Question

. A, B and C can complete a work in 10, 12 and 15 days respectively. All three of them starts together but

after 2 days A leaves the job and B left the job 3 days before the work was completed. C completed the

remaining work alone. In how many days was the total work completed?​

Answer 1

Step-by-step explanation:

A left after working 2days,

B leaves 3 days before completion of work----(x-3)

2/10+(x-3)/12+x/15=1

LCM of 10,12,15 is 60

6×2+5(x-3)+4x=60

12+5x-15+4x=60

9x-3=60

9x=60+3

x=63/9

x=7

Total work completed in 7 days

Answer 2

In 7 days, the total work was completed

Solution:

Given that,

A, B and C can complete a work in 10, 12 and 15 days respectively

Given that,

After 2 days, A leaves the job

So, A worked for 2 days alone

Let "x" be the days in which total work was completed

B left the job 3 days before the work was completed

Therefore,

B worked for (x - 3) days

C completed the  remaining work alone

Thus C worked for "x" days

Therefore,

$\frac{2}{10} + \frac{x-3}{12} + \frac{x}{15} = 1\\\\Make\ the\ denominators\ same\\\\\frac{2 \times 6}{10 \times 6} + \frac{(x-3) \times 5}{12 \times 5} + \frac{x \times 4}{15 \times 4} = 1\\\\12 + 5x - 15 + 4x = 60\\\\9x - 3 = 60\\\\9x = 63\\\\x = 7$

Thus, in 7 days, the total work was completed

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