<PRE><B>Question 1186104</B>
&lt;br&gt;
You can get come good mental exercise by solving the problem using logical analysis and a bit of mental arithmetic.&lt;br&gt;
(1) A year ago, the mother&#39;s age was twice Sally&#39;s age, so the mother&#39;s age was an even number.  So now the mother&#39;s age must be an odd number.&lt;br&gt;
(2) Since a year ago the mother&#39;s age was twice Sally&#39;s age, now her age is still very close to twice Sally&#39;s age.&lt;br&gt;
So the mother&#39;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:&lt;br&gt;
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!!!&lt;br&gt;
ANSWER: 73 and 37 now&lt;br&gt;
It&#39;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....&lt;br&gt;
Let the mother&#39;s current age be the 2-digit number &quot;AB&quot; (where AB does NOT indicate multiplication)
Then the daughter&#39;s current age is &quot;BA&quot;&lt;br&gt;
Algebraically, then, the mother&#39;s age is 10A+B and the daughter&#39;s age is 10B+A.  &lt;-- MAKE SURE you understand that; the VALUE of the 2-digit number &quot;AB&quot; is 10A+B&lt;br&gt;
We are told that a year ago the mother&#39;s age was twice Sally&#39;s age.&lt;br&gt;
Mother&#39;s age 1 year ago: 10A+B-1
Sally&#39;s age 1 year ago: 10B+A-1&lt;br&gt;
{{{10A+B-1=2(10B+A-1)}}}
{{{10A+B-1=20B+2A-2}}}
{{{8A=19B-1}}}&lt;br&gt;
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...&lt;br&gt;
ANSWERS:
mother: &quot;AB&quot; = 73
Sally: &quot;BA&quot; = 37&lt;br&gt;
</PRE>