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

How to Deal with Maximizing/Minimizing Strategies on the GMAT

BY KARISHMA, VERITAS PREP

The strategy to be used to deal with Maximizing/Minimizing questions varies from question to question. What works in one question may not work in another. You might have to think up on what to do in a question from scratch and you have only 2 mins to do it in. The saving grace is that once you know what you have to do, the actual work involved to arrive at the answer is very little.

Let’s look at some maximizing minimizing strategies in the next few weeks. We start with an OG question today with a convoluted question stem.

Question: List T consists of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum of the 30 decimals, E, is defined as follows. Each decimal in T whose tenths digit is even is rounded up to the nearest integer, and each decimal in T whose tenths digits is odd is rounded down to the nearest integer. If 1/3 of the decimals in T have a tenths digit that is even, which of the following is a possible value of E – S?

I. -16
II. 6
III. 10

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

Solution:

There is a lot of information in the question stem and a lot of variables are explained. Let’s review the given data in our own words first.

T has 30 decimals. The sum of all the decimals is S.

10 decimals have even tenths digit. They will be rounded up.

20 decimals have odd tenths digit. They will be rounded down.

The sum of rounded numbers is E.

E – S can take many values so how do we figure which ones it cannot take? We need to find the minimum value E – S can take and the maximum value it can take. That will help us figure out the values that E – S cannot take. Note that E could be greater than S and it could be less than S. So E – S could be positive or negative.

Step 1: Getting Minimum Value of E – S

Let’s try to make E as small as possible. For that, we need to do two things:

1. When we round up the decimals (even tenths digit), the difference between actual and estimate should be very small. The estimate should add a very small number to round it up so that E is not much greater than S. Say the numbers are something similar to 3.8999999 (the tenths digit is the largest even digit) and they will be rounded up to 4 i.e. the estimate gains about 0.1 per number. Since there are 10 even tenths digit numbers, the estimate will be approximately .1*10 = 1 more than actual.

2. When we round down the decimals (odd tenths digit), the difference between actual and estimate should be as large as possible. Say the numbers are something similar to 3.999999 (tenths digit is the largest odd digit) and they will be rounded down to 3 i.e. the estimate loses approximately 1 per number. Since there are 20 such numbers, the estimate is 1*20 = 20 less than actual.

Overall, the estimate will be approximately 20 – 1 = 19 less than actual.

Minimum value of E – S = -19

Step 2: Getting Maximum Value of E – S

Now let’s try to make E as large as possible. For that, we need to do two things:

1. When we round up the decimals (even tenths digit), the difference between actual and estimate should be very high. Say the numbers are something similar to 3.000001 (tenths digit is the smallest even digit) and they will be rounded up to 4 i.e. the estimate gains 1 per number. Since there are 10 even tenths digit numbers, the estimate will be approximately 1*10 = 10 more than actual.

2. When we round down the decimals (odd tenths digit), the difference between actual and estimate should be very little. Say the numbers are something similar to 3.1 (tenths digit is the smallest odd digit). They will be rounded down to 3 i.e. the estimate loses approximately 0.1 per number. Since there are 20 such numbers, the estimate is approximately 0.1*20 = 2 less than actual.

Overall, the estimate will be approximately 10 – 2 = 8 more than actual.

Maximum value of E – S = 8.

The minimum value of E – S is -19 and the maximum value of E – S is 8.

So E – S can take the values -16 and 6 but cannot take the value 10.

Answer (B) This question is discussed HERE
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Maximizing Your Efficiency on Min-Max Problems

On nearly every GMAT, you’ll see at least one of the “Min/Max” variety of word problems, a category that’s difficult for even the brightest quant minds largely for one major reason: these aren’t your typical word problems, and they don’t lend themselves very well to algebra. They tend to be every bit as “situational” as “mathematical” and in fact are labeled “scenario-driven Min/Max problems” in the Veritas Prep Word Problems lesson. Why? Because they’re almost entirely driven by the situation, including:

The figures almost always have to be integers. The problems use situations like “the number of people” or “the number of trees,” a subtle clue that algebra won’t quite work because you’re not using all real numbers, but instead nonnegative integers. But be careful (as you’ll see below).

The questions ask for a very specific value in a very specific way. You’ll often see them ask “did at least three” (3 or more means “yes”) or “was the number sold greater than 50″ (50 itself means “no” – to get “yes” it has to be 51 or more, provided you’re dealing with integers).

The rules of the game often dictate whether repeat numbers are allowed. Quite often you’ll find a stipulation that “no two could be the same” (but make sure you see that stipulation before you act on it!).

