When a GRE quantitative problem features multiple ratios, many of you suffer headaches. This is because the “math” way of solving the problem is brutal, and students who don’t use logic will dive head-first into a morass of ugly substitutions, mistakenly assuming that the GRE is a math test. Here’s the kind of problem I’m talking about:
In a particular mixed candy bag, the ratio of Skittles to M&M’s is 4 to 5, while the ratio of Reese’s Pieces to M&M’s is 9 to 7. What is the ratio of Skittles to Reese’s Pieces?
The “math” way to do this problem is to set up two equations, solve one for M&M’s, and plug that value into the other one. If that sounds painful, that’s because it is. Don’t do this. Make a simple table instead:
S | M | R
4 : 5
7 : 9
Take a moment to confirm that you understand where the numbers above are coming from. They’re just a translation of the information in the word problem.
The question asks for the ratio of S to R. Can you just say it’s 4 to 9? No way. The value connecting them — the M — is different. It’s 5 in one ratio and 7 in the other. So, rewrite the ratios to make the M term the same in both, creating a kind of “bridge.”
Multiply the first ratio by 7: 7×(4:5) = 28:35
Multiply the second ratio by 5: 5×(7:9) = 35:45
Next, check out your new table:
S | M | R
28 : 35
35 : 45
Now you can just “walk across the bridge,” as it were — the ratio of S to R is simply 28:45. Try this technique on your next multiple-ratios problem and let us know how it goes!
My fiancé and I have five pets: A dog, a kitten, and a trio of guinea pigs. We are normally excellent animal parents, but the one thing that we consistently disagree on is whether it’s okay to feed them outside of their mealtimes and occasional treats. I taught the dog that begging from the table gets her nowhere with me, but my fiancé is the king of slipping her bites of steak or “dropping” scraps of bacon onto the ground (or into her already-open mouth, as is more often the case). Our last conversation on the topic began when he, instead of putting his leftover pork into the fridge, put it into the dog’s bowl. It went something like this:
Me: “If you keep doing that, she’s going to get fat!”
Him: “Oh, please – how much can a few bites possibly matter? That much meat wouldn’t affect my weight at all.”
He made two crucial mistakes in that moment: He both triggered my inner GRE instructor, who loves percentages and data analysis, and made me determined to prove him wrong once and for all. So I decided to figure out the percentage by which that extra pork increases the dog’s caloric intake, and compare that to how much it increases his.
My fiancé eats anywhere from 2,500 calories to 2,800 calories per day, with the upper end being during football season – not because he plays, but because he and his friends spend every Sunday feasting on buffalo wings and sliders while watching NFL games. So what percentage of his daily intake does a 4-ounce pork loin comprise? We can use the percentage formula to find out. The percentage formula is (y/x) * 100%, where x is the total value and y is the smaller portion of the whole. In this case, since 4 ounces of pork has roughly 160 calories, our percentage would be (160/2,800) * 100%.
Now, you could very reasonably be asked to estimate this percentage on the GRE, so let’s simplify it without using the calculator. Let’s ignore the * 100% at the end of the equation for now, and just focus on the fraction. We can start reducing 160/2,800 by cancelling out one factor of 10 in the numerator and the denominator, and then factoring out a 4 from each term. That would look like this:
160/2,800 = 16/280 = 4/70
Now, if we add the * 100% back in, we have (4/70) * 100%, which gives us 400/70 when we multiply 100 times the 4 in the fraction’s numerator. Once again, the numerator and the denominator each have a factor of 10, and when we cancel it out we’re left with 40/7. We know that 42/7 equals 6, so 40/7 equals something just a bit smaller – we can estimate it to be 5.8%.
By comparison, our 30-pound dog only needs 900 calories per day to maintain her weight. So every time someone makes her day by giving her 160 calories worth of meat, he increases her intake by (160/900) * 100%. The fraction in this case doesn’t reduce quite as easily as the previous one did, so let’s look at the 100% and the 900 in the denominator – that reduces to 1/9. So we’re really dealing with 160/9, which again is not very easy to reduce. Let’s take a quick quiz to test your strategic estimation skills: Without using the calculator, can you determine which of these percentages is roughly equivalent to 160/9?
Again, this is another very GRE-like question, and your ability to quickly approximate will serve you well. 160/10 would give us 16%, and 160/9 is going to be just a bit larger than that – the only answer that can possibly match is (C), 18%.
So while that piece of pork would only increase my fiancé’s calories by less than 6%, it would increase our dog’s calories by nearly 20% – a significant difference that definitively proved me right. I presented my findings to my fiancé, who had no choice but to cede his case. When dinnertime rolled around, however, one of the ham steaks went missing – apparently tired of listening to us debate, the dog had taken matters into her own hands and stolen it off of the counter.
I know that I’m not the only one who uses GRE-like percentage, ratio, and arithmetic calculations in my everyday life – how do these types of math factor into your day-to-day activities? Let us know in the comments!
