Question

Susan has 4 times as many quarters as dimes and 16 less dimes than nickels. If she has $13.45 in coins. How many of each coin does she have?
Help pls

Answer 1

Using a system of equations, it is found that Susan had 44 quarters, 11 dimes and 27 nickels.

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We are going to use:

• x as the number of quarters.
• y as the number of dimes.
• z as the number of nickels.

A quarter is worth $0.25. A dime is worth $0.1, and a nickel is worth $0.05. Total of $13.45, then:

$0.25x + 0.1y + 0.05z = 13.45$

4 times as many quarters as dimes, thus:

$x = 4y \rightarrow y = 0.25x$

16 less dimes than nickels, thus:

$z = y + 16 = 0.25x + 16$

Replacing in the first equation, we find x:

$0.25x + 0.1y + 0.05z = 13.45$

$0.25x + 0.1(0.25x) + 0.05(0.25x + 16) = 13.45$

$0.2875x = 12.65$

$x = \frac{12.65}{0.2875}$

$x = 44$

Now, with x, we find y and z:

$y = 0.25x = 0.25(44) = 11$

$z = y + 16 = 11 + 16 = 27$

Susan had 44 quarters, 11 dimes and 27 nickels.

A similar problem is given at brainly.com/question/24342899