Question 950667
yeboah is 37 times as old as asseidu.


let a = age of yeboah.
let b = age of asseidu.


equation becomes a = 37 * b


in 16 years yeboah will be twice as old as asseidu.


equation becomes (a + 16) = 2 * (b + 16)


find their ages.


you have two equations that need to be solved simultaneously.


they are:


a = 37 * b
(a + 16) = 2 * (b + 16)


replace a in the second equation with 37 * b.


your second equation of (a + 16) = 2 * (b + 16) becomes:


(37 * b + 16) = 2 * (b + 16)


simplify to get


37 * b + 16 = 2 * b + 32


subtract 16 from both sides of the equation and subtract 2 * b from both sides of the equation to get:


37 * b - 2 * b = 32 - 16


simplify to get:


35 * b = 16


divide both sides of the equation by 34 to get:


b = 16 / 35 = .4571428571


since a = 37 * b, then a = 16.91428571


in 16 years, a will be equal to 16.91428571 + 16 = 32.91428571


in 16 years, b will be equal to .4571428571 + 16 = 16.4571428571.


in 16 years a will be 2 * b.


this becomes 32.91428571 = 2 * 16.4571428571.


2 * 16.4571428571 is equal to 32.91428571.


this confirms the solution is correct.


assuming the problem was stated correctly.


yeboah is 16.91428571 years old today.
asseidu is .4571428571 years old today.
16.91428571 / .4571428571 = 37
yeboah is 37 times as old as asseidu today.



yeboah is 32.91428571 years old in 16 years.
asseidu is 16.45714286 years old in 16 years.
32.91428571 / 16.45714286 = 2
yeboah is 2 times as old as asseidu in 16 years.


strange numbers, but that's the way it comes out, assuming the original problem is stated as given.