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Solutions
Solutions (1-5):
S1. Ans.(b) S2. Ans.(d) S3. Ans.(a) S4. Ans.(d) S5. Ans.(c) Solutions (6-10):
Number Box
8 U
7 C
6 R
5 Q
4 T
3 S
2 D
1 P
S6. Ans.(c) S7. Ans.(a) S8. Ans.(e) S9. Ans.(e) S10. Ans.(e) S11. Ans. (b) Sol. Only conclusion II follows. S12. Ans. (d) Sol. if neither conclusion I nor conclusion II follows S13. Ans. (d) Sol. if neither conclusion I nor conclusion II follows
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S14. Ans. (c) Sol. Either conclusion I or II follows. S15. Ans. (b) Sol. Only conclusion II follows. Solutions (16-20):
Month Student
January C
February A
March G
April E
June D
August F
October B
S16. Ans.(b) S17. Ans.(b) S18. Ans.(e) S19. Ans.(d) S20. Ans.(e) S21.Ans.(d) Sol.
S22.Ans.(e) Sol.
S23.Ans.(d) Sol.
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S24.Ans.(a) Sol.
S25.Ans.(b) Sol.
Solutions (26-30):
Row 1. ↓ P V S T R Q
Row 2. ↑ C F A E B D
S26. Ans.(d) S27. Ans.(a)
S28. Ans.(b) S29. Ans.(b) S30. Ans.(c)
Solutions (31-35):
Facing = ngi
With=snk Problem/health = mlp /hlt
On = sa
Rise = rtv
Every = lne
Challenge = riy
Each/day = nop/hus
S31. Ans.(c) S32. Ans.(a)
S33. Ans.(c) S34. Ans.(b) S35. Ans.(e)
S36. Ans.(d)
Sol. OX
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S37. Ans.(b) Sol.
S38. Ans.(d)
S39. Ans.(c)
Sol.4623957= 6842846
Digits will appear twice= 6, 8 and 4
S40. Ans.(c) Sol.
S41. Ans.(d)
Sol. Sum of differences = 20 + 10 + 20 + 20 + 20 + 20
= 110
S42. Ans.(a)
Sol. Total students in 2012 &2015 = 650 + 820 = 1470
Total students from A in all given years = 2310
Desired % = 1470
2310 x 100 = 63.6%
S43. Ans.(c)
Sol. No. of children for Class B in all years = 2240
No of children for class A in all years = 2310
Desired ratio = 2240
2310 = 32 : 33
S44. Ans.(a)
Sol. Total desired sum = (320 + 400) + (400 + 440)
= 1560
S45. Ans.(d) Sol. Class B = 2240
Class A = 2310) Total = 4550
Desired value = 4550−2240
4550× 100 ≈ 50.8%
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S46. Ans.(e)
Sol. × 0.5 + 1, × 1 + 1, × 1.5 + 1, × 2 + 1, × 2.5 + 1
27 × 2.5 + 1 = 68.5
S47. Ans.(b)
Sol.
S48. Ans.(a)
Sol. × 1 + 1, × 2 + 2, × 3 + 3, × 4 + 4, × 5 + 5
S49. Ans.(d)
Sol.
S50. Ans.(b)
Sol.
S51. Ans.(e)
Sol.
𝐈. 2x2 − 4x − x + 2 = 0 ⇒ 2x(x − 2) − 1(x − 2) = 0
⇒ (2x − 1)(x − 2) = 0
⇒ x =1
2, 2
||
𝐈𝐈. 2y2 − 9y + 7 = 0
⇒ 2y2 − 7y − 2y + 7 = 0
⇒ y(2y − 7) − 1(2y − 7) = 0
⇒ y =7
2, 1
∴ No relation
S52. Ans.(a)
Sol.
