Answer to question #2

Here's another problem that gives you an option between algebra and plug-in. Let's begin, however, with eliminating some of the answers without doing any work. Can you eliminate any answers using what you know about money?

Sure, the answer can't be even because an even number of quarters comes out as a value ending in "0" so when you add any number of dimes you can't come out with a 5 in the units place. Since the total value is $3.25, we have an odd number of quarters.

Did you do it algebraically? If so, you restated (Pillar II again) the information as .25n + .10(22-n) = 3.25 and solved for n. There are a couple of places you can make simple algebra errors so watch out when you multiply .10 times (22-n). You need to end up with 2.2 -.1n. Then you'll probably clear all the decimals. So be very careful.

Let's do it by substitution. Where do you start?

C. Nine quarters gives us $2.25. That means there has to be too many (13) dimes left because the total will be more than $3.25. If you noticed that there can't be an even number of quarters, you'd go directly to A. If not, you'd calculate, eight quarters is $2.00 and see that there isn't any number of dimes that can give you $3.25 leaving you with A. Be sure to check your calculations by doing (A). Seven quarters are $1.75 plus (22-7) 15 dimes is $3.25. Do you think this way was easier than algebra? Practice it. You'll like it.



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