Question 232563: Could you help me with this problem. I can't seem to figure it out correctly.
Amy has two more nickels than dimes and five more dimes than quarters. Her nickels, dimes, and quarters total $3.25. How many of each kind of coin has she?
Answer by rjd41(9)   (Show Source):
You can put this solution on YOUR website! First, I recommend changing everything to cents.
Dimes is listed on both sides of the problem, this is what you will focus on then.
You want to show all coins listed in the problem:
x - will represent dimes.
10(x) + 5(x+2) + 25(x-5), notice the three coins and how they follow the problem.
10x + 5x+10 + 25x-125 = 325
so you get 40x + 10 - 125 = 325
add 115 to each side to get 40x alone
40x + 115 - 115 = 325 + 115
40x = 440
divide each side by 40 to get x alone
x =11, the 11 is how many dimes you have. 11(10) = 110
Two more nickels than dimes, 11 + 2 = 13, 13(5)=65
five more dimes than quarters so 11 - 5 = 6, 6(25)= 150
check: 110 + 65 + 150 = 325
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