BATCH #1 (10 QUESTIONS) 1. Absolute values: 1.1 For how many prime numbers x is |x - 1| + |4 - x| 4 However, notice that we are only concerned with prime numbers x. Since prime numbers are positive integers and the least prime number is 2, we can ignore the first two ranges (since no prime numbers lie in those ranges) and only check the last two. 3. 1 -8. Since, we consider 1 4 For this range: x - 1 is positive, so |x - 1| = x - 1. 4 - x is negative, so |4 - x| = -(4 - x). x + 11 is positive, so |x + 11| = x + 11. Thus, |x - 1| + |4 - x| 4 range, then finally for it we get 4 < x < 16. The prime numbers in this range are 5, 7, 11 and 13. Thus, the prime numbers that satisfy the inequality are 2, 3, 5, 7, 11 and 13, which is a total of 6 prime numbers. Answer: A. Takeaway: When solving absolute value questions, remember the key property of absolute value: |x| = x when x ≥ 0, and |x| = -x when x < 0. This is what allows us to split the number line into ranges and rewrite the inequality correctly in each range. What this question tests: This question tests basic absolute value properties and their application in inequalities. In addition, it adds a number properties layer, since after solving the inequality, you must use the restriction that x is prime to identify the valid values.

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Problem Solving Questions

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Bunuel

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Updated on: 23 Apr 2026, 02:59

Originally posted by Bunuel on 15 Apr 2026, 06:04.
Last edited by Bunuel on 23 Apr 2026, 02:59, edited 1 time in total.

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BATCH #3 (10 QUESTIONS)

11. Mixture Problems

11.1. A jeweler had an alloy that was 24% silver by weight. After 200 grams of a metal containing no silver were mixed in, the percentage of silver in the alloy decreased by one-third. What was the weight of the original alloy?

A. 96 grams
B. 100 grams
C. 200 grams
D. 400 grams
E. 600 grams

Let the weight of the original alloy be x grams. Then the amount of silver in it is

0.24x

If the percentage decreased by one-third, then the new percentage is

24% * 2/3 = 16%

After 200 grams of a metal containing no silver are added, the total weight becomes

x + 200

Since no silver was added, the amount of silver stays the same. So we get

0.24x = 0.16(x + 200)

Solving:

0.24x = 0.16x + 32
0.08x = 32
x = 400

Therefore, the weight of the original alloy was 400 grams.

Answer: E

Takeaway
In mixture problems like this one, the key is to track the amount of the substance of interest. If something containing none of that substance is added, the amount of that substance stays the same, even though the total weight increases and the concentration falls.

What This Question Tests
This question tests mixture and concentration reasoning, together with algebraic translation. It also tests whether you can work with a fractional decrease in percentage and recognize that the amount of the key substance remains unchanged while the total weight changes.

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11. Mixture Problems

11.2. A container is full of a solution that is 55% salt. Part of the solution is removed and replaced with the same amount of a 15% salt solution. If the resulting solution is 27% salt, what fraction of the original solution was replaced?

A. 15%
B. 28%
C. 30%
D. 40%
E. 70%

Let x be the fraction of the original solution that was replaced, and assume the container originally held 1 liter.

Then x liters of the 55% salt solution were removed and replaced with x liters of the 15% salt solution.

After the removal, the amount of salt left from the original solution is

0.55(1 - x)

The replacement solution adds

0.15x

Since the final solution is 27% salt, the final amount of salt is 0.27 liters. So we get

0.55(1 - x) + 0.15x = 0.27

Now solve:

0.55 - 0.55x + 0.15x = 0.27
0.55 - 0.40x = 0.27
0.40x = 0.28
x = 0.70

So 70% of the original solution was replaced.

Answer: E.

Takeaway
In problem-solving questions where no actual quantities are fixed and the information is given only in percentages and fractions, assuming a smart convenient total is often better than working with abstract variables alone. It makes the setup cleaner and the relationships easier to track. In mixture problems, the key is to focus on how much of the substance of interest is being removed, kept, or added, and then translate that change into a clean equation using the given concentrations.

What This Question Tests
This question tests a specific type of mixture problem, namely a mixture with replacement, together with algebraic translation and manipulation.

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12. Multiples and Factors

12.1. How many integers are there between -1000 and 1000, inclusive, that are not divisible by either 12 or 27?

A. 1777
B. 1778
C. 1779
D. 1780
E. 1781

The total number of integers from -1000 to 1000, inclusive, is

1000 - (-1000) + 1 = 2001 (There are 1000 negative integers, 1000 positive integers, and 0.)

Now let us determine how many integers in this range are divisible by 12 or 27, and subtract that number from 2001.

The number of multiples of an integer within a range can be calculated using the formula

(last multiple in the range - first multiple in the range) / multiple + 1

Thus:

• The number of multiples of 12 in the given range is

(996 - (-996))/12 + 1 = 1992/12 + 1 = 166 + 1 = 167

To find the greatest multiple of 12 in the range, note that 12 = 3 * 4, so the number must be divisible by both 3 and 4. Since 1000 is divisible by 4 but not by 3, we look at the next smaller multiple of 4, which is 996. Since 996 is also divisible by 3, 996 is the greatest multiple of 12 in the range.

