Question

A, b and c can do a piece of work in 18 days, 27 days and 36 days respectively. they start working together. after working for 4 days. a goes away and b leaves 7 days before the work is finished. only c remains at work from beginning to end. in how many days was the whole work done?

Answer 1

Given:

A, B and C can do a piece of work in 18 days, 27 days and 36 days respectively.

All 3 work together for 4 days

After 4 days A leaves and B leaves before 7 days the work is finished

To find:

In how many days was the whole work done?

Solution:

So,

In 1 day, the amount of work A will do = $\frac{1}{18}$

In 1 day, the amount of work B will do = $\frac{1}{27}$

In 1 day, the amount of work C will do = $\frac{1}{36}$

∴ The amount of work done by A, B & C together in 4 days is,

= $4[\frac{1}{18} +\frac{1}{27} +\frac{1}{36} ]$

= $4[\frac{6+4+3}{108} ]$

= $4[\frac{13}{108} ]$

= $\frac{52}{108}$

= $\frac{13}{27}$

∴ The remaining amount of work to be done after A leaves,

= $1 - \frac{13}{27}$

= $\frac{27-13}{27}$

= $\frac{14}{27}$

Let "x" be the no. of days when B and C works together.

We know that B leaves 7 days before the work is finished, so C will do the remaining work in 7 days.

Now, we can form the equation as,

[Amount of work done by B and C in x days] + [Amount of work done by C in 7 days] = $\frac{14}{27}$

⇒ $x[\frac{1}{27} + \frac{1}{36} ] + [\frac{7}{36} ] = \frac{14}{27}$

⇒ $x[\frac{4 + 3}{108} ] + [\frac{7}{36} ] = \frac{14}{27}$

⇒  $x[\frac{7}{108} ] + [\frac{7}{36} ] = \frac{14}{27}$

⇒  $x[\frac{7}{108} ] = \frac{14}{27} - \frac{7}{36}$

⇒  $x[\frac{7}{108} ] = \frac{56 - 21 }{108}$

⇒ $x = \frac{35}{7}$

⇒ $x = 5$

∴ Total no. of days taken to complete the whole work = 4 + 5 + 7 = 16 days

Thus, in 16 days, the whole work was done.

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