SOLUTION: A woman is 3 times as old as her son. 8 years ago the product of their ages was 112. Find their present ages.

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Question 1194795: A woman is 3 times as old as her son. 8 years ago the product of their ages was 112. Find their present ages.
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
A woman is 3 times as old as her son. 8 years ago the product of their ages was 112.
Find their present ages.
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Let their present age be x for the son and 3x for the mother.


8 years ago they were (x-8) and (3x-8) years old, respectively.


An equation from the other part of the problem is

    (x-8)*(3x-8) = 112.


Simplify 

    3x^2 - 8x - 24x + 64 = 112

    3x^2 - 32x -48 = 0


Solve using the quadratic formula.


You will get two roots of the equation: one positive integer of 12, and the other 
negative rational  -11%2F3.


Use positive integer root and ignore negative one.


ANSWER.  The son is 12 years old; the mother is 36 years old.


CHECK.  (12-8)*(36-8) = 4*28 = 112.    ! Correct !

Solved.



Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

let woman’s age be w and son’s age s
if a woman is 3 times as old as her son, we have
w=3s..........eq.1
if +8 years ago the product of their ages was 112
%28w-8%29%28s-8%29=112
s+w+-+8+s+-+8+w+%2B+64+=+112
s+w+-+8+s+-+8+w+-+48+=+0........eq.2
from eq.1 and eq.2 we have
s+%2A3s+-+8+s+-+8%2A3s+-+48+=+0
3s%5E2+-+8+s+-+24s+-+48+=+0
3s%5E2+-+32s+-+48+=+0
3s%5E2+-+32s+-+48+=+0
%28s+-+12%29+%283+s+%2B+4%29+=+0
we need integer solution, so s=12
then
w=3%2A12..........eq.1
w=36

their present ages: woman is 36 and son is 12 years old
check the product of their ages +8 years ago
+8 years ago woman was 36-8=28 and son was 12-8=4
28%2A4=112 which confirms our result