Will you please answer question # 27, Test 3, section 4. Is it possible to use real numbers in your example? Thanks!

Sure! The thing you want to remember when you’re plugging in numbers is that you’re doing it to make your life easier, so pick numbers that will really make your life easier. In this case, that means I’m going to say that the original rectangle is a 10 by 10 square, so its area is 100. Obviously, it’s very easy to work with 100 when dealing with percents, and it’s also easy to take different percents of 10 to increase/decrease side lengths.

So, yeah. Here’s our original rectangle:


Now we need to increase its length by 10 percent (10 percent of 10 is 1, which means the length increases from 10 to 11), and decrease its width by p percent. What should we do for p? Well, why not backsolve?

Say p = 20, like answer choice C says. (I’m picking that because it’s in the middle, and also because 20 is the easiest choice to work with so why not try it first.) If p = 20, then we decrease the width by 2, taking it from 10 to 8. So here’s the new rectangle:


What’s the difference between the original area of 100 and the new area of 88? 12. And here’s where we pat ourselves on the back for picking numbers that gave us 100 as our starting area: 12 is what percent of 100? 12 percent! So the first choice we tried, C, is right. Awesome!

(Whenever I illustrate backsolving and the first choice I try is the right choice, I feel compelled to show what a wrong choice would look like. So let’s quickly look at choice B. We’re still increasing the length by 10 percent, so the length is still 11. The width is now decreased by 15 percent, making it 8.5. The new area is 11\times 8.5=93.5. Is that a 12 percent decrease from the original area of 100? No, 93.5 is too big! So if we didn’t already know the answer, we’d know that the next choice we should try should be smaller.)

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