Question 70726: 1. Find three consecutive numbers such that the sum of one-fourth the first and one-fifth the second is five less that one-seventh the third.
1/4n + 1/5n = (1/7n-5)
2. The present ages of a husband and wife are in the ratio of seven to six and five years ago the ratio was six to five. Find their ages now.
7/6=35/30 and 6/5=30/25 + 5/5=35/30
Husband is now 35 and wife is now 30.
3. An automobile's radiator has a capacity of fifteen quarters, and it currently contains twelve quarts of thirty per cent antifreeze solution.
a. How many quarts of pure antifreeze must be added to strengthen the solution to forty per cent?
b. If the radiator is filled to capacity with pure anti-freeze, what will be the strength (per cent ) of this solution?
30% of 12 = 30/100 x 12 = 360/100 = 3.6
a. 40% of 3.6 = 40/100 x 3.6 = 144
b. 40% of 15 = 40/100 x 15 = 600
4. Simplify the complex fraction n/3 - 3/n divided by 1 + n/3.
solve: ans. -3/n
Please help with these four problems. Thank you very much. It is urgent!
Answer by checkley75(3666)   (Show Source):
You can put this solution on YOUR website! THE THREE CONSECUTINE NUMBERS ARE X, (X+1), (X+2)
X/4+(X+1)/5=(X+2)/7-5 FIND A COMMON DENOMINATOR
(5X+4[X+1])/20=(X+2)/7-35/7
(5X+4X+4)/20=(X-33)/7
(9X+4)/20=(X-33)/7 NOW CROSS MULTIPLY
20(X-33)=7(9X+4)
20X-660=63X+28
20X-63X=28+660
-43X=688
X=688/-43
X=-16 ANSWER
X+1=-15 ANSWER
X=2=-14 ANSWER
PROOF
-16/4+-15/5=-14/7-5
-4-3=-2-5
-7=-7
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