SSC MAINS MATHS - 53 (SOLUTION)...MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA...

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Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD

1.(C) Let Price = ` 100After two successive discounts, price will be

= 100 × 10070

×10083

= ` 58.1

Single Discount (D) = 1001.58100

× 100

= 41.9%Shortcut:–Equivalent single discount (D)

=

100xyyx

= 30 + 17 –30 17

100

= 47 – 5.1= 41.9%

2.(C) Let the original price = ` x per kg

New price = `34x

2520 43x

–

2520x

= 15

3360 2520

x = 15

x = 15840

x = ` 56

New price = 34x

= `34

× 56

= ` 42 per kg.3.(B) Cost of 126 toffees = 64

after giving discount of 25%

Actual cost = 64 × 75

100 = ` 48

126 toffees cost = ` 48

1 toffee costs = `48

126

1170414 toffee cost = 48

126 ×1170414

= ` 445872

SSC MAINS MATHS - 53 (SOLUTION)

4.(B)S.P. C.P. 100

36

= C.P.

252 10036

= C.P.

C.P. = 700

S.P. = 136 700100

= ` 952

5.(C) Total present age = 26 × 7= 182 years

and5 years before total age = 152 – 5 × 7

= 147 yearsNumber of persons = 6

So, average age = 147

6= 24.5 years

6.(A) Total quantity of milk= 4 × 0.8 + 7 × 0.7 + 10 × 0.06= 3.2 + 4.9 + 6.0 = 14.1

Total quantity of water = 21 – 14.1 = 6.9

Milk : water = 9.61.14

= 69141

= 2347

7.(C) Let height of the tower = h cm AD = x cm

In ACD

tan 60º =hx 3 x = h

x = 3

h

In ACD

tan 30º = 30h

x

13 = 30

hx

Put the value of x

3h

+ 30 = 3 h

h + 30 3 = 3h

15 3 cm = h

Height of the tower = 15 3 cm

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8.(D)

C is mid point of AB

Co–ordinates of C = 2

2x

, 3

2y

22

x= 1 and

32

y= + 2

x = 0 and y =1 Coordinates of other end = (0,1)

9. (A) If x2 + 1 = 0 x2 = –1Put value in given polynomial.x4 + x3 + 8x2 + ax + b = 0(–1)2 + (–1)x + 8(–1) + ax + b = 01 – x – 8 + ax + b = 0x(a – 1) + b – 7 = 0

So, a = 1 and b = 7.10. (A) 2(cos2 – sin2 )= 1

cos2 – sin2 = 12

1 – sin2 – sin2 = 12

2sin2 =12

sin2 =14

sin2 = +12

For acute angle

sin = 12

= 30º11. (B) cos2 A – sin2 A = tan2 B

cos2 A – sin2 A+1 = tan2 B + 1 2cos2 A = sec2 B

Asec

22 =

Bcos1

2

2 cos2 B = sec2 A 2 cos2 B –1 = sec2 A –1 cos2 B – sin2 B = tan2 A

12. (A) Let the side of longer square = a unitSide of smaller square = m unit

ATQ,4a – 4m = 100 a – m = 25 .... (i)

Again from question:-a2 – 3m2 = 325 .... (ii)

From (i) and (ii):-m = 30, –5(side cannot be negative)

m = 3013. (B) Let OABCD be the right pyramid on a rect-

angular base ABCD

AB = CD = 32 cmBC = AD = 10cmSince it is a right pyramind,OXZ = 90º andOX = h = 12cmX is the midpoint of the base.Now, in OXZ , OXZ = 90ºOZ = 2 2OX XZ

= 2 212 16

(since XZ = 12 DC = 16)

= 20cmIn OXY , OXY = 90º

OY = 2 2OX XY

= 2 212 5 (XY = 12 BC = 4)

= 13 cmNow , since the base is a rectangle ,OBC = OAD and OAB = ODC

OBC =12 × BC × OZ =

12 × 10× 20 = 100cm2

OAB = 12 AB × OY =

12 × 32 ×13 = 208cm2

Slant surface of pyramid= 2 ( OBC+ OAB)= 2 (100 + 208) = 616 cm2

whole surface of pyramid= Slant surface + Base area= 616 + (32 × 10) = 936 cm2

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14. (C) Let A = (x1, y1) = (t, t – 2)B = (x2, y2) = (t + 2, t + 2)C = (x3, y3) = (t + 3, t)

The vertices of the vertices of the giventriangle.

