Question

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Answer 1

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Answer 2

Question:-

A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap takes to fill the tank completely ?

Given:-

• A tank fills completely in 2 hours if both the taps are open.

• The smaller tap takes 3 hours more than the larger one to fill the tank.

To Find:-

• How much time does each tap takes to fill the tank completely ?

Solution:-

• Let, the time required to fill the tank by large tap = x hours.

• Time required to fill the tank by smaller tap = x + 3 .

• Time required to fill the tank by smaller tap = (x + 3) hours.

• work done by large tap in one hour = 1/x.

• And work done by smaller tap in one hour = 1/x + 3.

According to the given condition:

work done in 1hour by small tap + work done by both

tap in 1hour.

$\sf \large \frac{1}{x + 3} + \frac{1}{x} = \frac{1}{2}$

$\sf \large \frac{x + x + 3}{x(x + 3)} = \frac{1}{2}$

$\sf \large \frac{2x + 3}{x {}^{2} + 3x } = \frac{1}{2}$

$\sf \large2(2x + 3) = x {}^{2} + 3x$

$\sf \large4x + 6 = x {}^{2} + 3x$

$\sf \large0 = x {}^{2} - 4x + 3x - 6$

$\sf \large x { }^{2} - x - 6 = 0$

$\sf \large x {}^{2} - 3x + 2x - 6 = 0$

$\sf \large x(x - 3) + 2(x - 3) = 0$

$\sf \large(x - 3)(x + 2) = 0$

${ \boxed{ \sf \large \red {x - 3 = 0 \: (or) \: x + 2 = 0}}} \\ { \boxed{ \sf \large \red{x = 3 \: (or) \: x = - 2}}}$

But x = – 2 is not acceptable because time cannot be negative.

∴ Time taken by larger tap = x = 3.

∴ Time taken by smaller tap = x + 3 = 3 + 3 = 6.

Answer:-

$\sf \large \pink{∴ Time \: taken \: by \: larger \: tap \: and } \\ \sf \large \pink{ smaller \: tap \: are \: \mathfrak{3} \: hours \: and \: } \\ \sf \large \pink{ \mathfrak{6} \: hours \: respectively.}$

Hope you have satisfied. ⚘