
A water tap A takes 7 minutes more than water tap B for filling up a tank with water. The tap A takes 16 minutes more than the time taken by both the taps together to fill the tank. find the time each tap would alone take to fill the tank.
Answer
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Hint: To solve this question first we have to assume that tap ‘A’ will fill in x minutes and from that x we have to use this to make equations from the given question statement. then simplify the equations and find the unknown value.
Complete step-by-step answer:
So water tap B will take (x-7) minutes to fill the tank.
So
In 1 minute water tap A will fill $\dfrac{1}{x}$part of the tank.
In 1 minute water tab B will fill $\dfrac{1}{{x - 7}}$part of the tank.
So in 1 min both tap together will fill $\dfrac{1}{x} + \dfrac{1}{{x - 7}}$part of the tank
$\dfrac{1}{x} + \dfrac{1}{{x - 7}} = \dfrac{{x - 7 + x}}{{x\left( {x - 7} \right)}} = \dfrac{{2x - 7}}{{x\left( {x - 7} \right)}} = \dfrac{1}{{\left( {\dfrac{{x\left( {x - 7} \right)}}{{2x - 7}}} \right)}}$ part of the tank is filled by both tap together in 1 min.
So the whole tank will be filled by both tap together in $\dfrac{{x\left( {x - 7} \right)}}{{2x - 7}}$min.
And it is given in question tap A takes 16 minutes more than the time taken by both the taps to fill together. So we have to subtract 16 minutes from time taken by tap A to equate with both tap together timing.
$
\dfrac{{x\left( {x - 7} \right)}}{{2x - 7}} = x - 16 \\
\Rightarrow {x^2} - 7x = 2{x^2} - 32x - 7x + 112 \\
\Rightarrow {x^2} - 32x + 112 = 0 \\
\Rightarrow {x^2} - 28x - 4x + 112 = 0 \\
\Rightarrow x\left( {x - 28} \right) - 4\left( {x - 28} \right) = 0 \\
\Rightarrow \left( {x - 28} \right)\left( {x - 4} \right) = 0 \\
$
Hence either x = 28 or x = 4.
If x = 4 then time taken by B will be 4-7 = -3 that is not possible.
So x =28min will be time taken by A and time taken by B will be 28 -7 = 21min.
Note: Whenever you get this type of question the key concept of solving is you have to assume timing of a tap and you have to proceed further using data given in question. Best way of solving is you have to find how much part fills in 1 min and from the unitary method the whole part in how much minute.
Complete step-by-step answer:
So water tap B will take (x-7) minutes to fill the tank.
So
In 1 minute water tap A will fill $\dfrac{1}{x}$part of the tank.
In 1 minute water tab B will fill $\dfrac{1}{{x - 7}}$part of the tank.
So in 1 min both tap together will fill $\dfrac{1}{x} + \dfrac{1}{{x - 7}}$part of the tank
$\dfrac{1}{x} + \dfrac{1}{{x - 7}} = \dfrac{{x - 7 + x}}{{x\left( {x - 7} \right)}} = \dfrac{{2x - 7}}{{x\left( {x - 7} \right)}} = \dfrac{1}{{\left( {\dfrac{{x\left( {x - 7} \right)}}{{2x - 7}}} \right)}}$ part of the tank is filled by both tap together in 1 min.
So the whole tank will be filled by both tap together in $\dfrac{{x\left( {x - 7} \right)}}{{2x - 7}}$min.
And it is given in question tap A takes 16 minutes more than the time taken by both the taps to fill together. So we have to subtract 16 minutes from time taken by tap A to equate with both tap together timing.
$
\dfrac{{x\left( {x - 7} \right)}}{{2x - 7}} = x - 16 \\
\Rightarrow {x^2} - 7x = 2{x^2} - 32x - 7x + 112 \\
\Rightarrow {x^2} - 32x + 112 = 0 \\
\Rightarrow {x^2} - 28x - 4x + 112 = 0 \\
\Rightarrow x\left( {x - 28} \right) - 4\left( {x - 28} \right) = 0 \\
\Rightarrow \left( {x - 28} \right)\left( {x - 4} \right) = 0 \\
$
Hence either x = 28 or x = 4.
If x = 4 then time taken by B will be 4-7 = -3 that is not possible.
So x =28min will be time taken by A and time taken by B will be 28 -7 = 21min.
Note: Whenever you get this type of question the key concept of solving is you have to assume timing of a tap and you have to proceed further using data given in question. Best way of solving is you have to find how much part fills in 1 min and from the unitary method the whole part in how much minute.
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