The post Distance Rate Time Formula Memorization Shortcut appeared first on Magoosh GRE Blog.

]]>When working on Distance-Rate-Time questions, you will inevitably use one of the following three formulas.

1. Distance = (Rate)(Time)

2. Rate = Distance/Time

3. Time = Distance/Rate

Did you know that there’s a quick and easy shortcut for recalling all three formulas?

The shortcut is similar to some of the shortcuts you learned in highschool science. It looks like this: D/RT (aka, the “DiRT” shortcut). As you may have guessed, D = distance, R = rate, and T = time.

Can you see all three formulas hiding in the simple fraction D/RT?

Here’s how it works. Let’s say, you need to recall the formula for finding the time it takes to complete a certain trip. Simply take the fraction D/RT and remove the T (for time). This leaves you with D/R. In other words, Time = Distance/Rate.

Similarly, if you need to recall the formula for finding the rate (speed), simply take the fraction D/RT and remove the R (for rate). This leaves you with D/T. In other words, Rate = Distance/Time.

Finally, if you need to recall the formula for finding the distance, take the fraction D/RT and remove the D (for distance). This leaves you with RT. In other words, Distance = (Rate)(Time).

So, instead of memorizing 3 formulas, just remember DiRT, which stands for D/RT.

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]]>The post Breakdown of the Most Commonly Tested GRE Quant Concepts appeared first on Magoosh GRE Blog.

]]>The 2^{nd} edition Official Guide has just been released. At the back of the book is a new test (GRE Practice Test 2). In general, the concept break down is not surprising, if you look at the concepts that pop up in Practice Test 1 (which is the same as the practice test in the first edition of the Official Guide).

Word problems abound. There is a fair amount of algebra. Even a few combinatorics problems rear their fearsome heads. And if you’re wondering how many statistics questions are on the GRE, a topic often given scant attention in many prep books – it can pop up in spades. Most interestingly, coordinate geometry only shows up once, and there is not a single rate problem.

The information below pertains to the 50 quantitative questions in Practice Test 2. I have lumped QC and Problem Solving questions together. I have not assigned only one category per question, but have ascribed multiple categories to the same question. For instance, a word problem can contain any number of concepts. To only call it a word problem is a vague. After all ten Word Problems on combinations is very different from ten word problems dealing with percents.

The breakdown of concepts detailed below will likely be very similar to what you see test day. That said you would probably see more than one coordinate geometry question and would probably see a rate question. And just because a concept did not show up on the test, does not mean you will not see that concept test day. Interest problems, both simple and compound, can very well show up.

What is the take away from all this? If you are weak in a certain area, say word problems that deal exclusively with adding a series of numbers, do not fret about this. You may only see one problem, if any, dealing with such a concept. On the other hand, if you are only somewhat comfortable with Statistics it is a good idea to strengthen this area, as you are far more likely to see it test day.

Word Problems (15)

Algebra (9)

Geometry (7)

Statistics (6)

Exponents (5)

Rates (0)

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]]>The post Compound Interest vs. Simple Interest on the GRE appeared first on Magoosh GRE Blog.

]]>For some this is the most “interest”ing concept on the Revised GRE (at least for those with a fondness for bad puns). For most, compound vs simple section can be a nuisance. Many think, what is the difference between the two, and/or how did that formula go again?

But remember, this concept involves money, and for many that means its practical (especially if you invest money yourself). But enough rambling…

Principal: The Amount of Money initially invested

Interest Rate: The amount return on an investment expressed as a percent of the principal.

Time: The length of time in which a principal is invested

1. John invests 100 dollars in account that yields 8% simple interest annually. How much money will John have in his account after one year?

(A) $4

(B) $8

(C) $104

(D) $108

(E) $110

2. Bob invests 100 dollars in a fund that yields 15% simple interest monthly. If Bob invests the principal in the middle of January, which is the first month will he have more than $200 total?

(A) June

(B) July

(C) August

(D) February

(E) March

3. In 2001, John invests x dollars in a special account that yields y% simple interest annually. If he has $250 in his account in 2006 and in 2008 he has $270 in his account, what is x + y?

(A) 5

(B) 25

(C) 200

(D) 205

(E) 210

Answers:

1. D

2. C

3. D

Okay, that was the easy part. Now for compound interest. In compound interest things become complicated. We no longer have a nice, clean linear increase. To illustrate:

If Mike invests $100 at 10% simple annual interest, he will have $110. After two years he will have $120. That is his money grows by $10 every year. After 10 years, Mike will have doubled his money.

Now, let’s say Mike’s friend Thomas invests $100 at a 10% rate that is compounded annually. After one year, Thomas will have made the same amount as Mike. But then things start diverging. Remember how Mike always gets 10% of the original 100 (the original 100 is called the principal)? Well, Thomas – because things are getting compounded annually – gets 10% of whatever the value of the account is at the end of the year. Let’s see how this plays out over time.

1^{st} year: 10% of 100 = 110

2^{nd} year: 10% of 110 = 121

3^{rd} year: 10% of 121 = 133.10

4^{th} year: 10% of 133.00 (rounding down) = 146.30…

After 10 years, Thomas will have made $260, which is $60 more than Mike.

Okay, that may all seem like chump change, but the same percent increase applies to numbers with a few more zeroes thrown in. How would $260,000 vs. 160,000 sound?

Of course the point of this lesson is to understand the conceptual difference between the two forms of interest—and not to have you running to the nearest ban, since the numbers above are very generous.

