Mart has $4.10 in nickels, dimes, and quarters. She has four more
quarters than dines. She has three times as many nickels as quarters.
How many of each coin does she have?
Notice that I changed only the numbers in your problem. That's because
I didn't want to do your homework for you. All you have to do is do
yours exactly as I did this one, step by step.
Mart has 4.10 in nickels, dimes, and quarters.
0.05N + 0.10D + 0.25Q = 4.10
She has four more quarters than dines.
Q = D + 4
She has three times as many nickels as quarters.
N = 3Q
So we have three equations and 3 unknowns:
0.05N + 0.10D + 0.25Q = 4.10
Q = D + 4
N = 3Q
Multiply first equation by 100 to remove decimals
5N + 10D + 25Q = 410
Q = D + 4
N = 3Q
Solve the second equation for D
Q = D + 4
Q - 4 = D
D = Q - 4
Substitute N = 3Q and D = Q - 4 into
5N + 10D + 25Q = 410
5(3Q) + 10(Q - 4) + 25Q = 410
15Q + 10Q - 40 + 25Q = 410
50Q - 40 = 410
50Q = 450
Q = 9
Substitute Q = 9 into
D = Q - 4
D = 9 - 4
D = 5
Substitute Q = 9 into N = 3Q
N = 3(9)
N = 27
So Mart has 9 quarters, 5 dimes and 27 nickels.
Checking:
Mart has $4.10 in nickels, dimes, and quarters.
9 quarters = 9($0.25) = $2.25
5 dimes = 5($0.10) = $0.50
27 nickels = 27($0.05) = $1.35
------------------------------
$4.10
That checks:
She has 4 more quarters than dines.
She has 9 quarters and 5 dimes, and 9 is 4 more than 5.
That checks.
She has three times as many nickels as quarters.
She has 27 nickels and 9 quarters, and 27 is three times 9.
That checks.
Now do yours the exactly same way, step by step.
Edwin
