ssc tier i – 2015 paper held on 16.08.2015 (evening shift) · pdf filessc tier-i –...

SSC TIER-I – 2015 PAPER held on 16.08.2015 (Evening Shift)

SOLUTIONS AND ANSWERSHEET

Qns. Ans. Qns. Ans. Qns. Ans. Qns. Ans. Qns. Ans.

1. 3 41. 3 81. 4 121. 2 161. 2 2. 1 42. 3 82. 3 122. 2 162. 4

3. 4 43. 4 83. 3 123. 2 163. 2 4. 2 44. 3 84. 3 124. 4 164. 1 5. 3 45. 1 85. 2 125. 2 165. 2

6. 2 46. 2 86. 3 126. 3 166. 4 7. 1 47. 4 87. 4 127. 4 167. 4 8. 4 48. 1 88. 2 128. 3 168. 2 9. 4 49. 4 89. 4 129. 1 169. 2

10. 3 50. 4 90. 1 130. 2 170. 1

11. 2 51. 3 91. 4 131. * 171. 4 12. 3 52. 2 92. 1 132. 2 172. 3

13. 1 53. 2 93. 4 133. 3 173. 3 14. 1 54. 3 94. 1 134. 3 174. 3 15. 4 55. 2 95. 1 135. 1 175. 1 16. 2 56. 2 96. 2 136. 1 176. 3

17. 4 57. 3 97. 2 137. 4 177. 1 18. 4 58. 4 98. 1 138. 4 178. 2 19. 4 59. 4 99. 4 139. 3 179. 4

20. 4 60. 3 100. 3 140. 1 180. 4 21. 3 61. 4 101. 3 141. 1 181. 1 22. 1 62. 2 102. 2 142. 4 182. 2 23. 1 63. 1 103. 1 143. 1 183. 4 24. 2 64. 2 104. 1 144. 4 184. 2 25. 3 65. 1 105. 1 145. 4 185. 4 26. 3 66. 1 106. 3 146. 2 186. 4 27. 2 67. 2 107. 4 147. 1 187. 1 28. 4 68. 4 108. 4 148. 2 188. 4

29. 3 69. 2 109. 4 149. 3 189. 4 30. 3 70. 1 110. 3 150. 4 190. 4 31. 4 71. 2 111. 4 151. 2 191. 1 32. 1 72. 4 112. 1 152. 4 192. 3 33. 2 73. 1 113. 4 153. 2 193. 4 34. 2 74. 3 114. 4 154. 4 194. 3 35. 3 75. 4 115. 2 155. 2 195. 4 36. 3 76. 2 116. 3 156. 3 196. 1

37. 4 77. 3 117. 4 157. 1 197. 3 38. 2 78. 3 118. 1 158. 1 198. 3 39. 2 79. 2 119. 2 159. 3 199. 4 40. 2 80. 4 120. 4 160. 3 200. 2

HINTS1. (3) Option (1)

8 – 7 + 3 × 5 = 35 7 + 8 – 3 × 5 = 35 7 + 8 – 15 ≠ 35 Option (2)

7 × 8 + 6 – 9 = 25 8 × 7 – 6 + 9 = 25 56 – 6 + 9 ≠ 25 Option (3)

6 + 8 × 2 - 7 = 0 6 – 14 + 8 = 0 14 – 14 = 0 Option (4)

8 × 2 + 7 – 6 = 9 8 × 2 – 8 + 6 = 9 14 – 14 ≠ 9

2. (1) The relation is : x : x

2 + 1

4 : (4)2 + 1

4 : 17 Similarly, 7 : (7)

2 + 1

7 : 50 3. (4) Nephron is the basic structural and functional unit of the

kidney. Similarly, neuron is the basic structural and functional unit of the Central functional System.

4. (2) the position of Y from the right end of the English alphabetical series is 2 and that of V is 5. (2)

2 = 4and (5)

2 = 25

5. (3)

Pairs of opposite letters. Similarly,

6. (2)Entomology is that branch of science which deals with

insects. Similarly, the scientific study of snakes is called ophiology.

7. (1)

Similarly

8. (4) English is different from Kannada. But both are included

in the class languages

9. (4) Some professors may be reserachers and vice-versa.

Some professors may be scientists and vice-versa. Some researchers may be seientists and vice-versa. Some professors who are researchers may be scientists. Some researchers who are scientists may be professors.

10. (3) Tiger is different from lion. But both are animals.

