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A can do a piece of work in 10 days, B in 12 days, and C in 15 days. All begin together but A leaves the work after 2 days and B leaves 3 days before the work is finished. How long did the work last?

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#### Solution

A's one day's work =`1/10`

B's one day's work =`1/12`

C's one day's work =`1/15`

A, B and C's one day's work =`1/10+1/12+1/15`

`=(6+5+4)/60=15/60=1/4`

Let the work completed in x days

∴ A's 2 days work + B's (x − 3) days work + C's x days work = 1

`⇒2xx1/10+("x"-3)xx1/12+"x"xx1/15=1`

`⇒1/5+("x"-3)/12+"x"/15=1`

`(12+5"x"-15+4"x"=60)/60` ...(L.C.M of 5, 12, 15 = 60)

⇒ 9x = 60 − 12 + 15

⇒ 9x = 63

`⇒"x"=63/9=7`

∴ Work will last for 7 days

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