ssc mains (maths) mock test-9 (solution)

Centres at:

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

1

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

SSC MAINS (MATH) MOCK TEST - 9 (SOLUTION)

1.(D) Let x = 8 2 8 2 8......... � � � , then

x = 8 2� x

squaring both sides, we get,

x2 = 8 + 2x

� x2 – 2x – 8 = 0

� (x – 4) (x + 2) = 0

� x = 4

2.(A)

2 2

2 2

3.07 .0193

.307 .00193

�

�

2 2

2 2

3.07 .0193

3.07 .0193

10 10

��

2 2

2 2

3.07 .0193100

3.07 .0193

��

= 100

3.(B) Men to be arranged = (6000 – 71) = 5929Number of men arranged in each row

= 2 5929 77�

4.(B) 7 5, 5 3, 9 7, 11 9� � � �

On rationalizing each term

=2 2 2 2

, , ,7 5 5 3 9 7 11 9� � � �

Smallest denominator = 5 3�

So largest value = 5 3�5.(C) Let the price of 1 book = ` x

and price of 1 pencil = ` y

According to the question:

4x + 3y = 8x + y

2y = 4x

y = 2x

Price of 4 books and 3 pencils

= 4x + 3y

= 4x + 6x = 10x

and if 6 books and 2 pencils = 6x + 2y

= 6x + 4x = 10x

Both are equal

6.(B)

So 47 is wrong

7.(C) 1000 is not a perfect square so we need

to make a perfect square.

we need 24 more plants

8.(A) 2557 –1 � 71–1 = 6 25 = 4×6+1��

9.(A) Let the five consecutive integers be x,

x + 1, x + 2, x + 3, x + 4, then

x+x+1+x+2+x+3+x+4 = a

� 5x + 10 = a..................(i)

Also, 5x + 35 = b..................(ii)

b – a 5 35 – 5 –10 1

100 100 4

��

x x

10.(A) I no. × II no. = L.C.M. × H.C.F.

(x2 +2x –3) × P = (x3–7x+6) × (x +3)

P =

3

2

( –7 + 6) ( +3)

2 – 3�

x x x

x x

P =

2

2

( +3)( –3 2)( +3)

2 – 3�

x x x + x

x x

=

2( +3) ( –2) ( –1)

( +3) ( –1)

x x x

x x

= (x +3) (x –2)

= x2 +x –6

11. (A)91 2 34 1 4 4 4� � �

= 914 1 4 16 64� � �

= 914 85

= 5 × 17 × 491

Divisible by 17.

12. (B) 1 + 2 + 3 +..........99 + 100 + 99+...3 + 2 + 1

(1 + 2 + 3+ ........100) + (99 + 98+....3+2+1)

By using the formula of sum of first 'n'

Natural Number = 1

2

�n n

We get, 100 100 1 99 99 1

2 2

� ��

= 5050 + 4950 = 10000

13.(A) Given

x = 0.037

Centres at:

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

2


x 1\3 = 1/3

0.037

= (37/999)1/3

= (1/27)1/3

= 1/3

= 0.3

14.(B) Average age of the couple is 25 years

The sum = 2 × 25

= 50 years

After 3 years, sum = 50 + 2 × 3

= 56 years

Age of baby = 2 years

Then average (X) = � x

n

= 56 2 1

58 /3 193 3

�� years

15.(C) Let minors be x

Consumption by adults = 8 × 15 = 120

Total Consumption = (x + 8) × 10.8Average consumption by minors

= 8 10.8 120

6 7� �

� � �x

xx

16. (C) Sum of 8 numbers = 20 × 8 = 160

1 115 2 21 3 4 7 160

2 3

� � � ��

x x x

31+ 64 + 3x + 11 = 160

3x = 160 – 106 � x = 54/3 � x = 18

8th Number = x + 7

= 18 + 7 = 25

17.(C) Total brass = 33 kg

copper =33 7

21 kg11

��

33 4zinc = 12 kg

11

��

copper 21 21 7 : 6

zinc 12 6 18� � �

�18.(A)

��1 1

I : II : 10 :18 5 : 918 10

� � �

Clearly, for 5l of I-Mixture required amount

of II mixture = 9l.

