Maximizing/Minimizing Strategies on the GMAT - All in One Topic!

Here, let’s look at another max-min question in which we apply the strategy of focusing on the extremes. The largest or the smallest values are often found at the extremes of a given range.

Question: If x and y are integers such that (x+1)^2 is less than or equal to 36 and (y-1)^2 is less than 64, what is the sum of the maximum possible value of xy and the minimum possible value of xy?

(A) -16
(B) -14
(C) 0
(D) 14
(E) 16

Solution: To get the sum of the maximum and minimum possible values of xy, we need to know the maximum and minimum values of xy. For those, we need to find the values that x and y can take. So first, we should review the information given:

\((x + 1)^2 \leq 36\)

\((y – 1)^2 < 64\)

We need to find the values that x and y can take. There are many ways of doing that. We can solve the inequality using the wave method discussed in this post or using the concept of absolute values. Let’s discuss both the methods.

Wave method to solve inequalities:

Solve for x: \((x + 1)^2 \leq 36\)

\((x + 1)^2 - 6^2 \leq 0\)

\((x + 1 + 6)(x + 1 - 6) \leq 0\)

\((x + 7)(x - 5) \leq 0\)

\(-7 \leq x \leq 5\) (Using the wave method)

Solve for y: \((y - 1)^2 < 64\)

\((y - 1)^2 - 8^2 < 0\)

\((y - 1 + 8)(y - 1 - 8) < 0\)

\((y + 7)(y - 9) < 0\)

\(-7 < y < 9\) (Using the wave method)

Or you can solve taking the square root on both sides

Solve for x: \((x + 1)^2 \leq 36\)

\(|x + 1| \leq 6\)

\(-6 \leq x + 1 \leq 6\) (discussed in your Veritas Algebra book)

\(-7 \leq x \leq 5\)

So x can take values: -7, -6, -5, -4, … 3, 4, 5

Solve for y: \((y - 1)^2 < 64\)

\(|y – 1| < 8\)

\(-8 < y - 1 < 8\) (discussed in your Veritas Algebra book)

\(-7 < y < 9\)

So y can take values: -6, -5, -4, -3, … 6, 7, 8.

Now that we have the values of x and y, we should try to find the minimum and maximum values of xy.

Note that the values of xy can be positive as well as negative. The minimum value will be the negative value with largest absolute value (largest negative) and the maximum value will be the positive value with the largest absolute value.

Minimum value – For the value to be negative, one and only one of x and y should be negative. Focus on the extreme values: if x is -7 and y is 8, we get \(xy = -56\). This is the negative value with largest absolute value.

Maximum value – For the value to be positive, both x and y should have the same signs. If x = -7 and y = -6, we get \(xy = 42\). This is the largest positive value.

The sum of the maximum value of xy and minimum value of xy is \(-56 + 42 = -14\)

Answer (B) This question is discussed HERE.

Try to think of it in terms of a number line. x lies in the range -7 to 5 and y lies in the  range -6 to 8. The range is linear so the end points give us the maximum/minimum values. Think of what happens when you plot a quadratic – the minimum/maximum could lie anywhere.

Example: A group of people were given 2 puzzles. 79% people solved puzzle X and 89% people solved puzzle Y. What is the maximum and minimum percentage of people who could have solved both the puzzles?

(A) 11%, 0%
(B) 49%, 33%
(C) 68%, 57%
(D) 79%, 68%
(E) 89%, 79%

Solution: The first thing to note here is that we do not know the % of people who could not solve either puzzle. All we know is that puzzle X was solved by 79% of the people and puzzle Y was solved by 89% of the people.



Let’s first try to maximize the % of people who solved both the puzzles. We want to make these two sets overlap as much as possible i.e. we need to get them as close to each other as possible. Region of overlap can be 79% at most since we know that only 79% people solved puzzle X. In this case, the venn diagram will look something like this.



