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20 coins, consisting of dimes and quarters.

if the values were switched, the difference would be 90 cents.

This is how i came up with my answer.

solving the system:

one equation describes the number of coins: x + y = 20

where x is the number of dimes and y is the number of quarters.

the other equation describes the amount of money:

(0.25x + 0.1y) - (0.1x + 0.25y) = 0.90

(0.25x + 0.1y) describes when the dimes are quarters and the quarters are dimes.

(0.1x + 0.25y) describes when the dimes and quarters are their original values.

The difference is supposed to be $0.90 or 90cents.

Therefore,

by the elimination method of solving a system of equations:

x + y = 20 #1

(0.25x + 0.1y) - (0.1x + 0.25y) = 0.9 #2

simplify #2 to become:

0.15x - 0.15y = 0.9 #3

x + y = 20 #1

0.15x - 0.15y = 0.9 #3

multiply #1 by 0.15:

0.15x + 0.15y = 3 #4

0.15x - 0.15y = 0.9 #3

add #4 and #3:

0.30x = 3.9

x = 13

therefore, the number of dimes that the man has is 13.

to find the number of quarters, x is evaluated in the 1st equation.

x + y = 20

13 + y = 20

y = 20 - 13

y = 7

This can be checked as follows.

13 dimes = 130 cents

7 quarters = 175 cents

total = 305 cents

13 quarters = 325 cents

7 dimes = 70 cents

total = 395 cents

The difference between the two is 90 cents.

Therefore, the man had 13 dimes and 7 quarters.

By the way, is this a question that you are doing in class, have the answer to and are seeing who can figure it out, or are you proposing it for another reason?

Addendum (by PatrickLMT): Actually, this question was featured in the February 26, 1974 comic strip "Peanuts." Peppermint Patty had this question in school and despaired of finding the answer.

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Q: A man has twenty coins consisting of dimes and quarters If the dimes were quarters and the quarters were dimes he would have ninety cents more than he has now How many dimes and quarters does he have?
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