Question 1170525: A person has 11 coins consisting of dimes and nickels. If the total amount of money is $0.75, how much of each coin are there? Found 3 solutions by josgarithmetic, greenestamps, ikleyn:Answer by josgarithmetic(39620) (Show Source):
You can put this solution on YOUR website! Since you have only eleven coins for seventy five cents, you could simply look at several of the possible combinations of nickels and dimes and find the combination that works.
You could use a little algebra instead.
d dimes
11-d nickels
Simplify and solve.
A quick informal way to find the solution, without formal algebra, but also without blind trial and error....
(1) Imagine you have 11 coins that are all nickels; the total value would be 55 cents.
(2) The actual total value is 75 cents, which is 20 cents more than that.
(3) Exchange nickels for dimes one at a time, thus keeping the number of coins the same but increasing the total value by 5 cents each time.
Since you need to increase the value by a total of 20 cents, you need to do that exchange 20/5 = 4 times. So you end up with 4 dimes and 11-4=7 nickels.
Let d be the number of dimes; then (11-d) is the number of nickels.
Next, you write the total money equation
10d + 5*(11-d) = 75 cents.
You cancel factor of 5 in both sides
2d + (11-d) = 15
and simplify
2d - d = 15 - 11, or
d = 4.
ANSWER. 4 dimes and 11-4 = 7 nickels.
You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.
A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.
Read them and become an expert in solution of coin problems.
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