SOLUTION: From his paper route, Brody collected $5.55 in nickels and dimes. The number of nickels was 6 more than the number of dimes. How many nickels were there?

Question 1118429: From his paper route, Brody collected $5.55 in nickels and dimes. The number of nickels was 6 more than the number of dimes. How many nickels were there?
Found 2 solutions by math_helper, greenestamps:
Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!

(1) N = D + 6 (number of nickels was 6 more than the number of dimes)
(2) 5N+10D = 555 (Brody collected $5.55 = 555 cents = 5*number of nickels + 10*number of dimes)

Substituting for N from (1) into (2):
5(D+6) + 10D = 555
15D + 30 = 555
15D = 525
D = 35 �> N = 41
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Ans: There were 41 nickels (and 35 dimes)
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Check:
41 = 35+6 (ok)
5(41) + 10(35) = 205 + 350 = 555 (=$5.55) (ok)

Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

Without algebra, using logical analysis....

(1) Take away the "extra" 6 nickels, so that the remaining coins are equal numbers of nickels and dimes. The amount left is $5.55 - 6($.05) = $5.55-$.30 = $5.25.
(2) Pair the remaining coins up into groups of 1 nickel and 1 dime. Each of those groups has a value of $0.15.
(3) The number of those groups required to make the total of $5.25 is $5.25/$0.15 = 35. So there are 35 nickels and 35 dimes.
(4) Bring back the "extra" 6 nickels, giving you the final answer of 41 nickels nd 35 dimes.

Of course, if an algebraic solution is required, then the solution from the other tutor is fine.
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