I need help with the working in a very hard math problem please

Hi Ben – I'm not hooked up to skype, but happy to answer Q's here on Y!A, and for you to email me links alerting me to the fact that you've posted something.

cheers – J

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btw – thx for the BA too!

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Let the high speed pump working alone take x hours to fill the pool

Then the low speed pump will take ( x+3) hours to fill the pool alone

Together in 1 hour they will fill 1 / x + 1 / ( x+3) of the pool

Since together they take 8 hours we get the equation

1/ x + 1 / ( x+3) = 1/8

MULTIPLY BOTH SIDES BY 8x ( x+3)

8( x+3) + 8x = x( x+3)

8x +24 +8x = x^2 +3x

x^2 + -13x -24 = 0

x= 1/2 ( 13 + \/265)

x = 14.64 hours

ANSWER Slow speed pump = 17.64 hours High speed pump = 14.64 hours

Let the time taken by low speed pump to fill swimming pool be 'x' hours.
Let the time taken by high speed pump to fill swimming pool be 'y' hours.

As per the first condition:
x+y=8 --------(1)
As per the another condition:
x=3+y
x-y=3--------(2)
Add (1) and (2)
x+y=8
+ x-y=3
2x =11
x =11/2
x =5.5 hours..
Substitute x=5.5 in equation (1)
x+y=8
5.5+y=8
y=8-5.5
y=2.5hours
therefore the time taken by low speed pump is 5.5hours and high speed pump is 2.5 hours


..........Cross check.........
x+y=8
5.5+2.5=8

x=3+y
5.5=3+2.5...........
Hence following is the soluion,working,and answer of u r vry hard maths problem........I hope u must have understood.............

assuming the pool has a capacity of X
if filled by pump1, it takes t1, if filled by pump2, it takes t2
t1 = X/p1 => p1 = X/t1
t2 = X/p2 => p2 = X/t2
t = X/(p1+p2) = 8 => p1+p2 = X/8
p1 + p2 = X/ (1/t1 + 1/t2) = X/8
1/t1 + 1/t2 = 1/8 => t1t2/(t1+t2) = 8
t1 - t2 = 3

t1*(t1-3)/(t1+t1-3) = 8
t1^2 - 3t1 = 16t1 -24
t1^2 - 19t1 + 24 = 0
t1 = 17.64
t2 = 14.64

high speed pump- 8 hours
low speed pump- 11 hours