ssc mains (maths) mock test-7 (solution)

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

1.(B) 30.000729

0.085184= 3

729

85184= 3

9 9 9

44 44 44

� �

� �=

9

44

2.(B)

3.(D) It is easy to solve this question by using option.

By option(d),

Total number of apples at the starting = 255

Number of apples sold to first customer

= 255

2+

1

2= 128

� Remaining apples = 225 � 128 = 127

Number of apples sold to second customer

= 127 1

2 2� = 64


Number of apples sold to third customer =

63 1

2 2� = 32


Number of apples sold to fourth customer =

31 1

2 2� = 16


i.e. condition satisfied

4.(B) Let time spend on each Mathematics

question = x min

A.T.Q,

Total time spent =

50x +100 × 2

x+ 50 ×

2

x = 3 ×160

� x ( 50 + 50 + 25) = 180 � x = 180

125

� Required time = 50 ×180

125= 72 min.

5.(D) Required remainder = 919 + 6 = (1)19 + 6 = 7

[� 9 = 8 +1, so 9 is replaced by 1]

6.(A) LCM of 3, 4, 5, 6 and 8 =120

Now, by option (a) 14400 is a least five digit

perfect square number which is divisible by

120.

7(B)

SSC MAINS (MATHS) MOCK TEST-7 (SOLUTION)SSC MAINS (MATHS) MOCK TEST-7 (SOLUTION)SSC MAINS (MATHS) MOCK TEST-7 (SOLUTION)SSC MAINS (MATHS) MOCK TEST-7 (SOLUTION)

8.(D) Required remainder = 19100 = ( � )100 = 1

[� 19 = 20-1, so 19 is replaced by ( � 1)]

9.(C) According to the question, total number of

toys is a perfect square number because the

toys were packed in n boxes containing n

toys each, without any remainder and

among the options given only 1444 is a per-

fect square. Hence, option (c) is the correct

answer.

10.(B) Ratio of total capital of A and B

= 20000 ×12 : 35000 × 12

= 240000 : 420000

Now C gives 220000 to both to make the capital

equal.

� A's captial : B's capital

= 240000 : 420000

� 220000 : 220000

20000 : 200000

� Required ratio of divided amount = 1 : 10

11.(D) Work done by A in 4 days = 4

20=

1

5part

� Remaining work = 1 �1

5=

4

5part

Let C while working alone can complete the

work in x days.

A.T.Q, 18 18 4

30 5x� �

� 18 (x + 30) × 5 = 4 × 30x

� 90x + 2700 = 120x

� 30x = 2700

� x = 90 days

12.(D) Let the number of minutes taken to empty

the cistern be x min.

A.T.Q,

5 50

6 12 15

x x x� ��

�5 5

6 12 15 12 15

x x x� � � �

�45

6 60

x� � x = 45 min.

13.(C) Let B takes x days to complete the

remaining job.

A.T.Q,

1 1 1

24A B� � and

1 1

32A�

�1

B=

1

24

1 1

32 96� �

� B = 96 days

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Now, 8 8

32 96

x�� =1

� 24 + 8 + x = 96

� x = 96 � 32 � x = 64

Hence, B completes the remaining job in 64 days

14.(B) Let extra hours per day are x.

By 2 2 21 1 1

1 2

M D HM D H

W W�

1 1 (6 4) 1 1 (6 6 )

11 12

x� � � � � � ��

�3

2×10 = 12 + x

� 15 = 12 + x

� x = 15 � 12

= 3

� Extra hours of work per day is 3 hrs.

15.(B) Suppose B left the work after x days from

the start.

