GMAT Focus Edition Mock test - 3 - GMAT MCQ

To determine how much consumers paid for the suit after the price increase and subsequent discount, follow these two steps:

1. Calculate the Increased Price:

• Original Price: $200
• Price Increase: 30% of $200
  Increase Amount = 200 × 0.30 = $60
• New Price After Increase:
  New Price = 200 + 60 = $260

2. Apply the 30% Discount Coupon:

• Discount: 30% off the increased price of $260
  Discount Amount = 260 × 0.30 = $78
• Final Price After Discount:
  Final Price = 260 - 78 = $182

Summary:

• Original Price: $200
• After 30% Increase: $260
• After 30% Discount: $182

Therefore, consumers who used the 30% off coupon on the increased price paid $182 for the suit.

Here, we have two categories: (a) with or without two-car garage, and (b) with or without an in-the-ground pool.  Houses can be members of either, both, or neither category.  We will use a two circle Venn diagram:

We know the total of the group is 80 —– A + B + C + D = 80.  We know the green circle, two-car garages, has 50 members, so A + B = 50.  We know the blue circle, in-the-ground pool, has 40 members, so B + C = 40.  We also know the crucial overlap region, B = 35.  If B = 35, in the green circle, we can deduce that A = 15, and in the blue circle, we can deduce that C = 5.  Then
A + B + C + D = 15 + 35 + 5 + D = 80
D = 25
Thus, 25 houses in this development have neither a two-car garage nor an in-the-ground swimming pool. 

First of all, notice that the shaded area, quite literally and visually, is the difference of two squares — Area = y² – x². We know from the Difference of Two Squares pattern, this factors into:
Area of ABCD = y² – x² = (y + x)(y – x)
Rectangle S is supposed to have the same area, and it has a length of y + x, which is one of the factors in the above expression. Therefore, the other factor y–x must be the width of Rectangle S.

Step 1: Key Data from the Question Stem

Set S has 6 elements.
The elements of set S are distinct.
x and y are prime numbers. Because 2 is already an element in S, both x and y have to be odd.
0 < x < 40 and 0 < y < 40

Step 2: Check Statement I

I. The maximum possible range of the set is greater than 33.
The key word in this entire statement is maximum. We have to determine whether the maximum value possible for the range exceeds 33.
We know x and y are prime numbers. The largest prime number less than 40 is 37.
If either x or y is 37, the largest number in the set will be 37 and the smallest number is 2.
Therefore, the maximum range of the set will be 37 - 2 = 35. It is greater than 33.

Statement I is true. So, eliminate choices that do not contain I.
Eliminate choice D

Step 3: Check Statement II

II. The median can never be an even number.
There are 6 numbers in the set. Therefore, the median is the arithmetic mean of the 3rd and the 4th term when the numbers are written in ascending or descending order.
The elements are {2, 4, 6, 8, x, y}, where x and y are prime numbers.
If x and y take 3 and 5 as values, the median is 4.5
If x = 3, y = 7 or greater, the median is 5.
If x = 5, y = 7 or greater, the median is 5.5
If x = 7, y = 11 or greater, the median is 6.5
If x = 11 or greater and y = 13 or greater, the median is 7.
It is quite clear that the median is either an odd number or is not an interger. So, the median can never be an even integer.

Statement II is true. Eliminate choices that do not contain II.
Eliminate choices A and C as well.

Step 4: Check Statement III

III. If y = 37, the average of the set will be greater than the median.
If y = 37, the set will be {2, 4, 6, 8, x, 37}, where x is a prime number greater than 2 and less than 37.
The average will be 
If x = 3, median = 5 and average = 10. Average > median.
If x = 5, median = 5.5 and average = 10.33. Average > median
If x = 7, median = 6.5 and average = 10.66. Average > medain
If x = 11 or greater, the median = 7. Average will be definitely greater than 10. So, Average > Median.
It is true that the average is greater than the median if y = 37.

Statement III is also true.
Statements I, II, and III are true.

Choice E is the correct answer.

Whatever we pick from the first set is independent with whatever we pick from the second set, so we can use the simplified AND rule.

