August 24, 2011
Try this advanced GMAT probability question, testing your knowledge of the ins and outs of how probability works.
The events A and B are independent. The probability that event A occurs is 0.6, and the probability that at least one of the events A and B occurs is 0.94. What is the probability that event B occurs?
In order to find the probability that event B occurs in this problem, we need to set up and equation that includes the probabilities we are given and allows us to solve for B. We are told that the probability that at least one of A or B occurring is 0.94. ‘At least one of A or B’ means that an outcome is desired if A occurs and B does not, B occurs and A does not or A and B both occur.
It is important to remember two rules of probability here. First, when you encounter an ‘or’ situation you add and when you see an ‘and’ situation you multiply. Second, the probability that an event does NOT occur is equal to one minus the probability it does occur.
Based on these rules, we can translate ‘at least one of A or B occurs is 0.94’ to the following equation:
.6B [the probability that both A and B occur] + .6(1-B) [the probability that A occurs and B does not] + B(1-.6) [the probability that B occurs and A does not] = .94
We can simplify this equation to .6B + .6(1-B) + .4B = .94 and solve for B.
.6B + .6(1-B) + .4B = .94
.6B + .6 – .6B + .4B = .94
.6 + .4B = .94
.4B = .34
B = .34/.4
B = 34/40 = .85
Thus, our answer is 0.85, or answer choice (E).
August 17, 2011
What should you do when you see a GMAT problem asking you for the average rate over an entire journey? Try your hand at this problem and let’s see.
A canoeist paddled upstream at 10 meters per minute, turned around, and drifted downstream at 15 meters per minute. If the distance traveled in each direction was the same, and the time spent turning the canoe around was negligible, what was the canoeist’s average speed over the course of the journey, in meters per minute?
In average rate problems many students forget that average rate means total distance divided by total time and not the average of the rates. This is especially true on problems, such as this one, that give the test-taker two rates, but no distances and no times. When this occurs, the most concrete strategy, which will be quickest for moth test-takers, is to pick numbers.
Keep in mind that when picking numbers, the numbers you choose must conform to any constraints in the problem. Here we are told that the distance was the same in both directions, so we should pick a number that is a multiple of both speeds. The lowest common multiple of our speeds, 10 and 15, is 30, so we will set our distance as 30 meters in each direction.
Next, we calculate the time in each direction using this distance. Going upstream we travel 10 meters per minute for 30 meters. Since distance/rate equals time, we know 30/10= 3 minutes. Returning we travel the same distance at 15 meters per minute. Thus, we know that 30/15 = 2 minutes.
Finally, we need total distance/total time to find our average rate. Our total distance is 30 + 30 = 60. Our total time is 2 + 3 = 5. 60/5 = 12, which is answer choice (B).
This can also be solved algebraically, and if you read the question and VERY QUICKLY and ACCURATELY can set up the appropriate equations and solve, that might be your best approach. However for the majority of test-takers, there is some uncertainty and hesitation when setting up the equations, and hence it can be riskier in case your equations or algebra are not perfectly accurate. Plus the equations will include dividing by variables and won’t be extremely straight forward, whereas picking ‘30’ for distance here made the scenario concrete and the test-taker who took that approach will often be done before the algebraic solver as they confidently solve. It’s great to practice both techniques as you study (algebra and picking numbers), so that you have multiple tools you can use on test day, which might serve you better for different problems.
July 20, 2011
Knowing how to use the distance = rate x time formula in various permutations will help you tremendously on the classic GMAT speed word problems. Just remember that when asked about average rate over an entire journey, you must think about total distance and total time over that journey….
A car drove from Town A to Town B without stopping. The car traveled the first 40 miles of its journey at an average speed of 25 miles per hour. What was the car’s average speed, in miles per hour, for the remaining 120 miles if the car’s average speed for the entire trip was 40 miles per hour?
To solve this problem, you must remember that average speed means total distance divided by total time over an entire journey, and is not the average of the speeds. The total distance in this problem is 160 miles, but we will need to express the total time in a more complex way.
For the first 40 miles, the car traveled at 25 miles per hour. Therefore, we can say that the time this took was 40/25, because distance divided by rate equals time.
For the last 120 miles, we are trying to solve for the rate, so we can call this speed R. Thus, the time for the final 120 miles is 120/R.
If we put all of this together, knowing that the average speed over the entire journey is 40 miles per hour, we get the equation 160/(40/25 + 120/R) = 40. We can now solve for R and reach our answer.
160/(40/25 + 120/R) = 40
4/(8/5 + 120/R) = 1
4 = 8/5 + 120/R
4R = 8R/5 + 120
20R = 8R + 600
5R = 2R + 150
3R = 150
R = 50
50 is choice (C) and is the correct answer.
July 13, 2011
As you may have noticed in prepping for the GMAT, in many cases the challenges you face in GMAT problems are less about the specific math skills, and more about translating word problems into mathematical equations in a fast and efficient way. Figuring out quickly HOW to approach the problem is one of the key skills the GMAT is testing (and, incidentally, a key skill in the business world as well). Try this typical word problem translation question, and be sure to practice GMAT-style word problems frequently, in addition to just practicing algebraic skills.