A.T.Q, 7

125 15

x x ��

�3 5 35

75

x x� �= 1

� 8x = 75 � 35

� x = 40

8 = 5

16.(C) According to the question,

1 1 1 1

3A B C� � �

�1 1 1 1

8 6 3C� � �

�1

C=

1

3�

1 1

8 6

� ��

� �=

1

24

� Ratio in shares of A, B and C = 1

8:1

6:

1

24

= 3 :4 : 1

� C's share = 1

3 4 1� �× 1200 = ` 150

17.(A) Let marked price = ` 100 and selling

price = ` 80

If the loss is 10%, the cost price of article =

80 100

90

�= `

800

9

� Required profit percentage =

80095

9 100800

9

��

= 55

8 = 6.9% (approx.)

18.(C) Given, selling price of an umbrella = ` 30

profit percentage = 20%

� Cost price of an umbrella = 30 100

120

�= ` 25

During the clearance sale, selling price of

an umbrella = 30 90

10

�= ` 27

� Required profit percentage = 27 25

25

�×100

= 8%

19.(D) Let cost price = ` 100

� marked price = 100 + 40 = ` 140

Let required discount be x %

So, A.T.Q, 140× 100

100

x��

=100

� 100 � x = 100 100

140

�

� x = 100 �100 100

140

�=

40 100

140

�= 28.5% (approx)

20.(B) Here, x = 15% , y = 10%

� Required profit percentage = 15 +10+15 10

100

�

= 26.5%

21.(D) Here, a = 10 L, n = 2 and x = 100L

� Quantity of wine in end = x 1

na

x

� ��

� �

=100

210

1100

� ��

� �= 81L

� Required ratio = 81: (100 � 81) = 81:19

22.(A) Let the smaller number = x

and the greater number = y

A.T.Q,

2

xy� �

� � �

= 4 2

xx� �

� � �

� y �2

x= 4

2

x� y = 2x +

2

x

� y = 5

2

x

� y : x = 5 : 2

23.(D) Let the numbers of men, women and

children are 3y, 2y and y and their wages

are 5x, 3x and 2x respectively.

Given, 3y = 90 � y = 30

Number of women = 60 and

Number of children = 30

� Now, A.T.Q,

Total daily wages = ` 10350

� 90 × 5x + 60 × 3x + 30 × 2x = 10350

� x (450 + 180 +60) = 10350

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� x = 10350

690= 15

� Daily wage of a man = 15 × 5 = ` 75

24.(C) Let the savings of A and B are 4x, 5x and

the share in cost of gift are 3y, 4y respec-

tively.

A.T.Q,

For A, 4x � 3y = 2

3×4x

� x = 9

4

y ------- (i)

For B, 5x � 4y = 145

� 5 ×9

4

y� 4y = 145 [from eq. (i)]

� y = 20

� Cost of gift = 3y + 4y

= 7 × 20 = ` 140

25.(D) Let the fixed charges = ` x

and the additional charges = ` y / km

According to the question,

x + 5y = 350 (i)

x + 20y = 800 (ii)

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

x = 200, y = 30

� Charge for a distance of 30 km

= x + 25y = 200 + 30 × 25 = ` 950

26.(D) Let the marks of A, B and C are 10x, 12x

and 15x respectively.

� Maximum marks of C = 15 × 6 = 90

and maximum marks of B = 12× 6 = 72

Hence, the marks of B cannot be in the range

of (80 � 90)

27.(D) According to the question,

70m + 91n = 80 ( m + n)

� 70m + 91n = 80m + 80n

� 10m = 11n

�n

m=

10

11

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


60x = 12 ×400 + 56 (x � 12)

� 60x � 56x = 4800 � 672

� 4x = 4128

� x = 1032

29.(D) Let the required score be x


80 90

80

x� �= 100

� 7920 + x = 8000

x = 80

30.(D) Let the monthly salary of a man = ` x

� The annual salary = ` 12x

A.T.Q,

Annually expenditure of a man

= 7 × 1694.70 + 5 × 1810.50 = ` 20915.40

� His monthly expenditure = 20915.40

12

= ` 1742.95

and monthly saving = 3084.60

12= ` 257.05

� His monthly salary = 1742.95 + 257.05

= ` 2000

31.(C) At present, total age of A, B and

C = 3 (27 + 3) = 90 yrs. and total age of B and

C = 2 ×(20 + 5 )= 50 yrs.

