A alone can do a piece of work in 12 days B alone can do the same work in 18 days C alone can do the...........

Quantitative Aptitude (MTS) Time and work

A alone can do a piece of work in 12 days. B alone can do the same work in 18 days. C alone can do the same work in 24 days.

In how many days will they complete 65% of the work, if all three work together?

This questions was previously asked in

SSC MTS 8th July 2022 Shift-1

3.6 days

3.9 days

4.2 days

4.5 days

- To solve this, we need to find out the combined work rate of A, B, and C.

- A's work rate is \( \frac{1}{12} \) of the work per day.

- B's work rate is \( \frac{1}{18} \) of the work per day.

- C's work rate is \( \frac{1}{24} \) of the work per day.

- Together, their combined work rate per day is \( \frac{1}{12} + \frac{1}{18} + \frac{1}{24} \).

- The least common multiple of 12, 18, and 24 is 72. Convert each fraction:

- \( \frac{1}{12} = \frac{6}{72} \)

- \( \frac{1}{18} = \frac{4}{72} \)

- \( \frac{1}{24} = \frac{3}{72} \)

- Together, they do \( \frac{6 + 4 + 3}{72} = \frac{13}{72} \) of the work per day.

- To complete 65% of the work:

- 65% of work is \( 0.65 \) of the complete work.

- Days required: \( \frac{0.65}{\frac{13}{72}} = \frac{0.65 \times 72}{13} \).

- Calculate the answer: \( \frac{46.8}{13} \approx 3.6 \) days.

Therefore, Option 1: 3.6 days is correct.

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