11. (2) First Column

1 + 8 + 27 = 36 36 – 1

2 = 35

Second Column 216 + 125 + 64 = 405 405 – 2

2 = 401

Third Column 343 + 512 + ? = 1575 + 3

2

855 + ? = 1584 ? = 1584 – 855 = 729

12. (3) First Column (2 × 4) + (4× 6) 8 + 24 = 32 Second Column (3 × 5) + (5 × 7) 15 + 35 = 50

Third Column (8 × 10) + (10 × 12)

80 + 120 = 200

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13. (1) First Row 4 × 3 × 2 + 8 24 + 8 = 32 Second Row 5 × 3 × 1 + 9 15 + 9 = 24 Third Row 7 × 3 × 3 + 7 63 + 7 = 70 Fourth row 2 × 9 × 4 + 12 72 + 12 = 84

14. (1) First Figure (11 × 12) – (6 × 9) 132 – 54 = 74

Second Figure (14 × 10) – (7 × 8)

140 – 56 = 84 15. (4) 5 = 2

2 + 1

10 = 32 + 1

26 = 52 + 1

50 = 72 + 1

122 = 112 + 1

16. (2)

17. (4)

18. (4)

19. (4)

B R O W N/B R O W N/B 20. (4)

21. (3) when paper is folded in the form of a cube, then

lies opposite

+ lies opposite

lies opposite 22. (1)

Option (1) 42 * 4 * 12 * 20 *9 42 – 4 ÷ 12 × 20 + 9 After changing the signs 42 × 4 + 12 ÷ 20 - 9

168+12 ÷ 20 – 9 180÷ 20 – 9=9-9=0

Option (2) 42 * 4 * 12 * 20 * 9

42 ÷ 4 + 12 – 20 × 9 After changing the signs 42 + 4 – 12 × 20 ÷ 9

42 + 4 -

42 + 4 -

46 -

Option (3) 42 * 4 * 12 * 20 * 9 42 + 4 – 12 ÷ 20 × 9 After changing the signs 42 – 4 × 12 + 20 ÷ 9

42 – 4 × 12 +

42 – 48 +

Option (4) 42 * 4 * 12 * 20 * 9 42 × 4 – 12 ÷ 20 + 9 After changing the signs 42 ÷ 4 × 12 + 20 – 9

23. (1)

24. (2)

25. (3)

36 – 2 = 34 34 – 4 = 30 30 – 2 = 28 28 – 4 = 24 24 – 2 = 22

26. (3)


Required distance AD = (25 + 5) km = 30 km

27. (2)

28. (4) Difference between the ratios of Ann = 5 – 2 = 3

: 3 21

∴: 1 =

29. (3) C is the father B. A is the wife of C. B, E and F are sons of A and C. D is a girl. Male members A, B, E AND F.

30. (3)

Similarly,

31. (4)

Similarly

32. (1)

B 01, 13, 20, 32, 44 E 56, 68, 75, 87, 99 A 03, 10, 22, 34, 41 K 57, 69, 76, 88, 95

33. (2)

34. (2) 428 4 × 2 = 8

338 3 × 3 = 9 326 3 × 2 = 6 339 3 × 3 = 9

35. (3) Kidnap is different from other three words. 36. (3)

37. (4) Except, Bristol, all others are cities of Switzerland. Berne

is the capital of Switzerland. 38. (2) 34 – 30

(3 + 4) – (3 + 0) 7 – 3 = 4

44 - 31 (4 + 4) – (3 + 1) 8 – 4 = 4

61 – 22 (6 + 1) – (1 + 2) 7 – 3 = 4

25 – 21 (2 + 5) – (2 + 1) 7 – 3 = 4

39. (2)

40. (2)

41. (3)

There is no ‘S’ letter in the given word. Therefore, the word CONSCIENCE can not be formed. I N C O N V E N I E N C E C O N V I N C E I N C O N V E N I E N C E C O N C E I V E I N C O N V E N I E N C E C O N C E I V E


42. (3) there is no ‘A’ letter in the given word. Therefore, the word SITUATION cannot be formed. D I S T R I B U T I O N D I S T U R B D I S T R I B U T I O N T U T I O N D I S T R I B U T I O N T R U S T

43. (4) There are no ‘C’ and ‘O’ letters in the given word. Therefore, the word DOCTOR cannot be formed. s u p e r i n t e n d e n t I N T E N S E s u p e r i n t e n d e n t N U R S E s u p e r i n t e n d e n t D E N T I S T

44. (3) Suppose the number of deer = d And, number of peacocks = p According to question d + P = 80 …………………………. (i) And, 4d + 2p = 200 Or, 2n + P = 100 ………………… (ii) From equations (i) and (ii) d = 20 therefore, number of peacocks = 80 – 20 = 60

45. (1) 46. (2) 47. (4)

AD = √( ) ( )

= √( ) ( )

= √ = √ = 5km 48. (1)First Premise is Particular Affirmative (I-type).

