ssc mains (maths) mock test-11 (solution)

Centres at: �� MUKHERJEE NAGAR �� MUNIRKA ��UTTAM NAGAR ��DILSHAD GARDEN �� ROHINI ��NARELA

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1

Mock Test - 11

1.(A)7

16 6 7 – 16 – 6 7�

�2 2

7

(3 7) – (3 – 7 )�

�7

3 7 – 3 7� ��

7

2 7=

1

2

2.(C) 5 , 3 4 , 5 2 , 7 3

L.C.M. of 2,3,5, and 7 = 210

5 = 1052 1055� = 210 1055

3 4 = 703 704� = 210 704

5 2 = 425 422� = 210 422

7 3 = 307 303� = 210 103

Largest number = 210 1055 = 5 Ans.

3.(C) Let the digit at Ten's place = z

N = 100x + 10z + y

N – 100x – y = 10z

N – 100x – y is divisible by 10.

4.(A) m = n2 – n

m = n (n – 1)

m2 – 2m = m(m – 2)

= n (n – 1) × (n2 – n – 2)

= n(n – 1) (n – 2) × (n + 1)

Now, for n = 3

m2 – 2m = (3 – 1) (3 + 1) (3 – 2)

= 24

m2 – 2m is divisible by 24. Ans.

5.(D) Let the two digits number be 10x + y.

two digits number in reverse order

= 10y + x

According to question,

� 10 × 2x + 2

y = 10y + x

� 20x – x = 10y – 2

y

� 19x = 2

19y

�y

x =

2

1

Unit's digit is two times the ten's digits. Ans.

6.(C) Let P & Q be any number say, P = 17, Q = 9.

Again, let divisor = 5

Divisor = 5 = R1 + R

2 – R

3

5 = 2 + 4 – 1 = 5 Ans.

7.(B) Let the two digits number be A & B


35% of A + B = 120% of B

� 35 A = 20 B

�A 20 4

B 35 7� � � 4 : 7

8.(D) Let the five numbers be x, y, z, u, and v.


� x + y + z + 6 = u + v ................(1)

� x + y + z = 2u v ...................(2)

From (1) & (2)

2 u + 6 = u + v

v – u = 6

Neither u nor v can be calculated with the

help of the above relation

��Data inadequate. Cases could be many.

9.(C) Number of different ways in which 4 boys &

2 girls can be seated.

= 6! = 720

When two girls seat together,

No. of arrangements = 5! × 2 = 240

No. of arrangements when girls do not seat

together = 720 – 240 = 480. Ans.

10.(B) Let the cost of 1 orange = Rs. x

Let the cost of 1 apple = Rs. 1.75x

Now,

�x75.1

40 +

x

16 = 14

� 40 + 16 × 1.75 = 14 × 1.75 × x.

� 40 + 28 = 24.5x

� x = Rs.5.24

68

cost of 1 apple = 5.24

6875.1 �


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2

No. of apples = 685.17

5.2440

��

= 8.24

= 8 (approx.) (B)

11. (B)

Atleast 67 electronic graduates should bethere to fulfill the conditions given in thequestion.

12. (C) Cost of painting on Monday = Rs. x

Cost of painting on Tuesday = Rs. x + 3y

Cost of painting on Wednesday = Rs. x + 2y

Cost of painting on Thursday = Rs. x + y Cost of painting on Friday = Rs. x + 2y

Average daily earning = 5

85 yx �

= yx5

8� Ans.

13. (B)

��20 ltr of 1st solution must be mixed withequal quantity of 2nd solution to make sugar10% in total mixture

��20 ltr. Ans.

14. (A)

�

15. (A) Sn =

1

)1(

��

r

ra n

a = first term

r = Common ratio

n = Number of terms

tn = last term

2 2 – 11022

2 – 1

n

�

511 = 2n–1

2n = 512

2n = 210

� n = 10 Ans.

