Minimum and maximum extremes

The min-max extremes strategy is one of the most useful timesaving strategies you can use in the GRE. The essence of the strategy is to find only the minimum and maximum possible values (i.e., the extreme values), instead of spending precious time solving for all the possible values.

This strategy is relevant for quantitative comparison questions as well as multiple-choice questions. The approach is similar, and we’ll walk through examples for 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 hints that the min-max strategy will be effective:

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

Applying the min-max strategy is simple for quantity comparison questions:

1. Determine the minimum for the unknown value
2. Determine the maximum for the unknown value
3. Compare the minimum and maximum extremes, together, to the known value
   • If Quantity B is less than both the Quantity A extremes, the answer is A
   • If Quantity B is greater than both the Quantity A extremes, the answer is B
   • If Quantity B is the same as both the Quantity A extremes, the answer is C
   • If Quantity B is within the Quantity A extremes, the answer is D

Keep in mind that there will be many other possible values of Quantity A aside from the minimum and maximum extremes! Nonetheless, because this is a quant comparison question, it’s only the minimum and maximum extremes that matter, since only these two values are relevant to determine which quantity must be greater.

Let’s solve the example fully:

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

We can solve for the maximum area using the maximum perimeter. The maximum double-digit perimeter that is also a multiple of 4 is 96. Since this is a square, each side will have a length of $96/4=24$, and the area will be $24\ast 24=576$.

We could stop right here and choose B as the answer, since if the maximum value for Quantity A is already less than Quantity B. The minimum value might be the same or less than the maximum value, but (obviously) it won’t be greater.

For completeness, we’ll double-check the minimum value. The minimum double-digit perimeter that is a multiple of 4 is 12. Each side will have a length of $12/4=3$, and the area will be only $3\ast 3=9$.

Quantity B is greater than both Quantity A’s maximum (575) and minimum (9) values. It follows that in every case, Quantity B will be greater than Quantity A.

Min-max for multiple choice questions

This strategy is also useful for the select all that apply multiple-choice questions. Simply solve for the minimum and maximum possible values, and then select those choices and any others that fall within the range. This can save a lot of time, since logically, anything within the range is guaranteed to be correct (but see the caveat below). This is only relevant for questions with numeric answers in ascending order - it typically won’t be helpful if the choices include variables or are shuffled.

Here’s an example question:

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 for this question!

Let’s find the maximum first. This question states that more than two-fifths of orders were late deliveries on Saturday, and only one out of 10 were late on Sunday. We have 65 total orders to work with, and since 2/5 is greater than 1/10, we want to allocate as few as possible into the 1/10 bucket to maximize the number of late deliveries. This would mean we have only 10 deliveries (including 1 late) on Sunday, and the other 55 happened on Saturday. Next, we want to maximize the “more than two-fifths late” portion. The maximum value for this, to the customers’ dismay, means all 55 orders were late. In total, the maximum extreme is 56 late orders.

Finding the minimum uses nearly the same approach. We want to allocate as many of the 65 orders as possible into the 1/10 bucket, which is 6 out of 60. This leaves us with 5 orders on Saturday, of which more than two-fifths are late. The smallest amount that fits the constraint is 3, giving us a minimum of $6+3=9$ late orders.

With a minimum extreme of 9 and a maximum extreme of 56, we just choose all choices that are equal to or within that range: 9, 26, and 56.

Note how one of the incorrect answers was just below the minimum extreme! ETS likes to include choices that are right near the edge to ensure you’re reading carefully. If you went a little too fast and mistakenly read the question as at least two-fifths are late instead of more than two-fifths are late, you would have ended up including the incorrect choice of 8.