### Question 9

SSC-CGL 2018 June 4 Shift 1

The ratio of efficiencies of A, B and C is 2 : 5 : 3. On working together, all three of them can complete work in 27 days. In how many days will both B and C together complete the 4/9th part of that work?

- 15 days
- 17-1/7
- 27
- 24

### Solution in Brief

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

$\displaystyle = 2k + 5k + 3k = 10k$

But A+B+C take a total of 27 days

$\displaystyle \therefore \text{ total work } = 270k$

B+C do (5k + 3k) = 8k in 1 day

$\displaystyle \therefore $ they do $\displaystyle 270k$ in $\displaystyle \bigg( \frac{1}{8k} \times 270k \bigg)$

$\displaystyle = 15$ days Ans.

### Solution 2

This is the classic, school math approach.

Total units of work are $\displaystyle 27$

Work to be done by B and C is $\displaystyle \frac 49 \times 27 = 12$ units

B + C can complete $\displaystyle \frac{5 + 3}{2 + 5 + 3} = \frac{4}{5}$ work in 1 day

Hence, they complete $\displaystyle 12$ units in $\displaystyle \frac{1}{4/5} \times 12 = 15$ 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, $\displaystyle M_1 = 2k + 5k + 3k = 10k$, $\displaystyle D_1 = 27, W_1 = 1$

When B+C work together, $\displaystyle M_2 = 5k + 3k = 10k$, $\displaystyle D_2 = x, W_2 = \frac 49$

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

Hence, $\displaystyle \frac{10k \times 27 }{1} = \frac{8k \times x}{\frac 49}$

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

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