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
Let x be the number of quarters Xavier had.
He had 124 - x dimes.
The total value of the quarters is 0.25x.
The total value of the dimes is 0.10(124 - x) = 12.4 - 0.10x.
The total value of the coins is 0.25x + 12.4 - 0.10x = 25.
0.15x + 12.4 = 25.
0.15x = 12.4.
x = 12.4 / 0.15 = 82.7.
Xavier had 82.7 quarters, which rounds down to 82.
# Answer
82
82 is not a choice
Apologies for the error. Let's correct the solution.
Let's say Xavior had q quarters and d dimes. We are given that he had a total of 124 coins, so we can write the equation:
q + d = 124 .......(1)
We are also given that he received $25. Since each quarter is worth $0.25 and each dime is worth $0.10, we can write another equation for the value of the coins:
0.25q + 0.10d = 25
Multiplying both sides of this equation by 100 to clear the decimal, we get:
25q + 10d = 2500 .......(2)
Now we have a system of equations (1) and (2) that we can solve simultaneously to find the values of q and d.
Multiplying equation (1) by 10, we get:
10q + 10d = 1240 .......(3)
Subtracting equation (3) from (2) we get:
(25q + 10d) - (10q + 10d) = 2500 - 1240
Removing the parentheses and simplifying:
15q = 1260
Dividing both sides by 15:
q = 84
So, Xavier had 84 quarters.
The correct choice is C. 84.