Question

A, b and c can complete a work separately in 24, 36 and 48 days respectively. They started together but c left after 4 days of start and a left 3 days before the completion of the work. In how many days will the work be completed?

Answer 1

$\bf\huge\underline\red{Answer}$

Let A take $\bf\frac{1}{24}$ days to complete a piece of work alone in 1 day.

Let B take $\bf\frac{1}{36}$ days to complete a piece of work alone in 1 day.

Let C take $\bf\frac{1}{48}$ days to complete a piece of work alone in 1 day

•°• Work done by A, B and C together,

A + B + C = $\bf\frac{1}{24}$ + $\bf\frac{1}{36}$ + $\bf\frac{1}{48}$

A + B + C = $\bf\frac{13}{144}$ $\bf\underbrace{LCM \:of\:24,\:36,\: and\: 48}$

$\begin{array}{r | l}</p><p></p><p></p><p>2 & 24,\:\: 36,\:\: 48 \\</p><p></p><p>\cline{2-2} 2 & 12, \:\: 18,\:\: 24 \: \\</p><p></p><p>\cline{2-2} 3 & 4, \:\: 6,\:\: 8, \\</p><p></p><p>\cline{2-2} 2 & 2, \:\: 3, \:\: 4\\</p><p></p><p>\cline{2-2} 2 & 1, \:\:3, \:\: 2 \\</p><p></p><p></p><p>\end{array}$

We calculated the denominator (144) via finding the LCM, to find the numerator (13) we just multiply the respective number by the numerator (in this case the numerators are 1) with the numbers present in the denominator (24,36,48) that would result in the product of 144.

Work done by (A + B +C) in 4 days

= $\bf\frac{13}{144}$×4

= $\bf\frac{13}{36}$

Work done by B in 3 days = $\bf\frac{1}{36}$ × 3

= $\bf\frac{1}{12}$

Remaining work

= 1 - $\bf\frac{13}{36}$ + $\bf\frac{1}{12}$ = $\bf\frac{5}{9}$

A + B 1 day's work = $\bf\frac{1}{24}$ + $\bf\frac{1} {36}$ = $\bf\frac{5}{72}$

$\bf\frac{72}{5} <strong>×</strong><strong> </strong>$ $\bf\frac{5}{9}$ = 8 days.

Hence total time taken = 4 + 3 + 8 = 15 days.