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In this blog we will solve three questions – an easy GMAT Data Insights Practice Question – A GMAT DS question testing concepts in quadratic equations and quadratic inequalities. What’s more? There are a couple of bonus DS questions that will help pick nuances and points that one has to watch out for when solving data sufficiency questions in algebra.
This data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether:
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
EACH statement ALONE is sufficient to answer the question asked.
Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
Is x = 3?
1. x^{2} – 6x + 9 = 0
2. x^{2} + 7x – 21 > 0
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The correct answer is Choice A. Statement 1 alone is sufficient; statement 2 is not.
The question to be answered is “Is x = 3?”.
An “Is” question is answered if you are able to provide a definite Yes or a definite No as an answer to the question. A “NO” is also a valid answer.
Let’s start by evaluating the two statements one at a time and determine whether we are able to get a definite answer. Else, let’s combine and evaluate. Finally, if we do not get a conclusive answer even after combining the statements, the given data is not sufficient to answer the question.
Step 1 – factorize the given quadratic equation
x^{2} – 3x – 3x + 9 = 0
Or (x – 3)(x – 3) = 0.
So, we get x = 3.
We have a unique value for x from the statement.
However, the question is not finding a unique value for x. The question “Is x = 3?”
From the solution to the given quadratic equation, we can decisively answer the question. We can say x is equal to 3. Yes is our answer.
Because we are able to find a conclusive answer to the question using statement 1, statement 1 alone is sufficient.
We can eliminate answer options B, C, and E. Answer narrows down to choices A or D.
This quadratic expression cannot be factorized. So, how do we proceed at this juncture?
There are two approaches at this stage. You could find the range of values that x takes using the formula to find the roots of the quadratic expression and then arrive at the answer.
Alternatively, let us determine whether the inequality holds good when x = 3.
Evidently, the second method is quicker.
When x = 3, x^{2} + 7x – 21 = 3^{2} + 7(3) – 21 = 9 > 0
At x = 3, the inequality holds good.
What sense can we make out of this finding?
The inequality holds good for x = 3. So, one of the values that x can take is 3.
What we need to establish is that x = 3 is the only value that will satisfy this inequality if statement 2 were to be sufficient.
Let’s look for a counter exam. A value of x other than 3 that will validate the inequality.
Let’s evaluate the inequality for x = 4
When x = 4, x^{2} + 7x – 21 = 4^{2} + 7(4) – 21 = 23 > 0.
The inequality holds good for 3 and also for 4. Essentially, it points to the fact that x could be 3 and it could take other values as well.
We are not able to answer the question conclusively using statement 2.
Statement 1 Alone is sufficient. Statement 2 is not sufficient.
In this variant of the given question, let us keep the question stem unchanged and evaluate it with two new statements.
Is x = 3?
1. x^{2} + 17x – 60 = 0
2. x – y^{2} > 0
The given quadratic equation can be factorized. So, let’s factorize the equation and compute the roots of the quadratic equation.
x^{2} + 20x – 3x – 60 = 0
x(x + 20) -3(x + 20) = 0
Or (x + 20)(x – 3) = 0
So, x = -20 or x = 3
What is the question? Is x = 3.
From statement 1, we can infer that x is one of the two values -20 or 3.
i.e., it may be 3 or it may not be.
We are not able to find a conclusive answer to the question using information in statement 1.
Statement 1 alone is not sufficient.
Let’s eliminate choices A and D. Choices narrow down to B, C, or E.
x and y can take infinite values that will satisfy the inequality.
One such possibility is x = 3 and y = 1. The second possibility is x = 4 and y = 1
Essentially, we arrive at the inference that x may or may not be 3.
So, statement 2 is NOT SUFFICIENT.
Eliminate choice B.
Let’s start from statement 2.
What we can glean from statement 2.
We can rewrite the statement as x > y^{2} .
For real values of y, y^{2} will never be a negative number.
The least value that y^{2} will take is 0. So, x is > 0 or x is positive.
From statement 1, we know x takes two values : -20 or 3; from statement 2, we know that x has to be positive.
Therefore, combining the two statements, we can narrow down the value that x can take to be 3.
The answer to the question “Is x = 3?” is a definite Yes.
In this question again, let’s not change the question stem. Modify the statements alone.
Is x = 3?
1. x^{2} + 10x – 39 = 0
2. x + 10y^{2} > 0
We can factorize the given quadratic equation.
x^{2} + 10x – 39 = 0
x^{2} + 13x -3x – 39 = 0
x(x + 13) -3(x + 13) = 0
(x + 13)(x – 3) = 0
x is either -13 or x = 3.
We end with the inference that x may or may not be 3.
So, statement 1 alone is not sufficient.
Eliminate choices A and D. Choices narrow down to B, C, or E.
Example: x = 3, y = 0. Satisfies the inequality. x = 3. Answer Yes.
Counter Example: x = -13, y = 2. Satisfies the inequality. x is not equal to 3. Answer No.
Because we are able to find a counter example, statement 2 alone is NOT sufficient.
Eliminate choice B.
From statement 1, we were able to narrow down the values that x takes to -13 and 3.
If x = 3 and y takes any real value, x + 10y^{2} > 0 will hold good. So, x = 3 holds good for both statements.
If x = -13 and y^{2} > 1.3, both statements will hold good.
For instance, if x = -13 and y = 2, both statements will hold good.
So, x = -13 and x = 3 are values that will satisfy statements 1 and 2 together.
Essentially, we are saying x may or may not be 3.
So despite combining the data in statement 1 and statement 2 together, we are not able to arrive an answer to whether x = 3.