Mathematics Test - 4 - SAT MCQ

Original inequality: 
Multiply by 3m (since m > 0, we don’t “flip” the inequality): 15 ≤ 2m
Divide by 2: 7.5 ≤ m
Therefore, the least possible value of m is 7.5.

That any equation in the form y = a(x - h)² + k has a vertex at (h, k) and is open up if a > 0 and down if a < 0. In the equation y = -(x + m)² + m; therefore, the vertex is (-m, m), and a = -1.
Since m > 1, this means that the vertex of the parabola has a negative x-coordinate and a positive y-coordinate, which means the vertex is in quadrant II. And since a < 0, the parabola is open down.
The only graph among the choices that is an open down parabola with a vertex in the second quadrant is the graph in choice (D).

Find AD with Pythagorean Theorem: (AD)² + 4² = 5²
Simplify: (AD)² + 16 = 25
Subtract 16: (AD)² = 9
Take square root: AD = 3
Or, even better, just notice that triangle ADB is a 3-4-5 right triangle.
Use triangle area formula to find AC:
Area = 1/2 bh = 1/2(AC)(4) = 19
Simplif y: 2(AC) = 19
Divide by 2: AC = 19/2
 or .615

Let’s call the 3 positive integers a, b, and c. If the average of these numbers is m, then

Multiply by 3: a + b + c = 3m
New average when 24 is included in the set: 
Substitute a + b + c = 3m: 

According to the histogram, 7 students received 15 credits, 1 student received 16 credits, and 1 student received 18 credits, for a total of 9 students who received 15 or more credits. This is 9/24 of the total, or 37.5%

(2 - i)(3 - 2i)
FOIL: 6 - 4i - 3i + 2i²
Substitute i² = -1: 6 - 4i - 3i + 2(-1)
Combine like terms: 4 - 7i

Let p = the initial price per share of the stock. After the first year, its price increased by 20%, so its price was (1.20)p. After the second year, this price declined 25%, so its price was (0.75)(1.20)p. After the second year, this price increased by 10% so its price was (1.10)(0.75)(1.20) p = 0.99p, which means that overall the price decreased by 1%.

On the drawing, we should first mark the areas of the three faces. The front and back faces both have an area of 3a. The left and right faces both have an area of 3b. The top and bottom faces both have an area of ab. We should now try to find integer values for a and b so that these areas match those given in the choices.
(A) 15, 18, and 30 This is possible if a = 5 and b = 6.
(B) 18, 24, and 48 This is possible if a = 6 and b = 8.
(C) 12, 15, and 24 This cannot work for any integer
values of a and b.
(D) 15, 24, and 40 This is possible if a = 5 and b = 8.

Because this polynomial has a degree of 3 (which is the highest power of any of its terms), it cannot have more than 3 zeros. These three zeros are given as -2, 3, and 6. We also know that g(0), the y-intercept of the graph, is negative. This gives us enough information to make a rough sketch of the graph.

This shows that the only values of x for which the function is negative are -2 < x < 3 and x > 6. Therefore the only negative value among the choices is (B) g(-1).

2^2n-2 = 32
When dealing with exponential equations, it helps to see if we can express the two sides of the equation in terms of the same base. Since 32 = 2⁵, we can express both sides in base 2:
2^2n-2 = 25
If X^a = X^b and x > 1, then a = b (if the bases are equal, the exponents are equal): 2n - 2 = 5
Add 2: 2n = 7
Divide by 2: n = 7/2 = 3.5

When faced with a question like this, we must analyze each statement individually.
(A) The average revenue per store increased by over 100% from 2005 to 2009. True or false? In 2005, according to the line of best fit, the average revenue per store was approximately $300,000. In 2009, the average revenue per store was approximately $500,000. This is a percent increase of 

FALSE
(B) The total number of retail stores increased by 50% from 2005 to 2012. True or false? According to the scatterplot, in 2005 there were 3 stores corresponding to the three dots above 2005. In 2012 there were 6 stores corresponding to the 6 dots above 2012. This is a percent increase of

FALSE
(C) The total revenue for all stores in 2012 is more than three times the total revenue from all stores in 2004. True or false? In 2004, there were 3 stores with an average revenue per store of approximately $250,000. Therefore the total revenue in 2004 was approximately 3 × $250,000 = $750,000. In 2012, there were 6 stores with an average revenue per store of approximately $650,000. Therefore the total revenue in 2012 was approximately 6 × $650,000 = $3,900,000. Since $3,900,000 is more than three time $750,000, this statement is TRUE.

