SOLUTION: victor has a coin collection worth $2.90 that contains nickels, dimes and quaters. if there are 3 times as many dimes as nickels, and the number of quarters is 2 more than the numb

Question 10785: victor has a coin collection worth $2.90 that contains nickels, dimes and quaters. if there are 3 times as many dimes as nickels, and the number of quarters is 2 more than the number of nickels. how many nickels, dimes and quarters are there?
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
Let x = number of nickels, since it is the last part at the end of the first phrase.
3x = number of dimes
x+2 = number of quarters

Total value of the coins = value of each coin times the number of coins. You can work in dollars or cents. I recommend that you use cents, so change $2.90 to 290 cents.

5(x) + 10(3x) + 25(x+2) = 290 cents
5x + 30x + 25x + 50 = 290
60x + 50 = 290
60x + 50-50 = 290-50
60x = 240
x= 4 Nickels
3x = 12 Dimes
x+2 = 6 Quarters

Check: Find the values of all the coins.
x= 4 Nickels = $.20
3x = 12 Dimes 1.20
x+2 = 6 Quarters 1.50

Total value = $2.90
It checks.

R^2 at SCC

See my Lesson Plans on Word Problems for Basic Algebra!!!

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