BOB can eat a cheesecake in THREE minutes. His sister JENNY can eat a cheesecake in TEN minutes. How long will it take the two of them, eating TOGETHER, to eat three cheesecakes?
If you’re anything like I used to be, these problems drive you nuts. But I say “used to be” because after I started teaching for Kaplan, I learned that there is actually a dirt-simple plug-and-chug formula that’ll spit out the answer for you. It’s call the combined work formula, and it’s this:
A student in my class last night asked, “Why weren’t we taught this in high school?” Why, indeed. When I learned this formula was real, I wanted to go find my high school math teachers and smack ‘em.
So Bob and Jenny’s cheesecake addiction problem isn’t a hair-puller after all. Just plug their speeds into the formula…
…and remember that this is the GRE, so there’ll always be a trap. Here, remember that the question asks for the time to eat THREE cheesecakes, so multiply the above by 3:
And there’s your answer. In case you missed it in high school, don’t miss it now: the combined work formula is a mighty addition to your arsenal on Test Day!
In my last entry, I showed you how to rejigger radical fractions to make them look the way the test makers want them to look. The example in that entry was a pretty easy one, though — all you had to do was multiply the top and bottom of the fraction by a radical. This time we’re going to crank it up a notch.
Imagine you worked out the math to a GRE quant problem and got this:
There’s a radical in the bottom of that fraction, so it will never be the credited answer to a GRE quantitative problem. As we saw last time, you’ve got to get the radical out from underneath the fraction. Some of you might perhaps be tempted to multiply the fraction through by the square root of 5, similar to what we did last time. Notice, though, what’ll happen to the bottom of the fraction if you do that:
Whoops! We got rid of the radical all right, but another one popped up in its place. What you have to do instead is multiply the top and bottom of the fraction by the complement of the denominator: that is, copy the denominator, but flip the sign. In this case, there’s a + to begin with, so switch it to a – and multiply by . Why do this? Well, you’re about to see!
Since this one’s more complicated, let’s do the two operations separately. First, on top, we’ve got:
Easy enough. The bottom is this:
And here’s why multiplying by the complement is so effective: this is a difference of squares. You absolutely have to know this property on Test Day:
Notice how this gets rid of the radicals cleanly, without making any new ones! All together, then:
Finally, every term in that fraction is a multiple of 4, so divide through by 4 to get the answer:
Click, confirm, and another seemingly impossible GRE challenge bites it. Got any questions or points of confusion about manipulating radicals? Let us know in the comments!
Few things are as unpleasant as that burst of terror you feel when you crunch all the math, get the answer — and it’s not in the choices. Oh no! What did I do wrong! Sometimes, the answer is … nothing! The GRE test makers will deliberately engineer problems to provide that moment of fright: you’ll get the right answer, but it won’t look like the right answer choice.
When this happens, don’t panic. There are a few tricks the test makers use to get your answer to look different from the credited choice. Instead of panicking, just assume that you do have the correct answer, and think back to the tricks. I’ll teach them to you in this post and in the next couple.
Test Maker Trick #1: Have the math work out to a fraction, then ask for a percent.
Sometimes the test makers will give you a bunch of fractions in the problem, but ask for a percent in the choices. For example, let’s say you work out the math and get an answer of:
But the choices are written as percents. What to do?
Well, one option is to pop the fraction into the on-screen calculator and match it up with the right answer. But that’s no fun; calculators are for wimps. Instead, exercise your cleverness and rewrite the fraction out of 100. Percents are out of 100, so any fraction with 100 on the bottom will convert naturally to a percent.
You can do this quickly — even more quickly than a calculator! — if you have the factors of 100 at your command. They are:
1, 2, 4, 5, 10, 20, 25, 50, 100
Play with the fraction to make the denominator one of these numbers, then multiply to reach 100. In this case, 40 is 20 × 2, so:
And now, since 100 = 20 × 5, multiply through the fraction by 5:
Note that while 8.5 × 5 may seem like something you’d need a calculator for, it’s not: eight times five is 40, and five times 0.5 is 2.5, so just add 40 and 2.5 to put the finishing touches on this GRE problem: 42.5%.