Second Premise is Universal Affirmative (A –type). All doctors are social workers.

Some social workers are politicians. A + I No Conclusion.

49. (4) Only conclusion II follows. It was expected that crop condition would improve after the rains.

50. (4)

Similarly,

101. (3) Let amount invested in each company be Rs. x.

S.I. =

According to the question,

= 1350

= Rs. 5000 102. (2)

∵ Market tax = Rs. 165 crores ∴ 33% = Rs. 165 crores

∴ 100 – 33 = 67%

= Rs. 335 crores 103. (1)

∵ 100%

∴ 35 + 10 = 45%

= Rs. 329. 85 crores 104. (1)

∵ 100%

∴ 1%

∴ 35%

105. (1) 1+cos2 = 3sin . cos

Dividing both sides by sin2 ,

cosec2 + cot

2 = 3 cot

1 + cot2 + cot

2

2 cot2 – 3 cot + 1 = 0

2 cot2 – 2 cot - cot + 1 = 0

2 cot2 (cot -1) – 1 (cot -1) = 0

(2cot2 - 1) (cot – 1) = 0

cot =

or 1

106. (3) Average units consumption in 2012

=

units

Required months July, August 107. (*) Average units consumption in the year 2013

=

=

units

108. (4) In the month of November, Difference = 500 – 200 =300 units In the month of August, Difference = 700 – 500 = 200 units

109. (*) Total consumption in 1012 = 2200 units Total consumption in 2013 = 2300 units

Percentage increase = (

)

=

110. (3) Let the numbers be 2x and 3x respectively. According to the questions,

9x + 24 = 8x + 32

9x – 8x = 32 – 24 = 8

x =8

Sum of numbers = 2x + 3x = 5x


= 5 x 8 = 40 111. (4) Let A, B, C, D and E in kg. represent their respective

weights. Then,

A + B + C = 84 x 3 = 252 kg. A + B + C + D = 80 x 4 = 320 kg. E = 68 + 3 = 71 kg. B + C + D + E = 79 x 4 = 316 kg. Now, (A+ B + C +D) – (B+C+D+E) = 320 – 136

A – E = 4 kg.

A = 4 + E = 4 + 41 = 75 Kg. 112. (1)

The ratio of the areas of two similar triangles is equal to the ratio of squares of any two corresponding sides.

Taking square roots of both sides,

16 x AC = 12 x 21

AC =

= 15.75 cm

113. (4) Expression = 3(sin4+cos

4) + 2 (sin

6+cos

6) + 12 sin

2 .

cos2

= 3 {(sin4 + cos

2)

2 – 2 sin

2 . cos

2}

+ 2{(sin2 + cos

2)3 – 3 sin

2 . cos

2 (sin

2 + cos

2) + 12

sin2 . cos

2

∵ a2+b

2 = (a+b)

2 – 2ab ; a

3 + b

3 = (a+b)

3 – 3ab (a+b)]

= 3 (1 – 2 sin2 . cos

2) + 12 sin

2 . cos

2 = 3 – 6 sin

2 .

cos2 + 2 – 6 sin

2 cos

2 + 12 sin

2 . cos

2

114. (4) Let the marked price of the camera be Rs. x. According to the question,

Rs 800

115. (2)

PQ = Tower A = 45 meter RS = Tower B = 15 meter,

QS = x metre (let)

PSQ = 600 ; RQS =

From PQS,

Tan 600 =

√

√

√ √ metre

From SRQ, tan =

√

tan = tan 300

= 300

sin= sin 300 =

116. (3) x = 4

equation of a line parallel to y – axis. y = 3

Equation of a line parallel to x-axis. Putting x =0 in the equation 3x + 4y = 12,

3 x 0 + 4y = 12 y =

= 3

Co-ordinates of the point of intersection on y-axis = (0, 3) Again putting y = 0 in the equation 3x + 4y =12,

3 + 4 x 0 = 12

Co-ordinates of the point of intersection on x–axis=(4, 0)

AC = 3 unit, BC = 4 units

Area of ABC

=

=

= 6 sq. units 117. (4)