• The number of multiples of 27 in the given range is

(999 - (-999))/27 + 1 = 1998/27 + 1 = 74 + 1 = 75

To find the greatest multiple of 27 in the range, note that 27 = 9 * 3. The largest multiple of 9 below 1000 is 999. Also, 999/9 = 111, and 111 is divisible by 3 since the sum of its digits is divisible by 3. Therefore, 999 is divisible by 27, so it is the greatest multiple of 27 in the range.

• The number of multiples of both 12 and 27 is the number of multiples of their least common multiple.

LCM(12, 27) = 108

To find the LCM, use prime factorization: 12 = 2^2 * 3 and 27 = 3^3. Then take the highest power of each prime that appears. So LCM(12, 27) = 2^2 * 3^3 = 108.

So the number of multiples of both 12 and 27, that is, the number of multiples of 108 in the range, is

(972 - (-972))/108 + 1 = 1944/108 + 1 = 18 + 1 = 19 (9 positive multiples of 108, 9 negative multiples of 108, and 0. Recall that 0 is a multiple of every integer.)

To find the greatest multiple of 108 in the range, note that 108 * 9 = 972, while 108 * 10 = 1080, which is greater than 1000. So 972 is the greatest positive multiple of 108 between -1000 and 1000.

Observe that these 19 numbers are included in both the count of multiples of 12 and the count of multiples of 27. To avoid double-counting, we subtract them once.

Therefore, the number of integers divisible by either 12 or 27 is

167 + 75 - 19 = 223

Consequently, the number of integers that are not divisible by either 12 or 27 is

2001 - 223 = 1778

Answer: B.

Takeaway
It is important to know how to find multiples of an integer in a given range. It is also important to know basic divisibility rules, such as divisibility by 2, 3, 5, and 9. You should also keep in mind basic properties of 0, such as the fact that 0 is a multiple of every integer and a divisor of none.

What This Question Tests
This question tests divisibility, counting multiples in a range, and the inclusion-exclusion principle. It also tests whether you can apply a systematic counting formula correctly when the interval includes negative integers, positive integers, and 0.

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12. Multiples and Factors

12.2. How many positive factors of 540 are divisible by 3?

A. 12
B. 16
C. 18
D. 20
E. 24

Must Know for the GMAT: How to Find the Number of Factors of an Integer

If n = a^p * b^q * c^r, where a, b, and c are prime factors, then the total number of positive factors of n is given by

(p + 1)(q + 1)(r + 1)

Notice that this count includes both 1 and n itself.

For example, if n = 72 = 2^3 * 3^2, then the total number of positive factors of 72 is

(3 + 1)(2 + 1) = 12

Now apply this to the question.

Since 540 = 2^2 * 3^3 * 5^1, the total number of positive factors of 540 is

(2 + 1)(3 + 1)(1 + 1) = 3 * 4 * 2 = 24

To count the factors that are not divisible by 3, we ignore the factor 3^3 and consider only 2^2 * 5^1. The number of positive factors of 2^2 * 5^1 is

(2 + 1)(1 + 1) = 6

So, among the 24 positive factors of 540, exactly 6 are not divisible by 3. Therefore, the number of positive factors of 540 that are divisible by 3 is

24 - 6 = 18

Answer: C.

Alternative approach:

Since 540 = 2^2 * 3^3 * 5^1, any positive factor of 540 must have the form

2^a * 3^b * 5^c

where
a = 0, 1, or 2
b = 0, 1, 2, or 3
c = 0 or 1

Because the factor must be divisible by 3, the exponent of 3 must be at least 1. So b can be 1, 2, or 3, which gives 3 choices.

For the exponent of 2, there are 3 choices: 0, 1, or 2.
For the exponent of 5, there are 2 choices: 0 or 1.

So the number of positive factors of 540 that are divisible by 3 is

3 * 3 * 2 = 18

Answer: C.

Takeaway
In factor-counting questions like this one, prime factorization is the key starting point. Once a number is written in prime-factorized form, you can count only the factors that satisfy a given condition by restricting the allowed exponents of the relevant prime factors.

What This Question Tests
This question tests prime factorization and counting factors under a divisibility condition. It also tests whether you can translate a requirement such as divisibility by 3 into a restriction on the exponent of a prime factor.

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13. Must or Could be True

13.1. If a^3 > b^6 > c^9, which of the following orderings of a, b, and c could be true?

I. a > b > c
II. c > b > a
III. a > c > b

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

First, notice that the question asks which of the following could be true, not which of the following must be true.

In must be true questions, the correct statement must hold for every valid set of numbers. So, if you can find even one valid example in which a statement fails, that statement cannot be correct. This means the plug-in method is useful for eliminating choices, but it is not enough by itself to prove that a statement must be true.

Could be true questions work differently. Here, a statement is correct if there exists at least one valid set of numbers for which it holds. So, once you find one valid example that satisfies both the given condition and the statement, you have proved that the statement could be true.

