Question

K taps numbered from 1 to K are fitted to a full cistern. The rate at which the kth tap (for k = 2, 3, ...,K) empties the cistern, equals the sum of the rates
of all the taps numbered before it. If the fifth tap can empty h in 240 minutes, then find the time in which the ninth tap can empty it.
5: A
O 7.5 minutes
B.
15 minutes
C. O 12 minutes
D. O 8 minutes​

Answer 1

Given :- K taps numbered from 1 to K are fitted to a full cistern. The rate at which the kth tap (for k = 2, 3, ...,K) empties the cistern, equals the sum of the rates of all the taps numbered before it. If the fifth tap can empty it in 240 minutes, then find the time in which the ninth tap can empty it. ?

Solution :-

Let us assume that, the time taken by the nth tap to empty the tank is kn.

we have given that,

• The rate of the kth tap is equal to the sum of the rates of all previous taps.

conclusion :-

• The time taken by the 3rd tap = sum of the time taken by 1st and 2nd tap.
• k3 = k2 + k1.
• k4 = k3 + k2 + k1 = (k2+k1) + (k2+k1) = 2(k2 + k1)
• k5 = k4 + k3 + k2+ k1 = 2(k2+k1) + (k2+k1) + (k2+k1) = 4(k2 + k1) = 2²(k2 + k1).
• kn = 2^(n -3)(k2 + k1) .

given that, 5th tap can empty the full cistern in 240 minutes .

So,

→ 2^(5 - 3)(k2 + k1) = 240

→ 2²(k2 + k1) = 240

→ 4(k2 + k1) = 240

→ (k2 + k1) = 60 .

therefore, 9th tap can empty the cistern in ,

→ 2^(9 - 3)(k2 + k1)

→ 2⁶(k2 + k1)

→ 64(k2 + k1)

→ 64 * 60

→ 3840 minutes.

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