16.(C)

Largest 3 digit numbers divisible by 7 = 994Smallest 3 digit number divisible by 7 = 105

umber of terms = 994 –105

17

�

= 127 + 1= 128

Sum = 2

n[first term + last term]

= 128

2[994 + 105]

� 70336 Ans.17.(A)

18.(B)

A's share of rent = Rs. 1020 = 60 unit

� 108 units = 1020

10860

�

C's rent = 1836 Rs. Ans.

19.(B)

=

2

2

2

3

3

3

100

Pr

100

3

100 ��

��

�

rrP

= 2

2

2

2

100

Pr

3100100

Pr�

��

�r


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3

�8

25 =

100

r + 3

�8

1 =

100

r

� r = 12.5%

= 122

1% Ans.

20.(C)100

11TPR =

100

22TPR

8n = 7 ��

��

��

2

1n

� n = 3.5

ATQ,

100

5.38��P + P = 2560

� P = 2000 Ans.

21.(A) P = 100 . .S I

R T

��

= 900 100

3 6

�

� = Rs. 5000 Ans.

Now:-

C.I. - S.I. =

3 2

3 2

3

100 100

r rP

��

� ��

=

3 2

3 2

6 3 65000

100 100

��

� ��

= Rs. 55

22.(A) 2a � 2r �

10

–1

–1

na rS

r� �

102 2 –1

2 –1

= 2(31) 2 1

2 –1 2 1

��

�

�2(31)( 2 1)

1

�� 31 2 2�

23.(A)

Let ‘a’ & ‘b’ be the two numbers

We have, 4

5 2

a bab

��

� �

5 2ab a b� � �

Squaring both sides

25 ab = 4a2 + 8ab + 4b2

� 4x2 – 17ab + 4b2 = 0

the factors of which are (a – 4b) (4a – b) = 0

Hence,

(i) a – 4b = 0 �4

1

a

b�

(ii) 4a – b = 0 �1

4

a

b�

�� a : b = 1 : 4 Ans.

24.(A) Relative speed = (60 + 6) km/hr

= 5

6618

� = 55

/sec.3

m

Distance = 110 m

� Time = 110

355

� = 6 sec. Ans

25(A) S.P. of scooter at Abhishek's shop

= Rs. [90% of 28000]

= Rs. 25,200

S.P. of Bhanu's shop

= Rs.[88% of 20,000 + 92% of 8000]

= Rs. [17,600 + 7360]

= Rs. [24, 960]

Required Difference = Rs. [25,200 – 24,960]

= Rs. 240

26(A) Let C.P of each Camera = Rs. 100

� Actual C.P. = 36

10020

� = Rs. 180 /each

27(A) Let x litres of water added to the solution.

According to question:-

4% of (10 + x) = 1

� x = 100

–104

= 15 litres Ans.

28(B) Let the total number of workers = x


20% of 75% of x + 80% of 25% of x = 126

�100100

7520

�� x

+ 80 25

10 100

x� �

� = 126

� x = 126 100 100

(1500 2000)

� ��

= Rs. 360 Ans.


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4

29. (B) Let single ticket = Rs. x

Return ticket = Rs.4

5x

105% of 4

5x= 84

x = 1055

4484

��

= Rs. 64

30 (D) Ratio of the students in the states A & C appearing for exam in 1998

= 3 : 6 = 1 : 2

As the increment in the next year is same for the student of both the states.

� The no. of students who appeared in the stateA in 1998 are = any one of them – 3, 6, 9, 12etc.

� Data inadequate. Ans.

31(C) Let the no. of children be n.

No. of note books each child have = 8

1n

ATQ,

2

1n × 16 = n ×

8

1n

8n = 8

2n

n = 64

No. of note books distributed

= 64 × 8

1× 64 = 512 Ans.

32. Data Insufficient

33. Total Price of component = Rs. 50,000

Expected rejection = 5% = Rs. 2,500

Remaining = Rs. 47,500 = C.P. + 25% Profit

C.P. = Rs. 38,000

But, 50% goods rejected, so he was paid 50%of 50,000 i.e. Rs. 25,000

So, Loss = Rs. 38000 – Rs. 25,000

= Rs. 13,000 Ans.

