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# Aptitude::Time and Work

Home > Quantitative Aptitude > Time and Work > Subjective Solved Examples

Example 1 / 11

Worker A takes 10 hours to do a job and worker B takes 6 hours to do the same work. How long should it take both A and B working together to complete the same job?
A's 1 hour work = $$\frac{1}{10}$$
B's 1 hour work= $$\frac{1}{6}$$
(A+B)'s 1 hour work will be = $$\left( \frac{1}{10} + \frac{1}{6} \right)$$= $$\frac{8}{30}$$
Therefore both A and B will finish the work in $$\frac{30}{8}$$ hrs. = 3 hrs. 45 min. Ans.

Always calculate 1(day/hr/minute)work done by a particular person.
Also if the units are different we should convert the units according to the options.

For Ex.
A would have taken 200 minutes and B the same 6 hours and options were in minutes then our equation would be:
=$$\left(\frac{1}{200}+\frac{1}{360}\right)$$ part together in 1 minute = 0.007
So they will complete the entire work in 128.57 minutes

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If in the given question, it is asking to calculate the work done by both together then the formula is:
= $$\frac{PRODUCT\text{ } OF \text{ }THE \text{ }TWO \text{ }WORK}{SUM \text{ }OF \text{ }THE \text{ }WORK}$$
= $$\frac{10\text{ }*\text{ }6}{10\text{ }+\text{ }6}$$
= $$\frac{60}{16}$$
= $$\frac{15}{4}$$ hours. = 3 hrs 45 min.

adelle abdallah7 months ago

3.75

|| Reply

Example 2 / 11

A and B together can complete a piece of work in 4 days. If A alone can complete the piece of work in 10 days then how long will it take for B alone to complete the same work?
(A+B)'s 1 day work = $$\frac{1}{4}$$.
A's 1 day work = $$\frac{1}{10}$$
So B's 1 day work will be = ($$\frac{1}{4}-\frac{1}{10}$$)=$$\frac{3}{20}$$
So B will be able to complete the entire work in $$\frac{20}{3}$$ = 6$$\frac{2}{3}$$ days. OR 6 hrs. 40 min.Ans.
It is similar to type 1. You know the total number of days in which the work is going to be complete so just calculate the work done in 1 day by both of them together and then subtract A's 1 day from it. You will get the answer.
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If the work done together is given and work done by one is given, we have to calculate the work done by another person alone, and then we have the formula as

=$$\frac{PRODUCT}{DIFFERENCE}$$
= $$\frac{4*10}{(10 - 4)}$$= $$\frac{40}{6}$$ days
= $$\frac{20}{3}$$ days
= 6 hrs. 40 min.

adelle abdallah7 months ago

6.66

|| Reply

Example 3 / 11

A can do a piece of work in 7 days of 10 hours each and B can do 6 days of 6 hours each. How long will they take to do it, working together 5 hours a day?
A can complete the work in (7*9) = 63 hours
B can complete the work in (6*6) = 36 hours
A's 1 hour = $$\frac{1}{63}$$ and B's 1 hour work = $$\frac{1}{36}$$
Together in 1 hour =($$\frac{1}{63} + \frac{1}{36}$$) = $$\frac{11}{252}$$
Both will finish the work in 25$$\frac{2}{11}$$ hrs.
Number of days for 5 hrs each is 25$$\frac{2}{11} * \frac{1}{5}$$ = 4.58 days Ans.

Calculate the total hours taken by them and then work done by them in 1 hour. It's easy when we get them in 1 single unit that is hours. Rest is same as Example Type 1
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adelle abdallah7 months ago

4.75 days

|| Reply

Example 4 / 11

A & B can do a piece of work in 18 days. B& C can do it in 24 days. A& C can do it in 36 days. In how many days will A,B and C finish it, working together and separately?
(A+B)'s 1 day work = $$\frac{1}{18}$$
(B+C)'s 1 day work = $$\frac{1}{24}$$
(A+C)'s 1 day work = $$\frac{1}{36}$$
Adding all we get = 2(A+B+C)'s 1 day's work = ($$\frac{1}{18} + \frac{1}{24} + \frac{1}{36}$$) = $$\frac{9}{72}$$ = $$\frac{1}{8}$$ ------ eqn1
Hence (A+B+C)'s 1 days work = $$\frac{1}{16}$$
Now A's 1 day work alone =[(A+B+C)'s 1 day's wor -(B+C)'s 1 day's work]
= ($$\frac{1}{16} - \frac{1}{24}$$) = $$\frac{1}{48}$$ ------- eqn2
Hence A alone can finish the work in 48 days
Similarly, B and C can do in 28$$\frac{4}{5}$$ days and 144 days respectively.