Area of ABC = 21

|x1(y2 – y3) + x2(y3 – y1)

+ x3(y1 – y2)|

Area of ABC = 21

|{t(t + 2 – t) + (t + 2)

(t – t + 2) + (t + 3)(t – 2 – t – 2)}|

Area of ABC = 21

[{2t + 2t + 4 – 4t – 12)}]

= |–4| = 4 sq. unitsClearly, area of ABC is independent of t.

15. (C) 22

11

nn

nn

ba

ba

is the AM between a & b.

2ba

= 22

11

nn

nn

ba

ba

for n = 0 LHS = RHS

Hence, n = 016. (D) Cannot be determined.17. (D) Cannot be determined.18. (B)

Only agriculture and commercial increasesby more than 50% during the same period.

19. (B) Electricity consumption in traction in

2001-2002 = 10013

× 100 = 13

Electricity consumption in traction in

2014-2015 = 1002

× 1300 = 26

% increase = 131326

× 100 = 100%

20. (B) Industrial + Agriculture will be more willbe more than 50%.

21. (A) 20%.

22. (B)

squaring both side squaring both side

2 2

2 2

a b c a b c

a c b a b c

a c ac b a b ab c

a b c ac a b c ab

2a b ca b c

=

222

abac

= bc

23. (A) (875321 × 875319 + a) is a perfect squareSo, (875320 + 1) (875320 – 1) + a

= (875320)2 – 1 + a

So, a – 1 = 0 a = 1

24. (B) sin3 (1 + cot ) + cos3 (1 + tan )

= sin3cos1sin

+ cos3

sin1cos

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

25. (B) Let speed of boat = x km/hrspeed of current = 4 km/hrA.T.Q.

104x +

104x =

8060

10(x + 4 + x – 4) = 43 (x2 – 16)

15x = x2 – 16 x2 – 15x – 16 = 0

So x = 16, –1So speed of boat in still water = 16 km/hr

26. (C) sin + sin = 2

So, = 90° and = 90°

So, cos 2

= cos 90°

= 027. (A)Rise in water level

= 50

100 ×

2 1000 1000100

100 10

= 10m

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28. (A) A.T.Q. ABC DEF

So, ALDM

= ABDE

= 12

29. (D) p = cos x – sin x

q = 31 sin

1 sinx

x

= 1 + sin2x + sinx

r = 31 cos

1 cosxx

= 1 + cos2x – cosx

p + q + r = cosx – sinx + 1+ sin2x + 1 + cos2x – cosx

p + q + r = 3squaring both side(p + q + r)2 = 9

30. (D) Cost price = `2400

Selling price = 2400 × 120100 = ` 2880

So, marked price = 2880 × 10090 = ` 3200

If discount on selling price = ` 288Required difference= `(320 – 288)

= ` 3231. (C) A.T.Q.

Area of BDG = 16 × Area of ABC

= 24 cm2

32. (C) Ratio of their profit

= 81

: 31

:

31

811

= 81

: 31

: 2413

= 3 : 8 : 13Now, for A & C

A × 4 : C × 8 = 3 : 13A × 4 : 1560 × 8 = 3 : 13

A = 134381560

= ` 720

For B & C,B × 6 : C × 8 = 8 : 13B × 6 : 1560 × 8 = 8 : 13

B = 136881560

= ` 1280

Contribution by A & B together= 720 + 1280= ` 2000

33. (B) Let the present age of my son = x yrs.Then, the present age of mine = 3x yrs.

