Question

A and B together can complete a task in 20 days. B and C together can complete the same task in 30 days. A and C together can complete the same task in 40 days. What is the respective ratio of the number of days taken by A when completing the same task alone to the number of days taken by C when completing the same task alone ?

• A. 2 : 5
• B. 2 : 7
• C. 3 : 7
• D. 1 : 5

Answer 1

The Correct Option is D

The task is to find the ratio of the number of days taken by A to complete the task alone to the number of days taken by C. Let's denote A's, B's, and C's respective times to complete the task alone as a, b, and c days. From the problem, we have the following daily work rates:

1. A and B together: \(\frac{1}{a} + \frac{1}{b} = \frac{1}{20}\)

2. B and C together: \(\frac{1}{b} + \frac{1}{c} = \frac{1}{30}\)

3. A and C together: \(\frac{1}{a} + \frac{1}{c} = \frac{1}{40}\)

We need to find \(\frac{a}{c}\). First, let's add the three equations:

\(\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{a}+\frac{1}{c}\right)=\frac{1}{20}+\frac{1}{30}+\frac{1}{40}\)

Simplifying, we get:

\(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{20}+\frac{1}{30}+\frac{1}{40}\)

Calculating the right side: \(\frac{1}{20}+\frac{1}{30}+\frac{1}{40}=\frac{6+4+3}{120}=\frac{13}{120}\)

Thus, \(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{13}{120}\) implies \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{13}{240}\)

Subtract each pair from \(\frac{13}{240}\) to find individual values:

\(1. \frac{1}{c}=\frac{13}{240}-\frac{1}{20}=\frac{13}{240}-\frac{12}{240}=\frac{1}{240}\) so \(c=240\)

\(2. \frac{1}{b}=\frac{13}{240}-\frac{1}{40}=\frac{13}{240}-\frac{6}{240}=\frac{7}{240}\) so \(b=240/7\)

\(3. \frac{1}{a}=\frac{13}{240}-\frac{1}{30}=\frac{13}{240}-\frac{8}{240}=\frac{5}{240}\) so \(a=48\)

Now, the ratio \(\frac{a}{c}=\frac{48}{240}=\frac{1}{5}\)

Thus, the respective ratio of days taken by A to C is 1 : 5.