Question

While A does 1/3 of a piece of work B does 1/4, while B does 1/5, C does 1/6; in how many
hours will C finish a piece of work which A finishes in 20 hours?
​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Solution :-

Case 1) :- While A does 1/3 of a piece of work B does 1/4 .

Since time is same in both .

→ (1/3) / A = (1/4) / B

→ A/B = (1/3) / (1/4)

→ A/B = (4/3)

So,

→ A : B = 4 : 3 { Efficiency ratio }

or,

→ A : B = 8 : 6

Case 2) :- while B does 1/5, C does 1/6;

Since time is same in both .

→ (1/5) / B = (1/6) / C

→ B/C = (1/5) / (1/6)

→ B/C = (6/5)

So,

→ B : C = 6 : 5 { Efficiency ratio }

then,

→ A : B : C = 8 : 6 : 5

Now ,Let efficiency of A , B and C is 8x units / hours , 6x units / hours and 5x units / hours respectively .

Then,

→ Time taken by A to finish the work = 20 hours

So,

→ Total work = Efficiency of A * Time taken = 8x * 20 = 160x units .

therefore,

→ Time taken by C to finish the same work = Total work / Efficiency of C = 160x / 5x = 32 hours (Ans.)

Hence, C will finish the same piece of work in 32 hours .

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