(3x + 5) = 2 21

(x + 5)

(3x + 5) × 2= 5(x + 5)6x + 10 = 5x + 25

x = 15 yrs.Father's age = 45 years

Son's age = 15 years

34. (B) Average area = n

n2222 ...321

= nnnn

6)12)(1(

= 6)12)(1( nn

=

312

21 nn

35. (A)

104 – 38 = 66 Students speak Marathi104 – 30 = 74 Students speak English104 – 40 = 64 Students speak HindiBy Venn diagram(n) + (44 – n + 50 – n + 40 – n)

+ (n – 18 + n – 20 + n – 26) = 104n = 34

No. of students who speak exactly twolanguage= 40 – n + 44 – n + 50 – n

= 134 – 3n= 134 – 3 × 34= 134 – 102= 32

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36. (C) Let the Ist part be ` x.Then, 2nd part = ` (4350 – x)

SI1 = SI2

10019 x

= 100210)4350( x

9x + 20x = 4350 × 20

x = 29204350

= ` 3000Ist part = ` 30002nd part = ̀ 1350

37. (A) Amount = `400 5 2

400100

= ` 440

Amount returned by Ramu to Arun= 2% of 440

= 2 440

100

= ` 8.838. (C) Radius of the semicircle AQC

= 2 21 14 142

= 7 2 cmArea of the shaded region= Area of semi-circle AQC

+ Area of triangle ABC– Area of Quarter circle APCB

=12 ×

722

× 27 × 27 + 21

× 14 × 14

– 41

× 722

× 14 × 14=154 + 98 – 154= 98 cm2

39. (C) xax + yb

y + zc

z

=axa

xa2.

+ byb

yb2.

+ czc

zc2.

= axczbyax

+ byaxczby

+ czbyaxcz

= czbyaxczbyax

= 140. (A) h = 45 cm, r1 = 28 cm and r2 = 7 cm

Volume of the bunket

= 13 (r1

2 + r1r2+ r22)×h

= 13 ×

227

(282 + 28 × 7 + 72) × 45

= 13 ×

227

(784 + 196 + 49) × 45

= 13 ×

227

×1029 × 45= 48510 cm3

41. (D) Total change = 25 + 10 + 1001025

= 37.5%

So, revenue increased by 37.5%.

42. (C)54

11 13 94

= 18 [On solving]

Time taken in completing 18 th = 10 mins

Time taken in completing 35 =10 × 8 ×

35

= 48 mins

43. (B) Time 1

cross-sectional area of plpe

Time 2

1

4d

Time 21

d

2

1

tt =

21

22

( )( )dd

t2 + t1

21

2

dd

t1 = 40 mins, d1 = d, d2 = 2d

t2 = 40 2

2dd

t2 = 40 21

2

t2 = 10 mins

44. (B) 1hr 40 min 48 sec = 1hr 484060

min

= 1hr 4405

min = 1hr 2045 min

= 2041300

hr = 504300

speed = 42

504300

= 25 kmph

57

× usual speed = 25

Usual speed = 25 7

5

= 35 kmph

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45. (B) Let speed of the stream = x km/hrATQ,

Related Speed = (12 + x) + (15 – x)= 27 km/hr

Required time = 27108

= 4 km/hr

46.(C) Surface area of one cube = 6a2 when 6cubes are fixed on the 6 faces of a cubethen only 5 faces of a cube are visible ofeach cube. Since, central cube iscompletely covered. So, the only 6 cubesare visible each with 5 faces. Hence, thetotal surface area of this solid = 5a2 × 6