34.(D) 10 M + 15 W = 6 days

� 60 M + 90 W = 1 days .......... (i)

100 M = 1 day .......... (ii)

60 M + 90 W = 100M

4M = 9 W

� IM = 9

W4

10 M = 45

W2

45W

2 + 15 W = 6 days

or, 75

2 W = 6 days

1 W = Rs. 225 days

35.(D) Suppose women take x hrs to complete the

work.

Then child will complete in (x + 15) hrs.


15

18

�xwork + �

�

��

�x

6work =

5

3

)15(

)15(618

�

��

xx

xx=

5

3

3x2 + 45x = 90x + 30x + 450

x2 – 30x + 5x + 180 = 0

x(x – 30) + 5(x – 30) = 0

(x + 5)(x – 30) = 0

x = 30

1 work is completed by a women in 30 hrs.

�5

2 work is completed by a women in

5

2× 30

= 12 hrs. Ans

36.(C)

� Time taken by A alone = 35 hrs Ans.

37.(D) Let the speed of boat = x m/sec

and speed of water = y m/sec

Speed downstream = (x + y) m/sec

Speed upstream = (x – y) m/sec

Thus,

yx �

20 +

yx �

20= 4 min 40 seconds

= 43

2min

But here we must have speed of water to

calculate speed of boat

Hence, data inadequate. Ans.

38.(A) Total Pipe = 6

1 outlet pipe takes = 6 hr.

1 Inlet pipe takes = 9 hr.

Tank is filled in only 9 hour which means

the efficiency of one inlet pipe is utilized and

rest became neutral which is possible only

in one case

3 inlet pipes = 2 outlet pipes

��total inlet pipe = 4 Ans.

39.(D) Speed ratio = 6 : 5

& Speed of Train is 20% faster than the car.

Let, the time taken = t min



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5

6(t – 12.5) = 5t

� t = 6 × 10

5.12= 75 min

t = 14

1 hr.

Speed of car = t

75=

4

5

75

Sc = 60 km/hr Ans.

40.(D)

20 M = 30 W = 60 B

2 M = 3 W = 6 B

Now, 2 M + 8 W + x B � 2 day � 60 units

Thus, 2 M + 8 W + x B � 1 day � 30 units

2 × 3 + 8 × 2 + x × 1 = 30

6 + 18 + x = 30

x = (30 – 22) = 8 boys

41.(A)

42. (B) A � B 100% pure milk

B � 2 : 5

C � 3 : 8

Thus, 16 litre of milk A is added.

43.(B)

44.(C) Let the population be x.


100% of x + 15% of x = 45 million (Jan, 2006)

In Jan 2005 :

�x = 45

100115

� million = 39 million

45.(A)

Vacant seats = 73

46.(D)

� AG = GF,

Let � GAF = � AGF = x

Now, ext angle � FGE = 2x

� � FGE = � FEG = 2x

� AB = BC

� CAB = � ACB = x

Now, ext angle � CBD = 2x

In � ADE,

x + 3x + 3x = 180º

x = 7

º180= 25º (Approx.) Ans


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6

47.(B)

Let BC = x, FB = y = EF = AE

� CD = 3y

Now:-

ar( � CBF) = 2

1xy

or, ar( � CBE) = 2

1x × 2y = xy

� ar( � CEF) = xy – 2

1xy

= 2

1xy

Now:-

Area of rectangle = 3xy

� = xy

xy

32

1

�

�= 1 : 6 Ans.

48.(A)

Give AB = 7 m & BG = 2m

Thus, AG = 5 m

AH = 2m (given)

Thus, HE = AD – AH – ED

= 12 – 2 – 2 = 8 m

and OH = 4 m

Area of required figure

= (l × b) – 2 ��

��

�� hb

2

1

= 12 × 7 – 2 × 2

1 × 5 × 4

= 64 m2.

49.(B)

Direct method:-

QSR� = 90º – 1

2QPR�

= 90º – 1

2 × 50

So,

QSR� = 65º Ans.

