(solved)Question 6 SSC-CGL 2020 March 3 Shift 1

A completes a job in 40 days. The efficiency of B is 25% more than that of A. C is 28% more efficient than B. The work can be completed by A, B and C together in 5 days. How long will B alone take to complete the same work?

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Parveen,

Question 6
SSC-CGL 2020 Mar 3 Shift 1

A completes a job in 40 days. The efficiency of B is 25% more than that of A. C is 28% more efficient than B. The work can be completed by A, B and C together in 5 days. How long will B alone take to complete the same work?

Solution in Brief

Let the efficiency of B = 100, then of C = 128, and of A = 80. Total work = 5 x (100 + 128 + 80) = 1540. B will complete it in 1540/100 = 15.4 or 15-2/5 days ans!

Solution in Detail

Take efficiency of B as $\displaystyle 100$

Why B as 100? and not A? This choice is possible only with practice. Mathematical intuition plays a role here. Honestly, I tried with A = 100. But calculations became complex. Then I tried B = 100, and things simplified.

B is $\displaystyle 100$, so C = $\displaystyle 100 + 28\% = 128$

Since $\displaystyle 80 + 25\% = 100$, so A = $\displaystyle 80$

DEFINITION: efficiency means "the amount of work done in 1 day"

Given A+B+C together do in 5 days

$ \begin{aligned} \therefore \text{Work} &= 5 \times (80 + 100 + 128)\\ &= 1540 \end{aligned} $

B can do $\displaystyle 100$ in $\displaystyle 1$ days

So, he completes $\displaystyle 1540$ in $\displaystyle \frac{1}{100} \times 1540$

Or, in $\displaystyle 15.4$ days, or $\displaystyle 15\frac 25$ days Ans.

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