�� For 3 l of I–mixture required amount of

II mixture = 9 27 2

3 55 5 5� � � l

19.(A) Students Failed in Hindi = 100% – 80% = 20%

Students Failed in mathematics

= 100%–75% = 25%

Students Failed in both subject = 18%

Students Passed in both subject

= 100–(25+20–18)

= 73%

Let total students be x

73438

100

��

x� x = 600

Total students is 60020.(D) Let C.P = x

100 1001

100 100

� ��

r rx

100 100

100 r 100 r

��

� �x

= 2 2

10000

100 r�

x = 2

10000

10000 r�21. (C) 100 × 68/100 = 68 Kg.

Dry Fruits = 32 Kg + 20 kg = 52 kg22.(D) Let, the total number of voters be x

�� Number of votes cast in the election = 92

100x

Number of votes obtained by winner = 48

100x

�� Number of votes obtained by the defeated

Candidate = (92 – 48)

100x

= 44

100x

From question,

48 441100

100 100�

x x–

� 4x = 110000

�� x = 27500

�� Total no. of voters = 27,500

23.(C) According to the question

Let the numbers are a and b

(a – b): (a + b) : ab = 1 :7 : 24

Numbers are a = 8 b = 6

so product = 8 × 6 =48

24.(A) Let the Length of candles be L

Centres at:

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

3


The rate of burn of first candle = L

4 per hour

The rate of burn of second candle = L

3per hour

Let after x hour the ratio be 2:1

L LL – 2 L –

4 3

� ��

x x

L 1 – 2 L 1 –4 3

� � � ��

x x

4 – 3 –2

4 3

� ��

x x

x = 2

25

hours = 2 hours 24 min.

25. (A) Cost of raw material = 4x

Cost of labour = 3x

Cost of miscellaneous = 2x

Total cost = 4x + 3x +2x

= 9x

Amount = 4 110 3 108 2 95

100 100 100

� � ��

x x x

= 9.54x

Percentage rise = 9.54 – 9100

9�

x x

x = 6%

26.(A) Let the number be x

(3–x ) (7–x ) = (5–x ) (6–x )

x = 9

27. (B)2 1 2 5 7 3

a : b : ,b : c : ,d : c :9 3 7 14 10 5

� � �

a : b = 2 : 3 b : c = 4 : 5

d : c = 7 : 6 c : d = 6 : 7

then,

a : b : c : d = 48 : 72 : 90 : 105

= 16 : 24 : 30 : 35

28.(D) Given,

Total earning of

A+B+C = 76000.................(i)

percentage of their saving are

30%, 25% & 20% respectively

Let, savings of A, B & C

be 4x, 5x & 6x respectively Now, 30% of A = 4x

or,A

30 4100

� � x

or,40

A3

� x ..................(i)

also, 25% of B = 5x

or,B

25 5100

� � x

or, B = 20x.......................(iii) also, 20% of C = 6x

or,C

20 6100

� � x

or, C = 30x...........................(iv)

on using (ii), (iii) & (iv) in (i), we get

40 +20 30 76,000

3

xx + x =

or, x = 1200

��40 40

A = 1200 160003 3

� � �x

B = 20x = 20 × 1200 = 24000

C = 30x = 30×1200 = 36000

�� (A + B) – C = (40000 – 36000) = ` 400029.(C) Let money be P.

