Question

A and B together complete a task in 24 days,C and B together complete the same task in 18 days,and A and C together complete the task in 12 days.How many days will A take to complete the task if he works alone?​

Answer 1

Answer:-

Given:-

Amount of work done by A & B together in 24 days = 1

⟹ Amount of work done by A & B together in 1 day = 1/24

⟹ A's efficiency + B's efficiency = 1/24 -- equation (1)

Similarly;

⟹ B's efficiency + C's efficiency = 1/18 -- equation (2)

⟹ A's efficiency + C's efficiency = 1/12 -- equation (3)

Subtract equation (2) from (1).

⟹ A + B - (B + C) = (1/24) - (1/18)

⟹ A + B - B - C = (3 - 4)/72

⟹ A - C = - 1/72 -- equation (4)

Adding equations (3) & (4) we get,

⟹ A + C + A - C = 1/12 - 1/72

⟹ 2A = (6 - 1)/72

⟹ A = (5/72)(1/2)

⟹ A = 5/144

Now,

Let,

Amount of work done by A in x days = 1

⟹ Amount of work done by A in 1 day = 1/x

⟹ 5/144 = 1/x

⟹ 5x = 144

⟹ x = 144/5

⟹ x = 28.8 days.

∴ A alone can complete the task in 28.8 days.

Answer 2

Answer:

Question :-

A and B together complete a task in 24 days,C and B together complete the same task in 18 days,and A and C together complete the task in 12 days.How many days will A take to complete the task if he works alone?

Full Solution :-

A and B together do (1/14)th of the task in 1 day.

B and C together do (1/8)th of the task in 1 day.

A and C together do (1/7)th of the task in 1 day.

Adding all the three we get

2A+2B+2C do (1/14)+(1/8)+(1/7) = (56+98+112)/784 = 266/784= 133/392, or

A+B+C = 133/784 = 19/112

C does (19/112)-(1/14) = (19–8)/112 = 11/112th of the work in 1 day.

A does (19/112)-(1/8) = (19–14)/112 = 5/112th of the work in 1 day.

B does (19/112)-(1/7) = (19–16)/112 = 3/112th of the work in 1 day.

The least efficient person is B, who will do the task, alone, in 112/3 or 37.33 days.

Next is A, who will do the task, alone, in 112/5 or 22.4 days.

C, the most efficient, will do the task, alone, in 112/11 or 10.18 days.