Exponent Errors on the GMAT
August 31, 2011
Exponent questions are among those that give GMAT-preppers the most difficulty. The key to answering exponent questions correctly is to remember all of the rules you must follow. For example, knowing that ab x ac = ab+c, and ab x cb = (ac)b will be essential to answering exponent questions correctly.
However, you also want to avoid common exponent mistakes. This is especially important because wrong answer choices in exponent questions, as with the rest of the GMAT, will be based on common test-taker errors.
The most common mistake test-takers make on exponent questions is to erroneously believe that ab + ac = ab+c. Remember, exponent rules refer to multiplication and division, not addition and subtraction. When you encounter an exponent question on addition or subtraction, you will usually need to factor out like terms in order to simplify. For example, if a problem states that 2210 + 225, rewrite it as (225)(225) + (225)(1). Then we can factor out 225, which gives us 225(225 + 1). Be sure to note that you cannot simplify it to 2215, as this is a completely different number.
By being aware of common exponent mistakes you can ensure that you will not be tempted by wrong answer choices and that you will complete such problem correctly. This, in turn, will lead to a higher score on test day.
Kaplan GMAT Sample Problem: Probability and “at least”
August 29, 2011
Probability questions can be among some of the more advanced and trickier problems you’ll face on the GMAT Quantitative section. Be sure to pay attention to the wording of word problems such as this one; in this case when asked about a scenario with “at least twice”, it will be more efficient to solve for that NOT happening and subtract from 1 (since the probability of something happening plus the probability of that same thing NOT happening should add up to 1, or 100%.)
Question:
A fair coin is tossed five times. What is the probability that it lands heads up at least twice?
(A) 1/16
(B) 5/16
(C) 2/5
(D) 13/16
(E) 27/32
Solution:
The key phrase to solving this sample GMAT problem is ‘at least twice.’ This means that out of our five flips, two, three, four and five heads are all desired outcomes. On problems such as this one it is important to remember that we can find the probability that something does NOT happen and subtract that from one, in order to find the probability it DOES happen. In this case, if we flip zero heads or one head, we will NOT have at least two heads. Finding these two probabilities, adding them together (remember, that ‘or’ becomes addition in probability) and subtracting from one will be the fastest way to get to the answer.
Because probability means desired outcomes over possible outcomes, we will need to find each of these for this problem. The first outcome we want to consider is zero heads. Only one way exists for this to happen: all of the flips come up as tails. Thus, we have ONE desired outcome in which zero heads appear. The second outcome we are looking for is exactly one head. Five outcomes would provide one head, as the head could be first, with all other flips coming up tails, the head could be second, with all other flips coming up tails, and so on. Therefore, in total we have six outcomes that do not give us at least two heads.
Next, we need to find the number of possible outcomes (our denominator). Each flip has two possible outcomes: heads and tails. In order to find total possible outcomes, we multiply the possible outcomes for each individual flip. As we have five flips, we get 2 x 2 x 2 x 2 x 2 = 32 possible outcomes.
When we put these together we have a 6/32, or 3/16, probability of not flipping at least two heads. Since we found the probability of what we do not want to happen, we still need to subtract our result from one to find the probability it does happen. The math for this is as follows:
1 – 3/16 = 16/16 – 3/16 = 13/16
13/16 is answer choice (D) and is the correct answer.
Kaplan GMAT Sample Problem: Probability Problem Solving
August 24, 2011
Try this advanced GMAT probability question, testing your knowledge of the ins and outs of how probability works.
Problem:
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?
(A) 0.34
(B) 0.65
(C) 0.72
(D) 0.76
(E) 0.85
Solution:
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).
Kaplan GMAT Sample Problem: Average Rate
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.
Problem:
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?
(A) 11.5
(B) 12
(C) 12.5
(D) 13
(E) 13.5
Solution:
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.
