Minimum and maximum extremes

The min-max extremes strategy is one of the most useful time-saving strategies on the GRE. The basic idea is to find only the minimum and maximum possible values (the extreme values), instead of spending time solving for every possible value.

This strategy works for quantitative comparison questions and for certain multiple-choice questions. The overall approach is the same, and you’ll see examples of both.

Min-max for quant comparison questions

Consider this example:

A square has a double-digit perimeter that is also a multiple of 4.

Quantity A: The area of the square
Quantity B: 600

There are a few clues that the min-max strategy will work well here:

• We have one unknown value (Quantity A) and one known value (Quantity B)
• The unknown value is restricted by a set of constraints

Here’s how to apply the min-max strategy to quantitative comparison questions:

1. Determine the minimum possible value of the unknown quantity
2. Determine the maximum possible value of the unknown quantity
3. Compare both extremes to the known quantity
   • If Quantity B is less than both Quantity A extremes, the answer is A
   • If Quantity B is greater than both Quantity A extremes, the answer is B
   • If Quantity B is the same as both Quantity A extremes, the answer is C
   • If Quantity B falls between the Quantity A extremes, the answer is D

Keep in mind that Quantity A can take many values besides the minimum and maximum. But in a quant comparison question, the extremes are enough: if the known value is above both extremes, below both extremes, equal to both extremes, or between them, you can determine the correct relationship without listing every case.

Let’s solve the example:

A square has a double-digit perimeter that is also a multiple of 4.

Quantity A: The area of the square
Quantity B: 600

Find the minimum and maximum possible values for Quantity A, then answer the question.

Start with the maximum possible area by using the maximum possible perimeter.

• The largest double-digit perimeter that’s also a multiple of 4 is 96.
• A square has 4 equal sides, so each side is $96/4=24$.
• The area is $24\times 24=576$.

At this point, you can already choose B: even the maximum possible value of Quantity A (576) is less than 600.

For completeness, check the minimum possible area.

• The smallest double-digit perimeter that’s a multiple of 4 is 12.
• Each side is $12/4=3$.
• The area is $3\times 3=9$.

Quantity B is greater than both extremes for Quantity A (576 and 9). Therefore, Quantity B is greater than Quantity A in every valid case.

Min-max for multiple choice questions

This strategy is also useful for select all that apply multiple-choice questions. Find the minimum and maximum possible values, then select the choices that match the extremes and any choices that fall within the range.

This can save time because anything outside the range is guaranteed to be impossible. (There’s an important caveat in the sidenote below.) This approach is most useful when the answer choices are numeric and listed in ascending order. It’s usually less helpful when choices include variables or are not ordered.

Here’s an example:

A pizza shop delivers pizza to its customers. On Saturday, more than two-fifths of the orders were delivered late because of traffic. On Sunday, only one out of ten orders were delivered late. How many deliveries might have been late if there were a total of 65 deliveries over the weekend? Select all that apply.

A. 8
B. 9
C. 26
D. 56
E. 65

Try applying the min-max strategy to this question.

Find the maximum

To maximize the total number of late deliveries:

• Put as few orders as possible on Sunday, because Sunday has a smaller late fraction ($1/10$).
• Put the remaining orders on Saturday, where the late fraction can be much larger (it’s only constrained to be more than $2/5$).

The smallest number of Sunday orders that still allows “one out of ten” to make sense is 10 orders (so 1 late). That leaves $65-10=55$ orders on Saturday.

Now maximize Saturday’s late deliveries. “More than two-fifths” allows as many as all of them to be late, so Saturday can have 55 late deliveries.

• Sunday late: 1
• Saturday late: 55
• Maximum total late: $1+55=56$

Find the minimum

To minimize the total number of late deliveries:

• Put as many orders as possible on Sunday, because only $1/10$ are late.

Use 60 orders on Sunday, which gives $1/10\times 60=6$ late deliveries. That leaves $65-60=5$ orders on Saturday.

On Saturday, more than $2/5$ of 5 orders must be late:

• $2/5\times 5=2$
• “More than 2” means at least 3 late deliveries

So the minimum total late is $6+3=9$.

Select choices within the range

The possible number of late deliveries must be between 9 and 56, inclusive. Select the answer choices that fall in that range:

• 9, 26, and 56

Notice that one incorrect choice (8) is just below the minimum. ETS often places choices near the boundary to test whether you’re reading the constraint carefully. If you misread “more than two-fifths” as “at least two-fifths,” you might incorrectly include 8.