When I wrote this post about inequalities, I thought the most interesting thing about it was the way it showed how inequalities mix with equations. Your notes in the comments proved that the actual point of interest was something else.
Here’s the short version if you didn’t read that post or the comments. If y < -12, which of the following must be true?
A) y < -10
B) y < -14
It took a lot of math to get to this point, so I thought the worst was over. I casually wrapped things up by saying, “If y is less than -12, then it’s certainly also less than -10. Click choice A, confirm it, and move on.”
Boy was I wrong.
Two different people posted in the comments to submit a “correction” to the entry, arguing that y < -14 was the correct answer. And I’m very glad they did, because this is exactly the kind of mistake I want you to make BEFORE Test Day — so that I can help you fix it before you it costs you GRE points! The more emphatic commenter wrote:
y<-14 MUST be true because it lies in the required range calculated which is: y<-12 … Any number chosen less than -14 will always lie in the range less than -12. This is a very basic math question and there is no confusion what so ever.
Basic and no confusion, indeed. The error in this reasoning is that it works backwards: the writer starts by assuming that y < -14, then confirms that y < -12 also. But that’s not how GRE problems work. On must-be/could-be questions, you have to work from the problem to the choices, not from the choices to the problem! You can’t just assume that y < -14 is true. It’s an answer choice. You have to CHECK whether it’s true. The only thing you can take as a premise is what’s in the problem, which is y < -12.
Here are some numbers that satisfy y < -12:
-13, -14, -15, -16, …
Now let’s CHECK choice B, y < -14. Is every value of y less than -14? No sir: -13 is greater than -14, and -14 equals -14 (never forget that a number is neither greater than nor less than itself). Since this is a “must be true” question, a statement would have to be true for every value of y in order to be correct. And since choice B is false for some values, you have to cross it out!
The math is basic. The logic isn’t. Always work from the problem to the choices on must-be/could-be problems!
When a question asks us to determine the probability of a certain event occurring, we need to ask ourselves two questions:
- How many possible outcomes are there?
- How many of those possible outcomes will fulfill the requirements given?
The answers to the above questions will allow us to determine probability using the formula:
Let’s look at a GRE question that requires us to apply this formula:
What is the probability of tossing a 1 or a 3 on a fair 6-sided die?
(Notice that the GRE does not leave you wondering about whether the die is fair or not—they remove that doubt right away. Don’t spend time entertaining “what ifs” that have already been resolved.)
If we toss a 6-sided die, how many possible outcomes are there?
Well, we could get a 1, 2, 3, 4, 5 or 6. There are 6 possible outcomes.
How many of those outcomes would fulfill the requirements given (i.e., be a 1 or a 3)? Only 2—a 1 or a 3.
Plugging those numbers into the probability formula, we get:
There is a 1/3 probability of tossing a 1 or a 3. That means we would expect a 1 or a 3 to show up in one out of every three tosses of the die.
That was a pretty easy example. Let’s look at a question that is a bit more complex:
A bag contains 3 red marbles and 5 blue marbles. What is the probability of randomly drawing a red marble, replacing the red marble back in the bag, then randomly drawing a blue marble?
In this problem, we have to calculate two probabilities—the probability of drawing a red marble and the probability of drawing a blue marble.
Since there are 3 red marbles and 5 blue marbles, we know there are 8 marbles in the bag. Therefore, the number of possible outcomes is 8. That will be true for both probabilities because we are replacing the red marble back in the bag before drawing a second marble. Also, because we are returning the red marble to the bag, these probabilities are independent—the color of the second marble drawn will not be determined in any way by the color of the first marble drawn.
Using the probability formula again, we find the following probabilities:
In order to find the probability of drawing a red marble, then drawing a blue marble, we have to multiply the two independent probabilities. Doing so, we find that the probability of red, then blue is:
Since 15 and 64 do not have any common factors, we cannot simplify the fraction any further. Thus, we would expect to draw a red marble, then a blue one, 15 out of every 64 selections of two marbles.
Of course, GRE probability questions can be much more complex than these examples. Be on the lookout for my next probability post, which will focus on the probability of flipping a coin or tossing a die.