� A's present age = 90 � 50 = 40 yrs.

32.(A) Total height of Team A = 20 (5 ×12 + 11)

= 1420 inches

and total height of Team B = 18 (6×12+2)

= 1332 inches

� Overall average height =

1420 133272.42

38

�� inches

33.(C) Let cost price of transistor = ` x

A.T.Q

C.P. of transistor = x × 105

100

and S.P. of transistor = 115

6100

x� ��

� �

� Profit percentage = S.P. C.P.

100C.P.

��

�

115 1056

100 100 100105

100

x x

x

� ��

� 10 = 10 600 100

105

x

x

�

� 105x = 100x + 6000 � 5x = 6000

� x = ` 1200

34.(C) Let Cost Price = ` 1000

� Actual cost price = ` 920

and selling price = 1000 115

100

�= ` 1150

� Actual profit percentage = 1150 920

920

�×100%

= 25%

35.(D) Here , x = 20%

� Total percentage loss = 2

100

x=

20 20

100

�= 4%

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36.(C) Required profit = (100 + 65)× 100 100

120 125� � 100

= 110 � 100 = 10%

37.(C) Let the price of shirt = ` x

A.T.Q, 500

12

x �=

350

10

x �

� 10x + 5000 = 12x + 4200

� x = 800

2

� x = 400

Hence, the price of shirt = ` 400

38.(D) Required profit percentage

=

11 10

10 11

10

11

� ��

� �×100 =

121 100

100

�×100 = 21%

39.(B) Let total number of men = x

and total number of women = y

� Number of married men =45

100

x

and number of married women = 25

100

y

A.T.Q,

45 25

100 100

x y� � y =

9

5

x(i)

also,

Total number of married adults = 45 25

100 100

x y�

= 9 9

20 20

x x� [from eq. (i)]

= 9

10

x

and total population in city = x + y

= x + 9

5

x[from eq. (ii)]

= 14

5

x

� Required percentage =

9

10014

5

x

x ×100 = 32.14%

40.(A) Let the individual ration be x


72 × 54 × x = 90 × D2 × x ×

90

100

� D2 =

72 54 100

90 90

� �

�= 48

Hence the required number of days = 48

41.(A) Let the total number of candidates = x

Number of candidates who answered all the

5 questions =5

100

x

also, Number of candidates who answered

not a single question =5

100

x

� Remaining students = x - 5 5

100 100

x x� ��

� �=

9

10

x

� Number of candidates who answered only one

question = 9 25

10 100

x� =

9

40

x

Number of candidates who answered four

questions

= 9 20

10 100

x� =

9

50

x

Gvien, number of candidates who answered

either two questions or three questions = 396

� x �5 5 9 9

100 100 40 50

x x x x� ��

� �= 396

� x �10 10 45 36

200

� � ��

x = 396

� x 200 101

200

��

= 396

� x = 396 200

99

�= 800

42.(D) B's marks = C's marks + 5% of 400

= 300 +20 = 320

Now, A's marks = B's marks + 10% of 400

= 320 + 40 = 360

43.(A) Let the length of the train be x m.


9

x= speed (i)

and 150

15

x �= speed (ii)

FromEqs. (i) and (ii), we get

9

x=

150

15

x �

� 5x = 3x + 450

� x = 225m

44.(A) Distance between Arun and Bhaskar at

7 : 30 am = 1

8 12

� = 12km

Time taken by Bhaskar to cover a distance

of 12 km = 12

(12 8)�3h

� Required time = 10:30am

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45. (A) Let x km/h be the speed of the watercurrent

A.T.Q,

24 24

10 10x x�

� �= 5

� 24 [10 � x + 10+x ] = 5 (102 � x2)

� 100 � x2 = 24 20

5

�= 96

� x2 = 4

� x = 4 = 2km/h

Hence the required speed = 24km/h.

