Question

There are three taps A, B and C in a tank. They can fill the tank in 10 hrs, 20 hrs, and
25 hrs respectively. At first, all of them are opened simultaneously. Then after 2
hours, tap C is closed and A and B are kept running. After the 4th hour, tap B is also
closed. The remaining work is done by Tap A alone. Find the percentage of the work
done by Tap A by itself.​

Answer 1

Step-by-step explanation:

A + B + C = 19%. In the first two hours they will do 38 % of the work. Further, for the next two

hours work will be done at the rate of 15% per hour. Hence, after 4 hours 68% of the work will

be completed, when tap B is also closed. The last 32% of the work will be done by A alone.

Hence, A does 40% (first 4 days) + 32% = 72% of the work.

Answer 2

Answer:

Work done by tap A = 72%

Step-by-step explanation:

(a) Tap A's work in %

work by tap A in percentage = $\frac{100}{10}$ = 10%

(b) Tap B's work in %

work by tap B in percentage = $\frac{100}{20}$ = 5%

(c) Tap C's work in %

work by tap C in percentage = $\frac{100}{25}$ = 4%

When all of them are opened for 2 hours -

A + B + C = 19% , that is work in one hour.

Therefore, work done by tap (A + B + C) in 2 hours = 19×2 = 38 % of the work.

After 2 hours, tap C is closed

Therefore, work done by (A + B ) in 2 hours = 15×2 = 30 % of the work.

total work done till now =work done by tap {(A+B +C) +(A+B)}

total work done till now =38 % + 30 %

total work done till now =68 %

After the 4th hour, tap B is also closed

The last 32% of the work will be done by A alone.

Hence, tap A does 40% (first 4 hours) + 32% = 72%

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