Choice B is correct. Since the angles marked y° and u° are vertical angles, y  = u.
Subtracting the sides of y = u from the corresponding sides of x + y = u + w gives x = w.
Since the angles marked w° and z° are vertical angles, w =  z.
Therefore, x = z, and so I must be true.
The equation in II need not be true. For example, if x = w = z = t = 70 and y = u = 40, then all three pairs of vertical angles in the figure have equal measure and the given condition x + y = u + w holds.
But it is not true in this case that y is equal to w. Therefore, II need not be true.
Since the top three angles in the figure form a straight angle, it follows that x + y + z = 180. Similarly, w + u + t = 180, and so x + y + z = w + u + t.
Subtracting the sides of the given equation x + y = u + w from the corresponding sides of x + y + z = w + u + t gives z = t. Therefore, III must be true. Since only I and III must be true, the correct answer is choice B.
Choices A, C, and D are incorrect because each of these choices includes II, which need not be true.

The correct answer is either 1 or 2. The given equation can be rewritten as x⁵ − 5x³ + 4x = 0. Since the polynomial expression on the left has no constant term, it has x as a factor: x(x⁴ − 5x² + 4) = 0. The expression in parentheses is a quadratic equation in x² that can be factored, giving x(x² − 1)(x² − 4) = 0. This further factors as x(x − 1)(x + 1)(x − 2)(x + 2) = 0. The solutions for x are x = 0, x = 1, x = −1, x = 2, and x = −2. Since it is given that x > 0, the possible values of x are x = 1 and x = 2. Either 1 or 2 may be gridded as the correct answer.

Choice B is correct. On Earth, the acceleration due to gravity is 9.8 m/sec². Thus, for an object with a weight of 150 newtons, the formula W = mg becomes 150 = m(9.8), which shows that the mass of an object with a weight of 150 newtons on Earth is about 15.3 kilograms. Substituting this mass into the formula W = mg and now using the weight of 170 newtons gives 170 = 15.3g, which shows that the second planet’s acceleration due to gravity is about 11.1 m/sec². According to the table, this value for the acceleration due to gravity holds on Saturn.
Choices A, C, and D are incorrect. Using the formula W = mg and the values for g in the table shows that an object with a weight of 170 newtons on these planets would not have the same mass as an object with a weight of 150 newtons on Earth.

Choice B is correct. For any value of x, say x = x₀, the point (x₀, f(x₀)) lies on the graph of f and the point (x₀, g(x₀)) lies on the graph of g. Thus, for any value of x, say x = x₀, the value of f(x₀) + g(x₀) is equal to the sum of the y-coordinates of the points on the graphs of f and g with x-coordinate equal to x0. Therefore, the value of x for which f(x) + g(x) is equal to 0 will occur when the y-coordinates of the points representing f(x) and g(x) at the same value of x are equidistant from the x-axis and are on opposite sides of the x-axis. Looking at the graphs, one can see that this occurs at x = −2: the point (−2, −2) lies on the graph of f, and the point (−2, 2) lies on the graph of g. Thus, at x = −2, the value of f(x) + g(x) is −2 + 2 = 0.
Choices A, C, and D are incorrect because none of these x-values satisfy the given equation, f(x) + g(x) = 0.

Choice C is correct. Let I be the initial savings. If each successive year, 1% of the current value is added to the value of the account, then after 1 year, the amount in the account will be I + 0.01I = I(1 + 0.01); after 2 years, the amount in the account will be I(1 + 0.01) + 0.01I(1 + 0.01) = (1 + 0.01)I(1 + 0.01) = I(1 + 0.01)²; and after t years, the amount in the account will be I(1 + 0.01)t. This is exponential growth of the money in the account.
Choice A is incorrect. If each successive year, 2% of the initial savings, I, is added to the value of the account, then after t years, the amount in the account will be I + 0.02It, which is linear growth. Choice B is incorrect. If each successive year, 1.5% of the initial savings, I, and $100 is added to the value of the the account, then after t years the amount in the account will be I + (0.015I + 100)t, which is linear growth. Choice D is incorrect. If each successive year, $100 is added to the value of the account, then after t years the amount in the account will be I + 100t, which is linear growth.