46.(C) Let V is the usual speed of the man andtime is hr.

A.T.Q,

V1 t

1 = V

2 t

2

Vt = 1

12

V 20

60t� ��

� �

� t = 3

2

1

3t� ��

� ��

3

2t � t =

1

2

� t = 1h

47. (C) Let A's share = ` x

then B's share = ` (15494 � x )

A.T.Q, x

920

1100

� ��

� � = (15494 � x )

1120

1100

� ��

� �

� x = (15494 � x )

220

1100

� ��

� �

� x = (15494 � x )

21

15

� ��

� �

� x = (15494 � x )36

25

� � � �

� 25x = 15494 ×36 � 36x

� 61x = 15494 ×36

� x = 15494 36

61

�= ` 9144

Hence is A'share = ` 9144

48.(A) Let the principal be ` x

Amount @ 4% after first year = 104

100

x �

Amount @ 4% after second year

= 104

100

x �×

4

100

A.T.Q,

104 4

100 100

x � �

�= 208

� x = ` 5000

49.(D) Let the amount be ` x and the rate of

interest = r% p.a.

A.T.Q,

Amount after Ist year = ` 1200

� x 1100

r� ��

� �= 1200 (i)

also, Amount after IIIrd year = 1587

� x

3

1100

r� ��

� �= 1587 (ii)

On dividing Eq. (ii) by Eq. (i), we get

2

1100

r� ��

� �=

1587

1200

� 1+23

100 20

r�

�3

100 20

r�

r = 15%

50.(A)

Required difference

= P

2

100

r� � � �

300

100

r��

, where 'r' is the rate

of interest

= 10000

25

100

� � � �

305

100

� � � �

= ` 76.25

51.(B) Let the height of circular cylinder = H� total volume of the solid = 3× volume of the

cone

�

�

2 2

2

1

31

3

r H r h

r h

�

= 3

�2 21

3r H r h � = 2r h � 2r H =

22

3r h

�2

3H h�

52. (A) radius of pipe = 5

20 cm, (given)

height of pipe = 1000 cm

radius of vessel = 20cm

and height = 24cm

Volume of water that flows in one minute

through cylindrical pipe =

25

20

� � � �

×1000

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= 125

2 cm3

Also, volume of conical vessel = 1

3 (20)2 ×24

= 3200 cm3

� Time taken by pipe to fill the vessel

= 3200 2

125

�

= 1

515

or 51 min 12s

53.(D) Side of an equilateral triangle

= 2

( )3

l m n� �

= 2

3(6 + 9 + 12)= 18 3

� Perimeter of a triangle = 3 × side

= 3 ×18 3

= 54 3 cm

54.(B) Given area of right-angled triangle = 24cm2

�1

2× AB × BC = 24

� AB = 24 2

6

�= 8cm

� AC = 2 2AB BC�

= 2 28 6� = 10cm

Now, area of triangle = 1

2×AC ×BD

24 = 1

2×AC ×BD

� 1

2×10×h = 24

� h =24 2

10

�= 4.8cm

55.(C) Area of field = 1215

135= 9 hectare or 90000m2

� Side of field = 90000 = 300m

Perimeter of field = 4× 300 = 1200m

Now, cost of putting a fence around the field

= 75

1200100

� = ` 900

56.(B) Let the sides of trapezium be 5x cm and

3x cm respectively.

A.T.Q,

�1

5 3 12 3842

x x� � � �

� 8x = 384 2

12

�

� x = 64

8= 8cm

� Length of the smaller of the parallel sides

= 8 × 3 = 24cm

57.(A) Let the sides of the base are 5x cm

12x cm and 13x cm respectively.

Given, perimeter of base = 60 cm

� 5x + 12x + 13x = 60

� x = 60

30= 2

The sides of base are 10cm, 24cm and

26cm.

