Quantitative Aptitude – Time & Work

In Quantitative Aptitude, Quants & Reasoning
December 18, 2015

Dear Aspirants,

Welcome to D2G’s Quantitative Aptitude concept on Time & Work. In this article we provide you the concept and shortcuts for Time and Work.

Now, What is time and work?? The time and work problems are actually practical problem and its knowledge will bring you power of taking correct decision. For example, in our day to day life we come across problem of building a bridge or digging the well in a village. These planning for work requires proper computation.

So lets see how time and work really relate each other as they both may directly or indirectly proportional to each other.

First lets learn few rules on which time and work rely..

Rule 1: Universal Rule
This rule can be used in almost every problem.
If M1 persons can do W1 work in D1 days and M2 persons can do W2 works in D2 days then we can say
M1×D1×W2 = M2×D2×W1
If the persons work T1 and T2 hours per day respectively then the equation gets modified to
M1×D1×T1×W2 = M2×D2×T2×W1
If the persons has efficiency of E1 and E2 respectively then,
M1×D1×T1×E1×W2 = M2×D2×T2×E2×W1

Rule 2:
If A can do a piece of work in n days, then The work done by A in one day = 1/n

If A can do a work in D1 days and B can do the same work in D2 days then A and B together can do the same work in days.

Rule 3:
If A can do a work in D1 days and B can do the same work in D2 days then A and B together can do the same work in (D1×D2)/(D1+D2) days.

Rule 4: If A is twice as good a workman as B, then A will take half of the time taken by B to complete a piece of work.

Rule 5: If A is thrice as good a workman as B, then A will take one third of the time taken by B to complete a piece of work.

Rule 6: If A and B together can do a piece of work in x days, B and C together can do in y days and C and A together can do in z days, then the same work can be done

By A alone in 2xyz/(xy+yz-zx) days.

By B alone in 2xyz/(yz+zx-xy) days.

By C alone in 2xyz/(zx+xy-yz) days.

By A, B and C together in 2xyz/(yz+zx+xy) days.

Rule 7: If A can do a piece of work in D1 days, B can do in D2 days and C can do in D3 days then they together can do the same work in D1×D2×D3/ (D1×D2+D2×D3+D1×D3) days.

Rule 8: If A and B together can do a piece of work in D1 days and A alone can do it in D2days, then B alone can do the work in D1×D2/ D2-D1 days.

Rule 9: If the number of men are changed in the ratio of m:n, then the time taken to complete the work will change in the ratio n:m

Shortcuts:-

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In indirect variation if no.of men increase then no.of workers also will increase.

While in indirect variation if no.of men increases no.of days decreases.

Now lets solve the frequently asked problems.

1. Each child from a certain school can make 5 items of handicraft in a day. If 1125 handicraft items are to be displayed in an exhibition then in how many days can 25 children make these items?

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M1×D1×W2 = M2×D2×W1

5×25×? = 1125×1×1

125×? = 1125

?= 9

2. Three men or eight boys can do a piece of work in 17 days. How many days will two men and six boys together take to finish the same work?

3m = 1/17, 8b = 1/17

1m = 1/3×17 , 1b = 1/8×17

2 men and 6 boys so,

(2/3×17)×(6/8×17) = (2/3×17)×(3/4×17) = (2×4/3×17)×(3×3/4×17)

(12+9)/12×17 = 17/12×17 = 1/12 = 12 days

3. 119 typists typed 1785 pages in 1/20 hour. The number of pages typed per minute by an average typist is

In 1/20 hour(3 minutes), 119 typists can type 1785 pages.

In 1 minute, 119 typists can type 1785/3 pages

In 1 minute, 1 typist can type 1785/3×119

= 5 pages

4. A is twice as good a workman as B. Together, they finish the work in 14 days. In how many days can it be done by each of them separately?

If the work done by B in 1 day is 1/x then by A is 2/x

According to the question, 14(1/x + 2/x) = 1 or, 14(1+2/x) = 1

or, x=42

B can do this work in 42/1 = 42 days.

A can do the work in 42/2 = 21 days.

5. A can do a piece of work in 28 days. If B is 40% more efficient than A then the number of days required by B to do the same piece of work is

A’s one day’s work = 1/28

B’s 1 day’s work = 140% of 1/28

= 140/100 × 1/28 = 1/20

Hence B can do the same piece of work in 20 days.

6. A can do a piece of work in 8 days. B can do it in 10 days. With the help of C they finish the work in 2 1/2 days. In how many days C alone can do the whole work?

Efficiency(%) of A = 100/8 = 12.5%

Efficiency(%) of B = 100/10 = 10%

Efficiency of A, B and C is = 100/2.5 = 40%

Efficiency of C = Efficiency of A, B and C – Efficiency of A and B

= 40 – (12.5 + 10) = 17.5%

No. of days = 100/17.5 = 1000/175 = 40/7 = 5 5/7

7. Suresh can complete a job in 15 hours. Ashutosh also can complete the same job in 10 hours. If suresh worked for 9 hours and then quit how many hours will Ashutosh alone take to complete the remaining work?

In 9 hours Suresh can finish 9/15 = 3/5 work

Suppose Ashutosh can finish the same work in x hours.

Then, 9/15 + x/10 = 1or, 18+3x/30 =1

or, x= 12/3 hr = 4 hours.

8. B is thrice efficient as A and A can do a piece of work in 21 days. A started working and after a few days B joined him. They completed the work in 15 days from the beginning. For how many days did they work together?

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