𝐈. 3x2 + 3x + 4x + 4 = 0 ⇒ 3x(x + 1) + 4(x + 1) = 0
⇒ x = −1, −43⁄
||
𝐈𝐈. y2 + 5y + 4y + 20 = 0
⇒ y(y + 5) + 4(y + 5) = 0⇒ y = −4, −5
∴ x > y
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S53. Ans.(a)
Sol.
𝐈. 3x2 + 3x + 2x + 2 = 0 ⇒ 3x(x + 1) + 2(x + 1) = 0
⇒ x = −1,−2
3||
𝐈𝐈. y2 + 9y + 3y + 27 = 0
⇒ y(y + 9) + 3(y + 9) = 0⇒ y = −3, −9
∴x > y
S54. Ans.(d)
Sol.
𝐈. x2 − 5x − 2x + 10 = 0 ⇒ x(x − 5) − 2(x − 5) = 0 ⇒ x = 2, 5
|
𝐈𝐈. y2 − 9y − 5y + 45 = 0
⇒ y(y − 9) − 5(y − 9) = 0⇒ y = 9, 5
∴x ≤ y
S55. Ans.(b)
Sol.
𝐈. (x − 16)(x − 16) = 0⇒ x = 16 |
𝐈𝐈.y = ±16
∴x ≥ y
S56. Ans.(b)
Sol.
Let, original CP be x
Then, original SP =115
100x
New SP =11
10×
115
100x
ATQ,11
10×
115
100x =
110
100(x + 100)
or,15x
100= 100
or, x = 666.66
S57. Ans.(c)
Sol.
B – A = 6 ….(i) {
Let A′s age is → A
Let B′sage is → B
Let C′s age is → C
B + 9
C=
9
8
Or, 9C –8B = 72 ….(ii)
And C = 2A …(iii)
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From (ii) & (iii)
⇒ 18A − 8B = 72 ⇒ 18(B − 6) − 8B = 72 [∵ A = B − 6 … (i)] Or, B = 18 year
After 5 years B‘s age = 23 years
S58. Ans.(b)
Sol. Akash scored in subject A = 73 marks
Akash scored in subject B = 56% of 150 = 150 × 56
100 = 84 marks
Akash scored in subject C = X marks
Maximum mark of all three subjects is 150.
∴ Total marks = 150 × 3 = 450
Now, according to the question
Marks obtained in subject A + Marks obtained in subject B + Marks obtained in subject C
= 54% of total marks
⇒ 73 + 84 + X = 450 × 54
100
⇒ X + 157 = 243
⇒ X = 243 – 157 = 86
⇒ X = 86
Hence, Akash scored 86 marks in subject C.
S59. Ans.(b)
Sol. Let, B takes ‘x’ days to do the work alone
Then A takes (x + 2)’ days to do the work alone
According to the question,
3
x+
(57
5− 3)
x + 2= 1
Solving, x = 10
Shortcut,
Solve through option for faster calculation
S60. Ans.(a)
Sol. Let speed of train B be x m/s
And length of train B be y m
Then length of train A is 2y m
Speed of train A = 84 ×5
18=
210
9 m/s =
70
3 m/s
A.T.Q, 2y+y
10=
70
3− x ………….(i)
and 2y+y
22.5=
70
3− 2x
solving (i) and (ii), y = 50 m
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S61. Ans.(b) Sol. Total number of balls = 7 + 5 = 12 Now, three balls are picked randomly Then, the number of sample space n(s)
= 12C3=10×11×12
1×2×3= 220
The number of events
n(E) = 7C2 × 5C1 =6×7
2× 5 = 21 × 5 = 105
∴ P(E) =n(E)
n(S)=
105
220=
21
44
S62. Ans.(b) Sol. Cost price of rice per kg
=320 × 17.6 + 160 × 16.4
320 + 160=
5632 + 2624
480
=8256
480= Rs 17.2
Now, he sells the mixture Rs 9.45 above the CP. ∴ Selling price = 17.2 + 9.45 = Rs 26.65 S63. Ans.(b) Sol. Ratio of Aman’s profit to Bharti’s profit = 56 × 12 : 48 × 7 = 2 : 1 Now, let Aman’s share in profit be 2x and that of Bharti be x. Given x = Rs 3250 ∴ Total Profit = 2x + x = 3250 × 3 = Rs 9750 S64. Ans.(d) Sol. 2πrh ∶ πr2h = 1 ∶ 7 2 ∶ r = 1 ∶ 7 ⇒ r = 14 ⇒ diameter : Height ⇒ 2r : h = 4 : 3 ⇒ h = 21
Total surface area of cylinder = 2 ×22
7× 14 (14 + 21)
= 88 × 35 = 3080 S65. Ans.(c) Sol.
