Let the number of quarters be \( q \) and the number of dimes be \( d \).
We are given two pieces of information:
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The total number of coins: \[ q + d = 124 \]
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The total value of the coins in dollars: \[ 0.25q + 0.10d = 25 \]
To eliminate the decimals in the second equation, we can multiply the entire equation by 100: \[ 25q + 10d = 2500 \]
Now we have the system of equations:
- \( q + d = 124 \)
- \( 25q + 10d = 2500 \)
We can solve the first equation for \( d \): \[ d = 124 - q \]
Now, substitute \( d \) into the second equation: \[ 25q + 10(124 - q) = 2500 \]
Expanding this gives: \[ 25q + 1240 - 10q = 2500 \]
Combining like terms results in: \[ 15q + 1240 = 2500 \]
Now, subtract 1240 from both sides: \[ 15q = 1260 \]
Next, divide by 15: \[ q = \frac{1260}{15} = 84 \]
Now, substitute \( q = 84 \) back into the equation for \( d \): \[ d = 124 - 84 = 40 \]
Thus, the number of quarters \( q \) is \( 84 \).
To verify, check the total value:
- Value from quarters: \( 84 \times 0.25 = 21.00 \)
- Value from dimes: \( 40 \times 0.10 = 4.00 \)
- Total value: \( 21.00 + 4.00 = 25.00 \)
Both conditions hold true:
- Total coins: \( 84 + 40 = 124 \)
- Total value: \( 25.00 \)
Therefore, the number of quarters Xavior had is \( \boxed{84} \).
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