P(first pick = C) = 1/5

P(second pick = M) = 1/6

P(C and M) = P(C)*P(M) = (1/5)*(1/6) = 1/30

x and y are integers and |x - y| = 12
Approach: Square both sides and solve.
Squaring both sides, we get (x - y)² = 144
x² + y² - 2xy = 144
Add, 4xy to both sides of the equation.
x² + y² - 2xy + 4xy = 144 + 4xy
x² + y² + 2xy = 144 + 4xy
Or (x + y)² = 144 + 4xy
(x + y)² will NOT be negative for real values of x and y.
i.e., (x + y)² ≥ 0
∴ 144 + 4xy ≥ 0
Or 4xy ≥ -144
So, xy ≥ -36
The least value that xy can take is -36.

Step 1: Formula to find the sum of first 'n' terms of a GP

The sum of the first n terms of a G.P. is given by  where 'a' is the first term of the G.P., 'r' is the common ratio and 'n' is the number of terms in the G.P.
Therefore, the sum of the first 6 terms of the G.P will be equal to 
And sum of the first 3 terms of the G.P. will be equal to 

Step 2: Use the ratio between these two sums to find 'r'

The ratio of the sum of the first 6 terms : sum of first 3 terms = 9 : 1

Or r³ + 1 = 9
r³ = 8
r = 2

Choice C is the correct answer.

In order to figure out the average velocity, we need to know both the total distance and the total time.
From the question, we know the total distance is 600 miles.
We need to figure out the time of each leg separately.
In the first leg, T = D/R = 300/30 = 10 hr. In the second leg, T = D/R = 300/60 = 5 hours.
The total time is 10 + 5 = 15 hours.
The average velocity, total distance divided by total time, is 600/15 = 40 mph.

Step one: find the midpoint of the region.
The midpoint, halfway between and 20 and 90, is 55.
In other words, 20 and 90 have the same distance from 55, a distance of 35.
These endpoints are not included, but the region includes all the points that have a distance from from x = 55 that is less than 35.
Translating that into math, we get the following:

|x – 55| < 35

Given the foregoing discussion, it may be obvious now the trap-mistake answer is (C), which results from simply adding and subtracting the percents.
We need multipliers.

multiplier for a 10% increases = 1 + 0.10 = 1.1

multiplier for a 20% decreases = 1 – 0.20 = 0.8

multiplier for a 50% increases = 1 + 0.50 = 1.5

Now, multiply these. First, multiply (0.8) and (1.5), using the doubling & halving trick. Half of 0.80 is 0.40, and twice 1.5 is 3

(0.8) * (1.5) = (0.4) * (3) = 1.2

Now, multiply this by 1.1

1.2 * 1.1 = 1.32

Thus, the three percent changes combined produce a 32% increase. Answer = (B).

Here, we have three categories, so we need three circles. Every student must take at least one of these three performing arts extracurricular activities, so there will be no one outside the three circles.

The sum of all seven = 120 (we never use this number in this question)

The totals for the band (70), the chorus (73), and the drama (45) each involve the sum of four discrete regions.  We will have to find other information before we can employ them.

“8 students are in all three groups”

N = 8.

“37 students are in both the Band and Chorus”

37 = K + N = K + 8 —> K = 29

“20 are in both the Band and the Drama”

20 = M + N = M + 8  —> M = 12

“twenty-five students are just in the chorus, not in anything else”

L = 25

We now have identified three of the regions in the Chorus circle, so we can solve for P.

chorus = 73 = K + L + N + P

73 = 29 + 25 + 8 + P

P = 11

Now, we have identified three of the regions in the Drama circle, so we can solve for Q.

drama = 45 = M + N + P + Q

45 = 12 + 8 + 11 + Q

Q = 14

This is precisely what the question was asking: how many students are only in drama? There are 14 students who take only drama.

Answer = C

This question is made for the discriminant! You won’t have to guess and check if you know how to use the formula appropriately. To find D, plug in the following to the formula:

a = 1
b = −10
c = m
D = b² – 4ac
D= (−10)² − 4(1)(m) = 100 – 4m

Now, we know that we’re looking for two real solutions, so we want D = 100 – 4m > 0.

100 – 4m > 0 ⟹ 100 > 4m ⟹ 25 > m

So, we get m < 25. Be careful NOT to choose 25 as your answer! The largest integer that is less than 25 would be 24.