Now for the fun part: Notice how, in the case of Thomas, I seemed to be doing mathematical wizardry. After all, how did I know that 10% compounded annually at 10 years is going to yield 160% of the principal? Well, let’s meet the formula:

V = Total Value

P = Principal

r = annual interest rate

n = number of times per year invested

t = number of years

Pretty unpleasant, no? Well, let’s try to put the formula to the test. And you may want to get your calculators out (this is the Revised GRE after all!)

If $10,000 is invested at 10%, compounded semi-annually, how much will the investment be worth after 18 months?

(A) 11,500

(B) 11,505

(C) 11,576.25

(D) 11,625.30

(E) 12,000.50

Now don’t worry about the semi-annual bit—it just means twice a year. And remember the *n* from the scary little formula above: the number of times per year invested. And that 18 months? That corresponds to t, the number of years, which translates to 1.5.

.

That was easy—once you know where to put everything (and provided you remember the formula)!

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]]>The post Math Basics – Distance, Rate and Time appeared first on Magoosh GRE Blog.

]]>The 4^{th} of July weekend is upon us, and many of us will be taking road trips, or even taking a plane somewhere. To commemorate this collective movement, let’s learn the most fundamental formula when dealing with movement over time. First, let’s meet Bob…

*Bob drives at an average rate of 50 mph from Berkeley to Los Angeles, a distance of 350 miles. How long does it take him to complete the trip?*

* *

*(A) **4 hrs*

*(B) **5 hrs*

*(C) **7 hrs*

*(D) **10 hrs*

*(E) **12 hrs*

When dealing with distance, rate and time, we always want to remember the nifty little formula, D = R x T, in which D stands for the distance, R stands for the rate (or speed), and T stands for the time.

With the problem above, the distance between Berkeley and Los Angeles is 350 miles. So D = 350. Bob is traveling at 50 mph, so that is his rate. The question is how long will it take him to complete this trip. Therefore, we have to solve for T. Let’s set up the equation, plugging in the values for D and R:

350 = 50T.

Solving for T, we get 7. Because we are dealing with miles per hours, the 7 corresponds to hours. So T = 7. Answer (C).

Now let’s try another problem:

*Charlie takes 2.5 hours to fly from Los Angeles to Mexico City, a distance of 1200 miles. What is the average speed of his plane in miles per hour?*

* *

*(A) **200 mph*

*(B) **240 mph*

*(C) **410 mph*

*(D) **480 mph*

*(E) **533 mph*

Setting up the equation, we get 1200 (Distance) = 2.5 (Time) x R; 1200 = 2.5R. Before solving (assuming this is on the current GRE, in which you do not have a calculator), let’s change 2.5 to 5/2, as it is much easier to do the math with fractions than with decimals. We get 1200 = 5/2R. Solving for R, we multiply both sides by the reciprocal, 2/5. This gives us R on the right-hand side of the equation, and 1200 x 2/5 on the left-hand side. 2400/5 = 480 (D).

Note you could have solved this problem back-solving, in which you put the answer choices back into the question. Or, even better, we could just plug-in 500 mph. Of course that is not one of the answer choices, but using 500 will make the math very easy. If the number is a little too high when we plug-in 500, then the answer must be (D). If it is a little low, then the answer must be (E) 533. Again, you never try to back solve with an ugly number like 533—it will take too long.

Using 500 mph, we get 1200 (Distance) = 500 (Rate) x 2.5 (Time). You can see that multiplying these two numbers gives us 1250. Meaning, we flew too fast and missed Mexico City by 50 miles. Therefore, we have to slow down the plane a little—the answer is 480.

So, this weekend, whether you are traveling abroad or just making a trip to a friend’s BBQ, you can figure out your average distance, rate, or time. Who said GRE doesn’t relate to the real world?

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]]>The post GRE Math Strategies – When to Plug-In appeared first on Magoosh GRE Blog.

]]>See if you can finish this rate problem in less than two minutes.

*A transcontinental jet travels at a rate of x – 100 mph with a headwind and x + 100 mph with a tailwind between Wavetown and Urbanio, two cities 3,200 miles apart. If it takes the jet 2 hr 40 minutes longer to complete the trip with a headwind, then what is the jet’s rate flying with a tailwind?*

*(A) 500*

*(B) 540*

*(C) 600*

*(D) 720*

*(E) Cannot be determined by the information given.*

**Did you Plug in?**

You may have been tempted to put together an algebraic equation. If you are adept at doing so, and can usually get the answer quickly, then I encourage you to go ahead and make an equation. Most students, however, find this strategy cumbersome and problematic. Even if they set up the right equation, which they can’t be quite sure of unless they get the answer, they may very well make a mistake in solving the equation. If this scenario describes what just happened when you attempted the problem above, know that there is a better strategy: Plugging-In.

**When and Where to Plug-In**

Once we’ve decided to plug in, where do we start? Do we plug in our own numbers, or the answer choices? First, we want to look at the answer choices. Are they numbers? If so, plug them in. If they are variables, you will need to come up with your own numbers, as long as those numbers conform to the information provided in the question.

Here we have answers, so let’s try plugging them in. The first place to start is the middle. The logic is if the number is too low (or too slow, in this case), you need to pick a larger number. Note: if the middle answer is a weird number like 625, then I would recommend plugging in (B) or (D).

Luckily, we can work with answer choice (C) 600. If the speed with a tailwind is 600 mph, then the speed with the headwind is 400. The distance between the two cities is 3200 miles. Using d = rt, where d stands for distance, r stands for rate, and t stands for time, we find that the time it takes to fly with a tailwind is 5 hr 20 min, and the time with a headwind is 8 hours. The difference in time is 2 hr 40 min. And there is our answer. Just like that.

If that seems too easy, that’s not a bad thing. Plugging-In can look like magic, in that it can make a seemingly intractable problem fall into place, just like that.

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