Required Ratio = 2

25 6

6a

a

=51

= 5 : 1

47.(B) 2p + 1p = 4

p + 1

2p = 2

3

12

pp

= p3 + 3

18p + 3p

12p

12

pp

8 = p3 + 31

8p + 32 × 2

p3 + 31

8p = 8 – 3

= 548.(B) Let the each side of cube be a

CD = 2 a

CQ = 2a

Let the radius of cone be x and height beh, then r = h 2In APO and CQO ( triangle)

APAO =

CQOQ =

rh

= / 2

( )ah a

/ 2

( )ah a = 2

a = 2 (h – 2)

h = 32a

r = 23a

× 2 and h = 32a

Volume of cone = 13

23 2

2a

× 32a

= 394

a

and volume of cube = a3

Required Ratio = 394

a : a3

= 94 :1

= 2.25 :149.(C) x2 + y2 – 4x – 4y + 8 = 0

x2 – 4x + 4 + y2 – 4y + 4 = 0 (x – 2)2 + (y – 2)2 = 0 x = 2 and y = 2 x – y = 2 – 2

= 0

50. (C) Volume of new ball = 3 44 3 (r1

3 + r23 + r3

3)

= (13 + 23 + 33)= 36 cm3

ATQ,43 r3 = 36

r3 = 27

r = 3 27 = 3 cm.

51. (A) Average of the batsman after 12th innings= 63 – 11 × 2= 41

52. (C) BDE = 115°,ADF = 65° and AED = 75°AED ~ ABC

DE AE ADBC AB AC

2 103 AB AB = 15 cm.

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53. (D) Go through options considering option (D)No. of sides = 5 : 10Ext. angle = 72° : 36°Int. angle = 180° – 72° : 180° – 36°

= 108° : 144°= 3 : 4

54. (A)

Now, in ABD,

tan 60° =hy 3 =

hy ...(i)

In ABC,

tan 30° =h

x y13

...(ii)

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

yx y

3 13

x = 2y y =

x2

Required time = 210

= 5 minutes

55. (C)

Required ratio = 8 : 7

56. (B) Required percentage = 20

100 20 × 100

=20

120 × 100

= 50 216 %3 3

57. (A) Let the CP of 1 orange = ` 1 SP of 10 oranges = `13

gain % = 13 10

10

× 100 = 30%

58. (B) Let the Principal is ` x

A = PTR

1100

8x = x3R

1100

23 =3R1

100

24 = 4R1

100

Required time = 4 years.

59. (C) Ratio = 1 : 1 1:3 6

= 6 : 2 : 1

Sum of the ratios = 6 + 2 + 1 = 9

Middle part = 29

× 78 = 523

= 1713

60. (B) Women = 8343

× 311250 = 161250

Men = 311250 – 161250 = 150000Total number of literature person

= 161250 × 1008

+ 150000 × 10024

= 4890061. (C) Amount Left = ` 1400

Amount before gift = `(1400 + 120)= ̀ 1520

Amount spent on transport= 951520

× 5

= ` 8062. (A) If distance between stations be x km.

Then, speed of train =

6045x

= 34

x km/hr

ATQ,

534

xx

= 6048

1543

xx

= 54

15x = 16x – 60 x = 60 km

63. (C) C.P. of article = ` x

100108x

– 10092x

= 28

10016x

= 28 x = 1610028

= ` 175

64. (A)

OE AB and OF CDAE = EB = 5 cmCF = FD = 12 cmAO = OC = 13 cm

From AOE

OE = 22 513 = 12 cmFrom CDF,

OF = 22 1213 = 5 cm

EF = OE + OF = 17 cm

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65. (D)

CBE = 50ºBAC + BCA = 90ºABE = 90º – 50º = 40º

ABE = ACE = 40º [ angle on same chord]ACE = DEC = 40º [ opposite angle]

66. (C)

