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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


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.


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


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%


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


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)


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


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


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


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


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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| No. 1 APP for Banking & SSC Preparation Website: …Solution...3 Adda247 | No. 1 APP for Banking & SSC Preparation Website:store.adda247.com | Email:[email protected] S14. Ans