SBI Clerk Mains Quantitative Aptitude (Day-47)

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Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk Mains 2020 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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Missing Number Series

Directions (01 – 05): What value should come in the place of (?) in the following number series?

1) I) 12, 30, 75, 187.5, 468.75

II) 2……..78.125. If 78.125 is the nthterm find the value of n

A) 5th term

B) 4th term

C) 6th term

D) 7th term

E) None of these

2) I) 8640, 1440, 288, 72, 24

II) 1560……13. If 13 is the nthterm find the value of n

A) 5th term

B) 4th term

C) 6th term

D) 7th term

E) None of these

3) I) 25, 26, 54, 165, 664, 3325

II) 7…..57. If 57 is the nthterm find the value of n

A) 4th term

B) 6th term

C) 5th term

D) 3rd term

E) None of these

4) I) 14, 52, 140, 290, 445

II) 2…..175. If 175 is the nthterm find the value of n

A) 4th term

B) 6th term

C) 5th term

D) 3rd term

E) 7th term

5) I) 8, 177, 298, 379, 428, 453

II) 11…..301. If 301 is the nthterm find the value of n

A) 4th term

B) 6th term

C) 5th term

D) 3rd term

E) 7th term

Data Interpretation

Directions (06 – 10): Study the following information carefully and answer the questions given below.

The given bar graph shows the speed of the boat in still water (in kmph) in five different days.

6) If the time taken by boat travel 120km on day1 in upstream is 8 hours which is 2 more than the time taken by the same boat on day1 downstream, then what is the distance travelled by boat on day1 in downstream?

A) 120 km

B) 150 km

C) 175 km

D) 125 km

E) None of these

7) If the ratio of the time taken by boat to cover distance in upstream to downstream on day2 is 3:2 and the distance travelled by boat in downstream and upstream is 160 km and 120 km respectively, then what is the speed of the stream on day2?

A) 20 kmph

B) 15 kmph

C) 10 kmph

D) 5 kmph

E) None of these

8) If the distance travelled by boat in downstream on day 3 and day 4 is 200 km and 240 km respectively and the ratio of the time taken by boat to cover the distances in upstream to downstream on day 3 and day 4 is 1: 1 and 1: 1 respectively, then what is the sum of the distance travelled by boat in upstream on day 3 and day 4 together if the time in downstream on day 3 and day 4 is 5 hours and 4 hours respectively?

A) 120 km

B) 110 km

C) 130 km

D) 140 km

E) None of these

9) The distance travelled by boat in upstream on day5 is 80 km. If the speed of the stream on day5 is 5 kmph and the ratio of the time taken by boat in upstream to downstream is 4:3, then what is the difference between the distances travelled by boat in upstream and downstream on day5?

A) 10 km

B) 15 km

C) 20 km

D) 5 km

E) None of these

10) On day2, distance travelled in downstream is 80 km and the time taken in downstream and upstream in the ratio of 2:3. Distance travelled in upstream on day3 is 60 km. If the time taken by downstream on day2 is 2 hours and the distance travelled in downstream on day3 is 30 km more than the distance travelled in upstream on day2. What is the ratio of the distance travelled in upstream in day2 to distance travelled in upstream in day3?

A) 1:2

B) 2:3

C) 3:4

D) 4:5

E) None of these

Answers :

Directions (1-5) :

1) Answer: A

Series I,

12 * 2.5 = 30

30 * 2.5 = 75

75 * 2.5 = 187.5

187.5 * 2.5 = 468.75

Series II,

2 is the 1st term

2 * 2.5 = 5

5 * 2.5 = 12.5

12.5 * 2.5 = 31.25

31.25 * 2.5 = 78.125

78.125 is the 5th term

2) Answer: B

Series I,

8640/6 = 1440

1440 / 5 = 288

288 / 4 = 72

72 / 3 = 24

Series II,

1560 is the 1st term

1560 / 6 = 260

260 / 5 = 52

52 / 4 = 13

13 is the 4th term

3) Answer: A

Series I,

25 * 1 + 1 = 26

26 * 2 + 2 = 54

54 * 3 + 3 = 165

165 * 4 + 4 = 664

664 * 5 + 5 = 3325

Series II,

7 is the 1st term

7 * 1 + 1 = 8

8 * 2 + 2 = 18

18 * 3 + 3 = 57

57 is the 4th term.

4) Answer: C

Series I,

14 * 3 + 10 = 52

52 * 2.5 + 10 = 140

140 * 2 + 10 = 290

290 * 1.5 + 10 = 445

Series II,

2 is the 1st term.

2 * 3 + 10 = 16

16 * 2.5 + 10 = 50

50 * 2 + 10 = 110

110 * 1.5 + 10 = 175

175 is 5th term.

5) Answer: D

Series I,

8 + 132 = 177

177 + 112 = 298

298 + 92 = 379

379 + 72 = 428

428 + 52 = 453

Series II,

11 is 1st term.

11 + 132 = 180

180 + 112 = 301

301 is 3rd term.

Directions (6-10) :

6) Answer: B

Speed of the stream = x

Time taken in upstream = 120/(20 – x) = 8

(20 – x) * 8 = 120

20 – x = 15

x = 5 kmph

Time taken in downstream = 8 – 2 = 6 hours

D/(20 + 5) = 6

D = 150 km

7) Answer: C

(160/(30 + x))/(120/(30 – x)) = 2/3

4 * (30 –x)/[3 * (30 + x)] = 2/3

60 – 2x = 30 + x

3x = 30

x = 10 kmph

8) Answer: C

Speed of the stream on day3 = x

Speed of stream on day4 = y

200/(25 + x) = 5

40 = 25 + x

x = 15 kmph

240/(40 + y) = 4

60 = 40 + y

y = 20 kmph

Distance travelled in upstream on day3 = (25 – 15) * 5 = 50 km

Distance travelled in upstream on day4 = (40 – 20) * 4 = 80 km

Required sum = 50 + 80 = 130 km

9) Answer: A

Time taken by boat in upstream = 80/(25 – 5) = 4 hours

Time taken by boat travelled in downstream = 3 hours

Distance travelled by boat in downstream = 3 * (25 + 5) = 90 km

Required difference = 90 – 80 = 10 km

10) Answer: B

Speed of the stream on day2= x

Time taken in downstream on day2 = 2 hours

80/(30 + x) = 2

60 + 2x = 80

2x = 20

x = 10 kmph

Time taken in upstream on day2 = 3 hours

Distance travelled in upstream on day2 = 3 * (30 – 10) = 60 km

Distance travelled in downstream on day3 = 60 + 30 = 90km

Distance travelled in upstream on day2 = 60 km

Required ratio = 60:90 = 2:3

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