Key Data from the Question Stem:

1. a ≤ b ≤ c
2. a, b, and c are positive integers.
3. Average of the three integers = 20
4. Sum of all the three integers = 60
5. Median = b = a + 11

Check for the possible values of c

Theoretically, the least value of c is when c = b.
Therefore, a + (a + 11) + (a + 11) = 60 (b and c are equal and b, the median, is a + 11)
Or 3a = 38 or a = 12.66
So, b = c = 12.66 + 11 = 23.66

However, we know that these numbers are all integers.
Therefore, a, b, and c cannot take these values.
So, the least value for c with this constraint is NOT likely to be when c = b.

Let us increment c by 1. Let c = (b + 1)
In this scenario, a + (a + 11) + (a + 12) = 60
Or 3a = 37. The value of the numbers is not an integer in this scenario as well.

Let us increment c again by 1. i.e., c = b + 2
Now, a + (a + 11) + (a + 13) = 60
Or 3a = 36 or a = 12.
If a = 12, b = 23 and c = 25.
The least value for c that satisfies all these conditions is 25.

Choice C is the correct answer.

Picking an M is not disjoint with picking a C — they both could happen on the same round of the game. We have to use the generalized OR rule for this:

P(C or M) = P(C) + P(M) – P(C and M)

What will be the form of such 4-digit numbers?

The 4-digit numbers should contain the block 25.
The required 4-digit numbers will be of the form:
a. 25_ _
b. _ 25 _
c. _ _ 25

What is the test of divisibility by 75?

If a number is divisible by 75, then it will be divisible by 25 and 3.

Count the number of 4-digit numbers for three possiblities

a. Numbers of the form 25 _ _ that are divisible by 75
A number of the form 25_ _ is divisible by 25 if its rightmost 2 digits are 00, 25, 50, or 75.

Check which of these numbers are also multiples of 3
Only one number, 2550 satisfies the condition.

b. Numbers of the form _ 25 _ that are divisible by 75
A number of the form _ 25 _ is divisible by 25 if its unit digit is 0.
The 4-digit number will be of the form _ 250

What options exist for the left most digit so that the number is also divisible by 3?
The sum of the right most 3 digits of the number = 2 + 5 + 0 = 7.
If the first digit is 2 or 5 or 8, the sum of the 4 digits will be divisible by 3.
There are three 4-digit numbers that match the form _ 25 _ and are divisible by 75.

c. Numbers of the form _ _ 25 that are divisible by 75
All numbesr of the form _ _ 25 is divisible by 25.

What options exist for the first 2 digits so that the number is also divisible by 3?
We already have a 2 and 5 whose sum is 7. 7 is a multiple of 3 plus 1.
We have a (3k + 1) with us. If we add a (3m + 2), the sum will be 3(k + m) + 1 + 2 = 3(k + m) + 3, which is divisible by 3.
The least 2 digit number that is of the form (3m + 2) is 11.
For example, if 11 takes the 1st 2 places, the number is divisible by 3

11 is not the only such number.
All 2-digit numbers of the form (3m + 2) will work

How many are there? The largest 2-digit number that is of the form 3m + 2 is 98.
And all of these numbers are in arithmetic progression with a common difference of 3
So, apply the arithmetic progression formula to compute the nth term: 98 = 11 + (n - 1)3
3(n - 1) = 87
(n - 1) = 29
Or n = 30
30 such 4-digit numbers exist

Add the count of all three possiblities

1 + 3 + 30 = 34
34 such 4-digit numbers exist

Choice C is the correct answer.

In the first leg, we know time and rate, so find distance: D = RT = (3)*(40) = 120 miles.
In the second leg, we know distance and rate, so find time: T = R/D = 300/60 = 5 hours.
Total distance = 120 + 300 = 420 miles. Total time = 3 + 5 = 8.
Average velocity = 420/8 = 210/4 = 105/2 = 52.5 mph.

Some folks might think this involves a sophisticated calculation, but much of this can be done with simple spatial analysis. Look at the second inequality, the more complicated one: |x – 8| > |x + 4|. All this says is that we are looking for points such that the distance to x = 8 is greater than the distance from x = –4; in other words, we want all the points that are closer to x = –4 and farther from x = 8.