� Volume of prism = 1

2×10 × 24 × 50 = 6000cm2

58.(C) Let the length, breadth and height of a

rectangular parallelopiped be l , b and h cm

respectively.

A.T.Q., l = 3b =5h = a (say)

� l = a, b = 3

a, h =

5

a

It is given that volume of parallelopiped

=14400cm3

� a × 3

a×

5

a= 14400

� a3 = 14400 ×15

� a = 3 14400 15� = 60 cm

Total surface area of parellelopiped

= 2 (lb +bh +hl )

= 2 (60 × 20 +20×12 × 60)

= 4320cm2

59.(A) Let the radius of hemispherical bowls

be r1 and r

2 respectively.


3

1

3

2

46.43

4 21.6

3

r

r

�

�

2

1

3

2

64

216

r

r� �

1

2

r

r=

4

6

� r1: r

2 = 2 : 3

� The required ratio

2

1

2

2

2

2

r

r

= 4 : 9

60.(D) Let the radius of solid spheres be r1and r

2

respectively.

A.T.Q

Surface area of B = 400% of surface area of A

7


=4×surface area of A

� 2

24 r = 2

116 r

�1

2

r

r = 1 1

4 2� or 1 : 2

Now, Volume of B × 100

100

k�= Volume of A

�3

2

4

3r

100

100

k��

= 4

33

1r

�

3

1

2

r

r

� � � �

= 100

100

k�

�1

8=

100

100

k�

� k = 700

8= 87.5

61.(A) Ratio of the areas of the circum circle and

incircle of a square =

2

2

(diagonal)

(side)

=

2

2

(side 2 )

(side)

�=

2

1or 2:1

62.(C) Perimeter of circular sheet = 2 r = 20 cm

� The perimeter of base of conical surface

= 20 × 100 40

100

�= 12

� Radius of base of conical surface = 12

2

= 6cm

& the height of conical surface = 2 210 6�= 8cm

So, The required ratio = 6 : 8 or 3 : 4

63.(D) A.T.Q.,

Area of circular shell

= Total surface area of cylinder

� (122 � 82) = 2 R1 (R

1+h)

� 80 = 2 ( 2

1R +R1h)

� 40 = 2

1R + R1h

� R1h = 40 � 2

1R

� h =

2

1

1

40 R

R

�

64.(A) Let, h m = height of platform

A.T.Q.,

400h = 22

7×(3.5)2 ×16

� h = 222 (3.5) 16

7 400

� �

�

= 1.54 m65.(C) Interior angle of the Ist polygon = 120º

Let, n1 = number of sides in Ist polygon

Then 1

1

2n

n

�×180º = 120º

� 3n1� 6 = 2n

1

� n1 = 6

� Sides of the IInd polygon = 6 × 2 = 12

� Interior angle of the IInd polygon

= 12 2

12

�×180º

= 150º

66. (A) x = 5 1 5 1

5 1 5 1

� ��

� �=

5 1

2

�

� x2 � x � 1 = 5 1 2 5 5 1

4 2

� � �� 1

= 6 2 5 2 5 2 4

4

� � � �=0

67.(B) On putting x = 3

2, we get

3 31 1

2 2

3 31 1 1 1

2 2

� ��

� � � �

=

2 3 2 3

2 2 2 3 2 2 2 3

� ��

� � � �

=2 3

2 4 2 3

�

� �+

2 3

2 4 2 3

�

� �

=2 3 2 3

2 1 3 2 3 1

� ��

� � � �

=2 3 2 3

3 3 3 3

� ��

� �

=2 3 3 3 2 3 3 3

3 3 3 3

� � � � �

� �=1

68.(A) Given, y = � 5

� y = x � 5 or, y = - x - 5

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Area of the bounded region

= 1

2× AC × OB =

1

2×10×5 = 25sq units

69.(D) By option (d), taking x = 4

4 9 4 9x x� � �

= 16 9 16 9� � � = 7 5�70.(C) � a3 + b3 + c3 � 3abc

= 1

2 (a + b + c ) 2 2 2[( ) ( ) ( ) ]a b b c c a� � � � �

=1

2 (225 +226 227) [1+1+4]