Jacob is now 12 years younger than Michael. If 9 years from now Michael will be twice as old as Jacob, how old will Jacob be in 4 years?
The first step to answering this question is translating the information in the problem into equations. If Jacob is 12 years younger than Michael, we can say that J = M – 12, where J is Jacob’s age and M is Michael’s age.
The second equation is a bit trickier to determine. You must keep in mind that it refers to the relationship between their ages in 9 years. Thus, Jacob will be J + 9 years old and Michael will be M + 9 years old. The equation we can then write if Michael will be twice as old as Jacob in 9 years is M + 9 = 2(J + 9).
Because the question wants us to solve for Jacob’s age in 4 years, we should next rewrite our first equation as M = J + 12. This allows us to substitute J + 12 for M in the second equation, which becomes (J + 12) + 9 = 2(J + 9). Then solve for J as follows:
J + 12 + 9 = 2(J + 9)
J + 21 = 2J + 18
3 = J.
However, you must remember that the question asks for Jacob’s age in 4 years. Since Jacob is 3 years old today, we know that he will be 7 years old in 4 years. Thus, the correct answer is choice (B).
June 20, 2011
As you have studied for the GMAT, you have probably heard over and over again that you need to make sure not to fall for trap answers. In no type of question is this truer than in quantitative questions asking about average rates.
When you encounter an average rate question you must immediately remind yourself that average rate means the total distance divided by the total time; it does not mean the average of the two rates. You can be assured that an answer choice that simply averages the two rates you have been given will be listed (this is the trap!), so make sure you always keep the definition of average rate in mind.
When you see these problems, use the rates you have been told in the problem, along with any other information you have, to calculate the total distance traveled, and the total amount of time it took. Then, simply divide distance over time as you would in any speed problem.
While this sounds fairly straightforward, the GMAT has found ways to tempt test-takers into simply averaging the speeds. The most common of these is to introduce a problem along the following lines: a truck driver drives from town A to town B at a speed of 40 miles per hour and returns from town B to town A at a speed of 60 miles per hour; what is the average speed for the journey? Here the test makers have only included information on speed, but do NOT choose 50 miles per hour, which will likely be listed as a choice. Instead, pick a number for the distance, which we know is the same each way, calculate the times in either direction and then divided total distance by total time (in this case, the answer is 48).
By keeping that formula in mind every time you see an average rate problem you will be on your way to answering them correctly on a consistent basis.
Try your hand at this sample GMAT problem focusing on a specific probability situation.
Each person in Room A is a student, and 1/6 of the students in Room A are seniors. Each person in Room B is a student, and 5/7 of the students in Room B are seniors. If 1 student is chosen at random from Room A and 1 student is chosen at random from Room B, what is the probability that exactly 1 of the students chosen is a senior?
In this problem we are asked to determine the probability of choosing exactly 1 senior. To be more specific, this means that we will need to select 1 senior and 1 non-senior.
There are two ways this can be done. We can select a senior from Room A and a non-senior from Room B or we can select a non-senior from Room A and a senior from Room B. In probability, ‘and’ means multiply while ‘or’ means add. In this case, we see 2 ‘ands’ indicating that we will multiply probabilities together and one ‘or’ telling us that we will add the resulting products.
Our first option is to get a senior from Room A, which has a probability of 1/6, and a non-senior from Room B, which has a probability of 2/7 (we found this number by taking the probability of selecting a senior in Room B and subtracting it from 1.) Because this is an ‘and’ situation, we multiply 1/6 x 2/7, which equals 2/42.
For our other option we must find the probability of selecting a non-senior from Room A, which has a probability of 5/6, and senior from Room B, which has a probability of 5/7. Again, we have an ‘and’ statement, so we multiply. 5/6 x 5/7 = 25/42.
Lastly, we must address the ‘or’ in our original scenario by adding the two probabilities we determined together. This gives us 2/42 + 25/42 = 27/42. This simplifies to 9/14, which is choice (C).
May 11, 2011
We invite you to give our GMAT Combinations Challenge question below a try! Aim for about 2 minutes and give it your best shot before reviewing the answer & explanation. Good luck!
Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?
The first step in this problem is to recognize that order does not matter. We must create a committee of four people, but we are not putting these people in any sort of order. Because order does not matter, we must use the combinations formula, which is n!/[k!(n-k)!], in order to determine the number of possible outcomes.
However, as is the case with most advanced combinations problems, the solution is not as simple as plugging a few numbers into the formula. Instead, we need to consider exactly what types of outcomes we are looking for.
In this case, we need to select Jane and 3 other people, none of whom are Thomas; Thomas and 3 other people, none of whom are Jane; or Jane, Thomas and 2 other people. Additionally, we need to note that this is an ‘or’ situation, which means that we need to add all of these possibilities together.
