Test: Min/Max Problems - GMAT MCQ

To find the smallest possible value of ab given that |a - b| = 8, we need to consider the different cases and their corresponding values of ab.

Case 1: a > b
If a is greater than b, then a - b = 8. To minimize the value of ab, we can let a be the smallest possible integer that satisfies this condition, which is a = b + 8. Substituting this into the expression for ab, we get:

ab = (b + 8)b = b² + 8b

Now, we want to minimize this expression. To do that, we can complete the square:

ab = b² + 8b = (b² + 8b + 16) - 16 = (b + 4)² - 16

Since we want to minimize the value of ab, we want to minimize the term (b + 4)². The smallest possible value for this term is 0, which occurs when b = -4. Substituting this back into the expression for ab, we get:

ab = (-4 + 4)² - 16 = 0 - 16 = -16

Case 2: b > a
If b is greater than a, then b - a = 8. Following a similar approach as in Case 1, we let b = a + 8 and substitute it into the expression for ab:

ab = (a + 8)a = a² + 8a

Again, we want to minimize this expression, and we can complete the square:

ab = a² + 8a = (a² + 8a + 16) - 16 = (a + 4)² - 16

Minimizing the term (a + 4)² gives us a = -4, and substituting it back into the expression for ab, we get:

ab = (-4 + 4)² - 16 = 0 - 16 = -16

Comparing the two cases, we see that the smallest possible value of ab is -16.

Therefore, the correct answer is A. -16.

Given that k is an integer, we can start by trying out some values of k and calculating the expression |129 - 17k|:

For k = 0: |129 - 17(0)| = |129 - 0| = 129
For k = 1: |129 - 17(1)| = |129 - 17| = 112
For k = 2: |129 - 17(2)| = |129 - 34| = 95
For k = 3: |129 - 17(3)| = |129 - 51| = 78
For k = 4: |129 - 17(4)| = |129 - 68| = 61
For k = 5: |129 - 17(5)| = |129 - 85| = 44
For k = 6: |129 - 17(6)| = |129 - 102| = 27
For k = 7: |129 - 17(7)| = |129 - 119| = 10
For k = 8: |129 - 17(8)| = |129 - 136| = 7

As we can see, the expression reaches its minimum value of 7 when k = 8.

Therefore, the correct answer is D. 8.

A candidate needs the votes of at least 2/3 of its 1,331 members
Number of votes the candidate receive is ≥ 2/3 * 1331
=> votes ≥ 2662/3
=> votes  ≥ 887.23
Since the number of votes must be an integer
=> votes  ≥ 888

The given problem asks us to determine the minimum number of marbles a blindfolded student would have to draw from a bag containing 18 marbles in order to be certain of having at least three marbles of the same color.

To solve this problem, we can consider the worst-case scenario for each color. In this case, we assume that the student draws the maximum number of marbles for each color without getting three marbles of the same color. Let's go through the options one by one:

Option A: If the student draws 5 marbles, it is possible to have at most 2 marbles of each color (2 red + 2 green + 1 black). This option does not guarantee that the student will have at least three marbles of the same color. Therefore, option A is not the correct answer.

Option B: If the student draws 6 marbles, it is possible to have at most 2 marbles of each color (2 red + 2 green + 2 black). Similar to option A, this option does not guarantee that the student will have at least three marbles of the same color. Therefore, option B is not the correct answer.

Option C: If the student draws 7 marbles, it is still possible to have at most 2 marbles of each color (2 red + 2 green + 2 black + 1 additional marble). However, since there are only three colors, the student must have at least three marbles of the same color. Therefore, option C is the correct answer.

Option D: If the student draws 13 marbles, they will definitely have at least three marbles of the same color. However, this is more than the minimum number of marbles required to guarantee this condition. Therefore, option D is not the correct answer.

Option E: If the student draws 14 marbles, they will definitely have at least three marbles of the same color. However, this is also more than the minimum number of marbles required to guarantee this condition. Therefore, option E is not the correct answer.

In conclusion, the correct answer is option C, which is 7 marbles. The student would need to draw at least 7 marbles to be certain of having at least three marbles of the same color.

Given that -5 ≤ a ≤ 1 and 5 ≤ b ≤ 8, we know that a and b are both even numbers.

To find the maximum possible value for ab, we need to find the largest possible values for both a and b within their given ranges.

Since a is even and -5 ≤ a ≤ 1, the largest even number within this range is 0. Therefore, a = 0.

Similarly, since b is even and 5 ≤ b ≤ 8, the largest even number within this range is 8. Therefore, b = 8.

Now we can calculate ab: ab = 0 * 8 = 0.

Therefore, the maximum possible value for ab is 0.

Hence, the answer is option D.

Available Numbers = -5, -4, -3, -2, -1, 0, 1, 2, 3, 4

We need the least possible value, so we have to take the least available value in conjunction with other values.

