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Question 1 |
Jim has 10 apples and he gives 7 to Bill. Now Bill has 3 times as many apples as Jim originally had. How many apples did Bill have to begin with?
20 | |
17 | |
23 | |
2 |
Question 1 Explanation:
The correct answer is (C). Rephrase the question into an algebraic expression: 3 times Jim’s original amount is:
3 × 10 = 30
This came after Jim gave 7 to Bill, so reduce this number by 7 to find Bill’s original amount:
30 − 7 = 23
3 × 10 = 30
This came after Jim gave 7 to Bill, so reduce this number by 7 to find Bill’s original amount:
30 − 7 = 23
Question 2 |
If $f(x) = 3x\,$ and $\,g(x) = 2x^²$, what is $\,g(f(−2))$?
24 | |
−72 | |
48 | |
72 |
Question 2 Explanation:
The correct answer is (D). First evaluate $f(−2)$:
$3(−2)$
$= −6$
Next evaluate $g(−6):$
$2(−6)^2$
$=2(36)$
$= 72$
$3(−2)$
$= −6$
Next evaluate $g(−6):$
$2(−6)^2$
$=2(36)$
$= 72$
Question 3 |
What is the value of: $i^2*i^{-3}$
1 | |
$−i$ | |
$\dfrac{1}{i}$ | |
−1 |
Question 3 Explanation:
The correct answer is (C). Simplify the expression using the rules of exponents:
$i^2*i^{−3}$
$=i^{2 − 3}$
$=i^{−1}$
$=\dfrac{1}{i}$
$i^2*i^{−3}$
$=i^{2 − 3}$
$=i^{−1}$
$=\dfrac{1}{i}$
Question 4 |
Which of the following is the same as $\log 85$?
$\log 35 + \log 50$ | |
$\ln \frac{1}{85}$ | |
$\log 255 − \log 3$ | |
$\ln 85$ |
Question 4 Explanation:
The correct answer is (C). Recall the rules for condensing and expanding logarithms:
$\log a + \log b = \log (a*b)$
$\log a − \log b = \log (\frac{a}{b})$
Answer choice (C) can be rewritten as: $\log (\frac{225}{3})$ which is equal to $\log 85$.
$\log a + \log b = \log (a*b)$
$\log a − \log b = \log (\frac{a}{b})$
Answer choice (C) can be rewritten as: $\log (\frac{225}{3})$ which is equal to $\log 85$.
Question 5 |
A man went to the store and bought pencils, pens, and erasers in the ratio of 2:4:3.
If he bought 6 erasers then how many items total did he buy?
9 | |
18 | |
12 | |
10 |
Question 5 Explanation:
The correct answer is (B). The ratio can be expressed as:
2x: 4x: 3x where 3x = 6 represents the number of erasers bought.
Solving for x: x = 2, and plugging this value into the ratio:
2(2): 4(2): 3(2) → 4: 8: 6
Find the total number of items purchased by combining these numbers:
4 + 8 + 6 = 18
2x: 4x: 3x where 3x = 6 represents the number of erasers bought.
Solving for x: x = 2, and plugging this value into the ratio:
2(2): 4(2): 3(2) → 4: 8: 6
Find the total number of items purchased by combining these numbers:
4 + 8 + 6 = 18
Question 6 |
How many solutions does the following equation have?
$\sqrt{2+x}-x=0$
3 | |
2 | |
1 | |
0 |
Question 6 Explanation:
The correct answer is (C). A solution to an equation is a value that when plugged in yields a true statement. Begin by isolating the square root and then squaring both sides to eliminate it:
$\sqrt{2+x}=x $ $ \to (\sqrt{2 + x})^2 = (x)^2 $ $ \to 2+x =x^2$
Move all of the terms to one side and factor to solve for $x$:
$x^2 − x − 2 = 0$
$→ (x + 1) (x − 2) = 0$
$→ x = −1, x = 2$
Check for extraneous solutions by plugging both solutions into the original equation:
Test $x = -1$:
$\sqrt{2+(-1)} = -1$?
$\sqrt{1} = -1$?
