A dairy farmer uses a storage silo that is in the shape of the right circular cylinder above. If the volume of the silo is 72π cubic yards, what is the diameter of the base of the cylinder, in yards?

Solution
The correct answer is 6. The volume of a cylinder is πr^{2}h, where r is the radius of the base of the cylinder and h is the height of the cylinder. Since the storage silo is a cylinder with volume 72π cubic yards and height 8 yards, it is true that 72π = πr^{2} (8), where r is the radius of the base of the cylinder, in yards. Dividing both sides of 72π = πr^{2} (8) by 8π gives r^{2} = 9, and so the radius of base of the cylinder is 3 yards. Therefore, the diameter of the base of the cylinder is 6 yards.
A local television station sells time slots for programs in 30minute intervals. If the station operates 24 hours per day, every day of the week, what is the total number of 30minute time slots the station can sell for Tuesday and Wednesday?

Solution
The correct answer is 96. Since each day has a total of 24 hours of time slots available for the station to sell, there is a total of 48 hours of time slots available to sell on Tuesday and Wednesday. Each time slot is a 30minute interval, which is equal to a ^{1}⁄_{2} hour interval. Therefore, there are a total of \(\frac{48\, hours}{\frac{1}{2}hours/time\, slot}\)= 96 time slots of 30 minutes for the station to sell on Tuesday and Wednesday.
Number of Portable Media Players Sold Worldwide Each Year from 2006 to 2011
According to the line graph above, the number of portable media players sold in 2008 is what fraction of the number sold in 2011 ?

Solution
The correct answer is ^{5}⁄_{8} or .625. Based on the line graph, the number of portable media players sold in 2008 was 100 million, and the number of portable media players sold in 2011 was 160 million. Therefore, the number of portable media players sold in 2008 is \(\frac{100\, million}{160\, million}\) of the portable media players sold in 2011. This fraction reduces to ^{5}⁄_{8}. Either ^{5}⁄_{8} or its decimal equivalent, .625, may be gridded as the correct answer.
The posted weight limit for a covered wooden bridge in Pennsylvania is 6000 pounds. A delivery truck that is carrying x identical boxes each weighing 14 pounds will pass over the bridge. If the combine weight of the empty delivery truck and its driver is 4500 pounds, what is the maximum possible value for x that will keep the combined weight of the truck, driver, and boxes below the bridge’s posted weight limit?

Solution
The correct answer is 107. Since the weight of the empty truck and its driver is 4500 pounds and each box weighs 14 pounds, the weight, in pounds, of the delivery truck, its driver, and x boxes is 4500 + 14x. This weight is below the bridge’s posted weight limit of 6000 pounds if 4500 + 14x < 6000. That inequality is equivalent to 14x ≤ 1500 or x < [latex]\frac{1500}{14}[/latex] = 107 ^{1}⁄_{7} . Since the number of packages must be an integer, the maximum possible value for x that will keep the combined weight of the truck, its driver, and the x identical boxes below the bridge’s posted weight limit is 107.
Wyatt can husk at least 12 dozen ears of corn per hour and at most 18 dozen ears of corn per hour. Based on this information, what is a possible amount of time, in hours, that it could take Wyatt to husk 72 dozen ears of corn?

Solution
The correct answer is any number between 4 and 6, inclusive. Since Wyatt can husk at least 12 dozen ears of corn per hour, it will take him no more than \(\frac{72}{12}\)=6 hours to husk 72 dozen ears of corn. On the other hand, since Wyatt can husk at most 18 dozen ears of corn per hour, it will take him at least \(\frac{72}{18}\)= 4 hours to husk 72 dozen ears of corn. Therefore, the possible times it could take Wyatt to husk 72 dozen ears of corn are 4 hours to 6 hours, inclusive. Any number between 4 and 6, inclusive, can be gridded as the correct answer.
Which of the following is an equivalent form of the equation of the graph shown in the xyplane above, from which the coordinates of vertex A can be identified as constants in the equation?