P 12 4 P 15 5– 1890

100 100

� � � ��

27P 1890 1001890 P P 7000

100 27

�� `

30.(A) Let initial amount = ` x

7 2 2 10 2 4 12 21430

3 100 5 100 15 100

� � � � � ��

�x x x

14 4 81430

300 50 125� � �

x x x

7 2 81430

150 25 125

x x x� � �

35 60 481430

750

� ��

x x x

143x = 1430 × 750

1430 750

143

��x = ` 7500

31.(C) CP of Horse = x

C.P S.P

100 25 100 32 115 60

100 100 100

� �� x x

x = 375

32.(C) Let cost of 100 Articles is ` 100

( 1Article 1)� �`

Centres at:

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

4


25%Profit S.Pif 100 articles

125��

`

� Diff = 5 unit = 100

1 unit = 20

33.(B) Let Marked Price = x

(100 –10)108

100�

x� x = ` 120

34.(A) Let C. P. = ` x

S.P. =108

100

�x

Again C. P. = 80

100

�x

S.P. = 80 140

100 100�

x =

112

100

x

��112 108

– 640100 100

�x x

x = ` 16000

35.(A) According to the Question

36.(D) According to the question

He should purchase = 400

320 50%�

= 5 shirts

37.(A) S. P. = 720 115

100

� = ` 828

Marked price = ` x

we get,

90

100

�x = ` 828

x = ` 920

38.(A) Let work done by a man in a day be x

and work done by a woman be y

�� From question,

4x + 6y = 1/8...................(i)

3x + 7y = 1/10...................(ii)

on solving (i) & (ii), we get

11

400�x

1

400�y

required ratio = 11 1

400 400� �

x

y = 11 : 1

39.(B) Given,

1/B = 2/A

or, A = 2B...........(i)

also, given 1/C = 3/A

or, A = 3C .....................(ii)

also, given 1 1 1 1

A B C 2� � � ................(iii)

on using (i) in (iii), we get

1 1 3 1

2B B 3C 2� � �

or,1 1 3 1

2B B 2B 2� � � [using (i) & (ii)]

or,1 2 3 1

2B 2

� ��

or, 2B = 12

or, B = 6 days

40.(A)2 21 1

1 2

M DM D

W W

��

2M 80100 16

1/7 6/7

��

2 2

100 16 6M M 120

80

� ��

Required labourers = 120 – 100 = 20

41.(A) Given,

A + B +C = 14,400............................(i)

Let savings of A, B & C are 8x, 9x & 20x

respectively.

Also given percentage of expenditure of A,

B & C are 80%, 85% & 75% respectively.

�� Percentage of savings of A, B & C be 20%,

15% & 25% respectively.

Now, 20% of A = 8x

or,20 A

8100

�� x

or , A = 40x..........................(ii)

Again, 15% of B = 9x

or,15 B

9100

�� x

or, B = 60x...........................(iii)

again, 25% of C = 20x

or,25 C

20100

�� x

or, C = 80x .............................(iv)

on using (ii), (iii) & (iv) in (i), we get

40x + 60x + 80x = 14,400

or, x = 80

�� A = 40x = 40 × 80 = 3,200

B = 60x = 60 × 80 = 4,800

C = 80x = 80 × 80 = 6,400

Centres at:

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

5


42.(A) Speed = 15 km per hour

= 5 25

15 m /s.18 6

� �

Water flow out in one second

= .2×.15×25/6 m3

Volume of tank = 150× 100 × 3 m3

Time taken = 150 100 3 6

.2 .15 25

� � �

� �

= 100 hours.

43.(D) Given,

1/A = 1/8.....................(i)

also, given 1/A – 1/B = 1/20

or, 1/B = 1/A – 1/20

or, 1/B = 1/8 – 1/20

or, 1/B = 3/40

�� B takes 40/3 min. to empty the tank

also, given rate of water flowing out = 6k l

�� Capacity of tank = 40

3 × 6k l = 80 k l

44.(C) Speed = 350 60

21 1000

�� km / hr

Total time taken = 84

13 621

� �

4 + 78 min = 5 hours + 18 min.

45.(A) Let total coaches be N

decrease in the speed = x

�� N�x

K N�x [K = constant]

4 K 4�K = 2

�� 24 K N�

24 2 N�

N = 24/2

N = 144 coaches

�� Number of coaches that can be exactly

pulled by the engine = 144 – 1 = 143 coaches

46.(A) Time taken by C = 114 min.