$ \to 1 \neq -1$, so $x=-1$ is extraneous.
Test $x = 2$:
$\sqrt{2+(2)} = 2$?
$\sqrt{4} = 2$?
$ \to 2 = 2$, so $x=2$ is a solution.
The equation has only 1 solution.
$\sqrt{2+x}=x $ $ \to (\sqrt{2 + x})^2 = (x)^2 $ $ \to 2+x =x^2$
Move all of the terms to one side and factor to solve for $x$:
$x^2 − x − 2 = 0$
$→ (x + 1) (x − 2) = 0$
$→ x = −1, x = 2$
Check for extraneous solutions by plugging both solutions into the original equation:
Test $x = -1$:
$\sqrt{2+(-1)} = -1$?
$\sqrt{1} = -1$?
$ \to 1 \neq -1$, so $x=-1$ is extraneous.
Test $x = 2$:
$\sqrt{2+(2)} = 2$?
$\sqrt{4} = 2$?
$ \to 2 = 2$, so $x=2$ is a solution.
The equation has only 1 solution.
Question 7 |
The following are the results of a class test:
65, 75, 87, 86, 78, 92, 77, 81, 82, 86, 21, 91, 3
Which of the following is the best to use to measure the central tendency of scores on the exam?
Mean | |
Median | |
Mode | |
Standard Deviation |
Question 7 Explanation:
The correct answer is (B). Since there are a few extreme outliers: 3 and 21, which differ from the norm, it is better to use the median, which is unaffected by outliers, rather than the mean, which is affected by outliers.
Question 8 |
What is the x-intercept of $y = e^{5x}+ 2$?
2 | |
0 | |
ln 2 | |
no x-intercept |
Question 8 Explanation:
The correct answer is (D). The x-intercept is where the function crosses the x-axis and can be found by setting the function equal to 0 and solving for $x$:
$0 = e^{5x} + 2$
$ \to−2 = e^{5x}$
To isolate the variable, take the natural log of both sides:
$\text{ln}(-2) = \text{ln}(e^{5x})$
$\text{ln}(-2) = 5x$
Because $\ln(−2)$ is undefined, this function does not cross the x-axis.
$0 = e^{5x} + 2$
$ \to−2 = e^{5x}$
To isolate the variable, take the natural log of both sides:
$\text{ln}(-2) = \text{ln}(e^{5x})$
$\text{ln}(-2) = 5x$
Because $\ln(−2)$ is undefined, this function does not cross the x-axis.
Question 9 |
What is 0.25% of $\frac{1}{3}$?
0.83 | |
0.083 | |
0.0083 | |
0.00083 |
Question 9 Explanation:
The correct answer is (D). Convert 0.25% to a fraction and multiply it by $\frac{1}{3}$.
Convert 0.25% to a fraction:
$0.25\% = \dfrac{0.25}{100} $ $ = \dfrac{25}{10000} $ $ = \dfrac{1}{400}$
Multiply 0.25% and $\frac{1}{3}$:
$=\dfrac{1}{400}*\dfrac{1}{3}$
$=\dfrac{1}{1200}$
$=0.00083$
Convert 0.25% to a fraction:
$0.25\% = \dfrac{0.25}{100} $ $ = \dfrac{25}{10000} $ $ = \dfrac{1}{400}$
Multiply 0.25% and $\frac{1}{3}$:
$=\dfrac{1}{400}*\dfrac{1}{3}$
$=\dfrac{1}{1200}$
$=0.00083$
Question 10 |
What are the critical points of $y = 4x^2 + 3x$?
0 | |
$\frac{-3}{8}$ | |
0 and $\frac{-3}{8}$ | |
0 and $\frac{3}{8}$ |
Question 10 Explanation:
The correct answer is (B). A critical point of a function is a point where the derivative of the function is equal to 0 or is undefined. Evaluate the derivative with respect to $x$, set it equal to 0, and solve for $x$:
$0 = 8x + 3$
$\to x = \dfrac{-3}{8}$
$0 = 8x + 3$
$\to x = \dfrac{-3}{8}$
Question 11 |
$\ln (e^7) = 1 - x$, What is $x$?