50.(A) Area = 1

2 × sum of �� sides × height

35 = 1

2 × 14 × h

Height = 5 cm

In �DFC,

DC2 = DF2 + FC2

DC2 = 52 + 22

DC2 = 29 ��DC = 29

51.(B)

Join AF & EC

Clearly, � AFC = � AEC

(angle on a semi-circle)

= 90º

� AFC � � AEC (RHS)

Thus, AF = EC

And, �AFD � �CEB (RHS)

� FD = EB = 2 cm

52.(B) ar(�AXB) = 2

1(area of parallelogram) ..(i)

[� Having same base]

Similarly,

ar(�BFC) = 2

1(area of parallelogram) ..(ii)

From (i) and (ii),

ar(�AXB) = ar(�BFC) Ans.


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7

53.(A)

BD = a, EF = 2

a

[� EF || BD and EF = 2

1BD]

� Area of equilateral (�EGF) =

2

24

3��

��

�a

= 16

3 2aAns.

54.(A)

AP = n AB

Now:-

tan� =

AP

AB

2 =

ABn

AB

.2

= n2

1... (i)

Now:-

� tan( � +� ) = AP

AB=

ABn

AB

.=

n

1

� ��

��

tan.tan1

tantan=

n

1

�

n

n

2

1.tan1

2

1tan

��

�� =

n

1

��

tan2

1tan2

n

n =

n

1

� nn ��tan2 2 = �� tan2n

� �� tantan2 2n = nn �2

� tan � [2n2 + 1] = n

� tan � = 12 2 �n

nAns.

55.(B)

C1 = Centre of small circle

C2

= Centre of bigger circle

AB = 2AC = 2 × 22 = 24

56.(A) cosec � – sin � m and sec � – cos � = n

Now,

m = cosec � – sin �

= �sin

1– sin �

m = ��

sin

cos2

n = sec � – cos �

n = �

��cos

cos1 2

= ��

cos

sin2

� (m2n)2/3 + (mn2)2/3

=

3/2

2

2

2

4

cos

sin

sin

cos��

��

�

�

��

�

�+

3/2

2

42

cos

sin

sin

cos��

��

�

�

��

�

�

= (cos3 � )2/3 + (sin3 � )2/3

= sin2 � + cos2 � = 1 Ans.

57.(A) a sec � + b tan � + c = 0

p sec � + 9 tan � + r = 0

�bpaqarcpcbr �

��

��

�

� 1tan

9

sec

� sec � = bpaq

cpbr

�

�


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8

and tan � = bpaq

arcp

�

�

Now:-

sec2 � – tan2 � = 1

�

2

��

��

�

�

�

cqaq

cqbr –

2

��

��

�

�

�

bpaq

arcp = 1

� (br – cq)2 – (cp – ar)2 = (aq – bp)2 Ans.

58.(D) �

��

�

�2

4

2

4

sin

sin

cos

cos = 1

� cos4 � .sin2 � + sin4 � .cos2 � = cos2 � .sin2 �

� cos4 � (1 – cos2 � ) + cos2 � (1 – cos2 � )2

= cos2 � (1- cos2 � )

� cos4 � – cos4 � .cos2 � – 2cos2 � .cos2 � +

cos4 � .cos2 � = cos2 � – cos4 �

� cos4 � –2cos2 � .cos2 � + cos4 � =0

� (cos2 � – cos2 � )2 = 0

� cos2 � = cos2 �

� 1 – sin2 � = 1 – sin2 �

� sin2 � = sin2 �

��

��

�

�2

4

2

4

sin

sin

cos

cos

��

��2

22

cos

coscos +

�

��2

22

sin

sinsin

� cos2 � + sin2 � = 1 Ans.

59.(B) �� 8cos2222

= )14cos2222 2 ��

[�cos2 � = 2cos2 � – 1]

= �� 4cos422 8

= �� 4cos222

= )1–2cos2(222 2 ��

= �� 2cos42 2

= �� 2cos22

= )1(cos22 2 ��

= �2cos4 = 2 cos � Ans.