AB = AC = tangents from the same point.OB = OC = 3 cmOA = 12 cmABO = 90º

AB = 22 312 = 153

OAB = 21

× 3 × 153 = 2159

cm2

So, area of ABOC = 159 cm2

67. (C) 1º360

n – 2º360

n = 6º

360

)2)(1(12

nnnn

= 6

(n – 1)(n + 2) = 180n2 + n – 182 = 0

(n + 14)(n – 13) = 0n = 13

68. (D) Area of canvas cloth required

= rl = 722

× 2.5 × 314

= 3110

m2

Length of canvas cloth

= 3110

× 25.11

= 388

m

Cost of canvas for tent = 388

× 33 = ` 968

69.(B) l = 15 +15 = 30 cmb = 15, h = 15

Total Surface Area = 2(lb + bh + hl )= 2[30 × 15 + 15 × 15 +

30 × 15]= 2(450 + 225 + 450)= 2250 cm2

70. (D)

Area of equilateral triangle = 234

a

16 3 = 34

a2 a = 8 cm

Height of the equilateral triangle

= 3 82

= 4 3 cm

Radius of the circumcircle

= 23 × 4 3 =

83

cm

Area of the circle = × 83 ×

83

= 643

= 1213 cm2

71. (C) Area of square = 121 m2

Perimeter of square = 1214 m

= 44 mSo, perimeter of circle = 44 m

Area of circle = 4

)2( 2r

= 22474444

= 154 m2

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72. (A)

Area of 4 isosceles triangle = 400 m2

Area of an isoscles triangle = 100 m2

ATQ, 21

× AP × AS = 100

AP = 210 [AP = AS]

Length of greater square = 220 m

Length of new square = 2

220

= 20 mArea of new circle= 400 m2

73. (D) Price of tea in 2014 = 280680

lakh/tonne

Price of tea in 2015 = 200500

lakh/tonne

= 2.5 lakh/tonne

Required percentage =

280600

2806005.2

× 100

= 600100

× 100

= 16 32

%

74. (B) Price of tea in 2013 = 240600

lakh/tonne

= 2.5 lakh/tonne Value of quantity will be 2010 = (250) × 2.5

= 625 lakhs

75. (C) Price of tea in 2010 = 200520

× 100100000

= ` 260076. (C) In 2010 price was maximum

i.e. ` 2600/kg77. (D) L.C.M. of 15, 20, 36 and 48 = 720

Required number = 720 + 3 = 723

78. (A)12a b b a

At a = 64 and b = 289

=1264 289 289 64

=128 17 17 8

=125 3 =

122

79. (A) L.C.M. = (x2 + 6x + 8)(x + 1)= (x + 1)(x + 2)(x + 4)

H.C.F. = (x + 1)First expression= x2 + 3x + 2

= (x + 1)(x + 2)ATQ,(second expression) × (x + 1)(x + 2)

= (x + 1)(x + 1)(x + 2)(x + 4)Second expression = (x + 1)(x + 1)

= x2 + 5x + 480. (B) Rate = 10%

Time = 2 years

C.I. = P 1100

tr

– P

525 = P2101

100

– P

P =525 100

21

= ` 2500

S.I.=

12500 (2 2) 102

100

= ` 50081. (C) Side of equilateral triangle = a

Height of prism = hRequired ratio= Volume of prism : Volume of new prizm

= 234

a

× h : 232

4a

× 2h

= 1 : 2

82. (B) Work done by A and B =35

Work done by C = 1 – 35 =

25

Amount paid to C = 30,000 × 25

= `12,000

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83. (C) (x + 2)2 = 9 (y + 3)2 = 25(x + 2)2 = (±3)2 (y + 3)2 = (±5)2

x = 1, – 5 y = 2, – 8

Possible value of xy

are 1 1 5 5, , ,2 8 2 8

Maximum Value of xy

= 58

84. (B)

ATQ,

PR = 2 2PQ QR = 2 25 12 = 13 cm

Length of median in right angle triangle

= PR2 =

132 cm

OQ = 23 × OM =

132 ×

23 = 4

13

85. (B)