Because this is a could be true question, it is enough to find one valid example for a statement to show that it could be true.

Also, since taking cube roots preserves the order of an inequality, we can simplify

a^3 > b^6 > c^9

to

a > b^2 > c^3

which is much easier to work with.

I. a > b > c

This statement presents the same order as the variables in the given inequality, which means that when we square b and cube c, the order can still remain consistent. To check whether this could be true, try a small positive value for c, a somewhat larger positive value for b, and a large positive value for a.
For example, let a = 10, b = 2, and c = 1. Then:

a > b^2 > c^3
10 > 2^2 > 1^3

So statement I could be true.

II. c > b > a

If we want c to be greater than b, but still want b^2 > c^3, then positive fractions less than 1 are a good choice, because raising such numbers to positive integer powers makes them smaller.

Take c = 1/2 and b = 2/5. Then c > b, and (b^2 = 4/25) > (c^3 = 1/8).

Now choose a number smaller than b = 2/5 but greater than b^2 = 4/25. Let a = 1/4.

Then:

(c = 1/2) > (b = 2/5) > (a = 1/4)

and

(a = 1/4) > (b^2 = 4/25) > (c^3 = 1/8)

So statement II could be true.

III. a > c > b

Here it is helpful to make b negative. That way b can be smaller than c, but b^2, which becomes positive, can still be large.

Take b = -2 and c = 1. Then (b^2 = 4) > (c^3 = 1)

Now choose a = 5. Then

(a = 5) > (c = 1) > (b = -2)

and

(a = 5) > (b^2 = 4) > (c^3 = 1)

So statement III could be true.

Answer: E.

Takeaway
It is important to know the difference between must be true questions and could be true questions, because the strategy is different. In a must be true question, you are testing whether a statement works in every valid case, so one counterexample is enough to eliminate it. In a could be true question, one valid example is enough to confirm it. It is also important to be comfortable with how numbers behave when raised to powers, since powers can change the relative size of numbers, especially when comparing positive integers, fractions between 0 and 1, 0, and negative numbers.

What This Question Tests
This question tests could be true reasoning, together with number behavior under exponents and inequality number sense. It also tests whether you can choose efficient values and use them correctly to check which orderings are possible under the given condition.

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13. Must or Could be True

13.2. If x ≠ 0 and (|x| - 6)/x > 0, which of the following must be true?

I. x > -10
II. -6 < x < 0
III. x < 10

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

If x > 0, then |x| = x, and we get (x - 6)/x > 0. Multiply by x, which is positive, and keep the sign, to get x - 6 > 0, which is equivalent to x > 6.

If x < 0, then |x| = -x, and we get (-x - 6)/x > 0. Multiply by x, which is negative, and flip the sign, to get -x - 6 < 0, which is equivalent to -6 < x. Since we are considering the case x < 0, then for this case we get -6 < x < 0.

So, we have that (|x| - 6)/x > 0 means that -6 < x < 0 or x > 6.

Now, let's evaluate the statements:

I. x > -10

Any x from the valid ranges (-6 < x < 0 or x > 6) will for sure be greater than -10. So statement I must be true.

To elaborate further, if x is say -3, which is in the range -6 < x < 0, then it is still greater than -10. Likewise, if x is say 100, which is in the range x > 6, it is also greater than -10. So, again, no matter what the value of x is, since it must be either in the range -6 < x < 0 or in the range x > 6, it will definitely be greater than -10.

II. -6 < x < 0

This is not always true. For instance, x could be 10.

III. x < 10

This is not always true. For instance, x could be 100.

Answer: A.

Takeaway
In questions like this one, you should be comfortable with inequality manipulation and absolute value. You should also be familiar with the logic of must-be-true questions. In particular, you need to clearly separate what is given as a fact from what is given as a statement to be evaluated.

What This Question Tests
This question belongs to a type of questions that tends to be very confusing for students: inequalities with confusing ranges together with must-be-true reasoning. It tests whether you can work through the inequality correctly, identify the full set of valid ranges, and then determine which statement must hold across all of those ranges.

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14. Number Properties

14.1. For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 24 is 4, since 24= 2 * 2 * 2 * 3. If x is the smallest positive two-digit integer with length 5 and y is the largest positive two-digit integer with length 3, what is the value of x - y?

A. -67
B. -66
C. -51
D. -6
E. 67

We are given a definition of a new term: the length of a positive integer. However, to solve the question, you do not need to have seen this term before. Just focus on what the definition says. The length counts prime factors with repetition, so the length of an integer is simply the sum of the exponents in its prime factorization. For example, since 24 = 2^3 * 3^1, the length of 24 is 3 + 1 = 4.

First, find x, the smallest two-digit integer with length 5.

To make a number as small as possible while having 5 prime factors, we should use the smallest prime each time:

x = 2 * 2 * 2 * 2 * 2 = 2^5 = 32

So x = 32.

Now find y, the largest two-digit integer with length 3.

Since 99 is the largest two-digit integer, first check whether it has length 3:

99 = 3 * 3 * 11

So the length of 99 is 3.

That means y = 99.