$6$ | |
$\ln 7 − e$ | |
$−6$ | |
$−\ln e − 7$ |
Question 11 Explanation:
The correct answer is (C). Recall that ln $e^7$ can be rewritten as $7 * \ln e$ and that $\ln e = 1$. Rewriting the original expression and solving for $x$:
$7 * 1 = 1 − x$
$ \to x = −6$
$7 * 1 = 1 − x$
$ \to x = −6$
Question 12 |
If a recipe calls for 3 parts sugar and 2 parts milk, and a total of 45 oz is made, how much milk is there?
18 oz | |
9 oz | |
27 oz | |
30 oz |
Question 12 Explanation:
The correct answer is (A). The ratio of sugar to milk is:
$3x:2x$
Combining these and setting the amount equal to the given total:
$5x = 45$
$→ x = 9$
Plugging this back in to find the total amount of milk:
$2 \ast 9 = 18 \text{ oz}$
$3x:2x$
Combining these and setting the amount equal to the given total:
$5x = 45$
$→ x = 9$
Plugging this back in to find the total amount of milk:
$2 \ast 9 = 18 \text{ oz}$
Question 13 |
$\int^{2π}_π \sin(x)dx$
$2 \int^{\frac{π}{2}}_0 \cos(x)dx$ | |
$\int^{2π}_π \cos(x)dx$ | |
$-2 \int^{\frac{π}{2}}_0 \cos(x)dx$ | |
$\int^{π}_0 \cos(x)dx$ |
Question 13 Explanation:
The correct answer is (C). Recall that the anti-derivative of $\sin(x) = −\cos(x)$, because the derivative of $\cos(x) = −\sin(x)$.
$\int^{2π}_π \sin(x)dx = [-\cos(x)]^{2π}_π$
$= -\cos(2π) - (-\cos(π))$
$= -\cos(2π) + \cos(π)$
$= -(1) + (-1) = -2$
Calculate the definite integral for each answer choice to find which evaluates to $-2$:
Choice A:
$2 \int^{\frac{π}{2}}_0 \cos(x)dx = 2[\sin(x)]^{\frac{π}{2}}_0$
$= 2(\sin(\frac{π}{2}) - \sin(0)) $ $ = 2(1 - 0) = 2$
Choice B:
$\int^{2π}_π \cos(x)dx = [\sin(x)]^{2π}_π$
$= \sin(2π) - \sin(π) $ $ = 0 - 0 = 0$
Choice C:
$-2 \int^{\frac{π}{2}}_0 \cos(x)dx $ $ = -2[\sin(x)]^{\frac{π}{2}}_0$
$= -2(\sin(\frac{π}{2}) - \sin(0)) $ $ = -2(1 - 0) = -2$
Choice D:
$\int^{π}_0 \cos(x)dx = [\sin(x)]^{π}_0$
$= \sin(π) - \sin(0) $ $ = 0 - 0 = 0$
Answer choice (C) has the same value and is the correct answer.
$\int^{2π}_π \sin(x)dx = [-\cos(x)]^{2π}_π$
$= -\cos(2π) - (-\cos(π))$
$= -\cos(2π) + \cos(π)$
$= -(1) + (-1) = -2$
Calculate the definite integral for each answer choice to find which evaluates to $-2$:
Choice A:
$2 \int^{\frac{π}{2}}_0 \cos(x)dx = 2[\sin(x)]^{\frac{π}{2}}_0$
$= 2(\sin(\frac{π}{2}) - \sin(0)) $ $ = 2(1 - 0) = 2$
Choice B:
$\int^{2π}_π \cos(x)dx = [\sin(x)]^{2π}_π$
$= \sin(2π) - \sin(π) $ $ = 0 - 0 = 0$
Choice C:
$-2 \int^{\frac{π}{2}}_0 \cos(x)dx $ $ = -2[\sin(x)]^{\frac{π}{2}}_0$
$= -2(\sin(\frac{π}{2}) - \sin(0)) $ $ = -2(1 - 0) = -2$
Choice D:
$\int^{π}_0 \cos(x)dx = [\sin(x)]^{π}_0$
$= \sin(π) - \sin(0) $ $ = 0 - 0 = 0$
Answer choice (C) has the same value and is the correct answer.
