### Question 3

SSC-CGL 2018 June 4 Shift 2

The ratio of competencies A, B and C is 4: 5: 3. On working together, all three of them complete work in 25 days. In how many days will both A and C together complete 35% of the work?

- 18
- 15
- 12
- 10

### Solution 1

Work that A+B+C can together do in 1 day is $\displaystyle 4k + 5k + 3k = 12k$.

Total work in 25 days = $\displaystyle 25 \times 12k = 300k$

Days taken by A+C to complete (4k + 3k) = 7k work is = 1 day

$\displaystyle \therefore $ days to complete $\displaystyle 35\% \times 300k$ will be $\displaystyle \frac{1}{7k} \times 35\% \times 300k = 15$ days Ans.

EXPLANATION: efficiency means "the amount of work done in 1 day". Ratios are given so we should say that A completes 4k work in one day, B completes 5k, and likewise C completes 3k work. They together complete (4k + 5k + 3k) work in a day. On the same lines, we can conclude that A and C together complete (4k + 3k) amount of work in a day.### Solution 2

This is the classic, school math approach.

Total units of work are $\displaystyle 25$

A + C can complete $\displaystyle \frac{4 + 3}{4 + 5 + 3} = \frac{7}{12}$ work in 1 day

Work to be done by A and C is $\displaystyle 35 \% \times 25$ units

Hence, they complete this work in $\displaystyle \frac{1}{7/12} \times 35 \% \times 25 = 15$ days Ans.

### Solution 3

Use the principle that $\displaystyle \frac{M \times D}{W}$ is constant. M = number of men, D = days they work, and W is the fraction, or amount of work they do.

When A+B+C work together, $\displaystyle M_1 = 4k + 5k + 3k = 12k$, $\displaystyle D_1 = 25, W_1 = 1$

When A+C work together, $\displaystyle M_2 = 4k + 3k = 7k$, $\displaystyle D_2 = x, W_2 = 0.35$

Now, $\displaystyle \frac{M_1 \times D_1}{W_1} = \frac{M_2 \times D_2}{W_2}$

Hence, $\displaystyle \frac{12k \times 25 }{1} = \frac{7k \times x}{0.35}$

Solving, $\displaystyle x = 15$ days, Ans.

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