One of the most uncomfortable aspects of the GMAT is that you are not allowed to use a calculator for the quantitative section. This is uncomfortable because, throughout your everyday life, you are never more than about 5 feet from a calculator (yes, even in Death Valley). Almost everyone has a cell phone, a laptop, a desktop or a GMAT guru nearby to compute difficult calculations for them. Even high school students are generally allowed their calculators on test day. However, the lack of a calculator allows the GMAT to test your reasoning skills and time management skills much more easily than if you had access to electronic help.
As an example, remember open-book tests. These tests always seemed easier when they were discussed in theory than when they were attempted in practice. An open book test must necessarily test you on more obscure and convoluted material, otherwise the test becomes too easy and everyone gets 100. Closed-book tests, by contrast, can concentrate on the core material and gauge how much preparation each student has put in. Adding more tools only serves to make the test more difficult in order to overcome these enhancements.
With a calculator, asking you to calculate the square root of an 8 digit number or the 9th power of an integer is trivial if you only have to plug in some numbers. However, if you need to actually reason out a strategic approach in your head, you have accomplished more than a thousand brute force calculations would. On the GMAT, the mathematics behind a question will always be doable without a calculator, but the strategy chosen and the way you set up the equations will generally be the difference between the correct answer in two minutes and a guess in four.
Let’s look at a question where the math isn’t too difficult, but can get tedious:
Alice, Benjamin and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8 and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
(A) 3 / 140
(B) 1 / 28
(C) 3 / 56
(D) 3 / 35
(E) 7 / 40
This is a probability question, and therefore we must calculate the chances of any one event occurring. However, the question is asking about several possibilities, specifically any occurrence where two players win and the third loses (think of any romantic comedy). This means that we have to calculate several outcomes and manually add these probabilities. This is entirely feasible, but it can be somewhat tedious. Let’s look at the best way to avoid getting bogged down in the math:
Firstly, the three players’ are suitably abbreviated as A, B and C (convenient, GMAT, convenient). We therefore want to find the probability that A and B occur, but that C does not occur (denoted as A, B, ⌐C). This represents one of our desired outcomes. However, this is not the only possibility, as any situation where two occur and the other doesn’t is acceptable as well. Thus we can have A and C but not B (A, ⌐B, C), or B and C but not A (⌐A, B, C). The sum of these three outcomes is the desired fraction, so only some math remains.
Let’s do them in order. For (A, B, ⌐C), we take the probability of A, multiplied by the probability of B, and then multiplied by the probability of 1-C. If the chances of C are 2 / 7, then the probability of them not occurring must be the compliment of this, which is 5 / 7. The calculation is thus:
1 / 5 * 3 / 8 * 5 / 7.
In a multiplication, we only care about multiplying the numerators together, and then multiplying the denominators together. There is no need to put these elements on common denominators. The math gives us:
(1* 3 * 5) / (5 * 8 * 7). This is 15 / 280.
There is a strong temptation to cancel out the 5 on the numerator and on the denominator to make the calculation easier, but you should avoid such temptation on questions such as these. Why? (I’m glad you asked). If you simplified this equation, you would get the equivalent fraction of 3 / 56, which is easier to calculate, but since we still have to execute two more multiplications, we will end up adding fractions that have different denominators. This is not a pleasant experience without a calculator, and likely will cause us to revert to our common denominator for all three fractions, which is 5 * 8 * 7 or 280. Additionally, now that we’ve calculated it once, we don’t need to worry about the denominator for the following fractions, it will always be the same. Let’s continue and hopefully this strategy will become apparent.
The next fraction is (A, ⌐B, C), which is equivalent to
1 / 5 * 5 / 8 * 2 / 7. Note that ⌐B is (1 – 3/8)
Executing this calculation yields a result of 10 / 280.
Finally, we need (⌐A, B, C), which is equivalent to
4 / 5 * 3 / 8 * 2 / 7. Note that ⌐A is (1 – 1/5)
Executing this last fraction gives us 24 / 280.
Once we have these three fractions, we must add them together in order to get the probability of any one of them occurring (“or” probability, as opposed to “and” probability”). This is simple because they’re all on the same denominator, so we get 15 / 280 + 10 / 280 + 24 / 280 which is 49 / 280.
Now that we have this number, we can try to simplify it. 49 is a perfect square that is only divisible by 1, 7 and 49, whereas 280 has many factors, but one of them fairly clearly is 7. We can thus divide both terms by 7, and get 7 / 40. Since the numerator is a prime number, there is no additional simplification possible. 7 / 40 is answer choice E, and it is the correct pick on this question.
Had we simplified each probability as much as possible, we would have ended up with 3 / 56, 2 / 56 and 3 / 35. While the addition would not be impossible, it would become much more difficult. In fact, to correctly add these numbers together, you’d have to put them on their least common multiple, which would be 280 again. There is usually no point in simplifying fractions in questions like this because they must usually be recombined at the end. Save time and don’t convert once only to convert back.
The math on this question is not difficult, but having to add together multiple fractions and simplifying expressions can be quite time-consuming. With a calculator, you could simply add the decimals together, regardless of their fractional equivalents. However, the GMAT doesn’t allow you that shortcut on test day (unless you approximate in your head), so you must find a better tactic. The difference between solving all the questions and running out of time on the math section is often the approach you take on each question. Keep up a consistent strategy and you’ll solve a large fraction of the questions you face on test day.
Ron Awad is a GMAT instructor for Veritas Prep based in Montreal, bringing you weekly advice for success on your exam. After graduating from McGill and receiving his MBA from Concordia, Ron started teaching GMAT prep and his Veritas Prep students have given him rave reviews ever since.