Quarter Wit, Quarter Wisdom: Evading Calculations Part II

Quarter Wit, Quarter WisdomLast week we discussed how to solve equations with the variable in the denominator. We also said that the technique generally works for PS questions but you need to be careful while working on DS questions. Today, let’s look at the reason behind the caveat.

Say, the question stem of a DS question asks you to find the value of n, the number of people in the room. Statement 1 of the question gives you the following equation:

60/(n – 5) – 60/n = 2

We can easily figure out that a value of n that satisfies this equation is 15. Now, is that enough to say that statement 1 is sufficient alone? No! It could be a trap! The equation, when manipulated, gives us a quadratic. It is important to find out whether the second solution of the quadratic works for us. When n is the number of people, it must be positive. So one extra step that we should take is re-arrange the equation to get the quadratic. If the constant term i.e. the product of the roots is negative, it means one root is positive and one is negative. Since we have already found the positive root, it is the only answer and hence we can say that the statement 1 is sufficient alone.

60/(n – 5) – 60/n = 2

60*n – 60*(n – 5) = 2*n*(n – 5)

n^2 – 5n – 150 = 0

The constant term, -150, is negative so the product of the roots must be negative. This means one root must be negative and the other must be positive. Since we have already found the positive root i.e. the number of people in the room, we can say that statement 1 is sufficient alone.

Let’s look at an example where we could fall in the trap.

Say statement 1 gives us an equation which looks like this:

60/(n +5) – 10/(n – 5) = 2

As discussed last week, we will easily see that n = 10 satisfies this equation. So should we move on now and say that statement 1 is sufficient alone? No, not so fast! Let’s try to manipulate the equation to get the quadratic.

60/(n +5) – 10/(n – 5) = 2

60*(n – 5) – 10*(n + 5) = 2*(n – 5)(n + 5)

n^2 – 25n + 150 = 0

n = 10 or 15

So actually, there are two values of n that satisfy this equation. In PS questions, since we have a single answer, there would be only one solution so once you get one, you are done. In DS questions, you need to be certain that only one value satisfies. There is a possibility that both values satisfy your constraints in which case your answer would change.

Therefore, it may not be necessary to solve the equation for the PS question, but it is certainly necessary to solve it for DS. That’s counter intuitive, isn’t it? We hope you understand the reason.

Another related trap in DS questions: Statement 1 gives you a quadratic and asks you for the value of x (no constraints that x must be an integer or positive number etc). You know that it is a quadratic and it will give you two values of x so you say that statement 1 is not sufficient alone and move on. But hold it! What if both the roots of the equation are same? It may not apparent to you when you look at the equation. When you solve it, you realize that the roots are the same. Hence, ensure that you solve the equation in DS questions before you decide on the sufficiency.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep in Detroit, Michigan, and regularly participates in content development projects such as this blog!

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