BAC = 40

0,

ABC = 650

AED = 1800 - 40

0 – 65

0 = 75

0

DE || BC

AED = ACB =750

CED = 1800 - 75

0 = 105

0

118. (1) x2 + y

2 + z

2 = 2(x+ z – 1)

x2 + y

2 + z

2 = 2x + 2z – 2

x2 – 2x + y

2 + z

2 – 2z + 2 = 0


x2 – 2x + 1 + y

2 + z

2 – 2z + 1 = 0

(x -1)2 + y

2 + (z -1)

2 = 0

[∵ a2 + b

2 + c

2 = 0 a = 0, b = 0, c =0]

x – 1 = 0 x = 1 y = 0

z – 1 = 0 z = 1

x3 + y

3 + z

3 = 1 + 0 + 1 = 2

119. (2) Let the average cost of each book bought (of 64 books) be Rs. x. According to the question, 64 x – 50 (x +1) = 76

64x – 50x – 50 = 76

14x = 76 + 50 = 126

Required average price = 9 + 1 = Rs. 10 120. (4)

APB = 110

0 = CPD

APD = 1800 - 110

0 = 70

0 = BPC

PCB = 1800 - 70

0 - 30

0 = 80

0

Angles subtended by same arcs at the circumference are equal.

ACB or PCB = ADB = 800

121. (2) x2 + x = 5 (Given)

Let, x + 3 = a

Now,

a +

( )

( ) ( )

=

=

( )

(

)

(

)

= (5)3 – 3 x 5 = 125 – 15 = 110

122. (2)

In ABC,

BAC = 850

BCA = 750

ABC = 1800 - 85

0 - 75

0 = 20

0

The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

AOC = ABC = 400

OA = OC = radii

In OAC,

OAC = OCA (The angles at the base of an isosceles triangle are equal)

OAC + OCA = 1800 - 40

0 = 140

0

OAC =

= 70

0

123. (2) sec + tan = 2 √

∵ sec2 - tan

2 = 1

(sec + tan ) (sec - tan) = 1

sec - tan =

√

=

√ √

√ √

√

sec + tan + sec - tan

= 2 + √ + √

2 sec = 2√

sec = √ …..(i) Again,

sec + tan - (sec - tan )

= 2 + √ √

2 tan = 4 tan = 2 ….(ii)

sin =

√

124. (4)

QPR = 50

0

PQR + PRQ = 180

0 - 50

0 = 130

0

PQR +

PRQ = 65

0

The point of intersection of internal bisectors of angles is in-centre.

OQR =

PQR;

ORQ =

PQE

In OQR,

OQR + QOR + ORQ = 1800

QRO = 1800 - 65

0 = 115

0

125. (2)

=

By componendo and dividendo,

=


sin =

126. (3) Expression = √

= √

127. (4) Volume of prism = Area of base x height

7200 = √

P

2 x 100√

7200 = 50 x 3 x 3 P2

P2 =

P = √

128. (3) Single equivalent discount

= (

)

= (30 – 2)% = 28%

C.P. of article = 100 – 28 = Rs. 72 Actual cost price of article

=

Rs. 9.2

Required S.P. =

Rs. 91.08

129. (1)

AB = 10 cm.

AF = FB = 5 cm. CD = 24 cm.

CE = DE = 12 cm. Let OE = x cm

OF = (17 –x) cm

From ODE,

OD = √

= √ …..(i)

From OAF,

OA = √

= √( ) …..(ii)

∵ OA = OD

√ = √( )

x2 + 144 = 289 – 34x + x

2 + 25

34x = 289 + 25 - 144 = 170

From equation (i),

OD = √ = √

= √ cm. 130. (2)

x = z = 225, y = 226 ∴ x + y + z = 225 + 226 + 225 = 676

∴ x3 + y

3 + z

3 – 3xyz

= ½ (x + y + z) [(x-y)2 + (y –z)

2 + (z – x)

2]

= ½ × 676 [(225 – 226)2 + (226 – 225)

2 + (225 – 225)

2]

= ½ × 676 × (1 + 1) = 676

131. (*) Let, a = 1 +

= 1 +

= 1 +

Again,

b = = 1 -

= 1 -

= 1 +

∴ Expression = (a

2-b

2) ÷ ab

= {(a+b)(a-b)} ÷ ab

= (

) (

) (

)

=

132. (2) Let 3 kg of first alloy and 4 kg of second alloy be mixed together. ∴ In 3 kg of mixture, Tin = 1 kg. Iron = 2kg. In 4 kg of mixture.

Tin =

Iron =

∴ (1 + 1.6) : (2 + 2.4) = 2.6 : 4.4 = 13 : 22

133. (3) Required mass of lead

= 800 ×

(

)

= 800 ×

= 4764 kg.

