SBI Clerk Mains Quantitative Aptitude (Day-09)

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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

Direction (1 – 5): What will come in place of question mark (?) in the following number series?

1) 12, 84, 432, ? , 5232, 10476

a) 1740

b) 1830

c) 2240

d) 2680

e) 3240

2) 23, 31, ?, 81, 131, 203

a) 48

b) 49

c) 42

d) 39

e) 43

3) 32, 176, ?, 5324, 29282, 161051

a) 968

b) 1240

c) 876

d) 764

e) 1459

4) 120, 649, 1010, 1299, 1468, 1589, ?

a) 1670

b) 1638

c) 1656

d) 1640

e) 1672

5) 992, 1056, 1122, 1190, ?, 1332

a) 1260

b) 1240

c) 1220

d) 1280

e) 1250

Caselet

Directions (6 – 10): Study the following information carefully and answer the question given below?

Amit can do a piece of work in 60 days. With the help of Rahul, he can complete the work in 180/7 days. Bhagat can do the same work in 15 days less than that of Rahul, when working alone. Amit started the work, Rahul and Bhagat join him every third day and thus time taken by them to complete the work is — (A) —days. Efficiency of Seema is half of the efficiency of Rahul. Amit and Bhagat started the work. Seema joined them after five days. Time taken by them to complete the remaining part of the work is —- (B) —– days. Seema started the work and left after 10 days. Reeta complete the remaining part of the work in 80/3 days. Amit and Reeta together can complete the work in —- (C) —– days. Garima and Rahul together can complete half of the work in 90/13 days. Garima and Reeta started the work and left after five days. Bhagat can complete the remaining part of the work in —– (D) —– days. If Reeta, Seema, Amit and Bhagat started the work together, they can complete the work in — (E) —days.

6) Find the place value of (A)?

a) 53(1/11)

b) 59(3/13)

c) 47(7/17)

d) 49(1/6)

e) None of these

7) Find the place value of (B)?

a) 23/2

b) 135/11

c) 133/10

d) 44/3

e) None of these

8) Find the place value of (C)?

a) 15

b) 18

c) 20

d) 25

e) None of these

9) Find the place value of (D)?

a) 35/2

b) 25/3

c) 18

d) 21/2

e) None of these

10) Find the place value of (E)?

a) 140/11

b) 180/17

c) 160/19

d) 150/13

e) None of these

Answers :

Directions (1-5) :

1) Answer: A

12 * 6 + 12 = 84

84 * 5 + 12 = 432

432 * 4 + 12 = 1740

1740 * 3 + 12 = 5232

5232 * 2 + 12 = 10476

2) Answer: B

23 + 22 * 2 = 31

31 + 32 * 2 = 49

49 + 42 * 2 = 81

81 + 52 * 2 = 131

131 + 62 * 2 = 203

3) Answer: A

32 * 5.5 = 176

176 * 5.5 = 968

968 * 5.5 = 5324

5324 * 5.5 = 29282

29282 * 5.5 = 161051

4) Answer: B

120 + 232 = 649

649 + 192 = 1010

1010 + 172 = 1299

1299 + 132 = 1468

1468 + 112 = 1589

1589 + 72 = 1638

5) Answer: A

31 * 32 = 992

32 * 33 = 1056

33 * 34 = 1122

34 * 35 = 1190

35 * 36 = 1260

36 * 37 = 1332

Directions (6-10) :

6) Answer: B

1/Amit = 1/60

1/Amit + 1/Rahul = 7/180

=> 1/60 + 1/Rahul = 7/180

=> 1/Rahul = 7/180 – 1/60

=> 1/Rahul = (7 – 3)/180

=> 1/Rahul = 4/180

=>1/Rahul = 1/45

1/Bhagat = 1/(45 – 15)

=>1/Bhagat = 1/30

3 days’ work = 1/60 + 1/60 + (1/60 + 1/45 + 1/30)

=>3 days’ work = (3 + 3 + 3 + 4 + 6)/180

=>3 days’ work = 19/180

=> 3 x 9 = 57 days’ work = 19/180 x 9 = 171/180

Remaining work = 1 – 171/180

= (180 – 171)/180

= 9/180

= 1/20

58 day’s work = 1/60

59 day’s work = 1/60

Remaining part of the work = 1 – 171/180 – 1/60 – 1/60

= (180 – 171 – 3 – 3)/180

= (180 – 177)/180

= 3/180

= 1/60

Let time taken by them to complete remaining part of the work = n

n x (1/60 + 1/45 + 1/30) = 1/60

=> n x (3 + 4 + 6)/180 = 1/60

=> n = 1/60 x 180/13

=> n = 3/13

Required time = 59 + 3/13 = 59 3/13

Place value of (A) = 59 (3/13)

7) Answer: B

1/Seema = ½ x 1/45

=>1/Seema = 1/90

Let time taken by them to complete remaining part of the work = n

5/60 + 5/30 + n x (1/60 + 1/30 + 1/90) = 1

=> 1/12 + 1/6 + n x (3 + 6 + 2)/180 = 1

=> 11n/180 = 1 – 1/12 – 1/6

=> 11n/180 = (12 – 1 – 2)/12

=> n = 9/12 x 180/11

=> n = 135/11

Place value of (B) = 135/11

8) Answer: C

Let, Reeta alone can complete the work in n days.

10/90 + 80/3n = 1

=> 1/9 + 80/3n = 1

=> 80/3n = 1 – 1/9

=> 80/3n = (9 – 1)/9

=> n = 80/3 x 9/8

=> n = 30

1/Reeta = 1/30

Let, Amit and Reeta together can complete the work in t days.

t x (1/60 + 1/30) = 1

=> t x (1 + 2)/60 = 1

=> t = 60/3

=> t = 20 days.

Place value of (C) = 20

9) Answer: A

1/Garima + 1/Rahul = 13/90 x 1/2

=>1/Garima + 1/45 = 13/180

=> 1/Garima = 13/180 – 1/45

=> 1/Garima = (13 – 4)/180

=> 1/Garima = 9/180

=>1/Garima = 1/20

Let, required number of days = n

5/20 + 5/30 + n/30 = 1

=> ¼ + 1/6 + n/30 = 1

=> n/30 = 1 – ¼ – 1/6

=> n/30 = (12 – 3 – 2)/12

=> n = 30 x 7/12

=> n = 35/2 days

Place value of (D) = 35/2

10) Answer: B

Let, required number of days = n

n x (1/30 + 1/90 + 1/60 + 1/30) = 1

=> n x (6 + 2 + 3 + 6)/180 = 1

=> n = 180/17 days

Place value of (E) = 180/17

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