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 taps together to fill tank. Find the time each tap alone would take to fill the tank?
Give answer I mark as a brainlist

Answer 1

let the time taken by tap a be x min. and by tap b be y min.
We know
x = 7+y...........(1)
and
x=16+(x+y)...........(2)
but,
for 2 objects doing one work,

time taken by both two object in whole= (time excess taken by one object than other+ time excess taken by same in whole)/2+1 /2
here,
x+y= (7+16)/2+1/2
=23/2+1/2
=24/2
x+y=12
put x+y=12 in (2)
x= 16+12
x=28
time taken by tap a is 28 min
now
put x=28 in (1)
so,
28-7=y
y=21
time taken by tap b is 21min.




hope it helps
plz mark as 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.

=================

@GauravSaxena01