Question

A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?​

Answer 1

Solution :-

Work done by A = $\dfrac{6}{24}$ = $\dfrac{1}{4}$

Work done by B = $\dfrac{(x-6)}{32}$

(where x is no. of days in which work is competed)

We know that $\dfrac{1}{4}$ + x – $\dfrac{6}{32}$ + $\dfrac{x}{64}$ = 1

We know 16 + 24 – 12 + $\dfrac{x}{64}$ = 1

We know 3x + 4 = 64

We know x = $\dfrac{60}{3}$ = 20 days

Answer 2

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Given

A----24

B----32

C----64

6day work of A+B+C=17*6=102

Remaining work=90

C's work fr 6dy=6*3=18

90-18=72

Done by B+C=72/9=8

6+6+8=20 days

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