In eqn 1 & 2 (A+B+C) is used cause we are simply summing the work done by each of them when they are paired up as given in the question. Now once we know in how many days the complete work can be done then we can simply subtract any pair's unit day work from the total unit day work to get an individual day's work.(eqn2)
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Example 5 / 11

A is twice as good as workman as B and together they finish a piece of work in 18 days. In how many days will A alone finish the work?
(A's 1 day's work : B's 1 day's work) = 2 : 1
(A+B)'s 1 day's work = $$\frac{1}{18}$$
Hence A's 1 day work = $$\frac{1}{18} * \frac{2}{3} = \frac{1}{27}$$
So, A can finish the whole work = 27 days.Ans
It's a simple question of ratio and proportion. Here 2/3rd is the part of work done by A according to the relation given in question.
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Example 6 / 11

A can do a certain job in 12 days. B is 80% more efficient than A. How many days does B alone take to do the same job?
Ratio of times taken by A and B = 180:100= 9:5
Suppose B alone takes x days to do the job
Then 9:5::12:x => 9x=60 => x = 20/3 days.Ans
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Example 7 / 11

A can do a piece of work in 100 days. He works at it for 10 days and then B alone finishes the remaining work in 40 days. In how much time will A and B working together will finish the same work?
Work done by A in 10 days = $$\frac{1}{100} * 10$$= $$\frac{1}{10}$$
Remaining work= 1-$$\frac{1}{10}$$ = $$\frac{9}{10}$$
Now $$\frac{9}{10}$$ work is done by B in 40 days
Whole work will be done by B in $$\left(40 * \frac{10}{9} \right)$$ = 44.44 days
A's 1 day's work = $$\frac{1}{100}$$ and B's 1 day's work = $$\frac{1}{44.44}$$
(A+B)'s 1 day's work = $$\left( \frac{1}{100} + \frac{1}{44.44} \right )$$ =.032
Hence both will finish the work in 31.25 days Ans.
Calculate the work left after A has worked for 10 days. Then that remaining work will be completed by B. From there you can get how many days B will take to complete the entire work. Similar to Example Type 1.
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Example 8 / 11

A and B undertake to do a piece of work for Rs 800. A alone can do it in 6 days and B alone can do it in 8 days. With the help of C, they finish it in 3 days. Find the share of each.
C's 1 day's work= $$\left(\frac{1}{3} - \frac{1}{6} + \frac{1}{8}\right)$$ = $$\frac{1}{24}$$
A:B:C= Ratio of their 1 day's work = $$\frac{1}{6}$$ : $$\frac{1}{8}$$ : $$\frac{1}{24}$$ = 4:3:1
Hence A's share = 800 * $$\frac{4}{8}$$ = 400 Rs.
B's share = 800 * $$\frac{3}{8}$$ = 300 Rs.
So C's share= $$\left(800 - 400 + 300\right)$$ = 100Rs. Ans.

Attempt such questions by calculating work done by them in 1 day.
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Example 9 / 11

A and B working separately can do a piece of work in 5 and 10 days respectively. If they work for a day alternately , A beginning in how many days the work will be completed?
(A+B)'s 2 day's work = $$\left(\frac{1}{5} + \frac{1}{10}\right)$$ = $$\frac{3}{10}$$
Work was done in 3 pairs of days = 3*$$\frac{3}{10}$$ = $$\frac{9}{10}$$
Remaining work= 1-$$\frac{9}{10}$$ = $$\frac{1}{10}$$
On the 11th day it's A's turn, $$\frac{1}{5}$$ work is done by him in 1 day
So, $$\frac{1}{10}$$ work will be done by him in $$5 * \frac{1}{10}$$ = $$\frac{1}{2}$$ day
Therefore total time taken = 6 + $$\frac{1}{2}$$ = 6$$\frac{1}{2}$$ days Ans.
It a tricky question. One individual will only work for 1 day that too alternately. Now when it is mentioned that A is beginning then after even number of days we need to calculate number of days taken only by A to complete the work. If B was asked then the same approach. Then after the 11th day we will only consider A.
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Example 10 / 11

50 men can complete work in 16 days. Six days after they have started working 30 more men joined them. How many days will they now take to complete the remaining work?
(50*16) men can complete the work in 1 day
Therefore 1 man's 1 day's work = 1/800
50 men's 6 day's work = $$\frac{1}{16} * 6$$ = $$\frac{3}{8}$$, Remaining work $$1 - \frac{3}{8}$$ = $$\frac{5}{8}$$
80 men's 1 day's work = $$\frac{80}{800}$$ = $$\frac{1}{10}$$
Now, $$\frac{1}{10}$$ work is done by them in 1 day
Therefore $$\frac{5}{8}$$ work is done by then in $$10 * \frac{5}{8}$$ = $$\frac{50}{8}$$ days.Ans.

It's a straight logic that if men are increasing then days required for doing that work will decrease. You can go by option verification provided that 3 options are more than 16 days.
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Example 11 / 11

2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 1 man and 1 boy do the work?
Let 1 man's 1 day,s work = x and 1 boy's 1 day's work = y
Then, 2x + 3y = $$\frac{1}{10}$$ and 3x + 2y = $$\frac{1}{8}$$
By Solving above both eq. , we get : x = $$\frac{7}{200}$$ and y = $$\frac{1}{100}$$
(1 man's + 1 boy's ) 1 day's work = $$1 * \frac{7}{200} + 1 * \frac{1}{100}$$ = $$\frac{9}{200}$$
So, 1 man and 1 boy together can finish the work in $$\frac{200}{9}$$ = 22.22 days. Ans
Same the unitary method approach and also using simultaneous equations solving you should know. Calculate for work done by each boy and each man in 1 day.
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