“After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?”

Step 1: Dissect the information in the word problem- There’s a lot of information here and it’s essential to make sure you get it all figured out. A sketch can often help. Even though I lack artistic skills, this diagram shows you that you don’t need to be an artist to organize information.

Step 2: Figure out what you need to solve for- In this case we’re asked to find how much further south Bob can run. That’s the section of the purple line that goes south to the turn around point.

Step 3: Set up an equation to solve- How could we isolate the variable we need to solve for in this situation? Here’s one approach:

Additional distance south = Total distance south – 3.25m.

Total distance south = 1/2 total distance

Step 4: Solve your equation- Now that we know we can find our desired variable if we can find the total distance or the total distance south, we can make a second equation and solve. Here we know that Bob runs at a constant rate of 8 minutes per mile, and that the purple section of our graph takes him 50 minutes to run. So, we get this equation:

1 mile/ 8 minutes * 50 minutes = 6.25 miles

That tells us that the purple segment of our line is 6.25 miles. All that’s left to do is isolate the piece of the line we want. We can eliminate the 3.25 mile segment at the top, leaving 3 miles. All that’s left is the turn around which is equal parts: south and north. So, we divide that 3 mile segment in two and we get our answer: 1.5 miles (A).

Keep practicing and good luck!

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“Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was $120,000. What was the median price of the three houses?

(1) The price of Tom’s house was $110,000.

(2) The price of Jane’s house was $120,000.”

Step 1: Identify your target and the given information- Here our goal is to find the median home price. We’re given the average price which tell us that the total of the three sales is $360,000.

Step 2: Find paths to sufficiency- in other words, figure out what you’d need to know in order to be able to find the median. When arranged in increasing order, the median is the middle term. In order to find the middle term we’ll either need to know the values of all three terms, or be guaranteed that one of our terms is the middle one.

Step 3: Assess the first statement- If Tom’s house was $110,000 we know that the two other houses sum to $250,000. If Jane’s house was $100,000 and Sue’s house was $150,000 the median would be $110,000. Alternatively if Jane’s house was $120,000 and Sue’s house was $130,000 the median would be $120,000. Since there is more than one possibility statement 1 is insufficient.

Step 4: Assess the second statement- If Jane’s house was $120,000 we know that the two other houses sum to $240,000. On the basis of this statement alone we can’t determine the exact prices of the other two houses. Many students will stop there, call this insufficient and get this question wrong. Remember, we could also have a sufficient amount of information if we could guarantee that $120,000 was the middle value. Since the other two houses sum to $240,000 we only have two options. The first option is that that one house is less than $120,000 and one is more. If that’s the case $120,000 is our median. The only other alternative is that all three houses cost $120,000. In that situation our median is still $120,000. Statement two is sufficient.

Follow all of these steps carefully and you’re well on your way to a great score. In data sufficiency questions you especially don’t want to jump ahead to combining both statements until you’ve fully evaluated each one.

Study hard!

]]>When you’re young, it’s simple: you get one candle for every year of the birthday you’re celebrating and all the candles tend to match. As you get older, the practicality of putting one candle per year decreases, and the candles that do make it to the cake are typically what’s left over in the drawer. At some point my family took the hodgepodge of candles as a mathematical challenge: what value could each color of candle take so that the sum equal’s the birthday person’s age?

As we got older that became too simple, so we added a twist that I have lately come to realize is exactly the kind of challenge that the GMAT test makers love to give: let each color of candle represent a unique prime number such that the sum equals the number of years being celebrated.

Here’s an example to get you started for my 31st birthday:

One red, two green, three purple = 31

Some hints:

1. Remember that 2 is the only even prime number. Whether you need to make an even or odd total will tell you how/if you can use 2.

2. Start with the small primes. A few small primes and one big one is usually the best approach.

My solution:

Red = 19

Green = 3

Purple = 2

Here’s what makes this such a great GMAT problem. It touches heavily on primes and number properties: two topics that the GMAT loves because they’re more focused on your pattern recognition and problem solving than your math skills. Let’s look at how logically applying those skills gets you to this solution.

