Two water taps together can fill a tank 9 and 3/8 hours. The tap of larger diameter takes 10 hour less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank?
Answers
Answer:
Let us consider the time taken by the smaller diameter tap = x
The time is taken by the larger diameter tap = x – 10
Totaltimeistakentofillatank=938
Total time is taken to fill a tank = 75/8
In one hour portion filled by smaller diameter tap = 1/x
In one hour portion filled by larger diameter tap = 1/(x – 10)
In one hour portion filled by taps = 8/75
⇒ 1x+1x−10=875 x−10+xx(x−10)=875 2x−10x2−10x=875
75 (2x-10) = 8(x2-10x)
150x – 750 = 8x2 – 80x
8x2 − 230x + 750 = 0
4x2−115x + 375 = 0
4x2 − 100x −15x + 375 = 0
4x(x−25)−15(x−25) = 0
(4x−15)(x−25) = 0
4x−15 = 0 or x – 25 = 0
x = 15/4 or x = 25
Case 1: When x = 15/4
Then x – 10 = 15/4 – 10
⇒ 15-40/4
⇒ -25/4
Time can never be negative so x = 15/4 is not possible.
Case 2: When x = 25 then
x – 10 = 25 – 10 = 15
∴ The tap of smaller diameter can separately fill the tank in 25 hours
Step-by-step explanation:
Let us consider the time taken by the smaller diameter tap = x
The time is taken by the larger diameter tap = x – 10
Totaltimeistakentofillatank=938
Total time is taken to fill a tank = 75/8
In one hour portion filled by smaller diameter tap = 1/x
In one hour portion filled by larger diameter tap = 1/(x – 10)
In one hour portion filled by taps = 8/75
⇒ 1x+1x−10=875 x−10+xx(x−10)=875 2x−10x2−10x=875
75 (2x-10) = 8(x2-10x)
150x – 750 = 8x2 – 80x
8x2 − 230x + 750 = 0
4x2−115x + 375 = 0
4x2 − 100x −15x + 375 = 0
4x(x−25)−15(x−25) = 0
(4x−15)(x−25) = 0
4x−15 = 0 or x – 25 = 0
x = 15/4 or x = 25
Case 1: When x = 15/4
Then x – 10 = 15/4 – 10
⇒ 15-40/4
⇒ -25/4
Time can never be negative so x = 15/4 is not possible.
Case 2: When x = 25 then
x – 10 = 25 – 10 = 15
∴ The tap of smaller diameter can separately fill the tank in 25 hours
Let the larger diameter tap fills the tank alone in (x – 10) hours.
In 1 hour, the smaller diameter tap can fill 1/x part of the tank.
In 1 hour, the larger diameter tap can fill 1/(x – 10) part of the tank.
Two water taps together can fill 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
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.