SOLUTION: please help me solve this word problems....AGE PROBLEMS 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

Question 348569: please help me solve this word problems....AGE PROBLEMS
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.
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?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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.
---------------
Equations:
A = 11B
(A+x)=5(B+x)
(A+x+5)=3(B+x+5)
-------
Substitute for "A" and solve for B and "x".
---
11B+x = 5B+5x
11B+x+5 = 3B+3x+15
---------------------------
Simplify:
6B = 4x
8B = 2x+10
---
B = (2/3)x
----
8[(2/3)x] = 2x+10
(16/3)x = 2x+10
(10/3)x = 10
x = 3 years
==================
B = (2/3)x = 2 yrs old now
---
A = 11B = 11(2) = 22 yrs old now
========================================

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?
------
Equation:
J = 0.3F
(J+20)= 0.58(F+20)
---
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)
-------
J = 0.3*30 = 9 yrs (John's age now)
===========================================
Cheers,
Stan H.

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