Question

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

Answer 1

Answer:

Step-by-step explanation:

Let there be two taps A and B and let A be the tap with longer diameter than tap B then

A.T.Q

tap A takes 2 hours less than tap B if tap B is taking X hours the. Tap A would take (X-2) hours

Now,

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

= x-2 + x / x^2-2x =8/15

= (2x-2 )15 = 8(x^2-2x)

= 30x - 30 = 8x^2 - 16x

8x^2-16x-30x+30= 0

8x^2 -46x +30 = 0

8x^2 -40x-6x +30 = 0

8x(x-5) -6(x-5) = 0

X=5 or X=6/8

If x=3/4....(as 6/8 = 3/4)

Then time taken by tap A that is x-2 would negative

Hence tap B take 5hours where as tap A takes ( 5-2 ) hours i.e 3 hours

Answers...... Hope this will help you all

Answer 2

Answer:

Step-by-step explanation:

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