= 678 × 3 = 230471.(B) We know that, AM > GM

�1

aa

� > 2

Here, 1

² 1² 1

x xx x

� � ��

> 2

� 2 � x2 > 2� x2 < 0

� x = 0 Hence, the given equation has only one solution

72.(D) Given x + a

x=1

� x2 + a = x� x2 � x = � a

Now, 2

3 2 2

1a

xx x a x

x x x x

� ��

��

= 2

a�

73.(A) 28 6 3� = 3 a + b

� (1 � 3 3 ) = 3 a + b

On comparing, we geta = -3, b = 1

� a + b = - 3 + 1 = - 274. (B) 232 � (2 +1) (22 +1) (24 +1)(28 +1)(216 +1)

= 232 � (2 � 1) (2+1) (22 +1)(24 +1)(28 +1) (216 +1)= 232 � (22 � 1) (22 +1) (24 +1)(28 +1) (216 +1)= 232 � (24 � 1) (24 +1) (28 +1)(216 +1)= 232 � (28 � 1) (28 +1) (216 +1)= 232 � (216 � 1) (216 +1)= 232 � (232 � 1) =1

75.(B) x + 809436× 809438= A square number

� x + (809437 � 1) (809437 + 1)= square number

� x + (809437)2 - 1 = A square numberIt is possible, when x = 1

76.(B) From figure,Sum of angles of quadrilateral ADOE

= 360º

� 80º + 90º + 90º + �DOE = 360º

� �DOE = 360º � (80º + 90º + 90º ) = 100º

� �BOC =�DOE = 100º

� �BOC =�DOE = 100º77.(C) From figure,

In �ONA, ON2 = r2 � a2 and

In �ONB, NP2 = OP2 � ON2

= c2 � (r2 � a2)=(c2 + a2) � r2

Now, In �OMD, OM2 = NP2 = r2 � b2

� c2 + a2 � r2 = r2 � b2

� 2r2 = a2 + b2 +c2

� r2 =2 2 2

2

a b c� �

� r = 2 2 2

2

a b c� �

78.(D) From figure

In �OCM, CM2 = 72 � 32 = 40

� CM = 2 10 cm

In � AOM

AM = 2 213 3� = 4 10

Now, AC = AM � CM

= 2 10 cm

79.(C) From figure,

PT2 = PA . PB

� (2x)2 = x (18+x )

� 4x2 = x (18+x )

� 4x = 18 +x

� x = 6

� PT = 2 × 6 = 12 units

�

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80.(C) Given, DE �BC

� � ADE � � ABC

then AB AC

AD AE�

�12

6

x=

16

2x� x2 =

16 6

12 2

�

�= 4

� x = 2cm

81.(A) From figure,

� AB + CD = BC + AD

� 6 + 5 = 7 + AD

� AD = 11 - 7

= 4cm

82.(C) The angle substended by an arc at the cir-

cumference of circle is half of the angle

substended by the same arc at the centre,

� PBO = 1

2× � POA

= 1

2× 120º = 60º

83.(B) From figure

�BIC = 2 × �BAC = 120ºand IB = IC

� � IBD =� ICD = 180º 120º

2

�= 30º

Now, �BID = 90º � 30º = 60º

84.(C)

In �OPM,

OM = 2 25 3� = 4cm

� MN = ON � OM = 1cm

In � POT,

PT2 = OT2 � OP2

= (5+x)2 � 52

= x2 + 10x (i)

In � PMT,

PT2 = PM2 + TM2

= 32 + (1 +x)2

= x2 + 2x + 10 (ii)