First, if we select Jane and three others, none of whom are Thomas, we must pick Jane plus 3 out of the remaining 6 people. Because we can pick any 3 out of the 6, we set n = 6 and k = 3. When we plug this into the formula it gives us 6!/(3!3!) = 20 (Quick review: 3! = 3x2x1, 6! = 6x5x4x3x2x1.)
Next we determine how many ways we can select three people, none of whom are Jane, to go along with Thomas. This is identical mathematically, thus we end up with 20 in this case too.
Lastly, we determine how many outcomes are possible in which we have Jane, Thomas and 2 others. We now select 2 out of 6, so n = 6 and k = 2. 6!/(4!2!) = 15.
All we need to do to get the answer is add up all of the possible outcomes. 20 + 20 + 15 = 55, which is the correct answer to this question (C).
The last time we looked at GMAT rate problems dealing with two trains, we discussed how to identify the correct strategy for a particular problem. Now we are going to take this a step further and look at an example of a GMAT quantitative problem in which these strategies can be implemented.
Let’s consider the following:
City A and City B are 140 miles apart. Train C departs City A, heading towards City B, at 4:00 and travels at 40 miles per hour. Train D departs City B, heading towards City A, at 4:30 and travels at 20 miles per hour. The trains travel on parallel tracks. At what time do the two trains meet?
In order to solve the above problem, we first must get the trains to leave at the same time. Train D leaves 30 minutes after train C, so we must find out how far train C has traveled in that first 30 minutes. Remembering that distance equals rate x time, we know that distance = 40 x .5 (note that 30 minutes needs to be expressed in hours.) This tells us that, at 4:30, train C has traveled 20 miles. Thus, our trains are now 120 miles apart. At this point all we need to do is add our rates together, to get the rate at which the trains are traveling towards one another, giving us 40 + 20 = 60 miles per hour, and plug into our three part formula of D = R X T: 120 = 60 x T, therefore T = 2 hours. If we started at 4:30 and traveled for two hours, it is now 6:30, which is the correct answer to this question.
The strategy we just employed will also work on problems in which the trains are traveling away from each other. On test day, rather than feel yourself getting anxious when you see a question that sounds similar to the “Two trains…” scenario, just take a deep breath, envision the scenario, and take it step by step.
Before students begin studying for the GMAT – before they even know anything about it beyond that it tests math and verbal – one question type worries students more than any other: a question about two trains traveling on parallel tracks.
Every time I start to go over a question about two trains, two cars or, in one rather quirky problem, two earthworms, students roll their eyes and prepare for the worst.
But these problems have received a bad rap. In order to solve them, students only need to be able to remember the basic rates formula, learn a two-step method and be able to differentiate between the two flavors in which this problem appears.
Learn the Rate Formula
First up, the rate formula. A rate is just something per something. It can be dollars per jobs, people per team or any other ratio. However, the most common rate on the GMAT is speed. Speed is equal to distance divided by time. This is the only formula students need to know in order to solve a classic two trains problem.
Two-Steps to Solve
Next, students need to remember a two-step process to solving these problems. Step 1 is to get the trains (or cars or earthworms) to start at the same time. Most GMAT problems of this type will have one of the trains starting earlier. Figure out how far this train has traveled by the time the second train starts and determine their distance apart at this time and you have completed step 1. Step 2 is to either add or subtract the train’s individual rates and, using the distance apart, calculate the missing piece (usually time) using the speed formula.
Know when to add and when to subtract
In order to implement this strategy effectively, the last piece of the puzzle is to know when to add the rates and when to subtract them. This is surprisingly straightforward. If the trains are coming towards each other or going away from each other, add their speeds. If one train is catching up to the other, subtract their speeds.
By remembering these three basic rules, you will be able to handle two train questions in under two minutes and save that time for truly time-consuming problems.
March 14, 2011
When a straightforward arithmetic problem appears on the GMAT, many test-takers treat it as a break from the more complicated problems that are on the GMAT. While arithmetic problems are often not as complicated as many of the other problems on the test, you should still be careful not to make careless math errors. The most common cause of such errors is a mistake in the order of operations.
Operations in an arithmetic problem need to be completed in a specific order. The best way to remember is via the acronym PEMDAS, which stands for parentheses, exponents, multiplication, division, addition and subtraction. Regardless of the actual order of the operations listed in the problem, they should always be completed in this order.
PEMDAS in Action
To see why order of operations is so important, let’s consider the following two expressions:
(5 – 3) + 4 x 6 ,
5 – (3 + 4) x 6.
Simplifying the first yields: (5 – 3) + 4 x 6 = 2 + 4 x 6 = 2 + 24 = 26
Simplifying the second yields: 5 – (3 + 4) x 6 = 5 – 7 x 6 = 5 – 42 = -37
Notice that all that changes between these two problems is the placement of the parentheses, but the results are completely different. The wrong answer choices on the GMAT will often be based on mistakes in order of operations, so it is essential to always follow the correct order in order to achieve the highest possible score.