Question

Two water taps together can fill a tank in one hour and twelve minutes. The tap
of smaller diameter takes 1 hour more than the larger one to fill the tank
separately. Find the time in which each tap can separately fill the tank.​

Answer 1

Answer:

Step-by-step explanation:

let the time taken by larger tap be x

And by smaller6p tap be [x+10]hrs

therefore....

larger tap can fill = 1/x par of the tank

smaller tap can fill = 1/[x +10] part of tank

TOGETHER THEY CAN COMPLETELY FILL THE TANK IN 9×3/8=75/8

1/x+10 + 1/x =1/75 upon 8

x+x+10/x*2+10x = 75/8

2x+10/x*2+10x = 8/75

HERE....,

150x+750=8x*2+80x

8x*2+80x-150x-750=0

8x*2+70x-750=0

4x(x-5)+25(x+15)=0

4x+25=0 OR x-15=0

x=-25/4 OR x=15

Since the time can't be in negative.

THEREFORE, x is not equal to-25/4,

so the answer is........

x=15

Answer 2

$\huge {\underline{\green{ SOLUTION}}}$

Let the tap of the larger diameter fills the tank alone in (x – 10) hours.

In 1 hour, the tap of the smaller diameter can fill 1/x part of the tank.

In 1 hour, the tap of the larger diameter can fill 1/(x – 10) part of the tank.

Two water taps together can fii a tank in 75 / 8 hours.

But in 1 hour the taps fill 8/75 part of the tank.

1 / x + 1 / (x – 10) = 8 / 75.

( x – 10 + x ) / x ( x – 10) = 8 / 75.

2( x – 5) / ( x2 – 10 x) = 8 / 75.

4x2 – 40x = 75x – 375.

4x2 – 115x + 375 = 0

4x2 – 100x – 15x + 375 = 0

4x ( x – 25) – 15( x – 25) = 0

( 4x -15)( x – 25) = 0.

x = 25, 15 / 4.

But x = 15 / 4 then x – 10 = -25 /4 which is not possible since time can't be negative

But x = 25 then x – 10 = 15.

Larger diameter of the tap can the tank 15 hours and smaller diameter of the tank can fill the tank in 25 hours.

$\red {\huge {THANK \;\: YOU}}$