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Thanks to Brandon for the suggestion! This is from a Peanuts comic that ran on February 21, 2000.
A man has 20 coins, consisting of dimes (10 cents) and quarters (25 cents). If the dimes were quarters and the quarters were dimes, he would have 90 cents more than he has now. How many dimes and quarters does he have?
Peppermint Patty just screams out, HELP! Can you figure it out?
As usual, watch the video for a solution.
Math Problem From Peanuts Comic
Or keep reading.
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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.
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Answer To Math Problem From Peanuts Comic
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
Let d be the number of dimes and q the number of quarters. For the number of coins to be 20, we have:
d + q = 20
Every dime is 10 cents and every quarter is 25 cents. Initially the man has:
10d + 25q
If dimes were quarters and vice versa, the number of each coin is swapped with the other. So there will be q dimes and d quarters, which make for a total monetary value of:
25d + 10q
Subtracting the initial amount leaves 90 cents more, giving the equation:
25d + 10q – (10d + 25q) = 90
15d – 15q = 90
d – q = 6
Adding to the first equation d + q = 20 eliminated the q variable to give:
2d = 26
d = 13
As there are 20 coins in total, it must be q = 20 – 13 = 7.
The man has 13 dimes and 7 quarters.
Reference
Peanuts comic February 21, 2000
https://www.gocomics.com/peanuts/2000/02/21
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