1). 102). $6\frac{1}{3}$3). $7\frac{1}{3}$4). 7

1.	1). 102). \(6\frac{1}{3}\)3). \(7\frac{1}{3}\)4). 7
Answer» A can complete the WORK in 10 days. ⇒ A’s 1day’s work = 1/10 B can complete the work in 20 days. ⇒ B’s 1 day’s work = 1/20 C can complete the work in 15 days. ⇒ C’s 1day’s work = 1/15 (A + B)’s 1 day’s work = (1/10) + (1/20) = 3/20(A + B implies A and B) (A + C)’s 1 day’s work = (1/10) + (1/15) = 1/6(A + C implies A and C) According to the QUESTION, A is assisted by B and C on ALTERNATE days. ⇒Their 2 day’s work = (3/20) + (1/6) = 19/60 [? (A + B) work on 1^st day and (A + C) work on 2^nd day] ⇒ Their (3 × 2 =) 6 day’s work = 3 × (19/60) = 57/60 ⇒ Remaining work = 1 - (57/60) = 3/60 = 1/20 On 7^th day (A + B) will work. (A + B) do (3/20)^th work in 1 day ⇒ Time TAKEN by (A + B) to do remaining work = (1/3) days ∴ The work will get completed in = $(6 + \frac{1}{3} \Rightarrow 6\frac{1}{3})$ days

Answer»

A can complete the WORK in 10 days.

⇒ A’s 1day’s work = 1/10

B can complete the work in 20 days.

⇒ B’s 1 day’s work = 1/20

C can complete the work in 15 days.

⇒ C’s 1day’s work = 1/15

(A + B)’s 1 day’s work = (1/10) + (1/20) = 3/20(A + B implies A and B)

(A + C)’s 1 day’s work = (1/10) + (1/15) = 1/6(A + C implies A and C)

According to the QUESTION, A is assisted by B and C on ALTERNATE days.

⇒Their 2 day’s work = (3/20) + (1/6) = 19/60

[? (A + B) work on 1^st day and (A + C) work on 2^nd day]

⇒ Their (3 × 2 =) 6 day’s work = 3 × (19/60) = 57/60

⇒ Remaining work = 1 - (57/60) = 3/60 = 1/20

On 7^th day (A + B) will work.

(A + B) do (3/20)^th work in 1 day

⇒ Time TAKEN by (A + B) to do remaining work = (1/3) days

∴ The work will get completed in = $(6 + \frac{1}{3} \Rightarrow 6\frac{1}{3})$ days

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