Question

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 alone would take to fill the tank

Answer 1

$\huge{\bold{\blue{A\:-28\:minutes}}}$

$\huge{\bold{\red{B\:-21\:minutes}}}$

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$\huge \boxed {\textt \fcolorbox \boxed \fcolorbox {\bold{\blue {Mark It Brainliest}}}}$

Answer 2

Solution :-

let time taken by A is x minutes and B is y minutes,

from the 1st statement we have,

x-y = 7 = > y = x-7......................... (I)

now in 1 min tank filled by A = 1/x

And  in 1 min tank filled by B = 1/y

Combined together both can fill = 1/x  + 1/y tank in one minute,

Hence the time taken by both the taps to fill the tank

= 1/(1/x + 1/y)

= xy/x+y

from the second statement,

x- xy/x+y   = 16

=> x² + xy - xy = 16(x+y)

=> x² -16x - 16y = 0

Putting the value of y from equation (I)

x²-16x - 16(x-7) = 0

=> x² - 32x + 112 = 0.

Solving the above quadratic equation, we get

x = 28 or 4

rejecting 4 as x can't be less than 7

x = 28

y = x-7 = 28-7 = 21

The time taken by both the taps are 28 and 21 minutes alone.

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@GauravSaxena01