Mathematics Test - 1 - SAT MCQ

Starting with the equation:

8x + 6 = 6m

Divide both sides by 2:

4x + 3 = 3m

Thus, the value of 4x + 3 is ^box(D).

Let’s choose a value, like b = 2, for our positive constant. This gives us an expression of 2n + 4 for the nth term of the sequence. Substituting n = 1, n = 2, n = 3, etc. gives us a sequence of 6, 8, 10, 12, 14, and so on. Choice (A) is clearly incorrect, because the first term of this sequence is not 2. Choice (C) is also incorrect because the average of the first three terms is (6 + 8 + 10)/3 = 8, not 2. Choice (D) is also incorrect because the ratio of the second term to the first is 8/6 = 4/3. Only choice (B), the difference between the fourth term and the third term, 12 - 10, gives us a value of 2.

First, we should notice that each choice can be interpreted as a distance between two points on the number line.
(A) |s - v| = t he distance bet ween s and v
(B) |s - t | = t he distance bet ween s and t
(C) |s + v| = |s - (-v)| = the distance between s and -v
(D) |u + v| = |u - (-v)| = the distance between u and -v
Thinking this way gives us a very straightforward way to solve the problem without doing any calculation. First we need to locate -v on the number line by just reflecting v over the origin at 0. (Recall that multiplication by -1 is equivalent to reflecting a point on the number line over the origin at 0.)
This makes it easy to see the distances the problem is asking us to compare:

Clearly, the greatest of these distances is (A).

• Original equation: 
• Substitute b = 4: 
• Simplify: 
•  2/3α = 3
• Multiply by 3/2: α = 9/2 or 4.5

• Since the sum of 55 and 35 is 90, which is 10 greater than 80, there must be at least 10 in the overlap between the two sets.
• Statement (B) is not necessarily true, because it is possible that all 35 students taking AP courses are also varsity athletes, which is more than half of 55.
• Statement (C) is not true because 80 - 55 = 25 students do not play varsity sports, and 80 - 35 = 45 students do not take at least one AP course.
• Statement (D) is not necessarily true because 35 students take at least one AP course and 25 students do not play a varsity sport, and this sum, 35 + 25 = 60, is less than the total number of students, so it is possible that there is no overlap between these two sets.

f (x) = x² - 1
Substitute x = 1/b: 
Simplify: 
Get common denominator: 
Subtract fractions: 
Factor numerator: 

The solutions to the system correspond to the points of intersection of the two graphs. The figure shows four such intersection points.

To find the slope of line l, we can find two points on l and then use the slope formula.
f(x) = 2x² - 4x +1
Plug in -1 for x: f(-1) = 2(-1)² - 4(-1) + 1
Simplify:    f(-1) = 2(1) + 4 + 1 = 2 + 4 + 1 = 7
Therefore line l intersects the function at (-1, 7).
Plug in 2 for x: f(2) = 2(2)² - 4(2) + 1
Simplify: f(2) = 2(4) - (8) + 1 = 8 - 8 + 1 = 1
Therefore line l intersects the function at (2, 1). Now we find the slope of the line containing these two points.

Given equation: y = (x + 2)²(x - 3)²
To find the y-intercept, set x = 0: y = (0 + 2)²(0 - 3)²
Simplify: y = (2)²(-3)² = (4)(9) = 36
Therefore the y-intercept is at (0, 36).
To find the x-intercepts, set y = 0: 0 = (x + 2)²(x - 3)²
By the Zero Product Property, the only solutions to this equation are x = -2 and x = 3, so there are two x-intercepts and a total of three x- and y-intercepts.

Add 2/5: 

Simplify: 5/x = 7/5
Cross multiply: 25 = 7x
Divide by 7: x = 25/7 or 3.57

Remember the definitions of the basic trigonometric functions: SOH CAH TOA. Since the “side of interest” (k) is the OPPOSITE side to the given angle (35°), and since we know the length of the HYPOTENUSE (12), we should use SOH.
sin x = opp/hyp
Plug in the values: sin35° = k/12
Substitute sin 35° = 0.574: 0 574 = k/12
Multiply by 12: (12)(0.574) = 6.88 = k

The graph of y = a(x - 2)(x + 1) is a quadratic w it h zeros (x-intercepts) at x = 2 and x = -1. The a x is of symmetry of this parabola is halfway between the zeros, at x = (2 + -1)/2 = 1/2. Since a < 0, the parabola is “open down,” and so we have a general picture like this:

If you trace the curve from x = 0 to x = 5, that is, from the y-intercept and then to the right, you can see that the graph goes up a bit (until x = 1/2), and then goes down again.
Alternately, you can pick a negative value for a (like -2) and graph the equation on your calculator.

Choice D is correct. By definition, for any positive integers m and n. It follows, therefore, that
Choice A is incorrect. By definition,  for any positive integer n.
Applying this definition as well as the power property of exponents to the expression is not the correct answer. Choice B is incorrect. By definition, for any positive integer n.
Applying this definition as well as the power property of exponents to the expression is not the correct answer. Choice C is incorrect. By definition, for any positive integer n. Applying this definition as well as the power property of exponents to the expression is not the correct answer.