Let's take -5

The next two values that will give the highest magnitude when multiplied is -4 * 4

Now, -5 * -4 * 4 is a positive value, so let's take the next least value is -3

So net product = -5 * -4 * -3 * 4 = -240

let's break it down step by step:

Step 1: Define variables
Let's denote the number of 5-cent stamps as x. Since Kim bought an equal number of 5-cent and 25-cent stamps, the number of 25-cent stamps is also x. Additionally, since she bought twice as many 10-cent stamps as 5-cent stamps, the number of 10-cent stamps is 2x.

Step 2: Calculate the total cost of stamps
The total cost of the stamps can be calculated by multiplying the number of each stamp denomination by its value and adding them together:
Total cost = (5 cents * x) + (25 cents * x) + (10 cents * 2x)
Total cost = 5x + 25x + 20x
Total cost = 50x

Step 3: Determine the number of 1-cent stamps
To find the minimum number of 1-cent stamps she could have bought, we need to determine the remainder when the total cost of the stamps is divided by 100 cents (equivalent to $1).

In this case, we have:
Total cost = 50x
Since the total cost is given as $2.65, we can convert it to cents: $2.65 = 265 cents.
So we have: 50x = 265

Step 4: Find the minimum number of 1-cent stamps
To determine the minimum number of 1-cent stamps, we need to find the remainder when 265 cents is divided by 100 cents (1 dollar).
265 cents = 2 dollars and 65 cents, which is equivalent to 2 * 100 cents + 65 cents.
The remainder when 265 cents is divided by 100 cents is 65 cents.

Therefore, the minimum number of 1-cent stamps she could have bought is 65.

Step 5: Select the answer choice
We need to determine which answer choice represents the minimum number of 1-cent stamps, 65.
Among the answer choices given, (C) 15 is the closest to 65.

Therefore, the correct answer is (C) 15.

Given that there are 30 marbles in total, and we stop the process as soon as 2 marbles of different colors have been selected, we can analyze the worst-case scenario for maximizing the number of selections.

To do this, we need to select as many marbles of the same color as possible before getting two marbles of different colors. This means we should select all the marbles of one color first, and then select marbles of a different color until we reach two marbles of different colors.

Given that there are 6 red marbles, 7 blue marbles, 8 yellow marbles, and the rest are green, we should start by selecting all the green marbles first since they make up the largest portion of the remaining marbles.

After selecting all the green marbles, we are left with 30 - (6 + 7 + 8) = 9 marbles, consisting of only red, blue, and yellow marbles. At this point, we can select any color marbles without stopping the process until we have two marbles of different colors.

Since we want to maximize the number of selections, we should select red marbles until we have only red marbles left or until we have selected two red marbles. This would require selecting all 6 red marbles.

Now we are left with 9 - 6 = 3 marbles, consisting of only blue and yellow marbles. Again, to maximize the number of selections, we should select marbles of the same color until we have two marbles of different colors.

Therefore, we can select 2 blue marbles, which will give us two marbles of different colors and stop the selection process.

To calculate the total number of selections needed, we add up the number of selections made for each color: 30 (green marbles) + 6 (red marbles) + 2 (blue marbles) = 38.

Therefore, the greatest number of selections that might be needed to stop the process is 38, which corresponds to answer choice (A).

To find the least possible value of x + y + z given the conditions 3x = 4y = 7z, we can start by finding the least common multiple (LCM) of 3, 4, and 7.

The prime factorization of 3 is 3,
The prime factorization of 4 is 2²,
The prime factorization of 7 is 7.

The LCM of 3, 4, and 7 is calculated by taking the highest power of each prime factor: 2² * 3 * 7 = 84.

Now, we need to express x, y, and z in terms of this LCM.

Since 3x = 84, we can solve for x: x = 84/3 = 28.
Similarly, 4y = 84, so y = 84/4 = 21.
And 7z = 84, which gives z = 84/7 = 12.

Finally, we can calculate x + y + z: 28 + 21 + 12 = 61.

Therefore, the least possible value of x + y + z is 61.

Hence, the correct answer is D.

The given problem states that a restaurant has a total of 16 tables, and each table can seat a maximum of 4 people. It is mentioned that 50 people are sitting at the tables in the restaurant with no tables empty. The question asks for the greatest possible number of tables that could be occupied by just 1 person.

To solve this problem, let's assume that the maximum number of tables occupied by just 1 person is x. Since each table can seat a maximum of 4 people, we can also say that the minimum number of tables occupied by more than 1 person is 16 - x.

Now, let's consider the total number of people sitting at the tables. We know that there are 50 people in total. Since each table can seat a maximum of 4 people, the total number of seats is 16 * 4 = 64.

We can create an equation based on the given information:

x + (16 - x) * 4 = 50

Simplifying the equation, we get:

x + 64 - 4x = 50
64 - 3x = 50
-3x = 50 - 64
-3x = -14
x = -14 / -3
x = 4.67

Since x represents the number of tables occupied by just 1 person, it cannot be a fraction or a decimal. Therefore, we need to round it down to the nearest whole number. The greatest possible number of tables that could be occupied by just 1 person is 4.

Hence, the correct answer is C.