= (5 1) + (4 1)= 5 + 4= 9

57. (A)

a

h

A

B

QPb

90-

here, h= height of tower AB

a

htan ---------- (i)

b

h )90tan(


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or,b

hcot

h

b tan ---------- (ii)

from (i) and (ii),

a

b

a

h abh

58.(A) CD = h metre, AB = 2h metre

h2h

OB = OD =x

2metre

From OCD,

tan =h

x

2

=h

x

2 (i)

From OAB,

tan (90

) =

AB

BO

cot =h

x

2

2

=h

x

4 (ii)

Multiplying both equations,

tan . cot =h h

x x

2 4

x2= 8h2

h=x

2 2meter

59. (C) tan2 . tan3 = 1

tan 3 =1

tan2= cot 2

tan 3 = tan (90 2 )

3 = 90 2 5 = 90

= 18

2cos22

5 1 = 2cos245 1

= 21

2 1 = 0

60.(B) sin 17 =x

y

cos 17 = 21 sin 17

=x y x

y y

2 2 2

2 21

=y x

y

2 2

2

sec 17 =y

y x2 2

sin 73 = sin (90 17)= cos 17

sec 17 sin 73

=y

y x2 2

y x

y

2 2

=

xyy

xyy

=

x

y y x

2

2 2

61.(B) X

YZ

7

5

62

XZ YZ = 2 (i)

XY2+ YZ2= XZ2

2

2 6 = XZ2 YZ2

24 = (XZ YZ) (XZ+YZ)

XZ + YZ = 12 (ii)Adding both the equations,2 Z = 14 XZ = 7 YZ = 7 - 2 = 5

sec X =7

2 6

tan X =5

2 6

sec X + tan X =7

2 6+

5

2 6=

12

2 6= 6

62. (A) Angles of triangle

(a d) , a , (a +d )

a d+ a+a +d = 180

3a =180 a = 60

a d

a d

60 60 1

180 3

d

d

60

60

=

1

3

180 3d= 60 + d

4d= 120 d = 30

a d= 60 30 = 30a= 60

a+ d= 60+ 30 = 90


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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINIBADARPUR BORDERSo, Angles of triangle are 30,60 and 90

63. (A)B (Balloon)

(Observer)

In the given figure,after leaving thepoint A, balloon

reach to point Bvertically upwardin 1.5 min

Here, O theobserver

So, BOA = 60

200m AO

60

tan 60 =OA

AB

AB = OA tan 60

= 200 3 m

So, speed of the ballon

=distance

time

=AB

time to reach from A to B

=sec

60

5.1

3200 m= 3.87 m/sec.

64. (C)A

House

Bstreet

C

D (window)

E

90

90

d

Here,

AB height of the house& CD height of the window

So, ATQ,

ADB = 90

Also,here, line AD makes as angle with the

vertical line DE.

ADE also, 90BDC

In BCD ,

)90tan( =CD

BC=

CD

dor,

CD

dcot

cot

dCD = tand

also,

In ADE ,

DE

d

DE

AEtan

cot

tand

dDE

So, the height of the house,AB= CD+DE

= )cot(tan

d

=

sincos

1

sin

cos

cos

sindd

= cosecsecd

65. (A)

BC a

bc

Let ABC is a

and a, b & c are the lengthsof BC, CA & AB respectively.

sin A : sin B : sin C = 1 : 1 : 2

By sine formula:

C

c

B

b

A

a

sinsinsin

a : b = sin A : sin B & b : c = sin B : sin C

a : b : c = 1 : 1 : 2

Let a= x, b= x& c = 2 x

ATQ,

c2

: (a2

+ b2

) = )(:)2(222

xxx = 2x2 : 2x2

= 1 : 1

66.(B) In s ACD and ABC.

CDA = CAB = 90

C is common.

ACD ABC

C

A

D

B

2

2

ACD AC

ACD BC

2

2

10 9

40 BC

BC2= 4 92

BC = 2 9 = 18 cm


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67.(B)A

OD

O

B

O D = 2 215 12

= 225 144

= 81 = 9

O D = 2 213 12

= 169 144 = 25 = 5

OO' = 9 + 5 = 14 cm

68. (B)A

CB

D E

DE BC

ADE = ABC

AED = ACB

ADE ABC

BDEC 1

ADE 1

BDEC

ADE

+1=1+1

ADE

ABC

= 2 =

2

2

AB

AD

AB

AD= 2

AB

AD 1= 2 1

BD

AD=

2 1

AD

BD=

1

2 1

AD : BD = 1 :1

2 1

69.(B) XZ = r+ 9 & YZ = r+ 2

90

X

B

A

Z

Y

2 2 2XY XZ ZY

172= (r+9)2+ (r+2)2

289 = r2+ 18r+ 18 + r2+ 4r + 4

2r2+ 22r+ 85 289 = 0

2r2+ 22r 204 = 0

r2+ 11r 102 = 0

r2+ 17r 6r 102 = 0

r ( r + 17) 6( r +17)= 0

( r 6) ( r +17)= 0

r = 6 cm

70.(A)