Some of the information in a Data Sufficiency version of a Min/Max is much more sufficient than it usually appears. This is largely because of the scenario, numbers, and question stem they’ve carefully crafted to sneak sufficiency past you.

Let’s consider an example so that you can see how one of these works:

Five friends recently visited a famous chocolatier, and collectively purchased a total of 16 pounds of fudge. Did any one friend purchase more than 5 pounds of fudge?

(1) No two friends purchased the same amount of fudge.

(2) The minimum increment in which the chocolatier sells fudge is one pound.

Look at the familiar symptoms of a min/max problem:

*The question stem asks a yes/no question about a very specific value (5 pounds)

*Statement 1 provides the caveat “no two can be the same”

*While the problem itself doesn’t dictate “integers” via the scenario – “pounds of fudge” can certainly come in fractions – Statement 2 comes in to limit the values to integers

Now, if you’re looking at the information from the question stem and statement 1, you could try to set up some algebra:

The given information: a + b + c + d + e = 16

Statement 1: a > b > c > d > e

The question, then: Is a > 5?

You should immediately see that this isn’t sufficient; with nonintegers in play, a could be 15.9 and the other four could add up to 0.1 (“yes”) or they could each be right around the average of 3.2, just a hair off to satisfy the inequality (“no”). But you should also see what makes problems like this tricky with algebra – there are a lot of variables and there’s a lot of inequality. Min/Max problems tend to require a lot more trial and error, and live up to their name because the technique that works best on them is to minimize and maximize particular values to figure out the possible range of the value in question. Eschewing algebra, let’s look at statement 2:

Given Information: 16 total pounds were purchased.

Statement 2: The purchases had to be in integer increments.

The question: Was one of those integers 5 or higher?

Here, to find the maximum value you can minimize the other values. What if four friends didn’t buy anything (0, 0, 0, 0) and the fifth bought all 16 pounds? That’s a resounding “yes”. But they could have split things much more easily – you’d do this by maximizing the smallest value(s). 3, 3, 3, 3, 3 would give you 15, allowing that one final pound to go to the highest making the highest value 4. So there’s your “no” and statement 2 is not sufficient.

When you take the statements together, however, you should see what really makes these problems tick. With algebra it’s still awful:

a + b + c + d + e = 16

a > b > c > d > e

a, b, c, d, and e are integers

Is a > 5?

But with an intent to minimize the highest value (by maximizing the others, sucking as much value away as possible) and maximize the highest value (by minimizing the others to drive all value toward the highest), you have a blueprint for trial and error.

Maximize the highest value / Minimize the others. To make sure you can get a “yes”, minimize the smallest values to see how high the highest can go. That means 0, 1, 2, and 3 – a total of 6 pounds leaving 10 for the highest. It’s easy to get a “yes”.

Minimize the highest value / Maximize the others. Since highest = 5 gives you “no”, see if you can then minimize that highest (5) and maximize the others (4, 3, 2, and 1). But notice that that only gives you a total of 15, and you need to account for 16. And here you cannot give that extra pound to any of the lower values without matching a higher one (add it to 1 and you match 2; add it to 2 and you match 3; etc.). So this guarantees that the highest value is 6 or more, and the answer is sufficient, C.

This question is discussed HERE.

More importantly, look at the technique – many great mathematical minds hate these problems because the “pure math” algebra is so ugly…but the GMAT loves these because they force you to think logically through a few situations. Since so many of these are Yes/No Data Sufficiency problems, keep in mind that your goals are to “prove insufficiency” looking for both a Yes and a No answer, by:

Minimizing the highest value by maximizing the others

Maximizing the highest value by minimizing the others

Minimizing the lowest value by maximizing the others

Maximizing the lowest value by minimizing the others

Essentially to ______ize one value, do the opposite to the others, and doing so will help you test the possible range. As you do so, make sure you consider:

-Can the values be nonintegers, negative numbers, or 0? (often the scenario dictates that the answer to a few of these is “no”)

-Can values repeat?

Min/Max Scenario problems can be a pain, as they maximize the amount of time you have to spend on them while minimizing your score. But if you know the game, you have an advantage – these problems are all about trial-and-error of Min/Max situations and about taking acute inventory of what is allowable for the values you do try. Play the game correctly, and you’ll be set up for maximal success with minimal (comparative) effort.
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Max-Min Strategies: Establishing Base Case

BY KARISHMA, VERITAS PREP

Continuing our discussion on maximizing/minimizing strategies, let’s look at another question today. Today we discuss the strategy of establishing a base case, a strategy which often comes in handy in DS questions. The base case gives us a starting point and direction to our thoughts. Otherwise, with the number of possible cases in any given scenario, we may find our mind wandering from one direction to another without reaching any conclusions. That is a huge waste of time, a precious commodity.