In my last post, I discussed how GRE math shows up in real life situations. My wife and I, while baking a carrot cake, decided to increase the recipe by adding more of each ingredient. This would, we surmised, create a Mega Carrot Cake (MCC). The problem we faced, though, was that we didn’t quite have enough flour to double the recipe. So instead of making a cake twice as big, we were forced instead to increase the amount of each ingredient proportionally, thus creating an Almost Twice As Big As Normal But Not Quite Twice As Big Mega Carrot Cake (ATABANBNQTABMCC).
Actually, you know what? That acronym is much too long. Let’s just stick with MCC.
Once we determined the amounts for each ingredient, we threw everything together in a bowl and began mixing. But then another thought hit me. Will this MCC fit in our cake carrier? We were making this cake for a friend’s housewarming party, and it was essential that when all was said and done, we be able to transport the MCC from our house to theirs. The cake pan we bought was deeper than our other pans, and the diameter was an inch larger. So not only would each layer of the MCC be thicker and wider, but stacking the layers would make the cake, as a whole, higher than a normal carrot cake. Our cake carrier was about 12″ tall, and previous iterations of this particular carrot cake stood about 9″. Would there be enough room in the cake carrier for our MCC?
If you remember back to my last post, you’ll remember that we increased the ingredients by 60% (going from 2.5 to 4 cups of flour is an increase of 1.5 cups, and 1.5/2.5 is 60%). That means the volume of our cake should increase by 60%, as well. So if I could find the volume of the old carrot cake, increase it by 60%, then divide by the area of the new pan, I should have the height of my new cake. Whew. I’m getting tired just thinking about it. All I wanted was to make a Giant Carrot Cake. Is that too much to ask?
Our old cake pan was advertised as having a diameter of 8 inches. If the height of the normal carrot cake was also 8 inches, what was the volume of the older cake?
If you’re having trouble finding that value, remember the formula for the volume of a cylinder:
p * r2 * h
In other words, if we can find the area of the round cake pan, then multiply it by the height of the cake, we should have our volume. The cake pan has a radius of 4 inches, so its area is 16p inches2. Multiply that by 9 and we end up with 144p inches2. Now, in order to determine the volume of the larger cake, we have to increase it by 60%. We can do that by multiplying by 1.6. When we do that, we find that our Mega Carrot Cake has a volume of 230.4p inches3.
That’s a big cake.
But we’re not done, are we? We still need to find the height of the MCC. (Please be less than 12″. Please be less than 12″.)
- Remember that the new cake pan actually has a larger diameter – this one is 9 inches across.
- If the diameter is 9 then the radius must be 4.5 inches, which means that the area of the base of the cake will be (4.5in)2p, or 20.25p in2.
- Divide 20.25p in2 inches into 230.4p inches3 and…YES!! 11.38 inches is the height of the MCC!
The giant cake will live! Well, live long enough to make it to the housewarming party, that is. After that, all bets are off.
As you can see, math in real life can be just as complicated as math on the GRE, and vice-versa. The concepts and principles that are being tested by the GRE test-makers are not relevant simply in the abstract. Ratios, proportions, percents, and volumes are mathematical concepts that have daily, real-world applications. And the more you practice your GRE math skills, the better you’ll get at navigating the waters of a math-filled world.
Now, where’s my fork? This MCC isn’t going to eat itself.
We’ve discussed GRE math in real life before. Yep, that’s right: real life situations that require real GRE math calculations. Now, you may think that the math section of the GRE focuses only on abstract math that couldn’t possibly be useful in your day to day life. Not true. Even though most of the math might seem abstract, the concepts that are tested largely focus on basic math that, on occasion, we actually do use in our daily lives. Don’t believe me? Keep reading.
The other day, I was baking a cake. Ok, I was actually helping my wife bake a cake. Same difference, right? I’m pretty sure that getting married means I get to take credit for my wife’s baking skills and she gets to take credit for my fantasy football prowess. Or at least, that’s what it said in the brochure. Anyway, we wanted to make a carrot cake, and we wanted to make a big one. You see, we had just purchased a fairly large cake pan, and we figured that we could fill it up by doubling the recipe. As we marched around the kitchen, pulling ingredients and utensils out of the cupboards, I took a look at our recipe and noticed something potentially problematic. The recipe called for 2.5 cups of flour, but we only had enough to squeeze out 4 full cups. That wasn’t enough to double the recipe. So, bummer. At that point, we had a decision to make: either run to the store and get more flour, or stick with the recipe as written and make a standard-sized cake. But wait a minute, I thought. Couldn’t we simply use 4 cups of flour, and bump up all of the other ingredients proportionally?
If you just made this face:
then you’re on the same wavelength as my wife. ”Proportions?” she said. ”I’m not sure about that. It sounds vaguely GRE-mathy.”