Solution
Any quadratic function q can be written in the form q(x) = a(x − h)^{2} + k, where a, h, and k are constants and (h, k) is the vertex of the parabola when q is graphed in the coordinate plane. (Depending on the sign of a, the constant k must be the minimum or maximum value of q, and h is the value of x for which a(x − h)^{2} = 0 and q(x) has value k.) This form can be reached by completing the square in the expression that defines q. The given equation is y = x^{2} − 2x − 15, and since the coefficient of x is −2, the equation can be written in terms of (x − 1)^{2} = x^{2} − 2x + 1 as follows: y = x^{2} − 2x − 15 = (x^{2} − 2x + 1) − 16 = (x − 1)2 − 16. From this form of the equation, the coefficients of the vertex can be read as (1, −16)
Choices A and C are incorrect because the coordinates of the vertex A do not appear as constants in these equations. Choice B is incorrect because it is not equivalent to the given equation.
For a polynomial p(x), the value of p(3) is −2. Which of the following must be true about p(x) ?

Solution
If the polynomial p(x) is divided by x − 3, the result can be written as \(\frac{p(x)}{x − 3}\) = q(x) + \(\frac{r}{x − 3}\), where q(x) is a polynomial and r is the remainder. Since x − 3 is a degree 1 polynomial, the remainder is a real number. Hence, p(x) can be written as p(x) = (x − 3)q(x) + r, where r is a real number. It is given that p(3) = −2 so it must be true that −2 = p(3) = (3 − 3)q(3) + r = (0)q(3) + r = r. Therefore, the remainder when p(x) is divided by x − 3 is −2.
Choice A is incorrect because p(3) = −2 does not imply that p(5) = 0. Choices B and C are incorrect because the remainder −2 or its negative, 2, need not be a root of p(x).
If the system of inequalities y ≥ 2x +1 and y > ^{1}⁄_{2}x − 1 is graphed in the xyplane above, which quadrant contains no solutions to the system?

Solution
To determine which quadrant does not contain any solutions to the system of inequalities, graph the inequalities. Graph the inequality y ≥ 2x + 1 by drawing a line through the yintercept (0, 1) and the point (1, 3), and graph the inequality y > ^{1}⁄_{2}x − 1 by drawing a dashed line through the yintercept (0, −1) and the point (2, 0), as shown in the figure below.
The solution to the system of inequalities is the intersection of the shaded regions above the graphs of both lines. It can be seen that the solutions only include points in quadrants I, II, and III and do not include any points in quadrant IV.
Choices A and B are incorrect because quadrants II and III contain solutions to the system of inequalities, as shown in the figure above. Choice D is incorrect because there are no solutions in quadrant IV.
A square field measures 10 meters by 10 meters. Ten students each mark off a randomly selected region of the field; each region is square and has side lengths of 1 meter, and no two regions overlap. The students count the earthworms contained in the soil to a depth of 5 centimeters beneath the ground’s surface in each region. The results are shown in the table below.
Which of the following is a reasonable approximation of the number of earthworms to a depth of 5 centimeters beneath the ground’s surface in the entire field?

Solution
The area of the field is 100 square meters. Each 1meterby1meter square has an area of 1 square meter. Thus, on average, the earthworm counts to a depth of 5 centimeters for each of the regions investigated by the students should be about \(\frac{1}{100}\) of the total number of earthworms to a depth of 5 centimeters in the entire field. Since the counts for the smaller regions are from 107 to 176, the estimate for the entire field should be between 10,700 and 17,600. Therefore, of the given choices, 15,000 is a reasonable estimate for the number of earthworms to a depth of 5 centimeters in the entire field.
Choice A is incorrect; 150 is the approximate number of earthworms in 1 square meter. Choice B is incorrect; it results from using 10 square meters as the area of the field. Choice D is incorrect; it results from using 1,000 square meters as the area of the field.
Katarina is a botanist studying the production of pears by two types of pear trees. She noticed that Type A trees produced 20 percent more pears than Type B trees did. Based on Katarina’s observation, if the Type A trees produced 144 pears, how many pears did the Type B trees produce?

Solution
Let x represent the number of pears produced by the Type B trees. Then the Type A trees produce 20 percent more pears than x, which is x + 0.20x = 1.20x pears. Since Type A trees produce 144 pears, the equation 1.20x = 144 holds. Thus x =\(\frac{144}{1.20}\)= 120. Therefore, the Type B trees produced 120 pears.
Choice A is incorrect because while 144 is reduced by approximately 20 percent, increasing 115 by 20 percent gives 138, not 144. Choice C is incorrect; it results from subtracting 20 from the number of pears produced by the Type A trees. Choice D is incorrect; it results from adding 20 percent of the number of pears produced by Type A trees to the number of pears produced by Type A trees.