Question : A and B work together to complete the rest of the job in 7 days. However, $\frac{37}{100}$ of the job was already 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 complete the entire work?
Option 1: 20
Option 2: 25
Option 3: 30
Option 4: 10
Correct Answer: 20
Solution :
Let the total work done be $100$ units.
Remaining work = $100 - \frac{37}{100}×100$ = $100-37$ = $63$ units
5 × efficiency of A = 4 × efficiency of B
$\frac{\text{Efficiency of A}}{\text{Efficiency of B}}$ = $\frac{4}{5}$
Efficiency of A = $4x$
Efficiency of B = $5x$
Total efficiency of A + B = $4x+5x$ = $9x$
Work done in 7 days = $7 × 9x$ = $63$ units
⇒ $63x$ = $63$
⇒ $x$ = 1
Efficiency of A = 4 and efficiency of B = 5
Time taken by B = $\frac{100}{5}$ = 20 days
Hence, the correct answer is 20.
Related Questions
Know More about
Staff Selection Commission Sub Inspector ...
Application | Eligibility | Selection Process | Result | Cutoff | Admit Card | Preparation Tips
Get Updates BrochureYour Staff Selection Commission Sub Inspector Exam brochure has been successfully mailed to your registered email id “”.