The midpoint between x = –4 and x = 8 is the point x = 2. This point is not included because it’s equidistant from both points, but everything to the left of this point on the number line is closer to x = –4 than it is to x = 8. That entire complicated inequality simplifies to x < 2.

Combine that with the first inequality, |x| < 20, which in the negative realm means that x must be greater than –20. Thus, the allowed region is –20 < x < 2.

Answer = (B)

A full deck of 52 cards contains 13 cards from each of the four suits. The probability of drawing a heart from a full deck is 1/4.  Therefore, the probability of “not heart” is 3/4.

P(at least three draws to win) = 1 – P(win in two or fewer draws)

Furthermore,

P(win in two or fewer draws) = P(win in one draw OR win in two draws)

= P(win in one draw) + P(win in two draws)

Winning in one draw means: I select one card from a full deck, and it turns out to be a heart. Above, we already said: the probability of this is 1/4.

P(win in one draw) = 1/4

Winning in two draws means: my first draw is “not heart”, P = 3/4, AND the second draw is a heart, P = 1/4. Because we replace and re-shuffle, the draws are independent, so the AND means multiply.

P(win in two draws) = (3/4) * (1/4) = 3/16

P(win in two or fewer draws) = P(win in one draw) + P(win in two draws)

= 1/4 + 3/16 = 7/16

P(at least three draws to win) = 1 – P(win in two or fewer draws)

= 1 – 7/16 = 9/16

Answer = B

The trap answer here would be to take the ratio of 80% and 20% — those don’t represent actually amounts that other person has, just the differences between amounts owned and the cost of the condo.  Think of this in terms of multipliers.  Use the variables:

B = amount Bert has in savings

R = amount Rebecca has in savings

P = price of the condominium

Then in terms of multipliers, the information given tells us that P = 1.8 * B, and P = 1.2 * R. Set these equal.

1.8 * B = 1.2 * R

P(at least one vowel) = 1 – P(no vowels)

The probability of picking no vowel from the first set is 3/5. The probability of picking no vowel from the second set is 5/6. In order to get no vowels at all, we need no vowels from the first set AND no vowels from the second set. According to the AND rule, we multiply those probabilities.

P(no vowels) = (3/5) * (5/6) = 1/2

P(at least one vowel) = 1 – P(no vowels) = 1 – 1/2 = 1/2

Answer = C

Key Data:

Total dollar value of all gift certificates in the initial configuration : 13 hampers × 12 dollars = $156
Total dollar value of all gift certificates in the new configuration = 12 hampers × 13 dollars = $156
No change in total value of gift certificates.

Susan started with 13 hampers.
So, she has 13 of $3, 13 of $4, and 13 of $5 certificates.

Key Question: What are the possible combination of certificates in the new hampers?

In the new configuration, $13 can be achieved as follows: {3, 3, 3, 4}, {4, 4, 5}, and {5, 5, 3}
The hampers that contain the configuration {3, 3, 3, 4} will not have $5 certificates.

Assign Variables and Frame Linear Equations

Let Susan make x hampers of {3, 3, 3, 4}, y hampers of {4, 4, 5}, and z hampers of {5, 5, 3}.
The answer to ‘x’ is the answer to the question.

Number of $3 certificates will be 3x + z = 13 ....(1)
Number of $4 certificates will be x + 2y = 13 ....(2)
Number of $5 certificates will be y + 2z = 13 ....(3)

Solve the 3 Equations

Multiply (3) by 2: 2y + 4z = 26
Subtract (2) from (3) × 2
2y + 4z = 26
-x - 2y = -13
------------------
4z – x = 13 ....(4)
------------------

Multiply (1) by 4: 12x + 4z = 52
Subtract (4) and (1) × 4
12x + 4z = 52
-4z + x = -13
----------------
13x = 39
-----------------
Or x = 3

'x' is the number of gift hampers without $5 certificates.
As x = 3, three gift hampers did not contain $5 gift certificates.

Choice D is the correct answer.

