A takes thrice the time taken by B and five times the time taken by C to complete a work. Working together, they can complete the work in 5 days. Find the number of days taken by A to complete the work alone.

Correct option is D

Given:

A takes 3 times the time taken by B

A takes 5 times the time taken by C

A, B, and C together complete the work in 5 days

We are to find how many days A alone will take to complete the work.

Solution:​
Let the time taken by A = x days
Then,

Time taken by B =$$\dfrac{x}{3}$$​​

Time taken by C =$$\dfrac{x}{5}$$​​
Their work rates:

​$$A: \dfrac{1}{x}, B: \dfrac{3}{x}, C: \dfrac{5}{x}$$​​

Total work per day $$= \dfrac{1}{x} + \dfrac{3}{x} + \dfrac{5}{x} = \dfrac{9}{x}$$​​

Since together they take 5 days:

​$$\frac{9}{x} = \frac{1}{5}$$​

x = 45

Alternate Method :

Let total work = LCM of the ratios = LCM of 3, 1, and 5 = 15 units of time
Let:

A takes 15 days

Then B =$$\frac{15}{3}$$​= 5 days

C =$$\frac{15}{5}$$​ = 3 days

Work per day:

​$$A = \frac{1}{15}, B = \frac{1}{5}, C = \frac{1}{3}$$​​

Together per day:

​$$\frac{1}{15} + \dfrac{1}{5} + \dfrac{1}{3} = \dfrac{1 + 3 + 5}{15} = \dfrac{9}{15} = \dfrac{3}{5}$$​​

Time to finish 1 work =$$\frac{1}{\frac35} = \frac{5}{3}$$days ( which is not equal to 5 days ).
So scale up total work to make total time = 5 days.

Let total work = LCM of 3, 5, 15 = 45 units

A’s rate = 1 unit/day

B = 3 units/day

C = 5 units/day
Combined = 9 units/day

​$$\text{Time} = \dfrac{45}{9} = 5 \text{ days }$$​ 

​$$​ \text{A alone takes } \dfrac{45}{1} = 45 \text{ days}$$​​