Question 785268: 1. presently, the ages of a father and his son are 10x and x years respectively. In 32 years time, the ratio of their ages will be 2:1. Fin the sum of their present ages.
2. In ten years time, a father will be as twice as old as his son. Ten years ago, he was times as old as his son. Find the sum of their ages in seven years time.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 1. presently, the ages of a father and his son are 10x and x years respectively.
In 32 years time, the ratio of their ages will be 2:1.
Find the sum of their present ages.
:
We know that is 11x, find x
:
 = 
cross multiply
10x+32 = 2(x+32)
10x + 32 = 2x + 64
10x - 2x = 64 - 32
8x = 32
x = 4
therefore
11(4) = 44 is the sum of their present ages
:
2. In ten years time, a father will be as twice as old as his son.
f + 10 = 2(s+10)
f + 10 = 2s + 20
f = 2s + 20 - 10
f = 2s + 10
Ten years ago, he was times as old as his son.
Assume the multiplier which is missing, = m
f - 10 = m(s-10)
f - 10 = ms - 10m
Now all you have to do is put in the missing multiplier, and solve it
:
I see one possibility, m=12 then Dad's 34, son's 12
:
Find the sum of their ages in seven years time. 60?
|
|
|
