Let's assume that Xavior had q quarters and d dimes.
We know that Xavior took a total of 124 quarters and dimes, so we have the equation q + d = 124. (Equation 1)
We also know that Xavior got exactly $25 back. Since the value of a quarter is $0.25 and the value of a dime is $0.10, we can write the equation 0.10d + 0.25q = 25. (Equation 2)
To solve this system of equations, we can use substitution.
We can rewrite Equation 1 as q = 124 - d.
Now, substitute this value of q into Equation 2:
0.10d + 0.25(124 - d) = 25
0.10d + 31 - 0.25d = 25
-0.15d + 31 = 25
-0.15d = 25 - 31
-0.15d = -6
d = -6 / -0.15
d = 40
So, Xavior had 40 dimes.
Now, substitute this value of d back into Equation 1:
q + 40 = 124
q = 124 - 40
q = 84
Therefore, Xavior had 84 quarters.
The correct answer is C) 84.
Xavior took a total of 124 quarters and dimes to trade in for cash at the bank. He got exactly $25 back. How many quarters did he have?
A.
40
B.
62
C.
84
D.
100
1 answer
1 answer