SOLUTION: Suppose that javier has a handful of coins, consisting of pennies, nickels, dimes, worth $2.63. The number or nickels is one less than the number of pennies, and the number of dime
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Question 7766: Suppose that javier has a handful of coins, consisting of pennies, nickels, dimes, worth $2.63. The number or nickels is one less than the number of pennies, and the number of dimes is 3 more than the number of nickels. How many coins of each kind does he have?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Let the number of pennies be "x".
Then number of nickels is x-1.
And number of dimes is x-1+3 = x+2
Value of pennies is (1)x cents
Value of nickels is (5)(x-1)= (5x-5) cents
Value of dimes is (10)(x+2) = (10x+20) cents
Equation:
Value of pennies + Value of nickels + Value of dimes = 263 cents
x + 5x-5 + 10x + 20 = 263
16x +15 = 263
16x = 248
x = 15.5
This would mean there are 15 1/2 pennies, which doesn't make
any sense. Check to see if you have properly posted the
problem.
Cheers,
Stan H.
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