134. (3) 4a -

On dividing by 4,

a -

∴a3 -

(

)

(

)

= (

)

= -

=

∴ a3-

=

135. (1) C.P of cycle = Rs. x (let)

∴ S.P. =

= Rs.

Case II,


New C.P. = Rs.

∴

= Rs.

= Rs. 2400

136. (1) x = √ √

√ √

y = √ √

√ √

∴ x + y = √ √

√ √

√ √

√ √

= (√ √ )

(√ √ )

(√ √ ) (√ √ )

= ((√ )

(√ )

)

= 5 + 3 = 8

xy = √ √

√ √ √ √

√ √

∴

= ( )

( )

=

137. (*) Expression = 2b2c

2 + 2b

2c

2 + 2a

2b

2 – a

4 –b

4

= 4b2c

2 – (2b

2c

2 – 2c

2a

2 – 2a

2b

2 + a

4 + b

4 + c

4)

= (2bc)2 – (a

2-b

2-c

2)

2

= (2bc + a2 –b

2 –c

2) (2bc –a

2+b

2 +c

2)

= (a2 – (b

2 + c

2 -2bc)) (b

2 + c

2 + 2bc –a

2)

= (a2 – (b-c)

2) ((b+c)

2 – a

2)

= (a –b + c) (a + b- c) (a + b + c) (b + c – a) If a + b – c = 0, ∴ Expression = 0.

138. (4) Rate downstream = (6 + 1.5) kmph = 7.5 kmph Rate upstream = (6 – 1.5) kmph According to the question

Time =

∴ Required time =

= 3 + 5 = 8 hours. 139. (3) Let the C.P. of article be Rs. 100 and the marked price

be Rs. x.

Case I

x =

= Rs.

Case II

S.P. =

= Rs. (

)

∴ Profit = Rs. (

)

= Rs. (

)

= Rs.

∴ Profit percent =

= 6

140. (1) Expression = cos 24 + cos 55 + cos 125 + cos 204 + cos 300 = cos24 + cos 55 + cos (180 - 55 ) + cos 55 + cos (180 + 24 ) + cos (360 – 60 ) = cos24 + cos55 – cos55 – cos 24 + cos 60

= cos 60

141. (1) 142. (4) 143. (1) Let time taken by A = x days

∴ time taken by B = 2x days Time taken by c = 3x days According to the question,

6x = 6 × 11

x =

= 11

∴ time taken by C alone = 3x = 3 × 11 = 33 days

144. (1) Part of tank filled by pipes A and B in 1 minute

=

part

Part of tank filled in 12 minutes

=

part

Remaining part

=

part

When pipe C is opened, Part of tank filled by all three pipes

=

Time taken in filling

part

=

minutes

Total time = 12 + 12 = 24 minutes

145. (4)

Let the radius of swimming pool be r meter.


Breadth of shaded part = 4 metre

OB = (r+4) metre According to the question,

(r + 4)2 – r

2 =

r

2

r2 + 8r + 16 – r

2

r

2

8r + 16 =

r

2

200r + 400 = 11r2

11r2 – 200r – 400 = 0

11r2 – 220r + 20r – 400 = 0

11r (r-20) + (r-20) = 0

(r – 20) (11r + 20) = 0

r = 20 metre because

r =

metre

146. (2) Let C complete the work in x days.

B’s 1 day’s work =

and A’s 1 day’s work

=

According to the question,

5 (

) (

)

(

)

days

147. (1) x +

x2 + 1 = x x

2 – x + 1 = 0

148. (2) tanA + cotA = 2

tanA +

tan2A + 1 = 2 tanA

tan2A – 2 tanA +1 = 0

(tan A -1)2 = 0

tan A – 1 = 0 tan A = 1

cot A = 1

tan10

A + cot10

A = 1 + 1= 2 149. (3) Here distance is constant.

Speed

Ratio of the speeds of A and

B=

∶

A’s speed = 7x kmph (let) B’s speed = 8x kmph

AB = 7 x 4 = 28x km.

Let both trains cross each other after t hours from 7 a.m. According to the question, 7x (t+2) + 8x t = 28x

7t + 14 + 8t = 28

15t = 28 – 14 = 14

t =

hours

= (

) minutes

= 56 minutes

Required time = 7 : 56 a.m. 150. (4) Radius of cylindrical verssel = r cm. (let).

Volume of conical piece of iron =

= (

) cu. cm.

Volume of raised water = x 6.4 cu. cm.

x 6.4

ssc tier i – 2015 paper held on 16.08.2015 (evening shift) · pdf filessc tier-i –...

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