GRE Math Refresher or Refreshing Math?
During class 2 of our program, we teach a lesson entitled Math Refresher. I thought it would be fun to share one of the problems with you and talk through two things: (1) the actual algebra (real math) we would need to solve the problem, and (2) the standardized testing skills that allow us to solve the problem much more easily.
First, try this problem (© Manhattan Prep). Note: there are no multiple choice answers: you have to come up with a number on your own!
Sarita and Bob together contribute to the cost of a $40 gift. If Bob contributes $12 more than Sarita does, how much does Bob contribute?
How did you do it? Did you write some equations? Try out some numbers? Find some other cool shortcut?
On this one, the best approach is a very neat shortcut that I’m going to show you “ but only after I show you the actual algebra. It’s important to know how to do the algebra, just in case you don’t know or can’t find any good shortcuts.
The first sentence tells us that the two people, Sarita and Bob, together pay for a $40 gift, so we can write an equation: S + B = 40. The second equation tells us how much more Bob contributes than Sarita, which gives us another equation: S + 12 = B.
For that second one: how do you know that the +12 should go with S and not with B? If you just know, think about how you’d convince a fellow student who wanted to put the +12 with B. What’s actually happening here?
I’m asking you to do this for a very important reason: if you can articulate the logic behind why you translate something from words to math in a certain way, then you’re much less likely to make a mistake “ and English-to-math translation is a very common source for errors on any standardized test.
Here’s how we know: which is greater, S or B? Bob contributes $12 more, so B represents the larger number. In order to get these two numbers to equal each other, then, we have to add to the smaller number, S.
Okay, whenever we have two equations, we can either line them up to add or subtract, or we can substitute and solve. Either way, we want to pay attention to the question. Which variable are we trying to find?
In this case, they’ve asked us for B. Try to set things up to solve for B, not for S. First, it’ll be more work if we solve for S because then we’ll still have to find B. Second, on a question with numerical answer choices, I can pretty much guarantee that the value for S will be in the mix “ just to see whether we’re all paying attention. We don’t want to fall into that trap!
Okay, we’ve got:
S + B = 40
S + 12 = B
It would be very easy to take the second equation and substitute into the first, but look what happens if we do:
S + (S + 12) = 40
We’re solving for S! That’s not what we want. Here’s the rule: you’re going to take one equation and substitute it into the other. Let’s say that we’ve decided to use the second equation, S + 12 = B. First, solve this equation for the variable that you do NOT want: S = B “ 12. Then, substitute into the first equation:
(B “ 12) + B = 40
Voilà ! Now we’re solving for the desired variable, B.
2B = 40 + 12
2B = 52
B = 26
One more note. I’m sure you noticed that I added parentheses both times that I substituted part of one equation into another. In these two equations, the parentheses were (GRE vocab word alert!) superfluous “ we didn’t actually need them. Nevertheless, this is a good habit to get into, every time, because sometimes it changes everything. What’s the difference between these two equations?
A “ (A + 12) = 40
A “ A + 12 = 40
Not sure? Try solving both.
That’s right “ you get two different answers. Why? Because, in the first equation, the subtraction sign gets distributed across both terms. The equation becomes A “ A “ 12 = 40. In the second equation, where we didn’t use the parentheses, the subtraction sign is applied only to the first substituted term, A.
Okay, so that’s how we solve this kind of question using algebra. It turns out, though, that there’s a fantastic shortcut that we can use. There are two things to learn: how the shortcut works, and when we can use it.
When a problem tells you that two people, or two machines, or two variables of some sort add up to a specific number, and then also tells you the difference between those two variables (how far apart they are), we can use this shortcut.
Take the add up to number, in this case 40. Divide by 2. Also take the difference number, in this case 12. Divide it by 2 also. Then, if you’re solving for the larger of the two numbers, add the two halved numbers together. If you’re solving for the smaller of the two numbers, subtract the smaller halved number from the larger halved number.
For example:
40 / 2 = 20
12 / 2 = 6
To solve for B, the larger number, add: 20 + 6 = 26
To solve for S, the smaller number, subtract: 20 “ 6 = 14
Isn’t that great? No algebra, just real numbers! Again, though, we can only use this when certain conditions apply: we’re told that two variables add up to some number and we also know the difference between the two variables.
Takeaways for wordy translation problems:
(1) Know how to translate from words to math accurately and efficiently. This can make the difference between solving on time and wasting time, getting a problem right and never finding the answer.
(2) Look for shortcuts! We discussed one in this article; there are plenty of others out there. You’ll discover these when reading explanations for problems, discussing with teachers or friends, and working on your own. After solving three or four very similar problems, you may start to notice patterns that can lead you to neat shortcuts “ but you’ll never notice if you’re not actually looking.