Length of transverse trangent

= 2 21 2XY (r r )

8 = 2 2XY 9

64 = XY2 - 81

XY2= 64 + 81 = 145

XY = 145

71. (C) AB is diameter ADB = 90

also DO AB at 'O', the centre of thecircle.

ADO BDO (by SAS cong. Rule) AD = DB (by CPCT)

DAB = ABD = 45But ACD = ABD (angles in the same

segment of a circle)

= 45

72. (A) CBDCAD (Angles iin the same

segment of a circle)

= 60

Now BAD = CADBAC = 30 + 60 = 90

Now 180 BCDBAD( ABCD is cyclic)

90 + BCD = 180

BCD = 180 90 = 90

73. (A) Perimeter of the rope

= 3 (3

1of circumference of a circle) +

3 diameter of a circle)


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= 3 3

1 2 + 3 2

= 2 + 6

74. (A) DC||AB (given)

AOB ~ COD (by AA similarity)

)(

)(

CODar

AOBar

= 2

2

DC

AB

= 2

2)3(

DC

DC= 2

29

DC

DC=

1

9

75. (B) In the given figure, ABC is a right angle

triangle, where 90BAE, BD and CF are the 3 mediansNow, AB = 12 cm, BC = 9 cm & AC = 15 cm

So, here,

41

21 ACBDACBD

4

5 ACCFAE

also,

4

5

4

1 ACCFAEBD

= 5.3372254

6

4

6AC cm

76. (A) Let xunit be the side of the square.

its diagonal = 2 xunit

A1: Area of the square = x2sq. unit

A2: Area of the equilateral described

on the diagonal of the square.

= 2)2(4

3x sq.unit

=22

4

3x sq. unit

1

2

A

A

=

2

3

x

x

= 2

3

A2: A

1= 3 : 2

77. (A) dm: diameter of the moon

de : diameter of the earth

Case I, dm= 4

1d

e

Let runit be the radius of the earth.

then, dm=

4

12r=

2

runit

Rm: radius of the moon =

2

2

r

=4

runit

m

eVV = 3

3

43

43

4

r

r

= 64 : 1

78. (B) Perimeter = 2(l+ b)P = 2(l+ w)

2

P w = l

its area = l b

k = wwP

2

2k = Pw 2w2

2w2 Pw+ 2k= 079. (C) Volume of the ice-cream in cylindrical

container = hr2 = 15667

22 cm3

Let rcm be the radius of the cone,its height = 4rcm

Volume of 1 cone with hemispherical top

=3

3

2

3

1rhr

=3

3

24

3

1rrr

=33

3

2

3

4rr

=3

3

6r = 32 r

V. of 10 such cones = 10 32 r cm

ATQ,

15667

22 = 3210 r

15667

22

=3

7

22

210 r

r3 =210

1566

=

222

666

r =2

6cm = 3 cm

80. (*) figure is not gettiva matched with the

question asked.


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81. (A) Area of rectangular field =1000

1

4

m

= 4000 m

breadth = 50 m

Length =4000

50= 80 m

New length of field = (80+20) m = 100 m

New area = 100 50 = 5000sq. m

Required exenditure = `1

50004

= ` 1250

82. (C) Increase in water level

=Volume of sphere

Area of base of cylinder

=

3

4r

r

= 3

145.33

4

3

4

r cm.

Required water level

= 7 14

3=

7

3cm

83. (A) Curved surface of cylinder = 2 rhCase II

Radius =1

r3

: height = 6h

Curved surface = 2 1

3r 6h = (2 rh)2

Increase will be twice.84. (A) BO = 4 units: OC = 3 units & BOC = 90

BC = 2 24 3 = 5 units

BC2= 25 sq. units

85.(A)2

33r = 19404

32 22

3 7r = 19404

r3 =19404 3 7

2 22

= 9261

r = 3 21 21 21 = 21cm

Total surface area = 3 r2

= 3 22

721 21 = 4158 sq. cm

86.(C)

DC

A B

BD = Diagonal = 16cm

Area of square =1

2 BD2

=1

21616 = 128 sq. cm.