Question: Four friends go to Macy’s for shopping and buy a top each. Three of them buy a pillow case each too. The prices of the seven items were all different integers, and every top cost more than every pillow case. What was the price, in dollars, of the most expensive pillow case if the total price of the seven items was $89? Statement 1: The most expensive top cost$16.

Statement 2: The least expensive pillow case cost $9. Solution: The first problem here is figuring out the starting point. There must be many ways in which you can price the seven items such that the total cost is$89. So we need to establish a base case (which conforms to all the conditions given in the question stem) first and then we will tweak it around according to the additional information obtained from our statements.

‘Seven items for $89’ means the average price for each item is approximately$12. But 12 is not the exact average. 12*7 = 84 which means another $5 were spent. A sequence with an average of 12 and different integers is$9, $10,$11, $12,$13, $14,$15.

But actually another $5 were spent so the prices could be any one of the following variations (and many others):$9, $10,$11, $12,$13, $14,$20 (Add $5 to the highest price)$9, $10,$11, $12,$13, $16,$18 (Split $5 into two and add to the two highest prices)$9, $10,$12, $13,$14, $15,$16 (Split $5 into five parts of$1 each and add to the top 5 prices)

$7,$9, $13,$14, $15,$16, $17 (Take away some dollars from the lower prices and add them to the higher prices along with the$5)

etc

Let’s focus on another piece of information given in the question stem: “every top cost more than every pillow case.”

This means that when we arrange all the prices in the increasing order (as done above), the last four are the prices of the four tops and the first three are the prices of the three pillow cases. The most expensive pillow case is the third one.

Now that we have accounted for all the information given in the question stem, let’s focus on the statements.

Statement 1: The most expensive top cost $16. We have already seen a case above where the maximum price was$16. Is this the only case possible? Let’s look at our base case again:

$9,$10, $11,$12, $13,$14, $15 (a further$5 needs to be added to bring the total price up to $89) Since the prices need to be all unique, if we add 1 to any one price, we also need to add at least$1 to each subsequent price. E.g. if we increase the price of the least expensive pillow case by $1 and make it$10, we will need to increase the price of every subsequent item by $1 too. But we have only$5 more to give.

If the maximum price is $16, it means the rightmost price can increase by only$1. So all prices before it can also only increase by $1 only and except the first two prices, they must increase by$1 to adjust the extra $5. Hence the only possible case is$9, $10,$12, $13,$14, $15,$16.

So the cost of the most expensive pillow case must have been $12. Statement 1 is sufficient alone. Statement 2: The least expensive pillow case cost$9.

A restriction on the lowest price is much less restrictive. Starting from our base case

$9,$10, $11,$12, $13,$14, $15, we can distribute the extra$5 in various ways. We can do what we did above in statement 1 i.e. give $1 to each of the 5 highest prices:$9, $10,$12, $13,$14, $15,$16

We can also give the entire $5 to the highest price:$9, $10,$11, $12,$13, $14,$20

So the price of the most expensive pillow case could take various values. Hence, statement 2 alone is not sufficient.

Answer (A) This question is discussed HERE.

Note that the answer is a little unexpected, isn’t it? If we were to read the question and guess within 20 secs, we would probably guess that the answer is (C), (D) or (E). The two statements give similar but complementary information. It would be hard to guess that one will be sufficient alone while other will not be. This is what makes this question interesting and hard too.

Our strategy here was to establish a base case and tweak it according to the information given in the statements. This strategy is often useful in DS – not just in max-min questions but others too.
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Max-Min Strategies: Focus on Extremes

BY KARISHMA, VERITAS PREP

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.
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A Sets Question That Upsets Many

BY KARISHMA, VERITAS PREP

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.

Attachment: Set1_Sep17_2012.jpg [ 14.41 KiB | Viewed 34053 times ]

Attachment: Set2_Sep17_2012.jpg [ 13.93 KiB | Viewed 34047 times ]

Attachment: Set3_Sep17_2012.jpg [ 13.75 KiB | Viewed 34060 times ]

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Questions to Practice:

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4 Questions To Ask Yourself On Min/Max GMAT Problems

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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_________________ Re: Maximizing/Minimizing Strategies on the GMAT - All in One Topic!   [#permalink] 01 Aug 2019, 04:33
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# Maximizing/Minimizing Strategies on the GMAT - All in One Topic!  