The correct answer is either 4 or 5. Because each student ticket costs $2 and each adult ticket costs $3, the total amount, in dollars, that Chris spends on x student tickets and 1 adult ticket is 2(x) + 3(1). Because Chris spends at least $11 but no more than $14 on the tickets, one can write the compound inequality 2x + 3 ≥ 11 and 2x + 3 ≤ 14. Subtracting 3 from each side of both inequalities and then dividing each side of both inequalities by 2 yields x ≥ 4 and x ≤ 5.5. Thus, the value of x must be an integer that is both greater than or equal to 4 and less than or equal to 5.5. Therefore, x = 4 or x = 5. Either 4 or 5 may be gridded as the correct answer.

Subtracting the sides of 3y + c = 5y − 7 from the corresponding sides of 3x + b = 5x − 7 gives (3x − 3y) + (b − c) = (5x − 5y). Since b = c − ¹⁄₂, or b − c = − ¹⁄₂, it follows that (3x − 3y) + ( −¹⁄₂) = (5x − 5y). Solving this equation for x in terms of y gives x = y − ¹⁄₄. Therefore, x is y minus ¹⁄₄.

Choices B, C, and D are incorrect and may be the result of making a computational error when solving the equations for x in terms of y.

Let x be the number of left-handed female students and let y be the number of left-handed male students. Then the number of right handed female students will be 5x and the number of right-handed male students will be 9y. Since the total number of left-handed students is 18 and the total number of right-handed students is 122, the system of equations below must be satisfied.

\(\left\{\begin{matrix} x + y = 18 & \\ 5x + 9y = 122 & \end{matrix}\right.\)

Solving this system gives x = 10 and y = 8. Thus, 50 of the 122 right-handed students are female. Therefore, the probability that a right-handed student selected at random is female is \(\frac{50}{122}\), which to the nearest thousandth is 0.410. Choices B, C, and D are incorrect and may be the result of incorrect calculation of the missing values in the table.

For the present population to decrease by 10 percent, it must be multiplied by the factor 0.9. Since the engineer estimates that the population will decrease by 10 percent every 20 years, the present population, 50,000, must be multiplied by (0.9)n , where n is the number of 20-year periods that will have elapsed t years from now. After t years, the number of 20-year periods that have elapsed is \(\frac{t}{20}\). Therefore, 50,000(0.9)\(^{\frac{t}{20}}\) represents the engineer’s estimate of the population of the city t years from now.

Choices A, B, and C are incorrect because each of these choices either confuses the percent decrease with the multiplicative factor that represents the percent decrease or mistakenly multiplies t by 20 to find the number of 20-year periods that will have elapsed in t years.

Let ℓ and w be the length and width, respectively, of the original rectangle. The area of the original rectangle is A = ℓw. The rectangle is altered by increasing its length by 10 percent and decreasing its width by p percent; thus, the length of the altered rectangle is 1.1ℓ, and the width of the altered rectangle is ( 1 − \(\frac{P}{100}\))w. The alterations decrease the area by 12 percent, so the area of the altered rectangle is (1 − 0.12)A = 0.88A. The  altered rectangle is the product of its length and width, so 0.88A = (1.1ℓ)( 1 − \(\frac{P}{100}\))w. Since A = ℓw, this last equation can be rewritten as 0.88A = (1.1)( 1 − \(\frac{P}{100}\))ℓw = (1.1)( 1 − \(\frac{P}{100}\)) A, from which it follows that 0.88 = (1.1)( 1 − \(\frac{P}{100}\)), or 0.8 = ( 1 − \(\frac{P}{100}\)). Therefore,\(\frac{P}{100}\) = 0.2, and so the value of p is 20.

Choice A is incorrect and may be the result of confusing the 12 percent decrease in area with the percent decrease in width. Choice B is incorrect because decreasing the width by 15 percent results in a 6.5 percent decrease in area, not a 12 percent decrease. Choice D is incorrect and may be the result of adding the percents given in the question (10 + 12).

The line passes through the origin, (2, k), and (k, 32). Any two of these points can be used to find the slope of the line. Since the line passes through (0, 0) and (2, k), the slope of the line is equal to \(\frac{k − 0}{2 − 0}=\frac{k}{2}\).

Similarly, since the line passes through (0, 0) and (k, 32), the slope of the line is equal to \(\frac{32 − 0}{k − 0}=\frac{32}{k}\). Since each expression gives the slope of the same line, it must be true that \(\frac{k}{2}=\frac{32}{k}\). Multiplying each side of \(\frac{k}{2}=\frac{32}{k}\) by 2k gives k² = 64, from which it follows that k = 8 or k = −8. Therefore, of the given choices, only 8 could be the value of k.

Choices A, B, and D are incorrect and may be the result of calculation errors.

The volume of the grain silo can be found by adding the volumes of all the solids of which it is composed. The silo is made up of a cylinder with height 10 feet (ft) and base radius 5 feet and two cones, each having height 5 ft and base radius 5 ft. The formulas V_cylinder = πr²h and V^cone = ¹⁄₃πr²h can be used to determine the total volume of the silo. Since the two cones have identical dimensions, the total volume, in cubic feet, of the silo is given by V_silo = π(5)²(10) + (2)(¹⁄₃)π(5)²(5) = (⁴⁄₃)(250)π, which is approximately equal to 1,047.2 cubic feet.

Choice A is incorrect because this is the volume of only the two cones. Choice B is incorrect because this is the volume of only the cylinder. Choice C is incorrect because this is the volume of only one of the cones plus the cylinder.

Let c be the number of students in Mr. Kohl’s class. The conditions described in the question can be represented by the equations n = 3c + 5 and n + 21 = 4c. Substituting 3c + 5 for n in the second equation gives 3c + 5 + 21 = 4c, which can be solved to find c = 26.

Choices A, B, and C are incorrect because the values given for the number of students in the class cannot fulfill both conditions given in the question. For example, if there were 16 students in the class, then the first condition would imply that there are 3(16) + 5 = 53 milliliters of solution in the beaker, but the second condition would imply that there are 4(16) − 21 = 43 milliliters of solution in the beaker. This contradiction shows that there cannot be 16 students in the class.

Since the angles are acute and sin(a°) = cos(b°), it follows from the complementary angle property of sines and cosines that a + b = 90. Substituting 4k − 22 for a and 6k − 13 for b gives (4k − 22) + (6k − 13) = 90, which simplifies to 10k − 35 = 90. Therefore, 10k = 125, and k = 12.5.

Choice A is incorrect and may be the result of mistakenly assuming that a + b and making a sign error. Choices B and D are incorrect because they result in values for a and b such that sin(a°) ≠ cos(b°).

One of the three numbers is x; let the other two numbers be y and z. Since the sum of three numbers is 855, the equation x + y + z = 855 is true. The statement that x is 50% more than the sum of the other two numbers can be represented as x = 1.5(y + z), or \(\frac{x}{1.5}\) = y + z. Substituting \(\frac{x}{1.5}\)for y + z in x + y + z = 855 gives x + \(\frac{x}{1.5}\) = 855. This last equation can be rewritten as x + \(\frac{2x}{3x}\) = 855, or \(\frac{5x}{3}\) = 855. Therefore, x equals ³⁄₅ × 855 = 513.

Choices A, C, and D are incorrect and may be the result of calculation errors.