Since 9 can be rewritten as 3², 9^3/4 is equivalent to Applying the properties of exponents, this can be written as 3^3/2, which can further be rewritten as 3^2/2(3^1/2), an expression that is equivalent to 3√3. 
Choices A is incorrect; it is equivalent to 9^1/3. Choice B is incorrect; it is equivalent to 9^1/4. Choice C is incorrect; it is equivalent to 3^1/2.

To solve the equation 2(p + 1) + 8(p − 1) = 5p, first distribute the terms outside the parentheses to the terms inside the parentheses: 2p + 2 + 8p − 8 = 5p. Next, combine like terms on the left side of the equal sign: 10p − 6 = 5p. Subtracting 10p from both sides yields −6 = −5p. Finally, dividing both sides by −5 gives p = 6/5 = 1.2. Either 6/5 or 1.2 can be gridded as the correct answer.

Because f is a linear function of x, the equation f(x) = mx + b, where m and b are constants, can be used to define the relationship between x and f(x). In this equation, m represents the increase in the value of f(x) for every increase in the value of x by 1. From the table, it can be determined that the value of f(x) increases by 8 for every increase in the value of x by 2. In other words, for the function f the value of m is 8/2, or 4. The value of b can be found by substituting the values of x and f(x) from any row of the table and the value of m into the equation f(x) = mx + b and solving for b. For example, using x = 1, f(x) = 5, and m = 4 yields 5 = 4(1) + b. Solving for b yields b = 1. Therefore, the equation defining the function f can be written in the form f(x) = 4x + 1. 
Choices A, B, and D are incorrect. Any equation defining the linear function f must give values of f(x) for corresponding values of x, as shown in each row of the table. According to the table, if x = 3, f(x) = 13. However, substituting x = 3 into the equation given in choice A gives f(3) = 2(3) + 3, or f(3) = 9, not 13. Similarly, substituting x = 3 into the equation given in choice B gives f(3) = 3(3) + 2, or f(3) = 11, not 13. Lastly, substituting x = 3 into the equation given in choice D gives f(3) = 5(3), or f(3) = 15, not 13. Therefore, the equations in choices A, B, and D cannot define f.

The DBA plans to increase its membership by n businesses each year, so x years from now, the association plans to have increased its membership by nx businesses. Since there are already b businesses at the beginning of this year, the total number of businesses, y, the DBA plans to have as members x years from now is modeled by y = nx + b.
Choice B is incorrect. The equation given in choice B correctly represents the increase in membership x years from now as nx. However, the number of businesses at the beginning of the year, b, has been subtracted from this amount of increase, not added to it. Choices C and D are incorrect because they use exponential models to represent the increase in membership. Since the membership increases by n businesses each year, this situation is correctly modeled by a linear relationship.

Adding the two equations side by side eliminates y and yields x = 6, as shown. 

If (x, y) is a solution to the system, then (x, y) satisfies both equations in the system and any equation derived from them. Therefore, x = 6.
Choices A, B, and C are incorrect and may be the result of errors when solving the system. 

The growth factor of a tree species is approximated by the slope of a line of best fit that models the relationship between diameter and age. A line of best fit can be visually estimated by identifying a line that goes in the same direction of the data and where roughly half the given data points fall above and half the given data points fall below the line. Two points that fall on the line can be used to estimate the slope and y-intercept of the equation of a line of best fit. Estimating a line of best fit for the given scatterplot could give the points (11, 80) and (15, 110). Using these two points, the slope of the equation of the line of best fit can be calculated as  The slope of the equation is interpreted as the growth factor for a species of tree. According to the table, the species of tree with a growth factor of 7.5 is shagbark hickory.
Choices A, B, and C are incorrect and likely result from errors made when estimating a line of best fit for the given scatterplot and its slope.

Choice C is correct. Using the distributive property to multiply 3 and (x + 5) gives 3x + 15 − 6, which can be rewritten as 3x + 9.
Choice A is incorrect and may result from rewriting the given expression as 3(x + 5 − 6). Choice B is incorrect and may result from incorrectly rewriting the expression as (3x + 5) − 6. Choice D is incorrect and may result from incorrectly rewriting the expression as 3(5x) − 6.
Alternatively, evaluating the given expression and each answer choice for the same value of x, for example x = 0, will reveal which of the expressions is equivalent to the given expression.

