Question 589118
THE FIFTH GRADER SOLUTION:
He obviously had more nickels than dimes. This must be a good mental math problem.
If I was told to swap all the nickels for dimes and all the dimes for nickels, I would not make all 20 swaps, because it's so much work.
Instead, I would just take the excess nickels and replace them with dimes.
So, if there were eight dimes, I would set aside all eight dimes and eight of the nickels, and I would only work on swapping the extra nickels for dimes. It's far less work.
As each nickel is replaced with a dime, Mongo gains 5 cents, so for him to gain 50 cents means that 10 coins had to be swapped.
The other 10 coins were left alone, because they were equal numbers of nickels and dimes, 5 of each.
It was only those 10 excess nickels that needed to be swapped.
Mongo had started with 5 dimes and 15 nickels.
 
THE ALGEBRA SOLUTION.
Let n be the number of nickels, and d be the number of dimes.
Mongo has 20 coins consisting of nickels and dimes means {{{n+d=20}}}
He now has (in cents) {{{5n+10d}}}.
If the nickels were dimes and dimes were nickels, he would have d nickels and n dimes, for a total amount of
{{{5d+10n}}} cents.
If the nickels were dimes and dimes were nickels, he would have 50c more than he now has means that
{{{5d+10n=5n+10d+50}}}
Let's simplify that.
We can subtract {{{5n}}} and subtract {{{10d}}} from both sides to get
{{{5d+10n-5n-10d=5n+10d+50-5n-10d}}} --> {{{5n-5d=50}}} --> {{{5(n-d)=50}}}
We can further simplify by dividing both sides by 5
{{{5(n-d)/5=50/5}}} --> {{{n-d=10}}}
Now we have a system of equations to solve:
{{{system(n+d=20,n-d=10)}}}
We could "eliminate" d by adding both equations to get
{{{2n=30}}} --> {{{2n/2=30/2}}} --> {{{n=15}}}
and then substitute {{{n=15}}} into {{{n+d=20}}} to get
{{{15+d=20}}} --> {{{15+d-15=20-15}}} --> {{{d=5}}}