Strategy#4 Comparing in Parts

To set up today’s Quantitative Comparison (QC) strategy, please solve the following question:

Attachment:

q1.png [ 9.63 KiB | Viewed 1445 times ]
There are several different ways to solve this question. For example, we could add the fractions in each column, and then rewrite both sums with a common denominator and then compare them (yikes!). Or, we could use the onscreen calculator to convert each fraction to a decimal and find the sum in each column (yeesh!).

Both of these approaches would take a long time.

Fortunately, we already know that, since the test-makers are reasonable, there MUST be an easier way to solve this question.

The fastest way to solve this question is to estimate and compare the columns in parts.

Here’s how it works.

First notice that 213/428 is approximately 1/2. Now, for this question, we need to be a little more accurate than that. Notice that, since 214/428 = 1/2, it must be the case that 213/428 is a little bit less than 1/2, which we’ll denote as 1/2-.

Using similar logic, we can see that, since 3007/9027 = 1/3, it must be the case that 3007/9101 is a little bit less than 1/3, which we’ll denote as 1/3-.

Next, since 731/1462 = ½, it must be the case that 741/428 is a little bit more than 1/2, which we’ll denote as 1/2+.

Using the same logic, we can see that 208/597 is a little bit more than 1/3, which we’ll denote as 1/3+.

At this point, we can simplify the two columns as:

Attachment:

q2.png [ 2.21 KiB | Viewed 1444 times ]
From here, we can compare the two sums in parts.

Since 1/2+is greater than 1/2-, and since 1/3+ is greater than 1/3-, the sum of 1/2+ and 1/3+ must be greater than the sum of 1/2- and 1/3-.

As such, the answer must be B.

_________________

My GRE Resources

Free GRE resources | GRE Prep Club Quant Tests

If you find this post helpful, please press the kudos button to let me know !