Time taken by B = 114/3 = 38 min.

Time taken by A = 38/2 = 19 min.

47.(A) Let correct time = t1�� From question,

1 1

11 540 t 50 t

60 60

� � � ��

1

19 19 t 60

60 60h� � � � = 19 mins

48.(C) Length of train = 120m.

Speed of train Ist = 120

12 m/s10

�

Speed of train IInd = 120

8 m/s15

�

time taken = 120 120

12 8

�

� =

24012

20� seconds

49.(D) Let the speed of the man = x km and the

speed of the current = y km/h

According to the question

D15�

x – y .......(i) (Where D is distance)

D10�

x + y........(ii)

Divide (i) by (ii)

�15 3

10 2� �

x + y

x – y

� 2x + 2y = 3x – 3y

� x = 5y

�5

5 :11

� �x

y

50.(A)1 1 1

3 3 3� �x y z

zyx �� 33131 )(

zyxyxyx �� )(3)()( 31313131331331

zzxyyx �� 3131)(3

31)(3 xyzzyx ��

xyzzyx 27)( 3 ��

027)( 3 �� xyzzyx

51.(C) Given,

or, x4 – 3x3 – 2x2 + 3x +1 = 0

or, x4 – 2x2 +1 – 3x3 + 3x = 0

or, (x2 –1)2 – 3x (x2 – 1) = 0

or, (x2 –1) (x2 – 1 – 3x) = 0

taking, x2 – 3x – 1 = 0

x = 2–(–3) (–3) – 4(1) (–1)

2 1

� �

�

or, x = 3 9 4

2

� �

or, x = 3 13

2

�

52.(D) (1+m2) x2 + 2mcx + c2 – a2 = 0

B = 2mc

A = (1+m2)

C = c2 – a2

Centres at:

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

6


�� roots are equal �� D = 0

B2–4AC = 0

(2mc)2 – 4 (1 + m2) (c2 – a2) = 0

4m2c2 – 4c2 + 4a2 – 4m2c2 + 4m2a2 = 0

–c2 + a2 + a2 m2 = 0

c2 = a2 (1 + m2)

53.(D) Given,

l x2 + nx + n = 0.................(i)

& / /� �� p q .................(ii)

Equation (i) � – /�� n l .........(iii)

& /�� n l .................(iv)

Equation (ii) � / /� �� p q .........(v)

& / /� �� q p .................(vi)

�� / / /� �p q q p n l

= �� (using (v), (vi) & (iv))

= 1

� ��

� �

= ( ) ( )

.

��

� �

= .� �

–n / l + n / l

= 0/ .� � = 0

54.(B) 3x2 + 2x +1 = 0

�� – 2/3

�� 1/3

Product of roots = 1 – 1 –

31 1

� ��

��

sum of roots = 1 – 1 –

21 1

� ��

��

Required equation is

x2 – (sum of the roots) x + product of roots = 0

x2 – 2x + 3 = 0

55.(B) 4x2 + 4x + 9

(2x)2 + 2.2x +1+ 8 make a perfect

square

(2x + 1)2 + 8

for minimum (2x +1)2 = 0 �� 0 + 8 = 8

We have minimun value of 4x2 + 4x + 9 is 8

56.(B) Given, x = 1 2 3� � ..........(i)

��1 1

–1 1 2 3 –1�

� �x

= 1

3 2�

= 1 3 – 2

3 2 3 – 2�

�

= 3 – 2

3 – 2

13 – 2

–1�

x

(i) + (ii) �1

1 2 3 3 – 21

� � � � ��

xx

= 1 2 3�57.(A) x 2+ y 2+ z 2+2 = 2(y – x )