Kaplan GMAT Sample Problem: Prime Factors
August 10, 2011
Dealing with prime numbers and prime factors is an essential GMAT skill. Technically it is a skill we learned as early as elementary school/early school age, however just because it was learned in childhood does not mean that these questions are necessarily easy for GMAT test-takers. First of all, dealing with prime factors may not be something we do on a daily basis, so you’ll want to be sure you practice during your studies. Secondly, sometimes the wording or steps along the way can be challenging on the GMAT, not to mention the time limit. Here is a fairly straightforward question, though it may take some time to work through.
Problem:
What is the smallest positive integer that is a multiple of 18, 20, 24, 25 and 30?
(A) 360
(B) 900
(C) 1,800
(D) 2,400
(E) 3,600
Solution:
While taking each answer choice, starting with the smallest, and dividing by each of the five numbers listed until you find one that divides evenly by all of them will get you to the right answer, you can avoid this time consuming process by employing prime factorization.
First, find the prime factors of the five numbers included in the question. These are as follows:
18 = 3 x 3 x 2
20 = 2 x 2 x 5
24 = 2 x 2 x 2 x 3
25 = 5 x 5
30 = 2 x 3 x 5
Second, we need to put these prime factors together in order to find our answer. We want to find the smallest number that is a multiple of all of our given numbers, so let’s start with the smallest one. The smallest multiple of 18 is 18, which equals 3 x 3 x 2. Next, our number must be a multiple of 20, which equals 2 x 2 x 5. Thus, we need 3 x 3 x 2 x 2 x 5 for our number to be a multiple of both 18 and 20, as this number includes all of the prime factors of both 18 and 20. Our third number is 24, which equals 2 x 2 x 2 x 3. However, the number we are building already includes two 2’s and a 3. Therefore, we only need to include one more 2 in order to make our number a multiple of 24. Now we are at 3 x 3 x 2 x 2 x 5 x 2. For our number to be a multiple of 25, we must add in another 5, giving us 3 x 3 x 2 x 2 x 5 x 2 x 5. Finally, to be divisible by 30 our number must include 2 x 3 x 5. However, as 2 x 3 x 5 is can already be found, we do not need to add in any additional numbers.
Lastly, to reach the answer, we calculate 3 x 3 x 2 x 5 x 2 x 5 in the following manner:
3 x 3 x 2 x 2 x 5 x 2 x 5 =
18 x 10 x 10 =
18 x 100 =
1800, which is answer choice (C).
Kaplan GMAT Sample Problem: Data Sufficiency Averages
August 3, 2011
GMAT Data Sufficiency questions can take simple concepts like averages and have test-takers pausing or falling into traps because of the way they are worded, and the fact that you have to keep in mind what your goal is with Data Sufficiency—to find out whether or not you have sufficient information to answer the question!
Problem:
Each of the 8 numbers s, t, u, v, w, x, y and z is positive. Is the average (arithmetic mean) of s, t, u, v, w, x, y and z greater than 46?
(1) The average (arithmetic mean) of s, t, u, v and w is greater than 74.
(2) The average of x, y and z is greater than 120.
Solution:
Before evaluating the statements, you should reword the question. We are asked if the average of a list of numbers is greater than 46. Since average is equal to the sum of the terms divided by the number of terms, we can write this question as: is sum/8 > 46? This can be simplified to: is sum > 46(8) or is sum > 368?
Statement 1 tells us that the average of the first five numbers in our set is greater than 74. At first, this may seem insufficient, as it tells us nothing about x, y and z. However, if five of our numbers have an average greater than 74, it means that those numbers must sum to a result greater than 74 x 5, which equals 370. If the sum of five numbers is greater than 370 and the other three numbers must all be positive, the overall sum must still be greater than 370. If the sum is greater than 370, then it is also greater than 368. Therefore, based on statement 1, we can answer the question as ‘always yes,’ which is sufficient.