Question 14 |
$i^3 \div i^{\frac{3}{2}} = \; ?$
$−i^{\frac{3}{2}}$ | |
$−i$ | |
$\sqrt i$ | |
$\sqrt {−i}$ |
Question 14 Explanation:
The correct answer is (D). Simplify the expression:
$i^3 \div i^{\frac{3}{2}} = i^{3 - \frac{3}{2}} = i^{\frac{6}{2} − \frac{3}{2}} = i^{\frac{3}{2}}$
$i^{\frac{3}{2}} = \sqrt {i^3}$
We can simplify $i^3$ as follows, recalling that $i = \sqrt {-1}$:
$i^3 = i^2 \cdot i = (\sqrt{-1})^2(i) $ $ = (-1)(i) = -i$
$i^3 = -i$
Substituting into our original equation:
$i^{\frac{3}{2}} = \sqrt{i^{-3}} = \sqrt{-i}$
You should consider committing the following to memory (or know how to derive):
$i = \sqrt{-1}$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$
$i^3 \div i^{\frac{3}{2}} = i^{3 - \frac{3}{2}} = i^{\frac{6}{2} − \frac{3}{2}} = i^{\frac{3}{2}}$
$i^{\frac{3}{2}} = \sqrt {i^3}$
We can simplify $i^3$ as follows, recalling that $i = \sqrt {-1}$:
$i^3 = i^2 \cdot i = (\sqrt{-1})^2(i) $ $ = (-1)(i) = -i$
$i^3 = -i$
Substituting into our original equation:
$i^{\frac{3}{2}} = \sqrt{i^{-3}} = \sqrt{-i}$
You should consider committing the following to memory (or know how to derive):
$i = \sqrt{-1}$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$
Question 15 |
John is running towards a finish line at 10 m/s and is 50m away. Bolt is running at 14 m/s but is 75m away. Assuming they keep their constant speed, who will reach the finish line first?
John | |
Bolt | |
It is a tie | |
Neither will finish |
Question 15 Explanation:
The correct answer is (A). Calculate how long it will take each runner to finish. The shorter time length indicates the first to finish. Recall that:
$\text{rate} = \dfrac{\text{distance}}{\text{time}}$
$\text{time} = \dfrac{\text{distance}}{\text{rate}}$
$ \dfrac{50 \text{ m}}{10 \text{ m/s}} = $ $ 5 \text{ seconds for John}$
$ \dfrac{75 \text{ m}}{14 \text{ m/s}} \approx $ $ 5.4 \text{ seconds for Bolt}$
John finishes before Bolt.
$\text{rate} = \dfrac{\text{distance}}{\text{time}}$
$\text{time} = \dfrac{\text{distance}}{\text{rate}}$
$ \dfrac{50 \text{ m}}{10 \text{ m/s}} = $ $ 5 \text{ seconds for John}$
$ \dfrac{75 \text{ m}}{14 \text{ m/s}} \approx $ $ 5.4 \text{ seconds for Bolt}$
John finishes before Bolt.
Question 16 |
What is the third derivative of $−3\sin(x)$?
$−3\cos (x)$ | |
$3\cos (x)$ | |
$−3\sin (x)$ | |
$3 \sin (x)$ |
Question 16 Explanation:
The correct answer is (B). Recall that the derivative of $\sin(x) = \cos(x)$ and the derivative of $\cos(x) = −\sin(x)$.
First derivative: $−3 * \cos(x)$
Second derivative: $−3 * (−\sin(x)) = 3\sin(x)$
Third derivative: $3 * \cos(x)$
First derivative: $−3 * \cos(x)$
Second derivative: $−3 * (−\sin(x)) = 3\sin(x)$
Third derivative: $3 * \cos(x)$
Question 17 |
After the vector $\lt a,3a \gt$ has been multiplied by 4, which statement is false?
