Question 163792: (1). How can $5,400 be invested, part of it at 8% and the rest of it invested at 10%, so that the two investments will produce the same amount of interest.
(2). Two cars start from the same place traveling in opposite directions. One car travels 4 miles per hours faster than the other car. Find the speed their speeds if after 5 hrs they are 520 miles apart.
(3). A tamk contains 50 gallons of a 40% solution of anitfreeze. How much solution needs to be drained out and replaced with pure anti-freeze to obtain 50% solution?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! How can $5,400 be invested, part of it at 8% and the rest of it invested at 10%, so that the two investments will produce the same amount of interest.
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Equation:
0.08x = 0.10(5400-x)
8x = 10(5400-x)
18x = 54000
x = $3000
5400-x = $2400
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(2). Two cars start from the same place traveling in opposite directions. One car travels 4 miles per hours faster than the other car. Find the speed their speeds if after 5 hrs they are 520 miles apart.
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Let speed of one of the cars be "x" mph; Speed of other is "x+4" mph.
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They are separating at x + x+4 = 2x+4 mph
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EQUATION:
d = rt
520 miles = (2x+4)*5
10x + 20 = 520
10x = 500
x = 50 mph (speed of 1st car)
x+4 = 54 mph (speed of 2nd car)
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(3). A tamk contains 50 gallons of a 40% solution of antifreeze. How much solution needs to be drained out and replaced with pure anti-freeze to obtain 50% solution?
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anti-freeze - antifreeze = antifreeze
0.40(50-x) + x = 0.50*50
40(50-x) + 100x = 50^2
2000 + 60x = 2500
x = 500/60
x = 8 1/3 gallons
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Cheers,
Stan H.
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