Choice C is correct. In the equation y = 0.56x + 27.2, the value of x increases by 1 for each year that passes. Each time x increases by 1, y increases by 0.56 since 0.56 is the slope of the graph of this equation. Since y represents the average number of students per classroom in the year represented by x, it follows that, according to the model, the estimated increase each year in the average number of students per classroom at Central High School is 0.56.
Choice A is incorrect because the total number of students in the school in 2000 is the product of the average number of students per classroom and the total number of classrooms, which would appropriately be approximated by the y-intercept (27.2) times the total number of classrooms, which is not given. Choice B is incorrect because the average number of students per classroom in 2000 is given by the y-intercept of the graph of the equation, but the question is asking for the meaning of the number 0.56, which is the slope. Choice D is incorrect because 0.56 represents the estimated yearly change in the average number of students per classroom. The estimated difference between the average number of students per classroom in 2010 and 2000 is 0.56 times the number of years that have passed between 2000 and 2010, that is, 0.56 × 10 = 5.6.

Choice D is correct. Starting with the original equation, h = −16t² + vt + k, in order to get v in terms of the other variables, −16t² and k need to be subtracted from each side. This yields vt = h + 16t² − k, which when divided by t will give v in terms of the other variables. However, the equation v = is not one of the options, so the right side needs to be further simplified. Another way to write the previous equation is  which can be simplified to v = 
Choices A, B, and C are incorrect and may be the result of arithmetic errors when rewriting the original equation to express v in terms of h, t, and k.

Choice C is correct. Since the angles are acute and sin(α°) = cos(b°), it follows from the complementary angle property of sines and cosines that α + 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 α + b and making a sign error. Choices B and D are incorrect because they result in values for a and b such that sin(α°) ≠ cos(b°).

If f is a function of x, then the graph of f in the xy-plane consists of all points (x, f(x)). An x-intercept is where the graph intersects the x-axis; since all points on the x-axis have y-coordinate 0, the graph of f will cross the x-axis at values of x such that f(x) = 0. Therefore, the graph of a function f will have no x-intercepts if and only if f has no real zeros. Likewise, the graph of a quadratic function with no real zeros will have no x-intercepts.
Choice A is incorrect. The graph of a linear function in the xy-plane whose rate of change is not zero is a line with a nonzero slope. The x-axis is a horizontal line and thus has slope 0, so the graph of the linear function whose rate of change is not zero is a line that is not parallel to the x-axis. Thus, the graph must intersect the x-axis at some point, and this point is an x-intercept of the graph. Choices B and D are incorrect because the graph of any function with a real zero must have an x-intercept.

Substituting x² for y in the second equation gives 2(x²) + 6 = 2(x + 3). This equation can be solved as follows:
2x² + 6 = 2x + 6 (Apply the distributive property.)
2x² + 6 − 2x − 6 = 0 (Subtract 2x and 6 from both sides of the equation.)
2x² − 2x = 0 (Combine like terms.)
2x(x − 1) = 0 (Factor both terms on the left side of the equation by 2x.)
Thus, x = 0 and x = 1 are the solutions to the system. Since x > 0, only x = 1 needs to be considered. The value of y when x = 1 is y = x² = 12 = 1. Therefore, the value of xy is (1)(1) = 1. Choices B, C, and D are incorrect and likely result from a computational or conceptual error when solving this system of equations. 

The graph of y = −f(x) is the graph of the equation y = −(2^x + 1), or y = −2^x − 1. This should be the graph of a decreasing exponential function. The y-intercept of the graph can be found by substituting the value x = 0 into the equation, as follows: y = −2⁰ − 1 = −1 − 1 = −2. Therefore, the graph should pass through the point (0, −2). Choice C is the only function that passes through this point.
Choices A and B are incorrect because the graphed functions are increasing instead of decreasing. Choice D is incorrect because the function passes through the point (0, −1) instead of (0, −2).

The value of c + d can be found by dividing both sides of the given equation by 3. This yields c + d = 5/3. 
Choice A is incorrect. If the value of c + d is 3/5, then however, 9/5 is not equal to 5. 
Choice C is incorrect. If the value of c + d is 3, then 3 × 3 = 5; however, 9 is not equal to 5.
Choice D is incorrect. If the value of c + d is 5, then 3 × 5 = 5; however, 15 is not equal to 5.

In 1908, the marathon was lengthened by 42 − 40 = 2 kilometers. Since 1 mile is approximately 1.6 kilometers, the increase of 2 kilometers can be converted to miles by multiplying as shown: 
Choices A, C, and D are incorrect and may result from errors made when applying the conversion rate or other computational errors.

According to the table, 74 orthopedic surgeons indicated that research is their major professional activity. Since a total of 607 surgeons completed the survey, it follows that the probability that the randomly selected surgeon is an orthopedic surgeon whose indicated major professional activity is research is 74 out of 607, or 74/607, which is ≈ 0.122.
Choices B, C, and D are incorrect and may be the result of finding the probability that the randomly selected surgeon is an orthopedic surgeon whose major professional activity is teaching (choice B), an orthopedic surgeon whose major professional activity is either teaching or research (choice C), or a general surgeon or orthopedic surgeon whose major professional activity is research (choice D).

