SOLUTION: Mary is younger than her husband, but their ages contain the same two digits. The sum of their age is 11 times the difference of their ages. How old is Mary?
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Question 852233: Mary is younger than her husband, but their ages contain the same two digits. The sum of their age is 11 times the difference of their ages. How old is Mary? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Mary is younger than her husband, but their ages contain the same two digits.
let a = husband 10s digit
let b = husbands unit digit
then
10a+b = husbands age
and
10b+a = M's age
:
The sum of their age is 11 times the difference of their ages.
(10a+b) + (10b+a) = 11[(10a+b)-(10b+a)]
11a + 11b = 11(10a - a - 10b + b)
11a + 11b = 11(9a - 9b)
simplify, divide thru by 11
a + b = 9a - 9b
b + 9b = 9a - a
10b = 8a
Simplify, divide by 2
5b = 4a
b = a
b = .8a
the only single digit solution is when a = 5
b = .8(5)
b = 4
Husband is 54, Mary is 45
:
:
See if that checks out in the statement:
"The sum of their age is 11 times the difference of their ages."
54 + 45 = 11(54-45)
99 = 11(9)