60.(C) cos� ��2

1��

��

��

aa

1

Squaring both sides:-

cos2� = 4

1

��

��

��

��

��

21

aa

2cos2� = 2

1

��

��

��

��

��

21

aa

Substracting 1 from both sides:-

2cos2� – 1 = 2

1

��

��

��

��

��

21

aa – 1

= 2

1��

��

�� 2

12

2

aa – 1

= 2

1��

��

��

2

2 1

aa + 1 – 1

LHS: = 2

1��

��

��

2

2 1

aa Ans.

61.(D) a = sin4

� =

2

1

b = cos 4

�=

2

1

c = – cosec 4

�= – 2

a + b + c = 2

1 +

2

1 – 2

= 2 – 2 = 0

a3 + b3 + c3 = 3abc [� a + b + c = 0]

L.H.S.= 3abc

= 3.2

1.

2

1(– 2 )

= 22

3�Ans.

62.(B)

� EF = 3AE = 3×3

5a


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9

EF = 5 a

Area of rectangle = 3

4a × 5 a ... (i)

Area of parallelogram = 2ar( � ADC)

= 2 × 2

1 × 2a ×

3

5a

= 3

52a2 ... (ii)

Required ratio = 3

52a2 :

3

54a2

= 1 : 2

63.(B) x cos � + ysin � = P and ..... (i)

x cos � + ysin � = P' ..... (ii)

Slope of equation (i):-

m1 = –

��

sin

cos

Slope of equation (ii):-

m2= �

�

sin

cos

Let the angle between line be � .

Then,

� tan � = 21

21

1 mm

mm

�

�=

��

��

�

��

��

�

sin

cos

sin

cos1

sin

cos

sin

cos

� tan � = ��

��

sinsincoscos

sincoscossin

� tan � = )cos(

)sin(

��

��

� tan � = tan( �� )

� � = ��

64.(A) By C & D.

)sin()sin(

)sin()sin(

yxyx

yxyx

��

��=

baba

baba

��

yx

yx

sincos

cossin=

b

a

y

x

tan

tan=

b

aAns.

65.(B) Putting x = 1

[Always put x = 1 in such questions]

� 2 Ans

66.(A) –1x y

y x� �

� x2 + y2 + xy = 0 ... (i)

� 2 (x3 – y3)

From (i)

� 2(x – y) (x2 + y2 + xy)

� 0 Ans.

67.(A) a(a +2) = a + b + c

� a + 2 = a

cba ��

b + 2 = b

cba ��

c + 2 = c

cba ��

Substituting:-

cba

a

�� +

cba

b

�� +

cba

c

��

�cba

cba

��

= 1 Ans.

68.(D) Multiplying

ax.ay.az = (x + y + z)x + y + z

a(x + y + z) = (x + y + z)x + y + z

� a = x + y + z

69.(A)aa �� 21

1 –

aa �� 21

1 – 421

2

aa

a

��

2 2

2 2 2

1 1

( )

a a a a

a a a

� � � � �

� � – 421

2

aa

a

��

241

2

aa

a

��+ 421

2

aa

a

��

= 0 Ans.

70.(A)cb

p

�=

ac

q

� =

ba

r

�

P = k(b – c), q = k(c – a) , r(a – b)

than,p + q + r

= k(b – c + c – a + a – b)

= 0 Ans.

71.(A) pqr = 1 ��1

pqr

� & 1

qrp

�

�

1 1 1

1 1 1 11 1

r qrq

qr q r

� ��

��1

1 1 1

qr r

qr q qr r qr r� �

� � � � � �

��1

1

qr r

qr r

� �� = 1 Ans.

10

72.(C)

From � BDC since � y = 90º – x

� � ADM = y

In � BDC,

y

x=

100

40.... (i)

In � ADM,

y

x=

AM

100.... (ii)

from (i) & (ii),

100

40=

AM

100

AM = 250 m

Now, AB = 250 + 40 = 290 m Ans.