Hence, the maximum % of people who could have solved both the puzzles is 79%.

Now, let’s try to minimize the % of people who solved both the puzzles. We want the sets to be as far apart as possible. In this case, the % of people who solved neither puzzle must be 0. Only then will the overlap of the sets be as little as possible.



In this case, 68% people must have solved both the puzzles.

Hence, the answer is (D) This question is discussed HERE.

Note: If the question instead gave you the % of people who did not solve either puzzle (e.g. by giving you that everyone solved at least one puzzle), then there is no question of maximizing/minimizing the % of people who solved both the puzzles. Consider this:

(For ease, let’s drop the percentage and work with just numbers.)

Total no. of people = No. of people who solved X + No. of people who solved Y – No. of people who solved both + No. of people who solved neither

100 = 79 + 89 – No. of people who solved both + No. of people who solved neither

No. of people who solved both – No. of people who solved neither = 68

We can maximize/minimize the two numbers by adjusting them against each other. If one increases, the other increases too. If one decreases, the other decreases too.

If no. of people who solve both = 68, no. of people who solve neither = 0

If no. of people who solve both = 69, no. of people who solve neither = 1

If no. of people who solve both = 79, no. of people who solve neither = 11

If you are given the number of people who solved neither, you have a fixed number of people who solved both. Hence, maximizing or minimizing becomes pointless.

You can also work on this concept logically –

Say, every solution of every puzzle was written down on a separate sheet of paper. Then there would be 168 sheets of paper in all (79 + 89). If everyone solved 1 puzzle, then we have accounted for 100 sheets. The other 68 sheets were made by 68 people who have already solved one puzzle each. Hence, 68 people would have solved both the puzzles. If instead, 99 people solve at least one puzzle and one person solves no puzzle, then 69 (obtained by 168 – 99) sheets would have been made by people who have already solved one. Hence 69 people would have solved both the puzzles. Note here that the number of people who solved neither and the number of people who solved both are not independent of each other. One number depends on the other. It will be good if you make a note of this in your log book.

* Data Sufficiency Statistics Questions
* Problem Solving Statistics Questions

Min/Max problems can be among the most frustrating on the GMAT’s quantitative section. Why? Because they seldom involve an equation or definite value. They’re the ones that ask things like “did the fisherman who caught the third-most fish catch at least 12 fish?” or “what is the maximum number of fish that any one fisherman caught?”. And the reason the GMAT loves them? It’s precisely because they’re so much more strategic than they are “calculational.” They make you think, not just plug and chug.

However…

There are three knee-jerk questions that you should plug (if not chug) into your brain to ask yourself every time you see a Min/Max problem before you ask that fourth question “What’s my strategy?”:

• Do the numbers have to be integers?
• Is zero a possible value?
• Are repeat numbers possible?

In the Veritas Prep Word Problems lesson we refer to these problems as “scenario-driven” Min/Max problems precisely because of the above questions. The scenario created by the problem drives the whole thing, related mainly to those three above questions. Consider these four prompts and ask yourself which ones can definitively be answered:

#1: “Four friends go fishing and catch a total of 10 fish. How many fish did the friend with the highest total catch?”

#2: “Four friends go fishing and catch a total of 10 fish. If no two friends caught the same number of fish, how many fish did the friend with the highest total catch?”

#3: “Four friends go fishing and catch a total of 10 fish. If each friend caught at least one fish but no two friends caught the same number, how many fish did the friend with the highest total catch?”

#4: “Four friends go fishing and catch a total of 10 pounds of fish. If each friend caught at least one fish but no two friends caught the same number, how many pounds of fish did the friend with the highest total catch?”

Hopefully you can see the progression as this set builds. In the first problem, there’s clearly no way to tell. Did one friend catch all ten? Did everyone catch at least two and two friends tied with 3? You just don’t know. But then it gets interesting, based on the questions you need to ask yourself on all of these.

