SOLUTION: one year ago sally's mom was twice sally's age. today they have the same digits in their ages but the numbers are reversed. how old are sally and her mom today?

Question 1186104: one year ago sally's mom was twice sally's age. today they have the same digits in their ages but the numbers are reversed. how old are sally and her mom today?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

You can get come good mental exercise by solving the problem using logical analysis and a bit of mental arithmetic.

(1) A year ago, the mother's age was twice Sally's age, so the mother's age was an even number. So now the mother's age must be an odd number.

(2) Since a year ago the mother's age was twice Sally's age, now her age is still very close to twice Sally's age.

So the mother's age is an odd number which is about twice as much as her age with the digits reversed. Now do some trial and error:

Now 31 and 13; a year ago 30 and 12. no....
Now 52 and 25; a year ago 51 and 24. no....
Now 73 and 37; a year ago 72 and 36. YES!!!

ANSWER: 73 and 37 now

It's also good practice solving the problem using formal algebra, which is what you probably want. This is not a particularly easy problem for a student just beginning to study algebra....

Let the mother's current age be the 2-digit number "AB" (where AB does NOT indicate multiplication)
Then the daughter's current age is "BA"

Algebraically, then, the mother's age is 10A+B and the daughter's age is 10B+A. <-- MAKE SURE you understand that; the VALUE of the 2-digit number "AB" is 10A+B

We are told that a year ago the mother's age was twice Sally's age.

Mother's age 1 year ago: 10A+B-1
Sally's age 1 year ago: 10B+A-1

With the requirement that A and B are both positive single-digit integers, the only solution to that equation is A=7 and B=3. So their ages are...

ANSWERS:
mother: "AB" = 73
Sally: "BA" = 37

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