From Eqs. (i) and (ii), we get

x2 + 2x + 10 = x2 + 10x

� x =10

8=

5

4

From Eq. (i)

PT = 2 10x x�

=

25 5

104 4

� � � ��

� � � � = 3.75cm

85.(C) From figure,

�AOB + �COD = 2 �ACB +2 �DBC� 15º + �COD = 2 (�ACB +�DBC)� 15º + �COD = 2 �APB� �COD = 2 ×30º – 15º

�COD = 60º � 15º = 45º

� tan2 �APB + cot2 �COD

= tan2 30º + cot2 45º = 1

13� =

4

3

86.(D)

In � ABC,

(x + 2)2 + x2 = (2 5 )2

� x2 + 4 + 4x +x2 = 20

Centres at:

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

10


� 2x2 + 4x - 16 = 0

� x2 + 2x - 8 = 0

� (x - 2) (x + 4) = 0

� x = 2

� AB = 4, BC = 2

� cos2 A - cos2C =

24

2 5

� � � �

-

22

2 5

� � � �

=3

5

87.(B)

Let AB = h km

� In �OAB, tan 45º = AB

OB

OB = h km

In �OLM, OM = 2 cos 30º = 3 km

� LN = BM = (h- 3 )km

In �OLM, sin 30º = LM

OL

LM = 2 sin30º = 1 km

BN = LM = 1km

In � ALN, tan 60º =AN

LN

or, 3 = AB BN

LN

�

or, 3 = 1

3

h

h

�

�

or, 3h - 3 = h -1

or, h = 2

3 1�×

3 1

3 1

�

�

� h = ( 3 +1)km

88.(A) f ( � ) = sin8 �+cos14 �

= sin8 �+ (1- sin2 � )7

� f (90º) = 1+ (1-1)7

[�Maximum value of sin2 �= 1]

= 1+ 0 =1

89.(D) l cos2 �+ m sin2 �

=2 2

2

cos (cos 1)

cos 1

ec

ec

� � �

� �=

2 2

2

cos (1 sin ) sin².

1 sin sin²

� � � �

� � �

=2 2

2

cos (1 sin )

cos

� � �

�= 1+ sin2 �

= cos2 �+ sin2 � +sin2 �

= cos2 � +2sin2 �

� (l -1) cos2 � = (2 - m)sin2 �

� tan2 �= 1

2

l

m

�

�

� tan �= 1

2

l

m

�

�

90.(A)

sin (10º 6'32'')= a

sin (90º - 79º 53' 28'') = a

� cos 79º 53' 28'' = a

� cos (79º 53' 28'') + tan (10º 6'32'')

= a + 21

a

a�=

2

2

(1 1 )

1

a a

a

� �

�

91.(C) sin �+ cossec �= 2

� sin �+1

sin�= 2

� sin2 �+1 = 2 sin �

� (sin² � -2sin �+1) = 0

� (sin � � 1)2 = 0

� sin �= 1 = sin90º

� � = 90º

So, sin7 �+ cosec7 �= 1 + 1 = 2

92.(D) tan 8

.tan

12

.tan

3

8

.tan

5

12

sin²

6

�

= tan8

.tan

12

.cot

3

2 8

� ��

� �.cot

5

2 12

� ��

� �

1

4�

= tan8

.tan

12

.cot

8

cot

12

1

4� = 1-

1

4=

3

4

93.(A) x sin3 a + y cos3 � = sin � cos �

� x sin � .sin2 � + y cos � .cos2 � =sin � cos �

� x sin � . sin2 � +x sin � .cos2�

= sin � cos � [�y cos � = x sin � ]

� x sin � (sin2 � +cos2 � ) = sin � cos �

� x = cos �also, y cos � = cos � sin �

� y = sin �

� x2 + y = sin2 � +cos2 � =1

94.(C) Given, sin � =4 3

9

x �

Taking �= 90º,1 = 4 3

9

x �

� 4x � 3 = 9 � 4x = 12

x = 3

95.(C) x = sin � + cos �

y = sec � + cosec �

Centres at:

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

11


= 1 1

sin cos�

� �

= sin cos

sin cos

��

� �

= sin cos

x

� �(ii)

� (sin �+cos � )2 = 1 +2 sin � cos �

x2 =1+2sin � cos �

sin � cos � =2 1

2

x �[From Eq. (ii)]

y = 2

2

1

x

x �

96.(D) Number of workers in scale V = 12% of

1500 = 180

Number of working male in scale V =12% of

800 = 96

Number of working female in scale V

= 180 � 96 = 84

97.(B) In scale VII.

Total number of workers = 8% of 1500 = 120

Number of male workers = 10% of 800 = 80

� Number of female workers = 120 - 80 = 40

� Required ratio = 80: 40 = 2:1

98.(A)

Number of females in scale I = 330-192=138

Number of females in scale VI = 210- 72 = 138

� Number of females are same in scale I and VI.

99.(d) Average of working females in all scales

=138 81 157 62 84 138 40

7

� � � � � �

=700

7=100

� Required no. of scales = 4 (II, IV, V, VII)

100.(D) Number of females in scale VII = 8% of

1600 - 10% of 800 = 128 - 80 = 48

� Required percentage =8

10040

� %

= + 20% (increase)

Corrections of SSC Mains Mock Test -6 Corrections of SSC Mains Mock Test -6 Corrections of SSC Mains Mock Test -6 Corrections of SSC Mains Mock Test -6

44.(D) part of tank filled till 11am

= 1 1

3 215 12

� � � ��

� � � �=

1 1

5 6� =

11

30part

Now, at 11 am, third pipe is also opened

Now, rate of filling or emptying the tank per

hour = 1 1

15 12�

1

4� =

4 5 15

60

� �=

6

60� =

1

10�

� Emptying of 1

10th part per hour.

So, Time required to empty the 11

30th part

= 11/30

1/30hr. =

23

3hour. = 3 hour 40 minutes

So, required point of time = 11am + 3 hour 40min.

= 2:40 pm

53.(*) � x = 1

32 1� � x = 2 1�

So, x33

1

x� = 2 1� �

1

2 1�=

2

2 1 1

2 1

� �

�

= 2 1 2 2 1

2 1

� � �

�=

2 2 2

2 1

�

�=

2 2 1

2 1

�

� = 2

Centres at:

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

12


1. B

2. B

3. D

4. B

5. D

6. A

7. B

8. D

9. C

10. B

11. D

12. D

13. C

14. B

15. B

16. C

17. A

18. C

19. D

20. B

21. D

22. A

23. D

24. C

25. D

26. D

27. D

28. B

29. D

30. D

31. C

32. A

33. C

34. C

35. D

36. C

37. C

38. D

39. B

40. A

41. A

42. D

43. A

44. A

45. A

46. C

47. C

48. A

49. D

50. A

51. B

52. A

53. D

54. B

55. C

56. B

57. A

58. C

59. A

60. D

61. A

62. C

63. D

64. A

65. C

66. A

67. B

68. A

69. D

70. C

71. B

72. D

73. A

74. B

75. B

76. B

77. C

78. D

79. C

80. C

81. A

82. C

83. B

84. C

85. C

86. D

87. B

88. A

89. D

90. A

91. C

92. D

93. A

94. C

95. C

96. D

97. B

98. A

99. D

100. D

SSC MAINS (MATHS) MOCK TEST-7 (ANSWER KEY)SSC MAINS (MATHS) MOCK TEST-7 (ANSWER KEY)SSC MAINS (MATHS) MOCK TEST-7 (ANSWER KEY)SSC MAINS (MATHS) MOCK TEST-7 (ANSWER KEY)

ssc mains (maths) mock test-7 (solution)

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