In PQSQPS = 180° – 90° – 60°

= 30°In PQR, if C is circumcentre than

QPR =12 × QCR = 65°

RPS = QPR – QPS= 35°

86. (C) Given, 4096 = 64

40.96 + 0.4096 + 0.004096

+ 0.00004096= 6.4 + 0.64 + 0.064 + 0.0064= 7.1104

87. (C) Expression

=

.....100 2 100 200100 1

100 99 98 ..... 3 2 1

=

(100 1).....(100 100)....(100 200)

100 99 98 .....3 2 1= 0

88. (C)

Required ratio = 4 : 389. (B) Let expenditure per student =` x

Total expenditure = ` yATQ,

40 × x = y ...(i)(40 + 8)(x – 2) = y + 4840x + 8x – 96 = y + 48

y + 8x – 96 = y + 488x = 144 x = 18

So, original expenditure of mess =40 × 18 = ` 720

90. (B) Cost price of land = ` 96,000

Loss on 25

th land = `2 6960005 100

= ` 2304

Overall profit = `1096000100

= ` 9600Required profit percentage

=9600 2304

3960005

× 100

= 2023 %

Short Cut :

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91. (C) Income = 9520 × 100 100 10080 85 70

= ̀ 20000

92. (A)

So, units of work done by A and C in a day= 4 – 1 = 3

Time required by A and C = 243 = 8 days

93. (B)1

1x +

22y +

10091009x

= 1 given

1x

x + 2

yy + 1009

zz

= 11

1x

+ 21

2y

+ 10091

1009z

= 1 + 1 + 1 – 1 2 1009

1 2 1009x y z

= 3 – 1 = 2

94. (A) sin = 2 2

2 2

m nm n

BC = 2 2 2 2 2 2( ) ( )m n m n = 2mn

cos = 2 2

2mnm n

95. (C)2sin68 2cot15cos22 5 tan75

–

3 tan20 .tan 40 tan 45 .tan50 .tan705

= 2cos22 2tan75cos22 5 tan75

–

3.tan20 .cot20tan40 .cot 40

5

= 2 – 25 –

35 =

10 2 35

= 1

96. (B)

AB = 108 m, CD = x m

From ABC, tan 60° = ABBC

3 = 108BC BC =

1083 = 36 3 m

From AED, tan 30° = AEED

AE = 13 × 36 3 = 36 m

Height of pole (x) = 108 – 36 = 72 m

97. (B)

AB = temple = 54 mCD = temple = h mBC = width of river = x m

From ABC,

tan 60° =ABBC 3 =

54x

x =543 = 18 3 m

From ADE,

tan 30° =AEDE

13 =

5418 3

h

54 – h = 18h = 54 – 18 = 36 m

98. (A) Greatest number that will divide 148, 246and 623 leaving remainders 4, 6 and 11respectively= H.C.F. of (148 – 4),(246 – 6) and (623 – 11)= H.C.F. of 144, 240 and 612= 12

Required number = 12

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

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

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

SSC MAINS MATHS - 53 (ANSWER KEY)

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

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

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

Note:- 1. For any issue related to RESULT Processing, kindly contact uson 8287200200.

2. If your opinion differs regarding any answer, please message themock test and question number to 8287500500.

99. (B) ATQ,15 th of the full tank + 22 litres =

34

th of full tank

3 14 5

th of the full tank = 22 litres

1120

th of the full tank = 22 litres

Capacity of the full tank =202211

litres

= 40 litres

100.(D) Let cost price = ` 100Profit earned = 26%

Selling price =126100100

= ` 126

Discount = 10%

Marked price = `10012690

= ` 140

Marked price is more then cost price by

= 140 100

100

=40%

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

71. (C)72. (A)73. (D)74. (B)75. (C)76. (C)77. (D)78. (A)79. (A)80. (B)

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

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

SSC MAINS MATHS - 53 (SOLUTION)...MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA...

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