Therefore,

x - y = 32 - 99 = -67

Answer: B

Takeaway
In number properties questions, it is important to know how to do prime factorization comfortably and accurately. Also, when the GMAT introduces a new term, as it sometimes does, such as “length of an integer,” do not panic. Check the definition carefully and look at the example given. Very often, the actual idea being tested is much simpler than it first appears.

What This Question Tests
This question tests prime factorization, including counting prime factors with repetition, together with optimization under constraints. All of this is hidden inside the introduction of an obscure new term, so it also tests whether you can look past unfamiliar wording, focus on the definition, and identify the underlying number-properties concept being tested.

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14. Number Properties

14.2. If n = p^3 * q^6 * r^10, where p, q, and r are different prime numbers, how many factors of n are cubes of an integer greater than 1?

A. 20
B. 21
C. 22
D. 23
E. 24

A factor of n will be a cube if each prime’s exponent in that factor is a multiple of 3. For example, p^3 * q^6 * r^9 = (p * q^2 * r^3)^3 and q^3 * r^6 = (q * r^2)^3 are cubes of an integer, since each exponent in these examples is a multiple of 3.

Now, since n = p^3 * q^6 * r^10, the exponent choices in a cube factor are:

For p: 0 or 3, so 2 choices
For q: 0, 3, or 6, so 3 choices
For r: 0, 3, 6, or 9, so 4 choices

So the total number of factors of n that are perfect cubes is

2 * 3 * 4 = 24

However, the question asks for cubes of an integer greater than 1, so we must exclude 1 itself. The factor 1 comes from choosing exponent 0 for all three primes.

Thus, the required number is

24 - 1 = 23

Answer: D

Takeaway
In number properties questions like this one, prime factorization is the key starting point. When a factor must be a perfect cube, each exponent in its prime factorization must be a multiple of 3. Similarly, for example, when a factor must be a perfect square, each exponent in its prime factorization must be a multiple of 2, that is, even.

What This Question Tests
This question tests prime factorization, factor counting under a restriction, and exponent rules for perfect cubes. It also tests whether you can translate a condition such as “cube of an integer greater than 1” into precise exponent choices and remember to remove the case that gives 1.

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15. Overlapping Sets

15.1. At a summer camp with 245 participants, 158 signed up for hiking and 127 signed up for kayaking. What is the difference between the greatest possible number and the smallest possible number of participants who could have signed up for both activities?

A. 72
B. 80
C. 87
D. 95
E. 118

First, notice that the number of participants who signed up for both activities cannot be greater than 127, since only 127 participants signed up for kayaking.

Therefore, the greatest possible number of participants who could have signed up for both activities is 127.

Next, use the relationship

Total = hiking + kayaking - both + neither

So,

245 = 158 + 127 - both + neither

245 = 285 - both + neither

both = 40 + neither

To make the number who signed up for both as small as possible, we make neither as small as possible. The smallest possible value of neither is 0.

So the smallest possible number who could have signed up for both is

both = 40

Therefore, the required difference is

127 - 40 = 87

Answer: C.

Takeaway
In overlapping-set questions like this one, the greatest possible overlap is the size of the smaller group, while the smallest possible overlap comes from making the number in neither group as small as possible. A standard relationship for overlapping-set questions to know is

Total = group 1 + group 2 - both + neither

What This Question Tests
This question tests overlapping sets, maximum and minimum possible overlap, and set-counting logic under constraints. It also tests whether you can use the standard overlap formula efficiently to find both an upper bound and a lower bound.

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15. Overlapping Sets

15.2. In a class of 55 students, 22 are in the drama club, 18 are in the debate club, and 13 are in the science club. Also, 8 are in both drama and debate, 5 are in both debate and science, and 6 are in both drama and science. If 19 students are in none of these three clubs, how many students are in exactly two of these clubs?

A. 10
B. 11
C. 12
D. 13
E. 14

Since 19 students are in none of the three clubs, the number of students in at least one club is

(total) - (none) = 55 - 19 = 36

Let x be the number of students in all three clubs.

Now use inclusion-exclusion:

(number in at least one club) = (drama) + (debate) + (science) - (drama and debate) - (debate and science) - (drama and science) + (all three)

So we get:

36 = 22 + 18 + 13 - 8 - 5 - 6 + x

Why do we add back x at the end?

Because students in all three clubs were counted once in each of the three individual club totals (22, 18, and 13), so they were counted 3 times at first. Then, when we subtract the three pairwise overlaps (8, 5, and 6), those same students are subtracted 3 times. So they disappear completely, even though they should be counted once in the final total. Therefore, we must add x back once.

Now solve:

36 = 53 - 19 + x
x = 2

So 2 students are in all three clubs.

Now find the number of students in exactly two clubs.

The sum of the pairwise overlaps is

8 + 5 + 6 = 19

But this total does not represent students in exactly two clubs, because each student in all three clubs is included in all three pairwise overlaps. So each such student is counted 3 times in the total of 19.

Therefore, to get the number of students in exactly two clubs, we subtract 3x:

19 - 3x = 19 - 3 * 2 = 19 - 6 = 13

Therefore, the number of students in exactly two clubs is 13.

