Strategy#3 - Number Sense
In previous posts (
Tip #1 – Dealing with Variables,
Tip #2: Looking for Equality), I have discussed two approaches when tackling Quantitative Comparison (QC) questions involving variables. Those approaches are:
1) Apply algebraic techniques
2) Plug in numbersIn those posts, we concluded that the algebraic approach is typically the faster and more reliable approach.
In today’s post, we’ll examine a third strategy that can sometimes be the fastest and easiest approach. We’ll call this the “logical approach.”
To set things up, please consider the following QC question:
Attachment:
q1.png [ 15.04 KiB | Viewed 4931 times ]
The algebraic approach to this question looks something like this:
First multiply both sides by 35 (the least common multiple of 5 and 7) to get:
Attachment:
q2.png [ 9.39 KiB | Viewed 4928 times ]
Then add 5x to both sides to get:
Attachment:
q3.png [ 8.59 KiB | Viewed 4931 times ]
Then add 42 to both sides to get:
Attachment:
q4.png [ 8.18 KiB | Viewed 4936 times ]
And, finally, divide both sides by 19 to get:
Attachment:
q5.png [ 7.25 KiB | Viewed 4921 times ]
So, now we’re comparing x and 3, and the given information tells us that x is greater than 3.
This means the correct answer must be A.
Now let’s take the original question and use logic to solve it (in about 5 seconds).
Attachment:
q6.png [ 9.59 KiB | Viewed 4926 times ]
Column A: If x > 3, then 2x > 6, which means that 2x-6 must be positive.
Column B: If x > 3, then 3-x must be negative.
So, the two columns can be rewritten as:
Attachment:
q7.png [ 11.67 KiB | Viewed 4923 times ]
From here, we can see that Column A is always positive and Column B is always negative. As such, Column A will always be greater than Column B. So, the correct answer is A.
Let’s try another one. See if you can solve it in your head.
Attachment:
q8.png [ 14.81 KiB | Viewed 4938 times ]
For this question, I’ll leave the algebraic approach to you.
Let’s apply some logic.
First, we’re told that . In order to apply some logic, let’s refer to the denominator as “something.” In other words, 18y divided by “something” equals 3. Well, we know that 18y divided by 6y equals 3, so that “something” must equal 6y.
In other words, it must be the case that 7y-x = 6y
Now consider the fact that 7y-x = 6y. If we now refer to “x” as “something,” we can see that 7y minus “something” equals 6y. Since we know that 7y-y=6y, we can see that “something” must equal y.
In other words x = y.
Now that we have concluded that x=y, we’ll return to the original question:
Attachment:
q9.png [ 9.19 KiB | Viewed 4901 times ]
If x=y, we can see that the answer here must be C.
So, although the algebraic approach is typically the superior approach for quantitative comparison questions involving variables, be sure to take a moment to see whether the problem can be solved by applying a little logic.
Attachment:
nervous-breakdown.png [ 28.82 KiB | Viewed 31737 times ]