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 $2(x + x)=4x$ (because twice as many quarters as nickels and dimes combined).

Step2: Calculate value of each coin type

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

Step3: Set up total value equation

The total value of all coins is $23$ dollars. So, $0.10x + 0.05x + x=23$.

Step4: Solve the equation

Combine like terms: $1.15x = 23$. Then, $x=\frac{23}{1.15}=20$.

Step5: Find value of quarters

We know the value of quarters is $x$ dollars (from Step 2), and $x = 20$. So the value of quarters is $20$ dollars? Wait, no, wait. Wait, in Step 2, the value of quarters is $0.25\times4x= x$. Wait, when $x = 20$, the number of quarters is $4x=80$, and the value of quarters is $0.25\times80 = 20$ dollars. Let's check:

Number of dimes: $x = 20$, value of dimes: $20\times0.10 = 2$ dollars.

Number of nickels: $x = 20$, value of nickels: $20\times0.05=1$ dollar.

Number of quarters: $4x = 80$, value of quarters: $80\times0.25 = 20$ dollars.

Total: $2 + 1+20 = 23$ dollars. Correct.

Answer:

$\$20$