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14 Nov 2007, 18:11
Kudos
Bookmarks
Please provide detailed answers 
OA are in Bold Face
Thanks!
1- Working alone, a small pump takes twice as long as a large pump takes to fill an empty tank. Working together at their respective constant rates, the pumps can fill the tank in 6 hours. How many hours would it take the small pump to fill the tank working alone?
(A) 8
(B) 9
(C) 12
(D) 15
(E) 18
2- On level farmland, two runners leave at the same time from the intersection of two country roads. One runner jogs due north at a constant rate of 8 miles per hour while the second runner jogs due east at a constant rate that is 4 miles per hour faster than the first runner’s rate. How far apart, to the nearest mile, will they be after 21hour?
(A) 6
(B) 7
(C) 8
(D) 12
(E) 14
3- If x = y + 4 and x = 20 – y, then (x^2) - (y^2)=
(A) 16
(B) 80
(C) 144
(D) 256
(E) 384
OA are in Bold Face
Thanks!
1- Working alone, a small pump takes twice as long as a large pump takes to fill an empty tank. Working together at their respective constant rates, the pumps can fill the tank in 6 hours. How many hours would it take the small pump to fill the tank working alone?
(A) 8
(B) 9
(C) 12
(D) 15
(E) 18
2- On level farmland, two runners leave at the same time from the intersection of two country roads. One runner jogs due north at a constant rate of 8 miles per hour while the second runner jogs due east at a constant rate that is 4 miles per hour faster than the first runner’s rate. How far apart, to the nearest mile, will they be after 21hour?
(A) 6
(B) 7
(C) 8
(D) 12
(E) 14
3- If x = y + 4 and x = 20 – y, then (x^2) - (y^2)=
(A) 16
(B) 80
(C) 144
(D) 256
(E) 384
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14 Nov 2007, 18:22
Kudos
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Ev123
The first one makes sense,
I'm not at all clear on the second. I make their rates 8 and 12 respectively. If they run at those rates North and East for 21 hours, I don't see how they can possibly be 7 miles apart.
I also dont get the answer to the third - I make it 80...
14 Nov 2007, 18:23
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Ev123
Solution:
If it takes twice as long to fill the tank, then the rate is half.
Solve for r, plugin in 6 hours as the time, 1 is 100% of the tank
(1/2)r(6) + r(6) = 1
r = 1/9
Remember this is the rate of the large pump, so 1/2*1/9 = 1/18 for the small pump's rate
solve for small pump t:
t = d/r
t = 1/(1/18) = 18
t = 1/(1/9)