Since we cannot say what was responsible for the increased academic standards, we can rule out any attempt to explain it. It could be that more residents of Merryweather began to attend the school, as opposed to a reduction in those from outside Merryweather, or vice versa, so avoid explanations as to why the population decreased.
The only thing we can infer from the information given is that there could be tuition increases if per capita revenue remained the same: Say students not from Merryweather used to account for 66% and were paying, for example, $100,000 in total for tuition, if they now account for 40%, they must pay a higher tuition to maintain a stable per capita revenue.

Because study participants who ate beets were also more likely to exercise regularly than those who did not eat beets, it is impossible to determine whether beets or regular exercise were more influential in preventing cancer. Thus this is the correct answer.

The conclusion states that the tonsils have no function in the body. To weaken this conclusion, we should try to show that the tonsils do in fact have a function, at least at some point in human life. Comparison with another organ is not really useful, as they might be very different from tonsils, and what happens in other animals is also not directly relevant. And discussing where the tonsils develop has nothing to do with their function. Finally, though an organ might develop as part of a system and yet have no function (like the appendix or muscles that wiggle the ear), the only answer left suggests there is a function even though this function can be taken over by another part of the body. This suggestion most weakens the argument posed above.

The conclusion of this argument is that when questions are ordered hardest to easiest, students perform more poorly on them. In order to refute this causal assertion, another source of the poor performance must be introduced. The fact that the student who performed more poorly has a lower grade point average than the other student is a significant intervening cause, and therefore the correct answer.

The title should capture the overall theme and content of the passage. The passage discusses the relationships between major tech players and startups, covering both the positive aspects of innovation alliances and the potential downsides. Option B, "Innovation Alliances: Tech Titans Unveiled," accurately reflects this dual perspective, making it the most appropriate title.

Option A is too focused on startups and does not encompass the broader scope of the passage. Option C focuses on the negative aspects only, overlooking the positive collaborations discussed in the passage. Option D is specific to one aspect (antitrust measures) and doesn't capture the overall theme. Option E introduces the idea of a balancing act but doesn't cover the range of topics discussed in the passage.

To infer the author's viewpoint, we need to examine the passage for clues. The passage discusses both the positive aspects of tech collaborations, such as fostering innovation, and the negative aspects, including the potential exploitation of startups. Option B, stating that collaborations lead to a more diverse tech ecosystem, aligns with the positive aspects discussed in the passage.

Option A reflects a negative viewpoint, and the passage suggests that acquisitions for the sole purpose of eliminating competition are detrimental. Option C is not supported by the passage, which advocates for antitrust measures to prevent concentration of power. Option D contradicts the passage's call for modern antitrust measures. Option E aligns with the passage's discussion on the negative impact of exploitative tactics.

To identify the question answered by the passage, we must refer to the information provided. The passage discusses a comparison between the current state of the tech industry and the corporate landscape of the early 20th century, suggesting a need for modern antitrust measures. Therefore, the correct answer is option A.
Options B and C are not addressed in the passage, as it focuses more on the relationships between tech giants and startups, rather than specific tactics or recent acquisitions. Option D is addressed in the passage, as it discusses the need for antitrust measures to prevent the concentration of power in the tech sector. Option E is not covered in the passage; it goes beyond the scope of the information provided.

In this argument, the flaw is framed in the following way:

X is always Y. A is Y. Therefore, A is X.

(Just because X is always Y does NOT mean that Y is always X.)

The flawed argument is therefore most like:

Except for 2, prime numbers are always odd. The number 13 is odd. Therefore, it is prime. 

Here

x = prime numbers

Y = odd

A = 13

The structure of this argument is the same as the structure argument in:

Eggplants are always purple. John bought a purple vegetable. Therefore, it was an eggplant. 

X= eggplants

Y = purple

A = purple vegetable

The advertisement's claim, that Coma Cola is the best tasting cola, is based on 99% percent of cola consumers choosing it over soda water. The flaw in the experimental setup is that just because a consumer chooses Coma Cola over soda water, it does not mean that she will choose Coma Cola over any other cola, let alone over all other colas. 

The correct answer recognizes that, in order to validly make such a claim, one would have to show a preference for Coma Cola over other colas, not over soda water (or even over other flavored soft drinks, as one of the incorrect answer choices indicates).