Answer: D.

Takeaway
In three-set overlapping questions, it is important to know the inclusion-exclusion formula and also to understand why it works. So do not just memorize the formula. Know what each term is counting. In particular, be careful about elements that belong to more than one set, because those are the ones that create overcounting and force you to adjust. The more clearly you track what has been counted once, twice, or three times, the easier these questions become.

What This Question Tests
This question tests three-set overlapping sets, inclusion-exclusion, and careful interpretation of what pairwise overlaps mean. It also tests whether you can distinguish between related but different ideas, such as being in at least one set, in exactly two sets, or in all three sets. Many mistakes in these questions come not from calculation, but from misunderstanding what a given count actually represents.

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BATCH #4 (10 QUESTIONS)

16. Percent and Interest Problems

16.1. A store listed a jacket at a price above its cost. During a sale, the listed price was reduced by 40%, yet the store still sold the jacket at an 8% profit on its cost. What was the original markup percentage on the cost before the sale?

A. 40%
B. 48%
C. 60%
D. 72%
E. 80%

Let the original marked price be p and the cost be c.

After a 40% discount, the selling price becomes

0.6p

We are told that even after this discount, the store still makes an 8% profit on cost. So that same selling price is also

1.08c

Thus,

0.6p = 1.08c

Solve for p:

p = 1.08c / 0.6 = 1.8c

So the original marked price was 180% of the cost.

Therefore, the original markup on cost was

180% - 100% = 80%

Answer: D

Takeaway
In profit and discount questions, it is important to keep track of the base for each percentage. A discount is usually taken on the marked price, while profit is measured on cost. The key is to connect both percentages through the actual selling price, since that is the same quantity viewed in two different ways.

What This Question Tests
This question tests percent reasoning, especially the relationship among cost price, marked price, selling price, discount, and profit. It also tests whether you can translate percentage statements into an equation and avoid mixing percentages that are based on different quantities.

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16. Percent and Interest Problems

16.2. Maria invested $1,000 for one year at r% annual interest, compounded semiannually. Elena invested $1,000 for one year at r% annual simple interest. If, after one year, Maria earned $3.60 more interest than Elena, what is the value of r?

A. 6
B. 10
C. 12
D. 24
E. 30

Since Maria’s investment is compounded semiannually, we should use the compound interest formula:

Final balance = Principal * (1 + r/(100C))^(Ct)

where C is the number of compounding periods per year and t is the number of years.

Here, Principal = 1000, C = 2 because the interest is compounded semiannually so twice a year, and t = 1. So Maria’s final amount after one year is:

1000(1 + r/200)^2

Therefore, Maria’s interest earned is:

1000(1 + r/200)^2 - 1000

Now consider Elena. Since her investment earns simple interest, her interest after one year is simply:

1000 * r/100 = 10r

We are told that, after one year, Maria earned $3.60 more interest than Elena. So:

[1000(1 + r/200)^2 - 1000] - 10r = 3.60

1000(1 + r/100 + r^2/40000) - 1000 - 10r = 3.60

1000 + 10r + r^2/40 - 1000 - 10r = 3.60

r^2/40 = 3.60

r^2 = 144

r = 12

Answer: C.

Takeaway
For interest problems, you should clearly understand the difference between simple interest and compound interest. You should also be careful not to confuse the interest earned over a period with the final balance after that period.

What This Question Tests
This question tests simple interest and compound interest, as well as the difference between interest earned and final balance. It also tests whether you can translate a verbal setup into an algebraic equation.

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17. Probability

17.1. Five fishermen go out to sea on a one-day swordfish expedition. For each fisherman, the probability of catching a swordfish during the expedition is 10%. What is the probability that exactly three of the five fishermen catch a swordfish?

A. 0.00081
B. 0.001
C. 0.0081
D. 0.01
E. 0.0729

We want exactly 3 out of 5 fishermen to catch a swordfish.

First, count how many different groups of 3 fishermen can catch a swordfish. The 3 fishermen who catch a swordfish can be chosen in

5C3 = 5!/(3! * 2!) = 10

ways.

Next, the probability of one specific arrangement is

0.1^3 * 0.9^2

because 3 fishermen catch a swordfish and 2 do not.

So the required probability is

10 * 0.1^3 * 0.9^2 =
= 10 * 0.001 * 0.81 =
= 0.0081

Answer: C.

Takeaway
In probability questions like this one, when the question asks for exactly k successes out of n trials, you need two parts: the number of valid arrangements and the probability of one such arrangement. Then multiply those two results.

What This Question Tests
This question tests basic probability with independent events, together with counting arrangements of successes and failures. It also tests whether you can distinguish between the probability of one specific pattern and the probability of all patterns that satisfy the required condition.

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17. Probability

17.2. A wildlife center has 12 parrots ready to be moved to new enclosures: 4 male-female pairs of green parrots and 2 male-female pairs of yellow parrots. If 3 parrots are selected at random, what is the probability that the 3 parrots include a male-female pair of the same color?

