Question

A, B and C working together completed a job in 10 days. However, C only worked for the first three days when 37/100 of the job was done. Also, the work done by A in 5 days is equal to the work done by B in 4 days. How many days would be required by the fastest worker to comlete the entire work?

• A. 25 days
• B. 20 days
• C. 30 days
• D. 40 days

Answer 1

Answer: (B)

20 days

Answer 2

A, B, and C together complete the job in 10 days.
$\therefore$ Their combined efficiency is $( \frac{1}{10} )$ of the job per day.
$\therefore$ A+B+C worked for the first 3 days and completed $( \frac{37}{100} )$ of the job.

After the first 3 days, the remaining work is $( 1 - \frac{37}{100} = \frac{63}{100} )$.

A and B complete this remaining work in ( 10 - 3 = 7 ) days.

Therefore, their combined efficiency is $( \frac{63}{100} \div 7 = \frac{9}{100} )$ of the job per day.

It is given that,
the work done by A in 5 days is equal to the work done by B in 4 days, $[ 5 \times \text{Work Rate of A} = 4 \times \text{Work Rate of B} ] [ \frac{\text{Work Rate of A}}{\text{Work Rate of B}} = \frac{4}{5} ]$

Now,
Let the efficiency of A be ( 4x ) and the work rate of B be ( 5x ).

$\therefore$ $( 4x + 5x = \frac{9}{100} \Rightarrow 9x = \frac{9}{100} \Rightarrow x = \frac{1}{100} )$.

efficiency of A is $( 4x = 4 \times \frac{1}{100} = \frac{4}{100} = \frac{1}{25} )$.

efficiency of B is $( 5x = 5 \times \frac{1}{100} = \frac{5}{100} = \frac{1}{20} )$.

Hence, B is the fastest worker and can complete the entire job in 20 days.