Question 179941
Joe has a collection of nickels and dimes that is worth $6.05. If the number of dimes is doubled and the number of nickels was decreased by 10, the value of the coins would be $9.85. How many dimes do we have?
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let the number of nickels and dimes be n and d respectively
:
.05n+.1d=6.05.........eq 1
.05(2n)+.1(d-10)=9.85.....eq 2
:
rewrite eq 1 to .05n=6.05-.1d--->n=121-2d and plug the value of n which is 121-d into eq 2
:
.1(121-2d)+.1(d-10)=9.85
:
12.1-.2d+.1d-1=9.85
:
-.1d=1.25
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d=12.5
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something is wrong with this problem. One of the amounts in the equation is off. there is no way to get a whole number for dimes in this scenario. Email me the correct numbers.  thanks Mathtut