A. 12/55
B. 18/55
C. 33/55
D. 37/55
E. 41/55

In many probability questions, it is easier to find the probability of the opposite event and subtract it from 1. Here, instead of finding the probability of getting a male-female pair of the same color, we first find the probability of getting no such pair.

To avoid getting such a pair, only the following cases are possible:

1. all 3 parrots are green and all 3 are the same sex
2. 2 parrots are green and the same sex, and the third parrot is yellow
3. 2 parrots are yellow and the same sex, and the third parrot is green

Case 1: GGG

P(GGG with no matched pair) = 8/12 * 3/11 * 2/10 = 2/55

Why?
• The first parrot can be any green parrot, so the probability is 8/12.
• After that, to avoid forming a green male-female pair, the second parrot must also be green and of the same sex as the first one. There are 3 such parrots among the remaining 11, so the probability is 3/11.
• Then the third parrot must again be green and of that same sex. There are 2 such parrots among the remaining 10, so the probability is 2/10.

Case 2: GGY

For one specific order, say green, green, yellow:

P(GGY) = 8/12 * 3/11 * 4/10 = 4/55

Why?
• The first parrot can be any green parrot, so 8/12.
• To avoid a green male-female pair, the second parrot must be green and of the same sex as the first one, so 3/11.
• Then the third parrot can be any yellow parrot, and there are 4 yellow parrots out of the remaining 10, so 4/10.

But this can happen in 3 different orders:

GGY, GYG, YGG

So the total probability for this case is

3 * 4/55 = 12/55

Case 3: YYG

For one specific order, say yellow, yellow, green:

P(YYG) = 4/12 * 1/11 * 8/10 = 4/165

Why?
• The first parrot can be any yellow parrot, so 4/12.
• To avoid a yellow male-female pair, the second parrot must be yellow and of the same sex as the first one. Since there are only 2 male yellow parrots and 2 female yellow parrots in total, only 1 such parrot remains among the 11 parrots left, so the probability is 1/11.
• Then the third parrot can be any green parrot, and there are 8 green parrots among the remaining 10, so 8/10.

This can also happen in 3 different orders:

YYG, YGY, GYY

So the total probability for this case is

3 * 4/165 = 12/165 = 4/55
_______________________

Therefore, the probability of getting no matched pair is

2/55 + 12/55 + 4/55 = 18/55

So the probability of getting at least one male-female pair of the same color is

1 - 18/55 = 37/55

Answer: D

Takeaway
In probability questions like this one, it is often easier to calculate the probability of the opposite event first and then subtract that result from 1. Also, when you calculate the probability of a specific case, do not stop after finding the probability of just one order of that case. If the same case can happen in different orders, each of those orders must be counted as well, so you need to multiply by the number of valid arrangements.

What This Question Tests
This question tests complementary probability, case counting, and careful organization of probability scenarios. It also tests whether you can identify all the ways the desired event can fail, count those cases correctly, and account for different valid arrangements within each case.

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18. Remainders

18.1. Positive integer m leaves a remainder of 7 when divided by 8 and a remainder of 5 when divided by 9. What remainder does m leave when divided by 18?

A. 5
B. 7
C. 13
D. 15
E. 17

Theory: Positive integer $a$ divided by positive integer $d$ yielding a remainder of $r$ can always be expressed as $a = qd + r$, where $q$ is the quotient and $r$ is the remainder. Note here that $0 \leq r < d$ (the remainder is a non-negative integer and always less than the divisor).

So, if m leaves a remainder of 7 when divided by 8, then m = 8p + 7. So possible values of m are:

7, 15, 23, 31, 39, 47, ...

If m leaves a remainder of 5 when divided by 9, then m = 9q + 5. So possible values of m are:

5, 14, 23, 32, 41, 50, ...

Now we can derive a general formula based on both pieces of information. In problems of this kind, the combined pattern has the form m = dx + r, where d is the least common multiple of the two divisors, and r is the first common value in the two patterns.

Here, the two divisors are 8 and 9, so d = LCM(8, 9) = 72, and the first common value is r = 23. So the general formula is:

m = 72x + 23

Thus, we need the remainder when 72x + 23 is divided by 18. Since 72 is a multiple of 18, only 23 affects the remainder. And since 23 leaves a remainder of 5 when divided by 18, the remainder is 5:

m = 72x + 23 = 18(4x) + 18 + 5

Answer: A

Takeaway
For remainder questions, you should know how to express a number in terms of divisor, quotient, and remainder. You should also know how to combine two or more remainder conditions and derive one general formula that satisfies all of them. Very often, once that general form is found, the final remainder becomes much easier to determine.

What This Question Tests
This question tests remainders, pattern recognition, and derivation of a general formula from multiple remainder conditions. It also tests whether you can use that general form to find a new remainder with respect to a different divisor.

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18. Remainders

18.2. Wat is the remainder when 67^67 is divided by 17?

A. 1
B. 12
C. 14
D. 15
E. 16

Notice that 67 is 1 less than 68, which is a multiple of 17. So we can rewrite

67^67 = (68 - 1)^67

If we expand the expression, every term except the last one will contain a factor of 68. Since 68 = 17 * 4, all of those terms are divisible by 17 and therefore leave remainder 0 when divided by 17.

