Question 1166932
<br>
The other tutor has shown one formal algebraic solution.<br>
I personally prefer to set up problems using a single variable if it is easy to do so.<br>
Let x be the number of pennies; the number of dimes is then x+11.<br>
Now we have a single equation to solve, instead of a system of two equations.<br>
{{{1(x)+10(x+11) = 1012}}}
{{{x+10x+110 = 1012}}}
{{{11x = 902}}}
{{{x = 902/11 = 82}}}<br>
ANSWER: x = 82 pennies and x+11 = 93 dimes<br>
If your mental arithmetic is good, you can get the answer informally, using virtually the same calculations as above.<br>
(1) Count the "extra" 11 dimes, with a total value of 110 cents.
(2) the remaining amount, 1012-110 = 902 cents, is made up of equal numbers of pennies and dimes.
(3) 1 penny and 1 dime together have a value of 11 cents.
(4) The number of groups at 11 cents each needed to make the remaining 902 cents is 902/11 = 82.
(5) So there are 82 pennies and 82+11=93 dimes.<br>