A, B and C can finish a work in 20, 25 and 30 days, respectively. They start working together but B quits after working for 3 days. In how many days from the start shall the work be completed ?
[amp_mcq option1=”$9\frac{8}{15}$ days” option2=”$10\frac{1}{15}$ days” option3=”$10\frac{14}{25}$ days” option4=”$11\frac{1}{10}$ days” correct=”option3″]
This question was previously asked in
UPSC CAPF – 2024
A can finish the work in 20 days, so A’s daily work rate is $\frac{1}{20}$.
B can finish the work in 25 days, so B’s daily work rate is $\frac{1}{25}$.
C can finish the work in 30 days, so C’s daily work rate is $\frac{1}{30}$.
They start working together (A, B, and C). Their combined daily work rate is $\frac{1}{20} + \frac{1}{25} + \frac{1}{30}$.
To add these fractions, find a common denominator (LCM of 20, 25, 30 is 300):
Combined rate = $\frac{15}{300} + \frac{12}{300} + \frac{10}{300} = \frac{15+12+10}{300} = \frac{37}{300}$.
They work together for 3 days. Work done in the first 3 days = $3 \times \frac{37}{300} = \frac{37}{100}$.
After 3 days, B quits. The remaining work is $1 – \frac{37}{100} = \frac{63}{100}$.
The remaining work is done by A and C. Their combined daily work rate is $\frac{1}{20} + \frac{1}{30}$.
LCM of 20 and 30 is 60:
Combined rate of A and C = $\frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12}$.
Time taken by A and C to finish the remaining work = $\frac{\text{Remaining Work}}{\text{Combined Rate of A and C}} = \frac{63/100}{1/12}$.
Time taken = $\frac{63}{100} \times 12 = \frac{63 \times 3}{25} = \frac{189}{25}$ days.
The question asks for the total number of days from the start.
Total time = Time A, B, C worked together + Time A and C worked together.
Total time = 3 days + $\frac{189}{25}$ days.
Total time = $\frac{3 \times 25}{25} + \frac{189}{25} = \frac{75+189}{25} = \frac{264}{25}$ days.
Converting the improper fraction to a mixed number: $264 \div 25$. $264 = 10 \times 25 + 14$.
So, $\frac{264}{25} = 10\frac{14}{25}$ days.
A’s daily units: 300/20 = 15 units/day.
B’s daily units: 300/25 = 12 units/day.
C’s daily units: 300/30 = 10 units/day.
In the first 3 days, (A+B+C) work = (15+12+10) * 3 = 37 * 3 = 111 units.
Remaining work = 300 – 111 = 189 units.
Remaining work is done by A and C. Their combined rate = 15 + 10 = 25 units/day.
Time taken for remaining work = 189 units / 25 units/day = $\frac{189}{25}$ days.
Total time = 3 days + $\frac{189}{25}$ days = $3 + 7\frac{14}{25} = 10\frac{14}{25}$ days.