Question

jay had as many nickels as dimes and twice as many quarters as his nickels and dimes combined. he had exactly $23. how much of it was in quarters ?

Explanation:

Step1: Define variables

Let the number of dimes be \( x \). Then the number of nickels is also \( x \) (since Jay had as many nickels as dimes). The number of quarters is \( 2(x + x)=4x \) (twice as many as nickels and dimes combined).

Step2: Calculate value of each coin

• Value of nickels: Each nickel is $0.05, so total value is \( 0.05x \).
• Value of dimes: Each dime is $0.10, so total value is \( 0.10x \).
• Value of quarters: Each quarter is $0.25, so total value is \( 0.25\times4x = x \).

Step3: Set up equation for total value

The total value is $23, so:
\[
0.05x + 0.10x + x = 23
\]
Combine like terms:
\[
1.15x = 23
\]

Step4: Solve for x

Divide both sides by 1.15:
\[
x = \frac{23}{1.15} = 20
\]

Step5: Find value of quarters

The value of quarters is \( x \) (from Step 2), and \( x = 20 \), so the value of quarters is $20. Wait, let's check again. Wait, in Step 2, value of quarters is LXI2? Wait, 4x is the number of quarters, each 0.25, so 0.25*4x = x. Then when x=20, value of quarters is $20? Wait, let's verify total:

• Nickels: 20 nickels, value = 20*0.05 = $1
• Dimes: 20 dimes, value = 20*0.10 = $2
• Quarters: 420 = 80 quarters, value = 800.25 = $20
• Total: 1 + 2 + 20 = $23. Yes, that works.

Answer:

\$20