Proportions aren’t just vaguely GRE-mathy, they are definitely GRE-mathy, and the GRE test-makers love asking questions about them. Luckily for us, they’re easy to spot and easy to rock on the test. So, what are proportions? Proportions are nothing more than two ratios set equal to each other. ”Ok, hot shot, what’s a ratio?” Ratios are pretty straightforward. They’re just a comparison between two different numbers.
In this case, let’s set up a ratio between the original amount of flour and the new amount of flour. That would be 2.5 cups to 4 cups (you might also see the ratio written as 2.5 : 4 ). Whenever possible, we want to write a ratio as a fraction. So let’s put the original amount over the new amount, which will give us a fraction of 2.5 / 4. We will then set this ratio equal to ratios of all the other ingredients, giving us a series of proportions. Setting up these proportions will help us determine how much sugar, carrots, and baking soda we’ll need to make our super big (but not quite double-sized) cake.
Let’s say that the original amount of sugar was 1.5 cups. Well, we need to know how much sugar to use in our new recipe, so we’ll set up a proportion where we insert 1.5 into the original amount and “x” as our unknown amount.
original 2.5 1.5
———- = —– = ——
new 4 x
Setting up a proportion this way allows us to “scale up” at the same rate for each ingredient. By whatever constant we multiplied 2.5 to get to 4, we have to multiply by the same constant to ramp up from 1.5 to our unknown “x”. So, let’s figure out what our x value is. Any guesses as to how we do that? (I’ll wait while your brain works its way back to 8th grade math.)
Yes, that’s right! When we have two fractions set equal to each other, and one of the values in one of the fractions is a variable, we should cross multiply to get rid of the fractions. Doing so would give us 2.5x = 4(1.5). If we keep going, we get 2.5x = 6. Divide both sides by 2.5 and we end with x = 2.4. So, we need 2.4 cups of sugar to stay consistent with the recipe and make our bigger (but not quite double-sized) cake.
Now that I’ve got you rolling with proportions, I’m going to turn it over to you to help me finish the rest of the cake. The original recipe said I needed 1 teaspoon of baking soda and 4 eggs. If we keep the proportions the same as above, then how much baking soda and how many eggs will I need for my new cake? Put your answers in the comments!
I’ll admit it: The topic of how to set up three-part ratios is one of my favorites to discuss with students. It represents the quintessential GRE math concept – something that can be done using a straightforward process, but that throws many test-takers for a loop because it’s not something they use in their daily lives. Think about it – how much time does the average person spend comparing or combining different ratios? Not much. Here’s an example of how three-part ratios can be tested in a Quantitative Comparison:
The quantities are relatively uncomplicated here – we’re asked to compare the smallest possible number of total employees to a fixed term, 336. However, in order to calculate any value for Quantity A, we’ll need to untangle a less-than-pithy paragraph in the centered information.
Let’s boil down the key pieces of information that we have: The ratio of employees in department A to those in department B is 7 to 12, and the ratio of employees in department B to those in department C is 5 to 24. We also know that each department employs at least 10 people.
So we’re given two relationships between the three departments. In order to calculate the minimum number of employees, we’ll need to compare all three departments to one another. In fact, we’ll need to use the information provided to create a three-part ratio.
We’ll start by lining up our variables horizontally, making sure that the one mentioned in both of the ratios given (in this case, B) is in the middle. Below that, we’ll line up the ratios that we already know:
Now that we have our information organized, how can we compare all three pieces? We certainly can’t determine the relationship between A and C at this stage, because the value for each is related to a different value of B. Before we can compare all the pieces of this ratio to each other, the number in the middle column needs to be the same. Once it is, then we will be able to see the ratio of A to C just as easily as we do the ratio of A to B and B to C.
The least common multiple of 5 and 12 is 60, so let’s multiply both of the ratios given so that the value in the B column equals 60. We’ll need to multiply the first ratio by 12, and the second one by 5.
Which gives us a result of:
Now that the value of B is constant, we can compare all three pieces to each other. The ratio of A to B to C is 35 to 60 to 228. We’ve found our three-part ratio!
But we’re not done with this problem yet. We still have to determine the smallest total number of employees in the three departments. To determine this, we need to add up the three values given: If there are 35 employees in department A, 60 in B, and 288 in C, then there are 383 employees in all – we’ve established that Quantity A is larger than Quantity B.
Be sure to note that a ratio ordinarily doesn’t tell you anything about the actual number of terms you’re dealing with – it merely represents the most-reduced relationship between the terms. The only reason that we could calculate a real value in this case is because we were looking for the minimum possible total – this will always be just the sum of the numbers in the ratio.
Now that you’ve seen one problem that requires you to set up a three-part ratio, you’ll be able to tackle any similar problems that the GRE throws at you with confidence and precision. Just remember the mantra “The common term must be the same”, and you’ll be set!