Test: Min/Max Problems - GMAT MCQ

•  Let x be the common multiple for the ratio of Books, Pens, and Pencils, which are in the ratio 2:7:9.
• Thus, the number of Books, Pens, and Pencils are 2x, 7x, and 9x respectively.
• The difference between the number of Pens and Pencils is 9x - 7x = 2x.
• For this difference to be a multiple of both 3 and 7, 2x must be a multiple of 21.
• The smallest integer value for x that satisfies this condition is 21.
• Therefore, the minimum number of Books is 2(21) = 42, Pens is 7(21) = 147, and Pencils is 9(21) = 189.
• The total minimum number of items is 42 + 147 + 189 = 378.
• Hence, the correct answer is option 2, 378.

To find the greatest possible value of r² + R, we need to consider the maximum remainders for r and R.

When x is divided by 4, the maximum remainder is 3 (since x is a positive integer). Therefore, the maximum value for r is 3.

When x is divided by 9, the maximum remainder is 8 (since x is a positive integer). Therefore, the maximum value for R is 8.

Substituting these values into the expression r² + R:

r² + R = 3² + 8 = 9 + 8 = 17

Therefore, the greatest possible value of r² + R is 17.

The correct answer is C: 17.

To determine the minimum amount of money the owner of the rental agency must spend to purchase enough oil, we need to calculate the total amount of oil required for all 6 machines.

Each machine requires 5 quarts of oil, so for 6 machines, the total number of quarts needed is 5 quarts/machine * 6 machines = 30 quarts.

Since there are 4 quarts in a gallon, the total number of gallons needed is 30 quarts / 4 quarts/gallon = 7.5 gallons.

Now, let's calculate the cost of purchasing 7.5 gallons of oil:

Cost of 7.5 gallons = 7.5 gallons * $12/gallon = $90

Therefore, the minimum amount of money the owner must spend to purchase enough oil is $90.

The closest answer choice is B: $94. However, the correct answer based on the calculation is $90.

To determine the greatest possible value of m for which 15!/3m is an integer, we need to consider the prime factorization of 15! (15 factorial).

15! = 15 * 14 * 13 * ... * 2 * 1

To simplify the expression 15!/3m, we need to determine the maximum number of times the prime factor 3 can be canceled out from the factorial.

The highest power of 3 that divides 15! is determined by the sum of the quotients obtained when dividing each of the numbers 15, 14, 13, ..., 2, and 1 by 3.

For each number greater than or equal to 3, the quotient obtained when dividing by 3 is 1 (except for 15, which gives a quotient of 5). Therefore, the sum of the quotients is 5 + 1 = 6.

So, the maximum value of m for which 15!/3m is an integer is 6.

Therefore, the correct answer is C: 6.

To find the minimum value of n for which the product of all positive integers from 1 to n is evenly divisible by 840, we need to determine the prime factorization of 840.

The prime factorization of 840 is: 2^3 * 3 * 5 * 7.

To ensure that the product of all positive integers from 1 to n is divisible by 840, we need to have at least three 2's, one 3, one 5, and one 7 among the factors.

The minimum value of n can be determined by considering the powers of prime factors:

n = max(power of 2, power of 3, power of 5, power of 7)

For 2: The power of 2 is 3.
For 3: The power of 3 is 1.
For 5: The power of 5 is 1.
For 7: The power of 7 is 1.

Therefore, the minimum value of n is max(3, 1, 1, 1) = 3.

Thus, the correct answer is B: 7.

Let's consider the weights of the three boxes of supplies. We'll call them A, B, and C.

We know that the average weight of the three boxes is 7 kilograms. This means that the sum of the weights of the three boxes is 7 * 3 = 21 kilograms.

We also know that the median weight of the three boxes is 9 kilograms. The median weight is the weight of the box that falls in the middle when the weights are arranged in ascending order. Since there are three boxes, the median weight must be the weight of the second box.

Let's consider the weights in ascending order: A, B, C.

Since the median weight is 9 kilograms, we have A ≤ 9 ≤ C.

Now let's analyze the maximum possible weight of the lightest box, A.

To maximize the weight of A, we need to minimize the weights of B and C. If we let B = 9 kilograms, then A + B + C = 21 kilograms.

From this, we can determine that A + 9 + C = 21, which simplifies to A + C = 12.

To maximize the weight of A, we need to minimize the weight of C. The minimum possible value for C is 3 kilograms (since A + C = 12). Therefore, the maximum possible weight of the lightest box, A, is 12 - C = 12 - 3 = 9 kilograms.

Thus, the correct answer is C: 3.

To find the least possible area of the rectangle, we need to consider the minimum values for the length and width.

Given that x ≥ 3, we can determine the minimum values for the expressions 2x+5 and 3x-7.

For the length (2x+5), when x is minimum (x = 3), the length is (2 * 3) + 5 = 11.

For the width (3x-7), when x is minimum (x = 3), the width is (3 * 3) - 7 = 2.

The area of a rectangle is calculated by multiplying the length and width, so the minimum area is 11 * 2 = 22.

Therefore, the correct answer is C: 22.

To find the lowest possible number of researchers who will not be assigned to the team they prefer, we need to determine the allocation of researchers to Team A and Team B based on the given percentages.

Out of the 50 researchers, 40% will be assigned to Team A, which is (40/100) * 50 = 20 researchers.

The remaining 60% will be assigned to Team B, which is (60/100) * 50 = 30 researchers.

However, 70% of the researchers prefer Team A, which is (70/100) * 50 = 35 researchers.

And 30% of the researchers prefer Team B, which is (30/100) * 50 = 15 researchers.

To minimize the number of researchers who will not be assigned to their preferred team, we need to assign the researchers who prefer Team A to Team A and the researchers who prefer Team B to Team B.

Since there are only 20 researchers assigned to Team A, which is fewer than the number of researchers who prefer Team A (35), we will have 35 - 20 = 15 researchers who will not be assigned to their preferred team.

Therefore, the lowest possible number of researchers who will not be assigned to the team they prefer is 15.

The correct answer is A: 15.

To find the maximum wage increase for a 40-hour week, we need to consider the highest possible increase per hour.

The maximum wage increase per hour is 87.5 cents.

To find the maximum wage increase for a 40-hour week, we multiply the maximum increase per hour by the number of hours worked:

Maximum wage increase = 87.5 cents/hour * 40 hours

Maximum wage increase = 3500 cents

Converting cents to dollars:

Maximum wage increase = $35.00

Therefore, the maximum wage increase for a 40-hour week is $35.00.

The correct answer is D: $35.00.