Asked by dama
Mary is now three times as old as her daughter. Four years ago the products of their age was 40 .calculate their present years
Answers
Answered by
Reiny
now:
daughter ---- x years
Mary -------- 3x years
4 years ago:
daughter : x-4
Mary : 3x-4
(x-4)(3x-4) = 40
3x^2 -16x + 16 = 40
3x^2 - 16x - 24 = 0
I was expecting it to factor, hoping for whole number solution, but ....
x = (16 ± √544)/6
= 6.55 or a negative
check:
if the daughter is appr 6.55
then Mary is 19.65 years old
Four years ago, Mary was 15.65 and the daughter was 2.55 , and 15.65(2.55) = 39.9 or
appr 40
The math works out, but Mary was 13 when the daughter was born. Somebody should be in jail.
daughter ---- x years
Mary -------- 3x years
4 years ago:
daughter : x-4
Mary : 3x-4
(x-4)(3x-4) = 40
3x^2 -16x + 16 = 40
3x^2 - 16x - 24 = 0
I was expecting it to factor, hoping for whole number solution, but ....
x = (16 ± √544)/6
= 6.55 or a negative
check:
if the daughter is appr 6.55
then Mary is 19.65 years old
Four years ago, Mary was 15.65 and the daughter was 2.55 , and 15.65(2.55) = 39.9 or
appr 40
The math works out, but Mary was 13 when the daughter was born. Somebody should be in jail.