The y-coordinate is 3 times the x-coordinate. | |
The difference between the y-coordinate and the x-coordinate is 8$a$. | |
The difference between the resulting y-coordinate and the original y-coordinate is 8$a$. | |
The difference between the resulting y-coordinate and the original y-coordinate is greater than that of the resulting x-coordinate and the original x-coordinate. |
Question 17 Explanation:
The correct answer is (C). Apply the scalar multiplication to the vector and evaluate the answer choices:
$4 * \lt a,3a \gt =$ $\lt 4a,12a \gt$
$4a \times 3 = 12a$, so (A) is true.
$12a - 4a = 8a$, so (B) is true.
$12a - 3a = 9a$, so (C) is false and is the correct answer.
$(12a - 3a) \gt (4a - a)$, so (D) is true.
$4 * \lt a,3a \gt =$ $\lt 4a,12a \gt$
$4a \times 3 = 12a$, so (A) is true.
$12a - 4a = 8a$, so (B) is true.
$12a - 3a = 9a$, so (C) is false and is the correct answer.
$(12a - 3a) \gt (4a - a)$, so (D) is true.
Question 18 |
What is the $\lim_{x\to \infty} \dfrac {1 − x^3}{x + 2}$?
$− \infty$ | |
$0$ | |
$\dfrac{1}{2}$ | |
$ \infty$ |
Question 18 Explanation:
The correct answer is (A). Begin by factoring out the largest power of $x$ in the denominator from both the numerator and denominator:
$\lim_{x\to - \infty} \dfrac {x( \frac{1}{x}-x^2)} {x(1+ \frac{2}{x})}$
Cancel out the $x$ and evaluate the limits:
$\lim_{x\to - \infty} \dfrac { \frac{1}{x}-x^2} {1+ \frac{2}{x}} $ $ \to \dfrac { 0-(- \infty)^2} {1+0} \to -\infty $
Recall that $\lim_{x\to - \infty} \dfrac {constant}{x}$ is $0$
$\lim_{x\to - \infty} \dfrac {x( \frac{1}{x}-x^2)} {x(1+ \frac{2}{x})}$
Cancel out the $x$ and evaluate the limits:
$\lim_{x\to - \infty} \dfrac { \frac{1}{x}-x^2} {1+ \frac{2}{x}} $ $ \to \dfrac { 0-(- \infty)^2} {1+0} \to -\infty $
Recall that $\lim_{x\to - \infty} \dfrac {constant}{x}$ is $0$
Question 19 |
If $f(x) = 2x^2 - x^3$, what is $f'(3)$?
15 | |
−9 | |
−15 | |
18 |
Question 19 Explanation:
The correct answer is (C). Recall the power rule:
If $f(x) = x^n, f'(x) = nx^{n-1}$
Here, $f'(x) = 2 * 2x^1 - 3x^2 $ $ = 4x - 3x^2$
$f'(3) = 4(3) − 3(3)^2$
$= 12 − 27$
$= −15$
If $f(x) = x^n, f'(x) = nx^{n-1}$
Here, $f'(x) = 2 * 2x^1 - 3x^2 $ $ = 4x - 3x^2$
$f'(3) = 4(3) − 3(3)^2$
$= 12 − 27$
$= −15$
Question 20 |
What is the probability of rolling a prime number in each roll in 4 consecutive rolls of a 6 sided die?
$ \dfrac{1}{4}$ | |
$ \dfrac{1}{8}$ | |
$ \dfrac{1}{16}$ | |
$ \dfrac{1}{32}$ |
Question 20 Explanation:
The correct answer is (C). The prime numbers in the range are: 2, 3, 5. On any roll, the probability of rolling a prime number is:
$ \dfrac{3 \text{ primes}}{6 \text{ total}}= \dfrac{1}{2}$
The probability of rolling this 4 times in a row is:
$ \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2} = \dfrac{1}{16}$
$ \dfrac{3 \text{ primes}}{6 \text{ total}}= \dfrac{1}{2}$
The probability of rolling this 4 times in a row is:
$ \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2} = \dfrac{1}{16}$
Question 21 |
$p = −3x + 5y$
$q = x + 2y$
What is the value of $p − 2q$?
