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SBICL-MN-A-0031.1 2.3 3.3 4.3 5.3

6.3 7.3 8.3 9. 4 10. 2

11. 3 12.2 13.5 14.1 15.2

16.5 17.2 18.4 19. 5 20. 4

21. 5 22. 5 23. 4 24. 2 25. 326. 3 27. 4 28. 5 29. 1 30. 3

31. 3 32. 1 33. 3 34. 4 35. 5

36. 2 37. 5 38. 3 39. 3 40. 4

41. 5 42. 1 43. 2 44. 1 45. 3

46. 1 47. 2 48. 4 49. 1 50. 2

51. 5 52. 2 53. 5 54. 3 55. 3

56. 2 57. 4 58. 3 59. 3 60. 4

61.2; Replace seem with seemed.

62.5; No error

63.1; Replace difference with line.

64.3; Replace as with but.

65.1; Replace forget with forgot.

(6670)C E D F H A G B

66.2 67.5 68.1 69.3 70.4

71.4 72.1 73.3 74.3 75.5

76.3 77.4 78.1 79.3 80.2

81.5 82.2 83.3 84.1 85.4

86.2 87.3 88.1 89.3 90.3

91. 5; I. 088p39p10 2

or, 088p55p16p10 2

or, 2p(5p - 8) + 11(5p - 8) = 0

or, (2p + 11) (5p - 8) = 0

p = 211

or5

8

II. 040q41q21 2

or, 040q15q56q21 2

or, 0)8q3(5)8q3(q7

or, (7q - 5) (3q + 8) = 0

q = 38

75 or

We can't get specific relationship between p and q.

92. 1; I. 020p43p14 2

or, 020p8p35p14 2

or, 7p(2p - 5) - 4(2p - 5) = 0

or, (7p - 4) (2p - 5) = 0

p = 74 or 2

5

II. 012q72q33 2

or, 012q66q6q33 2

or, 0)2q11(6)2q11(q3

or, (3q + 6) (11q + 2) = 0

q = -2 or 112

Hence, p > q.

93. 2; I. 042p97p56 2

or, 042p48p49p56

2

or, 7p(8p + 7) + 6(8p + 7) = 0

or, (7p + 6) (8p + 7) = 0

p = 87

76 or

II. 05q64q99 2

or, 05q55q9q99 2

or, 9q (11q - 1) - 5(11q - 1) = 0

or, (9q - 5) (11q - 1) = 0

or, (9q - 5) (11q - 1) = 0

q = 111

95 qor

Hence, p < q.

94. 1; I. 025p40p162

or, 025p20p20p16 2

or, 4p(4p - 5) - 5(4p - 5) = 0

or, (4p - 5) (4p - 5) = 0

p = 45

45 ,

II. 016q40q252

or, 016q20q20q25 2

or, 5q(5q - 4) - 4(5q - 4) = 0or, (5q - 4) (5q - 4) = 0

54

54 ,q

Hence, p > q.

95. 5; I. 024p38p15 2

or, 024p18p20p15 2

or, 5p(3p + 4) + 6(3p + 4) = 0

or, (5p + 6) (3p + 4) = 0

34

56 orp

II. 030q49q20 2

or, 030q25q24q20 2

or, 4q(5q + 6) + 5(5q + 6) = 0

or, (4q + 5) (5q + 6) = 0 q = 45 or 5

6

Hence, we can't get any specific relationship

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between p and q.

96.3; ? 269 + 1867 1346 262 = 528.

97.2; ? 100

62 6765 = 4194

98.5; ? = 13201001659604

98 13210

165 = 213444.

99.4; ? (29)2 (25)2+ (131 15)= 216 + 1965 = 2181

100.1; ? 100

13 714 + 101

= 92.82 + 101 101 93 = 8

101.3;

? = 20 + 3 = 23102.3;

? = 29 5 = 24103.2;

? = (7 7 6) + 76 = 119104.5;

? = 574 49 = 525105.1;

? = 187 + 192 = 379

106.2; Let the numbers be x and y.

x + y = y100

250

2x + 2y = 5y

2x = 3y

x : y = 3 : 2107.2; Let the two digit number be 10 x + y.

x + y = 11 ... (1)

10y + x = 10 x + y + 63

9 y 9x = 63

y x = 7 ... (2)

From (1) and (2)

2y = 18, y = 9 and x = 2

The number = 10x + y= 20 + 9 = 29

108.3; Let the number be x and y.

xy = 7 and x y = 6

(x + y)2 = (x y)2+ 4xy = 36 + 28 = 64

x + y = 8

x + x 6 = 8

2x = 14x = 7, y = 1

109.5; Let the number be 10 x +y

y = x + 3

y x = 3...................... (1)

(10y + x) (10x +y) = 27

9y 9x = 27

y x = 3 .................. (2)We cant get any conclusion from (1) and (2) both.

110.2;

2

x

1x

= 81 x2+

2

1

x

+ 2 = 81

x2+2

1

x

= 79

111.5; Let the salary be` x.