73.(C) Let angle of elevation for A, B & C are �, 2�and 3�. (According to given condition we

choose that)

From �PAB,

2� = ��+ �APB

��APB = ��PAB = �ABP = ��AB = BP =a

Similarly, in �BPC, �BPC = �from �OBP, sin 2� = h/a

��h = a sin 2�� h.2a sin��cos��

from �PBC, PB

sin(180–30) =

BC

sinθ

(by sine rule)

��sin3

a

�=

sin

b

��

a

b��

sin3

sin

�

�

�a

b��

33sin – 4sin

sin

� �

��

a

b�3–4sin2�

��4 sin2��3 –a

b��sin2��

3 –

4

b a

b

��sin��3 –

4

b a

b��cos2�=1–sin2�

��cos2��=3 –

1 –4

b a

b��

4

a b

b

��cos�=

4

a b

b

�

Putting value of sin� and cos��in (1), we get

h =3 –

2 .4 4

b a a ba

b b

�= ( )(3 – )

2

aa b b a

b�

74.(C)

Let the different objects are A and B.


In � APQ,

tan��c

y

In � AQB,

tan(�� = c

yx �

Now,

tan( � ) = tan[( � ��) –�� ]

= ��

��

tan)tan(1

tan)tan(

=

��

��

��

��

� ��

��

c

y

c

yxc

y

c

yx

1

= 22 yxyc

cx

��

� tan � = 22 yxyc

cx

��

� cot � ��cx

yxyc 22 ��


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11

� cx cot x = c2 + xy + y2 .... (i)

Similarly,

2 cx cot � = (2c)2 + xy + y2 .... (ii)

Substracting (i) from (ii):-

3 c2 = cx [2cot � – cot � ]

� x = �� cotcot2

3cAns.

75.(A)

= 7

22 ×

3

45 [(7)2 + (28)2 + 28 × 7]

= 7

22 ×

3

45× 7[7 + 4× 28 + 28]

��48510 cm2

76.(C) Volume = Area of trapezium × height

Area of Trepezium = 1

2 × (sum of parallel

sides)× height

= 1

2 × 22 × 8 = 88 cm2

Now,

Volume = Area of base × Height

� Height = 1056

88= 12 cm

77.(B) Each interior angle = n

n )2(180 �

Each exterior angle = n

360

ATQ,

n

n )2(180 � = �

�

��

�n

3602

� n = 6 Ans.

78.(C) Volume of pyramid

= 1

3 × Area of Base × Height

= 1

16 16 123

� � �

= 1024 cm3 Ans.

79.(C)

AB BD 8= = =2:3

AC DC 12

80.(*)3

77 ��y= 3,

3

93 ��x = 4

� y = –5, x = 6

� (x, y) = (6, – 5)

Area = )]()()([2

1213132321 yyxyyxyyx ��

= )75(9)57(3)77(62

1��

= 72 unit2.

81.(D) (a + b + c) = 5

Squaring both sides:-

[(a + b + c)]2 = 25

a2 + b2 + c2 + 2(ab + bc + ca) = 25

a2 + b2 + c2 = 25 – 20 (� ab + bc + ca = 20 )

= 5

Now,

a3+ b3 + c3 – 3abc

= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

= 5 (5 – 10)

= – 25

82.(C)

Area = 2

1 × b × h

= 2

1 ×

b

c ×

a

c

= 2

1

ab

c 2

83.(B) Ist 3-digit no. div. by 6 = 102

Last 3-digit no. div. by 11 : 996

AP = 102, ........., 996, (cd = 6)

99 = 102 + (n – 1)6

� n = 150 Ans.


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12

84.(C) 6th term = a + 5d

15th term = a + 14d and so on .....

ATQ,

a + 5d + a + 14d = a + 6d + a + 9d + a + 11d

19d = a + 26d

0 = a + 7d

� 8th term Ans.

85.(B) b2 = ac (G.P.)

2 2

2 2 2

–

1 1 1

a ac c

a b c

�

� �

��2 2

2 2

2 2

–

–

a ac c

c ac a

a c

�

�

��

��

�

��

��22

2222

caca

cacaca

= b4

86.(B) SP = (a – at2)

SQ = 2

aa

t�

2

2 2

1 1 1 t

SP SQ a at at a� � �

� �

2

2

(1 ) 1

(1 )

t

aa t

��

�

87.(A) x = 2

Then,

f(2)= (2)4 + 2(2)3 – 3(2)2 + 2 – 1

= 16 + 16 – 12 + 2 – 1

= 21 Ans.

