Question

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?​

Answer 1

Answer:

Step-by-step explanation:

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 = ¼

Let the work completed in x days

∴ A’s 2 days  work + B’s (x – 3) days work + C’s x days work = 1

⇒ 2 × (1/10) + (x – 3) × 1/12 + x × (1/15) = 1

⇒ 1/5 + (x -3)/12 + x/15 = 1

(12 + 5x – 15 + 4x = 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 day