(x2+2x+1)+ (y2–2y+1)+z2 =0

(x+1)2 + (y –1)2 +z2 = 0

x = –1, y = 1, z = 0

�� x 3+y 3+z 3 = (–1)3 + (1)3 + 03

= – 1+1+0 = 0

58.(A)1 1 – 1 1 –

1 – 1 – 1 1 –

� � � ��

� � �

x x x x

x x x x

21 1 – 1 1/2 3

3 /2

� ��

x

x

59.(C) 5 2 6� �x

5 2 6� �x

= 2 2

3 2 2. 3. 2� �

= 2

3 2�

x = 3 2�

1

x= 3 – 2

13 2 3 – 2� � � �x

x

= 2 3

60.(B)

According to the questionLet the height of the telegraph pole = h

so, in ABC�

Centres at:

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

7


tan30º20

�h

h = 20 20

3 AB33

� �

and cos 30º = 20

AC

AC = 20 2 40

3 3

��

so height of the pole= AB + AC

= 20 40 60

3 3 3� �

=60 3

20 33

�

61.(B) 14 20sin cos� �� xEven power means no negative value andmax value will be 1 and never be zero.

0 1� x

62.(A)

Here cos 10º > cos 80ºso, cos 10º – cos 80º > 0 or positive

63.(D) 1 + s inx + s in 2x .......� = 4 2 3�

�1 1

4 2 31 sin 1 sin

� � �� x x

�1

4 2 31 sin

� �� x

�1 4 2 3 3

1 sin 14 24 2 3

��

�x

�3 2

sin sin sin2 3 3

! !� � �x

so 2

or2 3

! !� �x

64.(A)

Let height of the tower = AB

So, In ABC�

ABtan45º

AC�

so, AB = 60 m. [�tan 45 = 1]

Also, In ADB�

60 1 60tan30º

60 603� � �

� �x x

� 60( 3 1) 60 (1.73 1)� � � � �x

� x = 60 × 0.73 = 43.8 m

Required Speed = 43.8 18

31.55 5

� � km/hr

65.(A)

In AFC�

tan30º 3 h� � �h

xx

.....(i)

In AEC�

tan60º 3160

� ��

h

x

� 3 (160- ) = x h

� 3 (160 – 3 )h = h From (i)

� 4 160 3�h � h = 160 3

3 ft = 340 ft

66.(A)tan3 sin3 .cos

tan cos 3 .sin�

x x x

x x x

=

3

3

(3sin – 4sin )cos

(4cos – 3cos )sin

x x x

x x x

or,

2

2

tan3 (3 – 4 tan )

tan (4cos 3)�

x x

x x –...........(i)

tan3 3 – 4 0 3Max 3

tan 4 – 3 1

��

x

x

(when x =0 º)

tan3 3 – 4 –1 1Min

tan –3 –3 3� � �

x

x

��tan3

tan

x

x vary from

1

3 to 3.

{clearly, tan3x /tan x never lies between –1/3and 0}

67.(B) tan 20º tan 80º cot 50º

= tan 20º tan 80º tan 40º

[�� cot (90º–40º) = tan 40º]

Centres at:

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

8


= tan [3×20º]

[�� tan tan (60º+) tan (60º–) = tan 3�]

= tan 60º

= 3

68.(C) sin 18º + cos 36º

= 5 –1 5 1

4 4

��

= 5 – 1 5 1 2 5 5

4 4 2

� ��

69.(B) We have,

2 4 6cos cos cos

7 7 7

! ! !� �

1 2 4 62sin cos 2sin cos 2sin cos

7 7 7 7 7 72sin

7

! ! ! ! ! !� �

!

��

1 3 5 3 5sin –sin sin –sin sin –sin

7 7 7 7 72sin

7

! ! ! ! !� � � !

!

��

1–sin sin

72sin

7

�!� � � !