Total expenditure

= 30 + 40 + 15 + 10 % of (70 +15)

= 85 + 8.5 = 93.5 %

6.5 % of x = 3900

x = 3900 6.5

100=`60,000

112.5; Number of way

= 12C10

12C8 12C

9 12C

12 12C

11

= 2

1112 12

23

101112

234

9101112

= 86248800

113.2; Required time =

2124

2421= 168 hours

114.5; Average runs for 11 innings = 47

Total runs in 11 innings = 41 11 = 517

Average runs after 12 innings = 47 + 9 = 56

Total runs = 56 12 = 672

Runs in last match = 672 517 = 155.115.3; Let the CP of each article be` 1

CP of 8 articles =` 8

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SP of 8 articles = CP of 17 articles =`17

gain % = 1008

817 = 100

8

9

=2

225= %

2

1112

116.4; The ratio of rupee, fifty paise and ten paise.

coins = 3 : 5

=45:40:24

9:8

10 paise coins value (Amount)

=109

45 3815

100

10=` 157.5

117.1; Height =2

3side

=2

3 322 = 11 3 = 33 cm

118.2; Ratio of profit = 1800 : 3000 : 4200 = 3 : 5 : 7

Ratio of investment = Ratio of profit = 3 : 5 : 7

Share of investments by second partner

= 000,60155

=` 20,000

119.3; Let there are x men originally.

80 x = (x 18) 100

4x =5x 90

x = 90 men

120.5; 1 boys one days work =12

1

15

1

4

1

=60

5415=

60

6=

10

1

A boy alone complete the work in 10 days.

121. 4; Speed of train =23

460= 20 km/hr

Speed of car =10065

20 =20

13 20 = 13 km/hr..

122.1;1002

617P

100

68P = 97.8

100

P[17 3 8 6] = 97.8

P (51 48) = 9780 P =`3260123.1; 2 r = 2(l + b)

)7656(7

22 r

r = 132 22

7= 42

Area of circle = 42427

22 = 5544 cm2

124.4;

A + B = C + 30 ... (i)And A = C 8 ... (ii)

From (i) and (ii): we get ,

C 8 + B = C + 30

B = 38 years.

125.5; Let the required distance be x km

60

20

7

x

6

x

42

x=

3

1

x =3

42= 14 km.

126.4; 100 % = 360 14 % =100

360 14 = 50.4

127. 1; 100 % = 1,60,000

15 % = 1,60,000 100

15=`24,000

128.3; Required ratio = (14 + 12) : (30 + 8)

= 26 : 38 = 13 : 19

129.2; 15 % = 40,000

Required difference = 5)(8

15

40,000 =`8000

130. 1; 100 % = 360

Angle formed by miscellaneous (16 %)

= 57.616100

360

60 % of miscellaneous = 6.57100

60= 34.56

131. 1; Subject Marks obtained

English 63

History 128

Physics 65

Maths 66 Science 134

Economics 34

Total marks: 490

132. 5;

200of%7

)71706764906570(

= 71% of 200 = 142

133. 2; Those two students are F and G.

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134. 2; Total marks obtained by the student E

= 64 + 134 + 64 + 64 + 134 + 32 = 492

Full marks

= 100 + 200 + 100 + 100 + 200 + 50 = 750

Hence, the required per cent = %6.65100750

492

135. 1; Overall percentage marks in the six subjects are asfollows:

Subject Overall percentage marks

(i) English: %637

66656463626160

(ii) History: %717

71706764906570

(iii) Physics:7

74756465886880 73.42%

(iv) Maths:7

78806466906890 76.57%

(v) Science:7

82856767807060 = 73%

(vi) Economics:7

86906468887270 76.85

136. 2; Grade B

137. 4; Grade E in finance.

138. 2; Average = 318~25.3184

1273

139. 3; Total number of hours = 1257 8 = 10056 hrs.

140. 4; Required average = 294

5

1470

(141-145):

141. 2

142. 3

143. 1

144. 5

145. 5

146.2; D E C E M B E R147. 3;

M O S Q U E

7 % 8 # 2

T E M P L E

$ 7 1 3

S T E M P

$ 7 18

148. 3; AFRICA : ACIRFA

INDIA : AIDNI

AUSTRALIA : AILARTSUA149. 2; CHIN, INCH

150. 5; Accompanied

(151 152):

Coolers

fans

Repairable

A.C.

151. 5;

152. 2;

153. 3;

Collegs

School

Coed

(Either I or II is true)

154. 4;

155. 3; Both the statements are Etype.

No conclusion follows from two negative statements.But complimentary pain. Hence, either I or II follows.

(156160):

156. 5; More than four

157. 1; Rohit

158.1; Third to the left.

159.1; Harish

160.3; Raju, Gautam161.2; 675 = 6 + 7 + 5 = 18

162.2; 534 3

435=145

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675 3

576= 192

163.1; After Interchanging

354 765 349 392 368

Hight Lowest

765 349 = 416

164.3;534 675 439 932 638

1 1 5 7 2

165.2; 435 576 934 239 836

Descending order : 934 836 576 435 239

166.5; 435 576 934 239 836

167.2;

Brother of M is uncle of O.168. 1;

169.3;

(170-174):

Name City Car

A Mumbai Polo

B Mumbai Honda

C Hyderabad Ford

D Hyderabad Swift

E Mumbai Beat

F Delhi Sentro

G Delhi Alto

170. 4 171. 3 172. 2 173. 1 174. 2

175.4; Question can be answered from I, II and III.

176. 5; From all three statements,

T > Q > P > R > S > U177. 2; A has two daughters.

178. 5; DOCUMENT = EOUCDMNT

Fifth from Right = C

179.3; 49, others are not perfect square.

180.1

181.2 182.1 183.4 184.1 185.2

186.2 187.4 188.1 189.2 190.3