The last term will be (-1)^67 = -1. So the question reduces to finding the remainder when -1 is divided by 17.

The remainder when -1 is divided by 17 is 16, since

-1 = 17 * (-1) + 16

Answer: D.

P.S. The process for finding the remainder when dividing a negative integer by a positive integer follows the same principles as when dividing a positive integer by a positive integer. For example:

• What is the remainder when dividing 26 by 8? We find the closest multiple of 8 that is less than 26, which is 24. Then we calculate

(dividend) - (closest multiple less than the dividend) = 26 - 24 = 2

So the remainder is 2.

• What is the remainder when dividing -29 by 9? Here, we find the closest multiple of 9 that is less than -29, which is -36. Then we calculate

(dividend) - (closest multiple less than the dividend) = -29 - (-36) = 7

So the remainder is 7.

What about dividing -74 by 12? The closest multiple of 12 less than -74 is -84. So the remainder is

(dividend) - (closest multiple less than the dividend) = -74 - (-84) = 10

So, for the question at hand, we do the same thing. We need the remainder when dividing -1 by 17, so we find the closest multiple of 17 that is less than -1, which is -17. Then we calculate

(dividend) - (closest multiple less than the dividend) = -1 - (-17) = 16

Alternatively, consider this method:

• What is the remainder when dividing -29 by 9? Dividing 29 by 9 gives a remainder of 2. To find the remainder for -29 divided by 9, subtract this 2 from the divisor. Thus, the remainder when dividing -29 by 9 is 9 - 2 = 7

• Similarly, what is the remainder when dividing -74 by 12? Dividing 74 by 12 gives a remainder of 2. Therefore, the remainder when dividing -74 by 12 is 12 - 2 = 10

So, for the question at hand, dividing 1 by 17 gives a remainder of 1. According to this rule, to find the remainder when dividing -1 by 17, we subtract that remainder from the divisor:

17 - 1 = 16

Takeaway
For remainder questions, it is important to understand how exponentiation can affect the remainder. It is also important to know how to find the remainder when a negative integer is divided by a positive integer, since that idea can be easy to mishandle if the underlying logic is not clear.

What This Question Tests
This question tests remainders, binomial expansion, and the effect of exponents on remainders. It also tests whether you can handle negative remainders correctly and use algebraic rewriting to simplify a large power expression before finding the final remainder.

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19. Roots

19.1. For which of the following values of x is $\frac{ \sqrt{3-\sqrt{7+\sqrt{x}}} }{\sqrt{2-\sqrt{4-\sqrt{x}}}}$ NOT defined as a real number?

I. 0
II. 4
III. 5

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

For an even root to be defined as a real number, the expression under the root must be nonnegative, because even roots of negative numbers are not real.

Also, for a fraction to be defined, its denominator must not be 0, because division by 0 is not allowed.

We are given

$\frac{\sqrt{3-\sqrt{7+\sqrt{x}}}}{\sqrt{2-\sqrt{4-\sqrt{x}}}}$

Now let’s check each option.

I. x = 0

Substitute 0 into the numerator:

$\sqrt{3-\sqrt{7+\sqrt{0}}}=\sqrt{3-\sqrt{7}}$

Since $\sqrt{7}<3$, the quantity $3-\sqrt{7}$ is positive, so the numerator is defined.

Now substitute 0 into the denominator:

$\sqrt{2-\sqrt{4-\sqrt{0}}}=\sqrt{2-\sqrt{4}}=\sqrt{2-2}=\sqrt{0}=0$

So the denominator is 0. Division by 0 is not allowed, so the expression is not defined for x = 0.

II. x = 4

Substitute 4 into the numerator:

$\sqrt{3-\sqrt{7+\sqrt{4}}}=\sqrt{3-\sqrt{7+2}}=\sqrt{3-\sqrt{9}}=\sqrt{3-3}=\sqrt{0}$

So the numerator is defined.

Now substitute 4 into the denominator:

$\sqrt{2-\sqrt{4-\sqrt{4}}}=\sqrt{2-\sqrt{4-2}}=\sqrt{2-\sqrt{2}}$

Since $\sqrt{2}<2$, the quantity $2-\sqrt{2}$ is positive, so the denominator is defined and is not 0.

Therefore, the expression is defined for x = 4.

III. x = 5

Substitute 5 into the numerator:

$\sqrt{3-\sqrt{7+\sqrt{5}}}$

Since $\sqrt{5}>2$, we have $7+\sqrt{5}>9$, so

$\sqrt{7+\sqrt{5}}>3$

That means

$3-\sqrt{7+\sqrt{5}}<0$

So the numerator is not defined as a real number.

Therefore, the expression is not defined for x = 5.

So the expression is not defined for I and III only.

Answer: D

Takeaway
For the GMAT, even roots of negative numbers are not defined, because they do not produce real numbers, and GMAT deals only with real numbers. You should also remember that division by 0 is never allowed, so the denominator of a fraction cannot be 0. More generally, you should know how to determine the domain of an expression by checking every restriction that comes from radicals and fractions.

