Friday Quiz, Take 2

Here's a word problem to keep you on your toes today. This one is coming to you from McDougal Littell. Yes, my textbook.

A man has 90ยข in quarters, dimes, and nickels. Half the coins are nickels, and a fourth of them are dimes. How many of each does he have?

For extra credit, and a little geometry, check out this Problem of the Week web site from St. Ann's, a creative, private school for gifted children in Brooklyn. A parent told me about this site, and I found some cool, animated questions, including this one.

Pretty please, send your favorite problems to chandlerm@washpost.com. Let me know your name, where you're from, and how you came about this particular problem.
Oh, and our math friendly readers are not only interested in algebra.

By Michael Alison Chandler  |  December 5, 2008; 7:43 AM ET  | Category:  Friday Quiz
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Comments



Good question - I like it because you can crank through the algebra, or you can be a little more creative and use the knowledge that the man probably doesn't have .3 quarters to skip forward a little bit.

You know that the total number of coins, n, has to be divisible by 4, since 1/4 of them are dimes. So, the problem can be reduced to checking multiples of 4.

We can throw out n < 0, because it doesn't make sense to have -1 quarters. We can throw out n = 0, because if you have 0 coins, you've got 0 cents. So the first multiple to try is n = 4. That's 2 nickels, 1 dime, and that leaves 1 quarter, which gives 45 cents. Now, you can try the next multiple of 4, which is n = 8, or you can recogize that 45*2 = 90, so we just need to double the number of each coin, which gives us n = 8. (This would be more useful if the man had, say, $1.80, because we would have to go through more multiples.)

You could take an even shorter cut if you recognize that you really have to have a multiple of 8, so that you don't wind up with an odd number of quarters, which would make the last digit of the total a 5. But at some point, you've got to actually solve the problem!

As for St. Ann's problem of the week, it seems to me that 1) one of the cubes is going to have to come out of the center of the cheese block, 2) you're going to have to separate that cube from the rest of the block on 6 faces, 3) that means 6 cuts, 4) 6 cuts is sufficient to create all 27 cubes (2 through each set of opposing faces). So my answer is 6.

Posted by: tomsing | December 5, 2008 9:10 AM | Report abuse

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