With #2, two big restrictions are in play. Fish must be integers, so you’re only dealing with the 11 integers 0 through 10. And if no two friends caught the same number there’s a limited number of unique values that can add up to 10. But the catch on this one should be evident after you’ve read #3. Zero *IS* possible in this case, so while the totals could be 1, 2, 3, and 4 (guaranteeing the answer of 4), if the lowest person could have caught 0 (that’s where “min/max” comes in – to maximize the top value you want to minimize the other values) there’s also the possibility for 0, 1, 2, and 7. Because the zero possibility was still lurking out there, there’s not quite enough information to solve this one. And that’s why you always have to ask yourself “is 0 possible?”.

#3 should showcase that. If 0 is no longer a possibility *AND* the numbers have to be integers *AND* the numbers can’t repeat, then the only option is 1 (the new min value since 0 is gone), 2 (because you can’t match 1), 3, and 4. The highest total is 4.

And #4 shows why the seemingly-irrelevant backstory of “friends going fishing” is so important. Pounds of fish can be nonintegers, but fish themselves have to be integers. So even though this prompt looks very similar to #3, because we’re no longer limited to integers it’s very easy for the values to not repeat and still give wildly different max values (1, 2, 3, and 4 or 1.5, 2, 3, and 3.5 for example).

As you can see, the scenario really drives the answer, although the fourth question “What is my strategy?” will almost always require some real work. Let’s take a look at a couple questions from the Veritas Prep Question Bank to illustrate.

Question 1:

Four workers from an international charity were selling shirts at a local event yesterday. Did one of the workers sell at least three shirts yesterday at the event?

(1) Together they sold 8 shirts yesterday at the event.

(2) No two workers sold the same number of shirts.

Before you begin strategizing, ask yourself the three major questions:

1) Do the values have to be integers? YES – that’s why the problem chose shirts.

2) Is zero possible? YES – it’s not prohibited, so that means you have to consider zero as a min value.

3) Can the numbers repeat? That’s why statement 2 is there. With the given information and with statement 1, numbers can repeat. That allows you to come up with the setup 2, 2, 2, and 2 for statement 1 (giving the answer “NO”) or 1, 2, 2, and 3 (giving the answer “YES” and proving this insufficient).

But when statement 2 says on its own that, NO, the numbers cannot repeat, that’s a much more impactful statement than most test-takers realize. Taking statement 2 alone, you have four integers that cannot repeat (and cannot be negative), so the smallest setup you can find is 0, 1, 2, and 3 – and with that someone definitely sold at least three shirts. Statement 2 is sufficient with really no calculations whatsoever, but with careful attention to the ever-important questions.

This question is discussed HERE.

Question 2:

Last year, Company X paid out a total of $1,050,000 in salaries to its 21 employees. If no employee earned a salary that is more than 20% greater than any other employee, what is the lowest possible salary that any one employee earned?

(A) $40,000

(B) $41,667

(C) $42,000

(D) $50,000

(E) $60,000

Here ask yourself the same questions:

1) The numbers do not have to be integers.

2) Zero is theoretically possible (but probably constrained by the 20% difference restriction)

3) Numbers absolutely can repeat (which will be very important)

4) What’s your strategy? If you want the LOWEST possible single salary, then use your answer to #3 (they can repeat) and give the other 20 salaries the maximum. That way your calculation looks like:

x + 20(1.2x) = 1,050,000

Which breaks out to 25x = 1,050,000, and x = 42000. And notice how important the answer to #3 was – by knowing that numbers could repeat, you were able to quickly put together a smart strategy to minimize one single value. Answer: C.

This question is discussed HERE.

The larger lesson is crucial here, though – these problems are often (but not always) fairly basic mathematically, but derive their difficulty from a situation that limits some options or allows for more than you’d think via integer restrictions, the possibility of zero, and the possibility of repeat values. Ask yourself these four questions, and your answer to the first three especially will maximize your efficiency on the strategic portion of the problem.

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