SOLUTION: An old jar of dimes and quarters contains 200 coins total, and the value of all the coins is $42.50. How many quarters are in the jar? A. 50 B. 130 C. 127 D. 150

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Question 1127278: An old jar of dimes and quarters contains 200 coins total, and the value of all the coins is $42.50. How many quarters are in the jar?
A. 50 B. 130 C. 127 D. 150

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
A QUICK FIFTH-GRADER ANSWER:
Using only quarters, you need 42.50%2A4=170 coins to make %22%2442.50%22 .
Using only quarters and dimes, you need an even number of quarters to make %22%2442.50%22 .
(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 %22%2442.50%22 ,
but the number of coins would increase by 5-2=3 to 170%2B3=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%2F3=10 such swaps.
Then you would be using 10%2A2=20 less quarters,
for a total of 170-20=highlight%28150%29 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%22=%22numberofquarters,
200-x%22=%22numberofdimes .
The total money amount, in cents, is
25x%2B10%28200-x%29=4250
You simplify and solve for x , maybe like this:
25x%2B10%28200-x%29=4250
25x%2B2000-10x=4250
15x%2B2000=4250
15x=4250-2000
15x=2250
x=2250%2F15
highlight%28x=150%29 .

ANOTHER SHOW-YOUR-WORK ANSWER FOR ALGEBRA CLASS:
With two variables:
x%22=%22numberofquarters,
y%22=%22numberofdimes .
The total number of coins is x%2By=200
The total money amount is
%22%240.25%22%2Ax%2B%22%240.10%22%2Ay=%22%2442.50%22
(or maybe, in cents, 25x%2B10y=4250 ).
With those two equations we have a system of linear equations to solve by the method of your choice.
Maybe, from system%28x%2By=200%2C25x%2B10y=4250%29 ,
you solve x%2By=200 for y to get y=200-x .
Then, you substitute 200-x for y in 25x%2B10y=4250 , get
25x%2B10%28200-x%29=4250
25x%2B2000-10x=4250
15x%2B2000=4250
15x=4250-2000
15x=2250
x=2250%2F15
highlight%28x=150%29 .
At that point you could also find y , the number of dimes,
but nobody asked for that.