100+ Time and Work Questions for SBI, IBPS Bank PO Exams - 1

Question: 1

A completes $${7} / {10}$$ of a work in 15 days. Then he completes the remaining work with the help of B in 4 days. The time required for A and B together to complete the entire work is

(A) $$8{1}/{4}$$

(B) $$10{1}/{2}$$

(C) $$13{1}/{3}$$

(D) $$14{1}/{4}$$

Ans: C

(A + B)’s 4 day’s work = $$(1 – {7}/{10})$$ = $${3}/{10}$$

(A + B)’s 1 day’s work = $$( {3/10} × {1/4})$$ = $${3}/{40}$$

Hence A and B together take $${40}/{3}$$

= $$13{1}/{3}$$ days to complete the entire work.

Question: 2

A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work is

(A) 1 days

(B) 4 days

(C) 6 days

(D) 8 days

Ans: B

Ratio of rates of working of A and B = 2 : 1.

So, ratio of times taken = 1 : 2.

∴ A’s 1 day’s work = $${1} / {6}$$; B’s 1 day’s work = $${1} / {12}$$.

(A + B)’s 1 day’s work = $$({1} / {6} + {1} / {12})$$ = $${3} /{12}$$ = $${1} / {4}$$

So, A and B together can finish the work in 4 days.

Question: 3

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) 20

(B) 30

(C) 40

(D) 50

Ans: C

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

(A + C)’s 4 day’s work = $$(1 - {7}/{10})$$ = $${3}/{10}$$

(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.

Question: 4

Two spinning machines A and B can together produce 3,00,000 metres of cloth in hours. If machine B alone can produce the same amount of cloth in 15 hours, then how much cloth can machine A produce alone in 10 hours?

(A) 50,000 metres

(B) 1,00,000 metres

(C) 1,50,000 metres

(D) 2,00,000 metres

Ans: B

Length of cloth produced by A and B in 10 hrs = 3,00,000 m.

Length of cloth produced by B in 10 hrs = $$({300000} /{15} × 10)$$m = 200000 m.

∴ Length of cloth produced by A in 10 hrs

= (300000 – 200000) m = 100000 m.

Question: 5

A tyre has two punctures. The first puncture alone would have made the type flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat?

(A) $$1{1} / {2}$$

(B) $$1{1} / {3}$$

(C) $$3{1} / {2}$$

(D) $$3{3} / {5}$$

Ans: D

1 minute’s work of both the punctures = $$({1/9} + {1/6})$$ = $${5}/ {18}$$.

So, both the punctures will make the tyre flat in $${18}/{5}$$ = $$3{3} / {5}$$ min.

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