Question 1127278
A QUICK FIFTH-GRADER ANSWER:
Using only quarters, you need {{{42.50*4=170}}} coins to make {{{"$42.50"}}} .
Using only quarters and dimes, you need an even number of quarters to make {{{"$42.50"}}} .
(Using an odd number of quarters you would make amounts ending in 5)
If you start with {{{170}}} quarters,
and swap {{{2}}} quarters for {{{5}}} dimes,
You would still have {{{"$42.50"}}} ,
but the number of coins would increase by {{{5-2=3}}} to {{{170+3=173}}}.
How many such swaps would it take to have a total of {{{200}}} coins?
{{{200}}} coins is {{{200-170=30}}} more coins than {{{170}}} .
So starting from {{{170}}} quarters, you would need
{{{30/3=10}}} such swaps.
Then you would be using {{{10*2=20}}} less quarters,
for a total of {{{170-20=highlight(150)}}} quarters.
If you can do that quickly, in your head,
you are on your way to a good SAT score,
and maybe a job as a casino dealer.
If you need pencil and paper, you could still do well on the SAT.
 
ONE SHOW-YOUR-WORK ANSWER FOR ALGEBRA CLASS:
With one variable:
{{{x}}}{{{"="}}}{{{number}}}{{{of}}}{{{quarters}}},
{{{200-x}}}{{{"="}}}{{{number}}}{{{of}}}{{{dimes}}} .
The total money amount, in cents, is
{{{25x+10(200-x)=4250}}}
You simplify and solve for {{{x}}} , maybe like this:
{{{25x+10(200-x)=4250}}}
{{{25x+2000-10x=4250}}}
{{{15x+2000=4250}}}
{{{15x=4250-2000}}}
{{{15x=2250}}}
{{{x=2250/15}}}
{{{highlight(x=150)}}} .
 
ANOTHER SHOW-YOUR-WORK ANSWER FOR ALGEBRA CLASS:
With two variables:
{{{x}}}{{{"="}}}{{{number}}}{{{of}}}{{{quarters}}},
{{{y}}}{{{"="}}}{{{number}}}{{{of}}}{{{dimes}}} .
The total number of coins is {{{x+y=200}}}
The total money amount is
{{{"$0.25"*x+"$0.10"*y="$42.50"}}}
(or maybe, in cents, {{{25x+10y=4250}}} ).
With those two equations we have a system of linear equations to solve by the method of your choice.
Maybe, from {{{system(x+y=200,25x+10y=4250)}}} ,
you solve {{{x+y=200}}} for {{{y}}} to get {{{y=200-x}}} .
Then, you substitute {{{200-x}}} for {{{y}}} in {{{25x+10y=4250}}} , get 
{{{25x+10(200-x)=4250}}}
{{{25x+2000-10x=4250}}}
{{{15x+2000=4250}}}
{{{15x=4250-2000}}}
{{{15x=2250}}}
{{{x=2250/15}}}
{{{highlight(x=150)}}} .
At that point you could also find {{{y}}} , the number of dimes,
but nobody asked for that.