Work Rate using Joint Variation | Arithmetic

Today, let’s discuss work-rate using joint variation. Though we will use the concepts of joint variation, we will solve the questions using a simple methodology.

Question 1: A contractor undertakes to do a job within 100 days and hires 10 people to do it. After 20 days, he realizes that one fourth of the work is done so he fires 2 people. In how many more days will the work get over?

(A) 60

(B) 70

(D) 80

(E) 100

Solution: Can we say that 10 people can finish the work in 100 days? No. If that were the case, after 20 days, only $\frac{1}{5}th$ of the work would have been over. But actually $\frac{1}{4}th$ of the work is over. This means that ‘10 people can complete the work in 100 days’ was just the contractor’s estimate (which turned out to be incorrect). Actually 10 people can do $\frac{1}{4}th$ of the work in 20 days. The contractor fires 2 people. So the question is how many days are needed to complete $\frac{3}{4}th$ of the work if 8 people are working?

We need to find the number of days. How is ‘no. of days’ related to ‘no. of people’ and ‘work done’?

If we have more ‘no. of days’ available, we need fewer people. So ‘no. of days’ varies inversely with ‘no. of people’.

If we have more ‘no. of days’ available, ‘work done’ will be more too. So ‘no. of days’ varies directly with ‘work done’.

Therefore,

$\frac{'no.\ of\ days' \times 'no.\ of\ people'}{'work\ done'} = constant$

$\frac{20 \times 10}{\frac{1}{4}} = \frac{'no.\ of\ days' \times 8}{\frac{3}{4}}$

No. of days = 75

So, the work will get done in 75 days if 8 people are working.

We can also do this question using simpler logic. The concept used is joint variation only. Just the thought process is simpler.

10 people can do $\frac{1}{4}th$ of the work in 20 days.

8 people can do $\frac{3}{4}th$ of the work in $x$ days.

Start with the no. of days since you want to find the no of days:

$x = 20 \times \frac{10}{8} \times \frac{3}{1} = 75$

From where do we get $\frac{10}{8}$ ? No. of people decreases from 10 to 8. If no. of people is lower, the no of days taken to do the work will be more. So 20 (the initial no. of days) is multiplied by $\frac{10}{8}$ , a number greater than 1, to increase the number of days.

From where do we get $\frac{3}{1}$ ? Amount of work increases from $\frac{1}{4}$ to $\frac{3}{4}$ . If more work has to be done, no. of days required will be more. So we further multiply by $\frac{\frac{3}{4}}{\frac{1}{4}}$ i.e. $\frac{3}{1}$ , a number greater than 1 to further increase the number of days.

This gives us the expression $20 \times \frac{10}{8} \times \frac{3}{1}$

We get that the work will be complete in another 75 days.

Answer (C)

Let’s take another question to ensure we understand the logic.

Question 2: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in 10 days. In the next 6 days, the company will need to run 9 cars for 12 hrs each so it rents 5 more cars which consume 20% less fuel than the company’s four cars. How many lts of fuel will be consumed in the next 6 days?

(A) 1200 lt

(B) 1555 lt

(D) 1728 lt

(E) 4800 lt

Solution: First let’s try to figure out what is meant by ‘consume 20% less fuel than the company’s cars’. It means that if company’s each car consumes 1 lt per hour, the hired cars consume only $\frac{4}{5}$ lt per hour. So renting 5 more cars is equivalent to renting 4 cars which are same as the company’s cars. Hence, the total number of cars that will be run for the next 6 days is 8 company-equivalent cars.