Question 348569
problem 1: A is eleven times as old as B. In a certain number of years A would be five times as old as B, and five years after that he would be three times as old as B. How old are they now.
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Equations:
A = 11B
(A+x)=5(B+x)
(A+x+5)=3(B+x+5)
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Substitute for "A" and solve for B and "x".
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11B+x = 5B+5x
11B+x+5 = 3B+3x+15
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Simplify:
6B = 4x
8B = 2x+10
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B = (2/3)x
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8[(2/3)x] = 2x+10
(16/3)x = 2x+10
(10/3)x = 10
x = 3 years
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B = (2/3)x = 2 yrs old now
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A = 11B = 11(2) = 22 yrs old now
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problem 2: At present John's age is 30 percent of his father's age. Twenty years from now, John's age would be 58 percent of his father's age. How old are they?
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Equation:
J = 0.3F
(J+20)= 0.58(F+20)
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Substitute for "J" and solve for "F":
0.30F+20 = 0.58F + 0.58*20
0.28F = 0.42*20
F = 30 (father's age now)
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J = 0.3*30 = 9 yrs (John's age now)
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Cheers,
Stan H.