Data Sufficiency Questions in MBA (CAT) Entrance Exams

Let’s review some problems. We will deliberately go step by step in the following solutions to help you imbibe the approach to solving Data Sufficiency problems. Take care not to get involved in lengthy solutions when you are writing the actual exams.

DIRECTIONS for question 1: Each question is followed by two statements, I and II. Answer each question using the following instructions:

Choose [1]      if the question can be answered by using one of the statements alone, but cannot      
                      be answered using the other statement alone.
Choose [2]      if the question can be answered by using either statement alone.
Choose [3]      if the question can be answered using both statements together, but cannot be
                      answered using either statement alone.
Choose [4]      if the question cannot be answered even by using both the statements together.

Problem 1

If are integers, is [a (b+1)] even?

I. a is even.

II. b is odd.

Solution

Taking the stem and statement I, we have the expression of the form:

Note that the result of the above expression will always be even. Statement I alone is sufficient to arrive at the answer. Therefore, options [1], [2] or [3] can only be the possible answers.

Now, take the stem and statement II, we have the expression of the form:

    , therefore this expression will, also, always be even based on similar reasoning as given above. Clearly, we are left with option [2] as the answer.

Problem 2
Let x be a real number. Is modulus of x, i.e.|x| necessarily less than 3?

I. x (x+3) < 0

II.x (x-3) > 0

Solution
We will consider the stem first and simplify it for better comprehension.

Considering |x| we get,

Note: the statements (i) and (ii) both pertain to the stem of the question.

It is clear from the statements that 0 ≤ x <3 or x€[0,3)

Now, taking the stem and statement I, we get

0≤x<3 and x(x+3)<0

Now, taking statement I we know that for x(x+3)<0, either 

x<0 and (x+3) > 0 OR x>0 and (x+3) <0

(i.e. one of the terms will be negative to get the multiplication result negative)

Considering each case one by one, we get the possible solution sets as:

x <0 and x>-3... this will give the result as x € (-3,0)         ...(iii)

From x>0 and x+3 < 0, we get the solution set as:

x €(-∞,-3)Ù(0,+∞)                                                         ....(iv)

That is with statement I we can arrive at possible values of x. Therefore, the possible answers are options [1], [2] and [3].

When we take statement II, we know that for x(x-3) >0, either

x > 0 and (x-3) > 0 OR x < 0 and (x-3) <0

(i.e. either both the terms will be-ve or both the terms will be +ve to get the multiplication result +ve). Considering each case one by one by one, we get the possible solution sets as:

x>0 and x>3... this will give that result as x>3, and

x>0 and x<3...this will give that result as x<0.

therefore the values permissible for x are x<0 and x>3.

The above is sometimes also written as: x € (-∞,0) Ù (3+∞)

This is with statement II we can again arrive at possible values of x for the question asked in the stem. Therefore, option [2] is the only possible answer.