Statement 2 we must approach in a similar manner. Now we know that the final three numbers in our set must have an average greater than 120. This means they must have a sum greater than 360. The other five numbers in the set can be equal to 1 at the smallest, therefore the total sum must be greater than 365. As 365 is smaller than 368, the average may or may not be greater than 46. ‘Sometimes yes, sometimes no’ is insufficient. So our final Data Sufficiency answer choice is (A) or choice (1); the first statement is sufficient to answer the question, the 2nd is not.
Multiplication Shortcuts on the GMAT
July 27, 2011
When working on the GMAT quantitative section, it is always important to remember that the questions are written so that they can be completed within about a two-minute timeframe. If you encounter a problem and the math seems as if it will take more than two minutes to do, it generally means that either you made an error or a faster way to solve exists. One of the most frequent cases in which the latter occurs is on problems that involve multiplication, since there are no calculators on the GMAT.
Unlike long division, which can be very useful on the GMAT, longhand multiplication is almost never necessary. Instead you should always look for shortcuts to solve. Not only will this be quicker, but it will also provide fewer opportunities for careless errors.
One such shortcut is prime factorization. If a problem asks you to multiply 525 by 16, you can break down 525 to 3 x 5 x 5 x 7 and 16 to 2 x 2 x 2 x 2 and multiply these strings of primes together, giving you 3 x 5 x 5 x 7 x 2 x 2 x 2 x 2. Since the order in which you multiply does not matter, recombine the primes into easier to use numbers. Especially keep an eye out for 5′s and 2′s you can make equal 10. Here we can rewrite our problem as (5 x 2) x (5 x 2) x (3 x 7) x (2 x 2) = 10 x 10 x 21 x 4. Again, look for multiplication you can complete quickly. 100 x 84 = 8400.
Another shortcut is to take a step back and consider what the multiplication problem really means. 64 x 9, is one fewer 64 than 64 x 10. So we can solve 64 x 10 = 640, then subtract 64, which equals 576.
Whenever you can find a faster way to reach the answer you should take it. The key is to remember, the right way to do the problem is the way that gets you to the correct answer most quickly.
Kaplan GMAT Sample Problem: Average Speed
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….
Problem:
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?
(A) 28
(B) 48
(C) 50
(D) 55
(E) 70
Solution:
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.
Kaplan GMAT Sample Problem: Algebraic Translation
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.
Problem:
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?
(A) 3
(B) 7
(C) 15
(D) 21
(E) 25
Solution:
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).
Commonly tested GMAT Number Properties Concepts
July 6, 2011
For almost every GMAT math problem, more than one way to solve it exists. On most problems, your options will be to do the straightforward algebra, or to use a strategy, such as picking numbers for unknowns. However, on number property questions, knowledge of the properties themselves will often allow you to solve without having to do time-consuming math. These number properties fall into three commonly-tested categories.
First, is integers and non-integers. An integer is any whole number, positive, negative or zero. If you add, subtract or multiply two integers together, the result will always be an integer. However, if you divide two integers, the result may or may not be an integer.
Second, is even and odd numbers. An even number is any integer that, when divided by two, produces another integer (and this includes zero.) An odd number is any integer that does not. If two even numbers are added or subtracted, the result is an even; if two odd numbers are added or subtracted, the result is an even; and if an even number and an odd number are added or subtracted, the result is an odd. If two even numbers are multiplied, the result is an even; if two odd numbers are multiplied, the result is an odd; and if an even number and an odd number are multiplied, the result is an even. Also, remember that no rules apply when dividing numbers.
Lastly, is positive and negative numbers. Keep in mind that a positive multiplied or divided by a positive is a positive, a positive multiplied or divided by a negative is a negative and a negative multiplied or divided by a negative is a positive.
By remembering all the rules listed above, you can often answer number property questions correctly more quickly than with any other method. This will leave you more time for problems that require you to work out all the math, which in turn leads to a higher score.