87. (A) Let 'r' be the radius of the semi-circle and 'x' be the side of the inscribed square.

A P O Q Bx

2

R

r

S

x

In OQR,

r2 = x2+

2

2

x

r2 =2

4

5x

x2 =2

5

4r

Area of the square =5

4 2rsq. unit

Again, let 'y' be side of the square inscribedin the circle of same radius.

D C

A B

r

r y

2

2

22

ry

yr

=

=

Area of the square = y2

=

2

2

2

rsq. unit

= 2r sq. unit

Required ratio =5

4 2r: 2r2

= 2r2

1:

5

2

= 2r2[2 : 5]= 2 : 5


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88. (A)

O

A B10

cm 10cm

Area of the minor segment

= sector area OABO area of OAB

= 10102

1

360

90101014.3

= 504

314 = 78.5 50 = 28.5 cm2

Area of the major segment= area of circle area of minor segment= 3.14 10 10 28.5

= 314 28.5 = 288.5 cm2 Required difference = 285.5 28.5

= 257 cm2

89. (C)

Total area of the square field= (44 44)m2= 1936m2

At the rate of Re. 1 per sq. mtr; the total

cost would be Rs. 1936,but the total cost = Rs. 3536Difference = Rs. 3536 Rs. 1936

= Rs. 1600 Rs. 1600 would be the extra cost on the

flower bed and as the extra cost on theflower bed is Rs. 1 per sq. mtr.

Area of flower bed = 1600 sq. mtr.

Side of flower bed = 1600 m2

= 40 m

So, width of the gravel path =2

4044

= 2 metre90. (B)

A B

CD

E F

26

cm

60 cm25cm

26

cm

ABCD is a trapezium. Draw CE || DA intersecting AB at E. ABCE is a ||gm.

DA = CE = 26 cm.

In BCE

S =2

262517 =

2

68=34

ar( BCE) = )2634)(2534)(1734(34 cm2

= 891734

= 2223317172

= 2 2 3 17

= 204 cm2

2

1 BE height = 204

or,2

1 17 CM = 204

CM =17

2204= 24 cm

ar(Trap. ABCD) =2

1 (60 + 77) 24

=21 137 24

= 1644 sq. cm.

91.(C) Required number of persons = 450 +250

+150 + 75 + 50 +25 = 100092.(B) Required answer = 250+150 = 400

93.(C) Required ratio = 250 : 75 = 10 : 3

94.(B) Age group 15 - 20 450

500=

9

10

95.(D) Required percentage =25

100500

= 5%

96. (D) Expenditure on clothing & miscellaneous= (20 + 30)% of 25000 = `12500

97. (C) Total expenditure = 100)2010(

15000

= `50,00098. (D) 360 = 100%

%15%6.3

5454

and among the options the two items having

difference of 15% (and hence 54) aremiscellanouns and Food

99. (B) Required % age = %10015

1015

%33.33%1003

1

100. (D) 90 = %25%6.3

90

and among the options the two itemsmaking central angle of 90, together, are

travelling and eutertainment.


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SSC MAINS (MATHS) MOCK TEST - 4 (ANSWER SHEET)

1. (D)2. (C)3. (A)4. (A)5. (D)6. (A)7. (C)8. (*)9. (C)10. (A)11. (C)

12. (B)13. (B)14. (C)15. (B)16. (A)17. (D)18. (A)19. (A)20. (C)

21. (C)22. (C)23. (A)24. (C)25. (A)26. (B)27. (D)28. (C)29. (B)30. (C)31. (A)

32. (D)33. (D)34. (C)35. (A)36. (D)37. (A)38. (C)39. (B)40. (C)

41. (B)42. (D)43. (A)44. (A)45. (C)46. (B)47. (C)48. (D)49. (B)50. (A)51. (A)

52. (A)53. (B)54. (D)55. (C)56. (C)57. (A)58. (A)59. (C)60. (B)

61. (B)62. (A)63. (A)64. (C)65. (C)66. (B)67. (B)68. (B)69. (B)70. (A)71. (C)

72. (A)73. (A)74. (A)75. (B)76. (A)77. (A)78. (B)79. (B)80. (*)

81. (A)82. (C)83. (A)84. (A)85. (A)86. (C)87. (A)88. (A)89. (C)90. (B)91. (C)

92. (B)93. (C)94. (B)95. (D)96. (D)97. (C)98. (D)99. (B)100. (D)

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Ssc Mains (Maths) Mock Test-4 (Solution)

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