What This Question Tests
This question tests the domain of a radical expression, especially a fraction containing nested square roots. It also tests whether you can apply the basic domain rules correctly: expressions under square roots must be nonnegative, and the denominator must be nonzero.

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19. Roots

19.2. Which of the following is equal to $\sqrt{18 + \sqrt{128}} - \sqrt{2}$ ?

A. 1
B. 2
C. 3
D. 4
E. 5

First simplify the expression:

$\sqrt{18 + \sqrt{128}} - \sqrt{2} = \sqrt{18 + \sqrt{64*2}} - \sqrt{2} = \sqrt{18 + 8\sqrt{2}} - \sqrt{2}$

Since all answer choices are rational numbers, the irrational part must cancel. That suggests rewriting $18 + 8\sqrt{2}$ as a perfect square.

Notice that

$18 + 8\sqrt{2} = 16 + 8\sqrt{2} + 2 = (4 + \sqrt{2})^2$

So,

$\sqrt{18 + \sqrt{128}} - \sqrt{2} = \sqrt{18 + 8\sqrt{2}} - \sqrt{2}$

$= \sqrt{16 + 8\sqrt{2} + 2} - \sqrt{2}$

$= \sqrt{(4 + \sqrt{2})^2} - \sqrt{2}$

$= (4 + \sqrt{2}) - \sqrt{2} = 4$

Answer: D.

Takeaway
You should know how to work with radicals and nested radicals. In addition, you should also know how to complete the square and recognize when completing the square helps simplify the expression.

What This Question Tests
This question tests radical simplification and pattern recognition. In particular, it tests whether you can recognize when completing the square under a square root helps simplify the expression.

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20. Sequences

20.1. The sequence $a_1$, $a_2$, $a_3$, ... is such that $a_n = \frac{a_{n-1}}{3}$ for every integer $n > 1$. If $a_7$ is between -0.243 and -0.027, then $a_9 * a_{10}$ is between

A. 0.0000003 and 0.0000243
B. 0.000003 and 0.000243
C. 0.000009 and 0.000729
D. 0.000027 and 0.002187
E. 0.000081 and 0.006561

Since $a_n = \frac{a_{n-1}}{3}$,

$a_8 = \frac{a_7}{3}$

$a_9 = \frac{a_8}{3} = \frac{a_7}{9}$

$a_{10} = \frac{a_9}{3} = \frac{a_7}{27}$

So

$a_9 * a_{10} = \frac{a_7}{9} * \frac{a_7}{27} = \frac{(a_7)^2}{3^5}$

We are given

$-0.243 < a_7 < -0.027$

Rewrite the endpoints as powers of 3 and 10:

$-\frac{3^5}{10^3} < a_7 < -\frac{3^3}{10^3}$

Since both endpoints are negative, squaring reverses their order:

$\left(\frac{3^3}{10^3}\right)^2 < (a_7)^2 < \left(\frac{3^5}{10^3}\right)^2$

$\frac{3^6}{10^6} < (a_7)^2 < \frac{3^{10}}{10^6}$

Dividing by $3^5$, we get

$\frac{3}{10^6} < \frac{(a_7)^2}{3^5} < \frac{3^5}{10^6}$

That is,

$0.000003 < a_9 * a_{10} < 0.000243$

Answer: B

Takeaway
For sequence questions such as this one, you should know how to use the given recursive formula to generate the terms of the sequence and to express one term in terms of another.

What This Question Tests
This question tests sequences, especially understanding a recursive formula, deriving later terms from earlier terms, and applying the recurrence rule correctly.

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20. Sequences

20.2. In a certain sequence of non-zero integers, each term after the first is obtained by multiplying the previous term by the same integer less than -1. If the sixth term of the sequence is between -800 and 800, what is the greatest possible number of integer values for the first term?

A. 47
B. 48
C. 49
D. 50
E. 51

Let the first term be x, and let the constant multiplier be r, where r is an integer less than -1.

Then the sequence is

x, xr, xr^2, xr^3, xr^4, xr^5

So the sixth term is xr^5

To make the number of possible integer values of x as large as possible, we want the absolute value of r to be as small as possible. Since r is an integer less than -1, the smallest possible absolute value is 2, so r = -2. Then the sixth term is:

x(-2)^5 = -32x

We are told that

-800 < -32x < 800

Now divide all parts by -32. Since we divide by a negative number, the inequality signs flip:

25 > x > -25

So

-25 < x < 25

Since x must be an integer and the sequence consists of non-zero numbers, x cannot be 0.

So the possible integer values of x are

-24, -23, ..., -1, 1, ..., 23, 24

That is 24 positive values and 24 negative values, for a total of 48 values.

Answer: B.

Takeaway
For sequence questions such as this one, you should know how to use the rule of the sequence to write a later term in terms of the first term. You should also pay close attention to what kinds of numbers are allowed, because that determines the domain of the sequence and can affect the final answer.

What This Question Tests
This question tests sequences, especially geometric sequences, along with minimizing and maximizing ideas.