# Aptitude::Time and Work

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Example 1 / 11

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.

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.

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

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

NA

If in the given question, it is asking to calculate the work done by both together then the formula is:

=

= \(\frac{10\text{ }*\text{ }6}{10\text{ }+\text{ }6}\)

= \(\frac{60}{16}\)

= \(\frac{15}{4}\) hours. = 3 hrs 45 min.

=

**\(\frac{PRODUCT\text{ } OF \text{ }THE \text{ }TWO \text{ }WORK}{SUM \text{ }OF \text{ }THE \text{ }WORK}\)**= \(\frac{10\text{ }*\text{ }6}{10\text{ }+\text{ }6}\)

= \(\frac{60}{16}\)

= \(\frac{15}{4}\) hours. = 3 hrs 45 min.

Example 2 / 11

(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.

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.

NA

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

=

= \(\frac{4*10}{(10 - 4)}\)= \(\frac{40}{6}\) days

= \(\frac{20}{3}\) days

= 6 hrs. 40 min.

=

**\(\frac{PRODUCT}{DIFFERENCE}\)**= \(\frac{4*10}{(10 - 4)}\)= \(\frac{40}{6}\) days

= \(\frac{20}{3}\) days

= 6 hrs. 40 min.

Example 3 / 11

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

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

NA

NA

Example 4 / 11

(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.

(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)

NA

NA

Example 5 / 11

(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.

(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.

NA

NA

Example 6 / 11

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.

Suppose B alone takes x days to do the job

Then 9:5::12:x => 9x=60 => x = 20/3 days.

**Ans**
NA

NA

NA

Example 7 / 11

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

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.

NA

NA

Example 8 / 11

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.

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.

NA

NA

Example 9 / 11

(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

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.

NA

NA

Example 10 / 11

(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.

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.

NA

NA

Example 11 / 11

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.

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.

NA

NA

Komal Mhaske1 month ago

3.15 hrs