Triangles ADB and CDB are congruent to each other because they are both 30°-60°-90° triangles and share the side  In triangle ADB, side  is opposite to the angle 30°; therefore, the length of  is half the length of hypotenuse  Since the triangles are congruent, AB = BC = 12. So the length of is 12/2 = 6.
Choice A is incorrect. If the length of  were 4, then the length of  would be 8. However, this is incorrect because is congruent to  which has a length of 12. Choices C and D are also incorrect. Following the same procedures as used to test choice A gives  a length of 12√2 for choice C and 12√3 for choice D. However, these results cannot be true because AB is congruent to  which has a length of 12.

Choice A is correct. Substituting –1 for x in the equation that defines f givesSimplifying the expressions in the numerator and denominator yieldswhich is equal to 10/-2 or –5.
Choices B, C, and D are incorrect and may result from misapplying the order of operations when substituting –1 for x.

Choice B is correct. In general, a binomial of the form x + f, where f is a constant, is a factor of a polynomial when the remainder of dividing the polynomial by x + f is 0. Let R be the remainder resulting from the division of the polynomial P(x) = αx³ + bx² + cx + d by x + 1. So the polynomial P(x) can be rewritten as P(x) = (x + 1)q(x) + R, where q(x) is a polynomial of second degree and R is a constant.
Since –1 is a root of the equation P(x) = 0, it follows that P(–1) = 0.
Since P(–1) = 0 and P(–1) = R, it follows that R = 0. This means that x + 1 is a factor of P(x).
Choices A, C, and D are incorrect because none of these choices can be a factor of the polynomial P(x) = αx³ + bx² + cx + d. For example, if x – 1 were a factor (choice A), then P(x) = (x –1)h(x), for some polynomial function h. It follows that P(1) = (1 – 1)h(1) = 0, so 1 would be another root of the given equation, and thus the given equation would have at least 4 roots. However, a third-degree equation cannot have more than three roots. Therefore, x – 1 cannot be a factor of P(x).

Choice D is correct. Factoring out the coefficient 1/3, the given expression can be rewritten as 1/3 (x² − 6). The expression x² – 6 can be approached as a difference of squares and rewritten as (x − √6)(x + √6). Therefore, k must be √6.
Choice A is incorrect. If k were 2, then the expression given would be rewritten as 1/3 (x − 2)(x + 2), which is equivalent to  − 2. Choice B is incorrect. This may result from incorrectly factoring the expression and finding (x – 6) (x + 6) as the factored form of the expression. Choice C is incorrect. This may result from incorrectly distributing the 1/3 and rewriting the expression as 1/3 (x² − 2).

The correct answer is 6632. Applying the distributive property to the expression yields 7532 + 100y² + 100y² − 1100. Then adding together 7532 + 100y² and 100y² − 1100 and collecting like terms results in 200y² + 6432. This is written in the form αy² + b, where α = 200 and b = 6432. Therefore α + b = 200 + 6432 = 6632.

Choice C is correct. Substituting (1, 1) into the inequality gives 5(1) − 3(1) < 4, or 2 < 4, which is a true statement. Substituting (2, 5) into the inequality gives 5(2) − 3(5) < 4, or −5 < 4, which is a true statement. Substituting (3, 2) into the inequality gives 5(3) − 3(2) < 4, or 9 < 4, which is not a true statement. Therefore, (1, 1) and (2, 5) are the only ordered pairs that satisfy the given inequality.
Choice A is incorrect because the ordered pair (2, 5) also satisfies the inequality. Choice B is incorrect because the ordered pair (1, 1) also satisfies the inequality. Choice D is incorrect because the ordered pair (3, 2) does not satisfy the inequality.

Choice D is correct. There are 60 minutes in one hour, so an 8-hour workday has (60)(8) = 480 minutes. To calculate 15% of 480, multiply 0.15 by 480: (0.15)(480) = 72. Therefore, Lani spent 72 minutes of her workday in meetings.
Choice A is incorrect because 1.2 is 15% of 8, which gives the time Lani spent of her workday in meetings in hours, not minutes. Choices B and C are incorrect and may be the result of computation errors.

Choice C is correct. If Mosteller’s and Current’s formulas give the same estimate for A, then the right-hand sides of these two equations are equal; that is,

Multiplying each side of this equation by 60 to isolate the expressiongives= or = 2(4 + w).
Therefore, if Mosteller’s and Current’s formulas give the same estimate for A, then is equivalent to 2(4 + w).
An alternate approach is to multiply the numerator and denominator of Current’s formula by 2, which gives
Since it is given that Mosteller’s and Current’s formulas give the same estimate for A, 
Therefore, = 2(4 + w).
Choices A, B, and D are incorrect and may result from errors in the algebraic manipulation of the equations.