Choice C is correct. Ken earned $8 per hour for the first 10 hours he worked, so he earned a total of $80 for the first 10 hours he worked. For the rest of the week, Ken was paid at the rate of $10 per hour. Let x be the number of hours he will work for the rest of the week. The total of Ken’s earnings, in dollars, for the week will be 10x + 80. He saves 90% of his earnings each week, so this week he will save 0.9(10x + 80) dollars. The inequality 0.9(10x + 80) ≥ 270 represents the condition that he will save at least $270 for the week. Factoring 10 out of the expression 10x + 80 gives 10(x + 8). The product of 10 and 0.9 is 9, so the inequality can be rewritten as 9(x + 8) ≥ 270. Dividing both sides of this inequality by 9 yields x + 8 ≥ 30, so x ≥ 22. Therefore, the least number of hours Ken must work the rest of the week to save at least $270 for the week is 22.
Choices A and B are incorrect because Ken can save $270 by working fewer hours than 38 or 33 for the rest of the week. Choice D is incorrect. If Ken worked 16 hours for the rest of the week, his total earnings for the week will be $80 + $160 = $240, which is less than $270. Since he saves only 90% of his earnings each week, he would save even less than $240 for the week.

Choice D is correct. The numerator of the given expression can be rewritten in terms of the denominator, x – 3, as follows: x² − 2x − 5 = x² − 3x + x − 3 − 2, which is equivalent to x(x – 3) + (x – 3) – 2. So the given expression is equivalent 
to . Since the given expression is defined for x ≠ 3, the expression can be rewritten as 
Long division can also be used as an alternate approach.
Choices A, B, and C are incorrect and may result from errors made when dividing the two polynomials or making use of structure.

The correct answer is 4. There are π radians in a 180° angle. A 720° angle is 4 times greater than a 180° angle. Therefore, the number of radians in a 720° angle is 4π.

Choice C is correct. Multiplying each side of 1 meter = 100 cm by 6 gives 6 meters = 600 cm. Each package requires 3 centimeters of tape. The number of packages that can be secured with 600 cm of tape is 600/3, or 200 packages.
Choices A, B, and D are incorrect and may be the result of incorrect interpretations of the given information or of computation errors.

Choice C is correct. According to the line of best fit, a planetoid with a distance from the Sun of 1.2 AU has a density between 4.5 g/cm³ and 4.75 g/cm³. The only choice in this range is 4.6.
Choices A, B, and D are incorrect and may result from misreading the information in the scatterplot.

Choice B is correct. Theresa’s speed was increasing from 0 to 5 minutes and from 20 to 25 minutes, which is a total of 10 minutes. Theresa’s speed was decreasing from 10 minutes to 20 minutes and from 25 to 30 minutes, which is a total of 15 minutes. Therefore, Theresa’s speed was NOT increasing for a longer period of time than it was decreasing.
Choice A is incorrect. Theresa ran at a constant speed for the 5-minute period from 5 to 10 minutes. Choice C is incorrect. Theresa’s speed decreased at a constant rate during the last 5 minutes. Choice D is incorrect. Theresa’s speed reached its maximum at 25 minutes, which is within the last 10 minutes.

Choice B is correct. The charity with the greatest percent of total expenses spent on programs is represented by the highest point on the scatterplot; this is the point that has a vertical coordinate slightly less than halfway between 90 and 95 and a horizontal coordinate slightly less than halfway between 3,000 and 4,000. Thus, the charity represented by this point has a total income of about $3,400 million and spends about 92% of its total expenses on programs. The percent predicted by the line of best fit is the vertical coordinate of the point on the line of best fit with horizontal coordinate $3,400 million; this vertical coordinate is very slightly more than 85. Thus, the line of best fit predicts that the charity with the greatest percent of total expenses spent on programs will spend slightly more than 85% on programs. Therefore, the difference between the actual percent (92%) and the prediction (slightly more than 85%) is slightly less than 7%.
Choice A is incorrect. There is no charity represented in the scatterplot for which the difference between the actual percent of total expenses spent on programs and the percent predicted by the line of best fit is as much as 10%. Choices C and D are incorrect. These choices may result from misidentifying in the scatterplot the point that represents the charity with the greatest percent of total expenses spent on programs.