(solved)Question 14 SSC-CGL 2020 March 4 Shift 1

A can do a work in 30 days. Efficiency of B is 25% more than A and efficiency of C is 20% more than B. They work together for 3 days. In how many days B will complete remaining work?
(Rev. 20-Jan-2024)

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Parveen,

Question 14
SSC-CGL 2020 Mar 4 Shift 1

A can do a work in 30 days. Efficiency of B is 25% more than A and efficiency of C is 20% more than B. They work together for 3 days. In how many days B will complete remaining work?

Solution in Short

Efficiency of A be = 100, then of B is 125, and of C is 150. Ratio A : B : C = 4 : 5 : 6. Total work is LCM (4, 5, 6) = 120. Already done in 3 days = (4 + 5 + 6) x 3 = 45. Remaining is 120 - 45 = 75. B alone will take 75/5 = 15 days ans!

Solution in Detail

Let efficiency of A = 100

"A" completes the work in 30 days

$\displaystyle \therefore$ work = $\displaystyle 30 \times 100 = 3000$

given B is 25% more efficient

so efficiency of B = 100 + 25% = 125

given C is 20% efficient than B

so, of C = 125 + 20% = 150

A+B+C work for 3 days. Work done:

$\displaystyle 3 \times (100 + 125 + 150) = 1125$

work left: $\displaystyle 3000 - 1125 = 1875$

B does 125 in 1 day

$\displaystyle \therefore $ B does 1875 in $\displaystyle \frac{1}{125} \times 1875$

$\displaystyle = 15 \text{ days }\underline{Ans}$

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