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We know that ‘Easy C’ is a common trap of DS questions – have you wondered whether there could be trap called ‘Easy A/B’ such that the answer would actually be (C)? Such questions also exist! The point is that whenever you feel that the question was way too simple, you might want to take a step back and review. GMAT will try every trick in the trade to delineate you. Let us show you a question which looks like an easy (A) but isn’t:

Question: 25 integers are written on a board. Are there at least two consecutive integers among them?

Statement 1: For every value in the list, if the value is increased by 1, the number of distinct values in the list does not change.

Statement 2: At least one value occurs more than once in the list.

**Solution**: Let’s first review the information given to us here:

25 integers are written on the board – we don’t know whether they are all distinct. We want to know if there is any pair of consecutive integers among them.

Let’s look at the statements:

**Statement 1:** For every value in the list, if the value is increased by 1, the number of distinct values in the list does not change.

It is easy to fall for statement 1 and think that it is sufficient alone. Say, if any single value is increased by 1 and it doesn’t match any other value already there in the list, it means that there are no consecutive integers, doesn’t it? Well, no! But we will talk about that in a minute. Let’s first look at why we might think that statement 1 is sufficient.

Say, the numbers are: 1, 5, 8, 10, 35, 76 …

If you increase 1 by 1, you get 2 and the list looks like this:

Now the numbers are 2, 5, 8, 10, 35, 76 …

Note that the number of distinct integers is the same.

Had there been two consecutive integers such as 1, 2, 8, 10, 35, 76 …

If we increase 1 by 1, the list would have become 2, 2, 8, 10, 35, 76 … – this would have decreased the number of distinct integers.

You might be tempted to say here that statement 1 alone is sufficient. What you might forget is that when you increase a number by 1, one distinct integer could be getting wiped out and another taking its place! It may not occur to you that the case might be different when one value occurs more than once, but statement 2 should give you a hint.

Obviously, statement 2 alone is not sufficient but let’s analyze what happens when we take both statements together.

Since statement 1 doesn’t tell you that all values are distinct, statement 2 should make you think that you need to consider the case where one value occurs more than once in the list. In that case, is it possible that number of different values in the list does not change even though there is a pair of consecutive integers?

Say the numbers are 1, 1, 2, 8, 10, 35, 76 …

Now if you increase 1 by 1, the list would look like 1, 2, 2, 8, 10, 35, 76 …

Here, the number of distinct integers stays the same even when you increase a number by 1 and you have consecutive integers! In this case, if there were no consecutive integers, the number of distinct integers would have increased. Hence if the numbers are not all distinct and the number of distinct numbers needs to stay the same, there must be a pair of consecutive integers.

This tells you that statement 1 is not sufficient alone but both statements together answer the question with a ‘Yes’.

Answer **(C)**.

**Takeaway** – Just as when you get an easy (C), you must check whether the answer could be (A) or (B), when you feel that the answer is an easy (A) or (B), you might want to check whether the other statement gives some relevant data and is necessary.