!� � �

1 1– sin 0 –

7 22sin

7

�!� � � �

! � � �

70.(C)

BL is bisecter of ABC"

so, MBL LBC" � " � x (let)

ML BC�

LBC MLB" � " � x

MLB BML" � "Can't be 90º then triangle does not exist

71.(C)

In ADC�

DAC 180 – 60 – 30 90º" � �

In DAC�

ACtan60º 3

AD� � �

a

x

3�

ax

72.(C) It is true that congruent triangles will be

similar but opposite is not true. Also III

will be true.

73.(C)

Given OA = 3 cm

and OM AC#

Let AOM" ��

In AOM�

AMsin

OA��

AC 3 1sin

20A 2 3 2��

�

sin sin30º��

30º��

74.(C)

We know that tangents from an external

point on the circumference of the circle

will be equal.

AP = AS

BP = BQ

CR = CQ

DR = DS

� AP+BP+CR+DR = AS+BQ+CQ+DS

� AB+CD = AD + BC

75. (A) According to the questions sides are 5xand 12x

=1

5 12 = 2702� x × x

x2 = 9x = 3

= Hypotenuse = 13x = 13 × 3 = 39 cms.

Centres at:

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

9


76.(A)

Radius of circle APBQ = 5 4 9

2 2

��

Area of circle APBQ = �R2 =

29 81

sq. cm.2 4

!! � ��

77.(C) Area of square > Area of rectangle

78.(A)

ABC 50º" �DAB ABC" � " [alternate angle]

�DAB = 50ºIn �ADB

A D B 180º" � " � " �50º ADB 65º 180º�" � �

ADB 180º –115º" �ADB 65º" �

79.(A)

� PAQ 68º" �

�� PAO 68 /2 34º" � �

In APD

APD PAD ADP 180º" � " � " �

APD 34º 90º 180º" � � �

APD 56º" �

APD APQ 56º" � " �

APQ 56º" �

80.(A)

Area of the rhombus = 21

1d d

2� �

= 1

55 48 13202� � � cm2

Area of the rhombus = base × height= DC×AE

� so DC × AE = 1320

2 255 48

P 13202 2

� � � ��

5329P 1320

4� �

1320P 36.16

36.5� �

so, 36. < P < 37 cm

81.(B)

Area of circlular segment = 23

360º

�� !r

= 260

3 (4)360º

� �!

= 8! cm2

Area of the 2 23

ABC (8) 16 3 cm4

� � � �

�� Area of the portion included between

them = (16 3 – 8 )! cm2

82.(B) Let the radius of the circle be rand side of the square be a.

so, 2 r 4! � a � given

2

!�

ra

Area of the square =

2r

2

� �!� ��

= 2.46r2......(i)

Area of the circle = 2r!= 3.14r2.......(ii)

so (ii) > (i) option (B) is correct

10


83.(B) GivenAC2 = AB × CB

x2 = 2 (2–x)

� x2 +2x–4 = 0

� x = 2 4 16

–1 52 1

� � ��

�

� BC = 2– ( –1 5� )

= 3 – 5 [leaving (3 5)� because

it is greater than 2]

84.(C) By the property of triangle

85.(C) According to the question

86.(D)

Let height of the pillar = h metres.

In DAB, tan 30º = /3

50

h

1

3 503�

�

h� h = 50 3 metres

87.(D) Given,

l + b + h = 19............(i)

& diagonal of cuboid = 11

or, 2 2 2 11� � �l b h

or, l 2+ b 2+ h 2 = 121........(ii)

we know that surface area of cuboid

= 2 (l b + b h + h l )

Now, (l + b + h )2 = l 2+ b 2+ h 2 + 2(l b + b h + h l )

� 2(l b + b h + h l ) = (l+b+h)2 – (l 2+ b 2+ h 2)

= (19)2 – (11)2

= 361 – 121

= 240 cm2

�� surface area of cuboid = 240 cm2

88.(C) Let height be h m.

