Teen's Age + Dad's Age = ?
A teenager wrote the digits of his fathers' age and then wrote the digits of his own age after that of his father.
From this four digit number he subtracted the difference of their ages and the answer was 4289.
Find the sum of the father's and the son's ages.
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1 solution
Let the portion of the 4 digit number representing father's age be XY and the son's be ZW.
The father's age is 10X +Y and the son's age is 10Z + W. But since the son is a teenager, then Z=1 and W must be between 3 and 9 inclusive.
XYZW represents a 4 digit number calculated as 1000X +100Y +10 +W.
The difference in their ages is (10X + Y) - (10 + W)
If we subtract the difference in their ages from the 4 digit XW1W, that becomes
(1000X +100Y +10 +W) - ((10X + Y) - (10 + W)) = 990X + 99Y + 2W +20 = 4289
While this is 1 equation with 3 unknowns, it can be solved quite easily because:
a. Since both ages are two digit numbers, we know that (4289 + difference in their ages) must exceed 4312 (XYZW) because Z= 1, W is greater than 2, and of course the son must be younger than the father..
b. So, we can conclude that X=4 and Y=3 and the father is 43 yrs old
c. Plugging the values into the equation above, 990(4) + 99(3) +2W +20 = 4289 ==> 4257 + 2W + 20 = 4289 ==> W=6
Since Y = 1 and W = 6 the son is 16 yrs old.
To verify, note that the difference in their age is 27. The XYZW number is 4316. 4316 - 27 = 4289 as required.
Therefore, the sum of their ages is 43 + 16 =
59