Question

two water taps together can fill a tank in 15 by 8 hours .the tap with longer diameter takes 2 hours less than the smaller one to fill the tank separately .find the time in which each tap can fill the tank separately

Answer 1

Answer:

The two pipes will take 5 hour and 3 hours respectively.

Step-by-step explanation:

Let the smaller dia tap takes x hours to fill the tank

=> time taken by the larger dia pipe = x - 2

Tank filled in one hour by smaller dia pipe = 1/x

Tank filled in one hour by larger dia pipe = 1/(x-2)

Time taken by both the pipes together = 15/8

=> Tank filled in one hour by both pipes = 8/15

=> 1/x + 1/(x-2) = 8/15

=> (2x-2)/(x²-2x) = 8/15

=> 30x - 30 = 8x² -16x

=> 8x² - 46x +30 = 0

=> x = [46 ± √(46² - 4x8x30)]/2x8

= [46 ± √(2116 - 960)]/16

= [46 ± √1156]/16

= [46 ± 34]/16

= 80/16 or 12/16

= 5 or 0.75

neglecting 0.75 as it can not be less than 15/8

x = 5

and,

x-2 = 3

Hence the two pipes will take 5 hour and 3 hours respectively.

Answer 2

Let,

time taken by smaller diameter pipe to fill tank separately be x hours.

=> time taken by larger diameter pipe to fill tank separately = x - 2 hours.

Thus,

part of tank filled in 1 hour by smaller diameter pipe = ¹/x

and,

part of tank filled in 1 hour by larger diameter pipe = ¹/(x-2)

given, time taken by both pipes to fill tank = ¹⁵/8 hours

=> part of tank filled in 1 hour by both pipes = ¹/(15/8) = ⁸/15

now,

part of tank filled in 1 hour by smaller diameter pipe + part of tank filled in 1 hour by larger diameter pipe = part of tank filled in 1 hour by both pipes

=> ¹/x + ¹/(x-2) = ⁸/15

=> [(x-2) + x] / (x²-2x) = ⁸/15

=> 8x ² - 16x = 30x - 30

=> 8x ² - 46x +30 = 0

thus, you got your quadratic equation. :)