### Question 2

SSC-CGL 2020 Mar 3 Shift 1

A, B and C can do a work in 24, 12 and 15 days respectively. All start a work together. After 3 days B and C leave the work. How many days will A take to complete the work?

### Solution in Brief

Suppose the total work is LCM(24, 12, 15) = 120. Efficiency of A = 120/24 = 5, similarly of B is 10 and of C is 8. Work completed in 3 days is (5 + 10 + 8) x 3 = 69. Remaining work is 120 - 69 = 51. Time by A to complete this will be 51/5 = 10-1/5 days answer!

### Solution in Detail

The approach in our basic math books is to assume total work as 1. But calculations can be simplified if we assume total work as LCM of the days taken by each of the workers to complete a given work.

Total work = LCM$\displaystyle (24, 12, 15) $

$\displaystyle \implies$ Total Work $\displaystyle = 120 \text{ units}$

[given] A takes 24 days for 120 units

[1]$\displaystyle \therefore $ work by A in 1 day = $\displaystyle \frac{120}{24} = 5$

likewise, work by B = $\displaystyle \frac{120}{12} = 10$

and, work by C in 1 day = $\displaystyle \frac{120}{15} = 8$

A+B+C work for 3 days

Work done = $\displaystyle 3 \times (5 + 10 +8) = 69$

Remaining work is 120 - 69 = 51

But A completes 5 units in 1 day

$\displaystyle \therefore $ he completes 51 in $\displaystyle \frac{1}{5} \times 51 $ days

Or, in $\displaystyle 10\frac15$ days answer!

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