Volume of remaining field = Volume of

soil

[18×15–7.5×6]. h = 7 . 5×6×8

7.5 6 8 .16m 16 cm

270 – 45

� �� h h

89.(C)

Let slant height be h cm

Pyramid height = h/2

In EDF22 2

2 22 4 4

2 4 3 33

� ��

h hh h h

height = 4 1 2

3 2 3� �

Volume of pyramid = 1

3 (Area of base) × height

= 21 3 2

43 4 3

� ��

= 38

3 cm9

90.(C)

In ABC

2 2AC 10.5 14� �

AC = 17.5m

l = 17.5 m

Total surface area = r l + 2 r h

= 22 22

14 17.5 2 14 37 7� � � � � �

= 1034 m2

The cost of painting= 1034 × 2

= ` 2068

91.(A)

Let length be x

breadth be y

(x+14) ( y–6) = xy

xy – 6x +14y – 84 = xy

14y – 6x = 84 ................(i)

(x–14) (y+10) = xy

Centres at:

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

11


xy + 10x – 14y – 140 = xy

10x – 14y = 140...............(ii)

solving (i) & (ii) we get,

x = 56, y = 30

92.(D) From graph, in standard (viii) the failure

of girls is minimum.

93. (*) Note :Read more in place of less in question

then, the solution is

Average result of the girls =

80 80 40 90 70 70

6

� � � � �=

430

6 = 71.67

clearly from graph in standard (ix) the

results of boys is more than the average

result of the girls.

94. (A) From graph,

The ratio of results of girls and boys in the

standard (vi)

= 80:70

again,

the ratio of results of girls and boys in the

standards (ix)

= 70:80

clearly, in pair (vi), & (ix) of standard the

results of girls and boys are in inverse

proportion.

95.(*) Note : Read less in place of more in question

then, the solution is

Average result of the boys

= 60 70 60 60 80 60

6

� � � � �= 390/6 = 65

Clearly from graph, in standard (vii) the

results of girls is less than the average

result of the boys.

96.(A) No. of students come by bus = 7200 50º

1000360º

��

97.(D) Total angle = 20º + 50º = 70º

required no. = 7200 70º

1400360º

��

98.(B) The ratio of number of students who come byfoot to the students who used two-wheeler

20º 1 1 : 6

120º 6� � �

99.(B) Ratio of students who come by car to that bycycle

= 1200

6 :112200

�

100.(*) Number of students come by (cycle + bus)

= (110º 50º) 160º

7200 7200 3200360º 360º

��

Number of students come by

Two-wheeler = 120º

7200 2400360º

� �

�� Exceed % = 3200 – 2400100

2400�

= 800 100 1

100 33 %2400 3 3

� � �

Centres at:

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

12


SSC MAINS (MATHS) MOCK TEST - 9 (ANSWER SHEET)

1. (D)2. (A)3. (B)4. (B)5. (C)6. (B)7. (C)8. (A)9. (A)10. (A)11. (A)12. (B)13. (A)14. (B)15. (C)16. (C)17. (C)18. (A)19. (A)20. (D)

21. (C)22. (D)23. (C)24. (A)25. (A)26. (A)27. (B)28. (D)29. (C)30. (A)31. (C)32. (C)33. (B)34. (A)35. (A)36. (D)37. (A)38. (A)39. (B)40. (A)

61. (B)62. (A)63. (D)64. (A)65. (A)66. (A)67. (B)68. (C)69. (B)70. (C)71. (C)72. (C)73. (C)74. (C)75. (A)76. (A)77. (C)78. (A)79. (A)80. (A)

81. (B)82. (B)83. (B)84. (C)85. (C)86. (D)87. (D)88. (C)89. (C)90. (C)91. (A)92. (D)93. (*)94. (A)95. (*)96. (A)97. (D)98. (B)99. (B)100. (*)

41. (A)42. (A)43. (D)44. (C)45. (A)46. (A)47. (A)48. (C)49. (D)50. (A)51. (C)52. (D)53. (D)54. (B)55. (B)56. (B)57. (A)58. (A)59. (C)60. (B)

ssc mains (maths) mock test-9 (solution)

Documents