# 1000+ Quantitative Aptitude Questions with Answers Pdf for Bank Exams - 1

Question: 1

Four examiners can examine a certain number of answer papers in 10 days by working 5 hours a day. For how many hours a day would 2 examiners have to work in order to examine twice the number of answer papers in 20 days?

(A) 4 hours

(B) 10 hours

(C) 12 hours

(D) $$8{1}/{2}$$ hours

Ans: B

Required number of hours = $${4 × 2 × 10 × 5} /{2 × 1 × 20}$$ = 10 hours.

Question: 2

A can do a piece of work in 4 hours, B and C can do it in 3 hours; A and C can do it in 2 hours. How long will B alone take to do it?

(A) 8hours

(B) 10 hours

(C) 12 hours

(D) 16 hours

Ans: C

A can do a piece of work in 4 hours

A and C can do it 2 hours.

∴ C alone can do it in $$({4 × 2} / {4 - 2})$$ = 4 hours.

B and C can do it in 3 hours.

∴ B alone can do it $$({4 × 3} / {4 - 3})$$ = 12 hours.

Question: 3

A can do a work in 12 days. When he had worked for 3 days, B joined him. If they complete the work in 3 more days, in how many days can B alone finish the work?

(A) 4 days

(B) 6 days

(C) 8 days

(D) 12 days

Ans: B

Work done by A for 3 days = $${3}/{12}$$ = $${1}/{4}$$

∴ Remaining work = $$1 - {1}/{4}$$ = $${3}/ {4}$$

∴ Work done by (A + B) for 1 day = $${3}/{4} × {1/3}$$ =$${1}/{4}$$

∴ Work done by B for 1 day = $${1/4} - {1/12}$$ = $${2}/{12}$$ = $${1}/{6}$$.

Question: 4

Raj and Ram working together to do a piece of work in 10 days. Raj alone can do it in 12 days. Ram alone will do the work in

(A) 20 days

(B) 40 days

(C) 60 days

(D) 70 days

Ans: C

(Raj + Ram)’s 1 day’s work = $${1}/{10}$$

Raj’s 1 day’s work = $${1}/{12}$$

∴ Ram’s 1 day work = $${1}/{10} - {1}/{12}$$ = $${6 - 5} / {60}$$ = $${1}/{60}$$

∴ Required time = 60 days.

Question: 5

A and B can separately do a piece of work in 20 and 15 days respectively. They worked together for 6 days, after which B was replaced by C. If the work was finished in next 4 days, then the number of days in which C alone could do the work will be

(A) 30

(B) 35

(C) 40

(D) 60

Ans: C

(A + B)’s 6 day’s work = $$6({1}/{20} + {1}/{15})$$ = $${7}/{10}$$;

(A + C)’s 4 day’s work = $${3}/{10}$$;

(A + C)’s 1 day’s work = $${3}/{40}$$.

A’s 1 day’s work = $${1}/{20}$$

∴ C’s 1 day’s work = $$({3/40} - {1/20})$$ = $${1}/{40}$$.

Hence, C alone can finish the work in 40 days.

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