First of all, we need Total Red + Total Green + Total Purple = 31. Let’s look at that in terms of evens and odds. Total Red + Even + Total Purple = Odd. Since there are two green, no matter what value we put in that spot, the total will be even. So, moving some things around we get the following:

Total Red + Total Purple = Odd – Even

And, since Odd – Even is always Odd, we get Total Red + Total Purple = Odd.

If the sum of two integers is odd, that always means that one of them is even and the other is odd. We have an odd number of red and purple candles, and the only way to get an even product from Odd* Integer is for that integer to be even. That’s where our use of 2 comes in. Since there is only one even prime number, two must be the value for either red or purple. Once we plug that in we only need to test a few values until we find a combination that works! (There may be others… list them in the comments if you find a good one)

Here’s a few celebrity birthdays for March 24th and their candles. See if you can figure these out! I’ll post suggested solutions at the bottom, but feel free to find your own!

Jim Parsons 41- One yellow, three red, two blue, two green

Peyton Manning 38- Four red, two green

Chris Bosh 30- One yellow , one blue, one green

Alyson Hannigan 40- Two pink, two orange

And the challenge question… today Harry Houdini would have turned 140- One blue, one red, two green

Jim Parsons 41- One yellow (19), three red (2), two blue (5), two green (3)

Peyton Manning 38- Four red (7), two green (5)

Chris Bosh 30- One yellow (2), one blue (5), one green (23)

Alyson Hannigan 40- Two pink (7), two orange (13)

Harry Houdini140- One blue (5), one red (131), two green (2)

]]>This year’s tournament has a little extra excitement as businessman Warren Buffett has offered $1,000,000,000 to anyone who can complete a perfect bracket. But Buffett didn’t luck into his fortune. Let’s use basic probability to determine just how likely it is that someone will be collecting that billion dollar check!

First of all, we need to determine whether we’re working with independent or dependent events. In this case, our outcomes are independent. The result of one game does not alter the odds in the next game. Dayton beating Ohio State doesn’t change the likelihood of Harvard beating Cincinnati. In contrast, a dependent event would be something like pulling marbles out of a bag without replacement. If we pull a blue marble out of the bag there is one less blue marble to pull out the next time, so the odds will change.

With independent events, we can compute the probability of all n events in a series happening by the following formula:

P(1)* P(2)* P(3)*…*P(n) = Probability of all events occurring

In this case, we need to know the probability of predicting all 63 events correctly. In our first case, let’s assume we know nothing about college basketball. In that case, we’ll randomly guess a winner from each matchup. Since we know nothing about the teams, we will estimate our likelihood of getting each guess correct at 1/2 or 0.5. In that case our likelihood of predicting a perfect bracket will be 1 in 2^63 or 1 inĀ 9,223,372,036,854,775,808. That’s more than 9 billion billions! So without any knowledge it’s safe to say that your chances are not good.

But what if you know a little bit more? What if you know that a 1 seed has never lost to a 16 seed? That makes four games slam dunks and just leaves 59 left to figure out. Maybe you’re a college basketball expert and you can predict 60% of outcomes correctly. Your chances of getting a perfect bracket are now 1/(0.6)^59 or 1 in 12,275,963,663,147. We’re down to one in twelve million billion, but Mr. Buffett’s money is still looking pretty safe.

Let’s try one other example. Let’s say you’re a college basketball expert with a time machine that allows you to look two days into the future. That will allow you to know the first 32 results with absolute certainty, and just leave the final 31 games that you can predict correctly 60% of the time. Now your chances are 1/(0.6)^31 or 1 in 7,538,956. Even with expertise and a time machine your chances are just one in seven and a half million!

Three quick takeaways:

1. When events are independent, multiply the probabilities to find the total probability of all events happening

2. Exponential growth makes numbers get really big (or really small) really fast

3. Warren Buffett’s money is probably safe… until we build a better time machine!

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