SOLUTION: I have 48 coins, all nickels, dimes and quarters, worth $4.80. I have 8 more nickels than dimes. How many quarters do I have?

Question 523224: I have 48 coins, all nickels, dimes and quarters, worth $4.80. I have 8 more nickels than dimes. How many quarters do I have?
Answer by Maths68(1474) (Show Source): You can put this solution on YOUR website!

Let
Number of Dime = x
Number of Nickel = x+8
Number of Quarters = 48-(x+x+8) = 48-(2x+8)= 48-2x-8=40-2x

Given
Total coins = 48
Total Amount = $4.80 = 4.80*100 = 480 cents

5*(Nickels)+10*(Dimes)+25*(Quarters)=Total Amount
5(x+8)+10x+25(40-2x)=480
5x+40+10x+1000-50x=480
-35x+1040=480
-35x=480-1040
-35x=-560
-35x/-35=-560/-35
x=16

Dime = x = 16
Nickel = x+8 = 16+8 = 24
Quarters =40-2x = 40 -2*16 = 40-32= 8

Check
=====
5*24+10*16+25*8=480
120+160+200=480
480=480
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