88. (A) y = x

� x – y = 0

& y = x + 3

��x – y +3 = 0

�1 1 1

2 2 2

a b c

a b c� !

� y = x + 3 & y = x are parallel

89. (*)

14y = 56

� y = 4

� x = 6, y = 4

Required co-ordinate = (6,4)

90.(B)

2 2(7 5 ) – (7 5 )

49 – 5

� �

49 5 14 5 – 49 – 5 14 5

44

� � �

75

11�

75

11a b� �

��a = 0 b=1

91.(D) Half male employee promoted from IT deptt.

= 1200100

13� = 156

Total male employee of IT deptt.

= 2040100

20� = 408

Required % = 100408

156� = 38.25% ~ 38% Ans.

92. (*) Total employee (Prod. + Mark.) – Male Employee (Prod. + Mark.)

= 3600(35 + 18)% + 2040 (50 + 10)%

= 1908 – 1224

= 684 Ans.

93.(A) Female worker = Total employee in A/c deptt – Male employee in A/c

deptt.

= 20% of 3600 – 5% of 2040

= 10

45� [4 × 90 – 51]

= 618 Ans.

94.(D) Required no. of employee

= 1003600

1200� = 33.3% or 33 Ans.

95.(B)

=

100

36002100

12001

�"

�"

×100

�56

1100 = 30.555 = 30.56 Ans.

96.(A) Expenditure =

ATQ,

1100

351

�

I =

1100

402

�

I

2

1

I

I =

140

135

� I1 : I

2 = 27 : 28 Ans.


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

13

97.(D) Given

(Income – Expenditure = 1.5 lakh)

� Profit % = Exp

ExpIncome � × 100

= 10

100

exp

5.1� = 40

� Expenditure = 15

4 =3.75 lakh Ans.

98.(D) Taking the approx. values

= 6

453035404540 ��

# 39 (same it is very close to 37)

� It's approx. value can also be seen from

graph itself.

99.(D) � Profit % = �

��

�1

.Exp

Income × 100

Income = Exp.

ATQ,

Exp. A �

��

�1

100

50 = Exp. B �

��

�1

100

30

B.Exp

A.Exp =

150

130

Exp A : Exp B = 13 : 15 Ans.

100.(D)BCompany

ACompany =

45

30

Profit :-

Com A : Com B = 2 : 3 Ans.


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

14

Mock Test - 11 (Answer Key)

1. (A)

2. (C)

3. (C)

4. (A)

5. (D)

6. (C)

7. (B)

8. (D)

9. (C)

10. (B)

11. (B)

12. (C)

13. (B)

14. (A)

15. (A)

16. (C)

17. (A)

18. (B)

19. (B)

20. (C)

21. (A)

22. (A)

23. (A)

24. (A)

25. (A)

26. (A)

27. (A)

28. (B)

29. (B)

30. (D)

31. (C)

32. (D)

33. (A)

34. (D)

35. (D)

36. (C)

37. (D)

38. (A)

39. (D)

40. (D)

41. (A)

42. (B)

43. (B)

44. (C)

45. (A)

46. (D)

47. (B)

48. (A)

49. (B)

50. (A)

51. (B)

52. (B)

53. (A)

54. (A)

55. (B)

56. (A)

57. (A)

58. (D)

59. (B)

60. (C)

61. (D)

62. (B)

63. (B)

64. (A)

65. (B)

66. (A)

67. (A)

68. (D)

69. (A)

70. (A)

71. (A)

72. (C)

73. (C)

74. (C)

75. (A)

76. (C)

77. (B)

78. (C)

79. (C)

80. (*)

81. (D)

82. (C)

83. (B)

84. (C)

85. (B)

86. (B)

87. (A)

88. (A)

89. (*)

90. (B)

91. (D)

92. (*)

93. (A)

94. (B)

95. (B)

96. (A)

97. (D)

98. (D)

99. (D)

100. (D)

ssc mains (maths) mock test-11 (solution)

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