# help 2 algebra questions

• Sep 19th 2005, 06:41 PM
u317dwang
help 2 algebra questions
1. 2 pumps of different sizes, working together, can empty a fuel tank in 5 hours. The larger pump can emty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?

2. How much water must be evaporated from 32 ounces of a 4% salt solution to make a 6% salt solution?
• Sep 19th 2005, 09:45 PM
MathGuru
Sorry not a full solution
Quote:

Originally Posted by u317dwang
1. 2 pumps of different sizes, working together, can empty a fuel tank in 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?

Pump A flowrate = A
Pump B flowrate = B
Volume of fuel tank = T
Number of hours it takes smaller pump(B) to empty tank = x

eq1) 5A + 5B = T
eq2) (x)B =T
eq3) (x-4)A = T

Deleted my bad work:o (thanks rgep and ticbol)
• Sep 19th 2005, 10:28 PM
rgep
If (4x-4)/(x+4) = -5 then multiplying up 4x-4 = -5x-20 so 9x = -16 which doesn't look right. The original equations were 5A + 5B = T, xB = T, (x-4)A = T which give B = T/x, A = T/(x-4) so T = 5A + 5B = 5T/(x-4) + 5T/x. Dividing through, 1 = 5/(x-4) + 5/x. Multiplying up, x(x-4) = 5x + 5(x-4), that is, x^2 - 14x + 20 = 0, x = 7 +- sqrt(49-20) = 1.6 or 12.4 approximately. Since x must be greater than 4 for the problem to make sense, the answer is 7+sqrt(29) ~ 12.4.
• Sep 20th 2005, 01:14 AM
ticbol

Let
L = rate of larger pump
S = rate of smaller pump
F = total volume of fuel tank
t = time the smaller pump can do the job alone.

Volume = (rate)*(time) --------***

"2 pumps of different sizes, working together, can empty a fuel tank in 5 hours."
F = (L+S)*5 ---(1)

" The larger pump can emty this tank in 4 hours less than the smaller one."
If F = S*t ---------(2)
then, F = L(t-4) ---(3)

F from (1) = F from (2),
5(L+S) = t(S)
5L +5S = t(S)
5L = t(S) -5S = S(t-5)
L = S(t-5)/5 ---------------(4)

F from (2) = F from (3),
t(S) = (t-4)L
L = t(S)/(t-4) -------------(5)

L from (4) = L from (5),
S(t-5)/5 = t(S)/(t-4)
Divide both sides by S,
(t-5)/5 = t/(t-4)
Cross multiply,
(t-5)(t-4) = 5t
t^2 -9t +20 = 5t
t^2 -9t -5t +20 = 0
t^2 -14t +20 = 0
t = {-(-14) +.-sqrt[(-14)^2 -4(1)(20)]} / (2*1)
t = {14 +,-10.77}/2
t = 12.39 or 1.62

Since 1.62 cannot be in (t-4), then t = 12.39 hrs.
That means, if left alone, the smaller pump can do the job in 12.39 hours. ---answer.

===============
2.) About some salty solutions ...

"How much water must be evaporated from 32 ounces of a 4% salt solution ..."
Salt content = 4% of 32 oz = (0.04)(32) = 1.28 oz. salt.
So, water content = 32 -1.28 = 30.72 oz. water.

"...to make a 6% salt solution?"
Salt content = 1.28 oz. salt also.
Total weight of new solution = (1.28 / 0.06) = 21.33 oz. 6% solution.
Water content = (21.33 -1.28) = 20.05 oz. water

Therefore, (30.72 -20.05) = 10.67 oz. of water must be evaporated. ---answer.