Downstream UpstreamSpeed → 7x km/hr : 3x km/hr
(D + 11
3x) = 3 (
D − 3
7x)
⇒ D = 52
7x =D + 18
2.5
⇒ x=4 Speed of current = 2x = 8 km/hr
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S66. Ans.(a)
Sol. Total spending on Tennis =45
360× 100 = 12.5%
S67. Ans.(c)
Sol. Desired value = (70−25)°
360× 144 = 18 𝑐𝑟.
S68. Ans.(b) Sol.
Increase in Hockey = 144 ×70
360×
120
100= 33.6
Decrease in football = 144 ×80
360×
70
100= 22.4
Difference in spending = 33.6 – 22.4 = 11.2 S69. Ans.(b) Sol. Amount spent on Tennis, Football and Hockey
= 195°
360× 144 = 19.5 × 4 = 78 cr.
Desired ratio = 78
84=
13
14
S70. Ans.(d) Sol. Amount spend on others
=10°
360× 144 = 4 cr.
Difference = 17cr – 4 cr = 13 cr. S71. Ans.(e) Sol. ≈ x2 − 900 = 30 × 3 − 29 ≈ x2 = 990 − 29 = 961 ≈ x = 31 S72. Ans.(c) Sol.
≈55
100× 400 +
50
100× 600 − 500 = √x
≈ 55 × 4 + 50 × 6 − 500 = √x
≈ √x = 20 or, x = 400 S73. Ans.(a) Sol.
≈ 32 + 8 + 31 + x =25
100× 700
≈ 71 + x = 25 × 7 ≈ x = 104
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S74. Ans.(d)
Sol.
≈350
50+ (10)2 × 20 = x3 −
19
100× 1000
≈ 7 + 2000 = x3 − 190
≈ x3 = 2007 + 190
≈ x = 13
S75. Ans.(e)
Sol.
≈ 729 + 256 + √x =20
100× 5000
≈ √x = 15
≈ x = 225
S76. Ans.(a)
Sol. No. of girls playing cricket in 2001 = 700 × 12
25×
1
4 = 84
No. of boys playing football in 2002 = 825 × 3
5×
5
9 = 275
Difference = 275 – 84 = 191
S77. Ans.(b)
Sol. Average no. of boys playing cricket in 2000, 2002 & 2003
=150 + 220 + 150
3
=520
3
Average no. of girls playing football in 2000, 2002 & 2003
=200 + 210 + 90
3
=500
3
Sum = 500 + 520
3=
1020
3 = 340
S78. Ans.(c)
Sol. No. of students in 2005 = 550 + 550 × 800
1100
= 550 + 400
= 950
No. of girls playing football in 2005 = 10
19 × 950 ×
60
100
= 300
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S79. Ans.(d)
Sol. Such years are 2000 & 2004
Average no. of boys in these two years = 250 + 250
2 = 250
S80. Ans.(a)
Sol. No. of girls playing cricket in 2000, 2002 & 2004
= 600 ×7
12×
3
7+ 825 ×
2
5×
4
11+ 550 ×
6
11×
8
15
= 150 + 120 + 160
= 430
No. of boys playing cricket in 2001 & 2003
= 700 ×13
25×
3
4+ 650 ×
7
13×
3
7
= 273 + 150
= 423
Required Ratio = 430 : 423
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