$5x + y$ | |
$-5x + y$ | |
$-y + 5x$ | |
$5 + y$ |
Question 21 Explanation:
The correct answer is (B). Substitute the values of $p$ and $q$ into the expression:
$(−3x + 5y) − 2(x + 2y)$
Distribute and combine like terms to simplify:
$= −3x + 5y − 2x − 4y$
$= −5x + y$
$(−3x + 5y) − 2(x + 2y)$
Distribute and combine like terms to simplify:
$= −3x + 5y − 2x − 4y$
$= −5x + y$
Question 22 |
A lab requires 0.76 L of a solution. How many 100 mL vials of the solution should the lab acquire?
7 | |
8 | |
6 | |
10 |
Question 22 Explanation:
The correct answer is (B). Recall that 100 mL = 0.1 L. Because the smaller vials only come in amounts of 100 mL, 8 vials will be needed to provide 0.76 L (760 mL).
Question 23 |
A lab has 17 black rats and 11 white rats. One black rat and one white rat are needed for today’s experiment. The lab technician plans to select the first rat at random. He will then select the second rat at random, but if it is the same color as the first rat, he will continue the process of selecting a second rat until it is a different color than the first. What is the probability that the second rat will be white?
$\dfrac{11}{28}$ | |
$\dfrac{11}{17}$ | |
$\dfrac{17}{11}$ | |
$\dfrac{17}{28}$ |
Question 23 Explanation:
The correct answer is (D). In order to satisfy the condition of choosing a white rat second, the first rat chosen must be black. The probability of selecting a black rat is:
$\dfrac{17 \text{ black rats} }{28 \text{ total rats}} = \dfrac{17}{28}$
$\dfrac{17 \text{ black rats} }{28 \text{ total rats}} = \dfrac{17}{28}$
Question 24 |
At what time will an object be at rest if its position function is given by: $x(t) = t^3 − 9t^2 + 24t$
$t = 0$ | |
$t = 3$ | |
$t = 4$ | |
$t = 8$ |
Question 24 Explanation:
The correct answer is (C). Recall that the derivative of the position function is the velocity function and that an object is at rest when its velocity is equal to 0. Evaluate the derivative, set it equal to 0 and solve for $t$:
$v(t) = x'(t) $ $ = 3t^2− 18t + 24$
$\to 0 = 3(t^2 - 6t + 8)$
$ \to 0 = t^2 − 6t + 8$
Factor the expression to solve for $t$:
$(t − 4) (t − 2) = 0$
$ \to t = 2, t = 4$
Only $t = 4$ is an answer choice.
$v(t) = x'(t) $ $ = 3t^2− 18t + 24$
$\to 0 = 3(t^2 - 6t + 8)$
$ \to 0 = t^2 − 6t + 8$
Factor the expression to solve for $t$:
$(t − 4) (t − 2) = 0$
$ \to t = 2, t = 4$
Only $t = 4$ is an answer choice.
Question 25 |
Evaluate: $\int \sin^2 (x) \cos (x) dx$
$\dfrac{\sin^3(x)}{3} + C$ | |
$\sin (x) \cos (x) + 3 + C$ | |
$\dfrac{\sin^2(x)}{3} + C$ | |
$3 \cos^2 (x) + C$ |
Question 25 Explanation:
The correct answer is (A). Evaluate the integral using a u-substitution:
$u = \sin (x)$, $du = \cos (x) dx$
The integral becomes:
$\int u^2 du$
Evaluate:
$\dfrac{u^3}{3} + C$, and replacing $u$ with $\sin (x)$,
$\dfrac {\sin^3 (x)}{3} + C$
$u = \sin (x)$, $du = \cos (x) dx$
The integral becomes:
$\int u^2 du$
Evaluate:
$\dfrac{u^3}{3} + C$, and replacing $u$ with $\sin (x)$,
$\dfrac {\sin^3 (x)}{3} + C$
Question 26 |
The mean of 17 exam scores is 26. After one exam is dropped, the mean drops to 25.625. What is the value of the exam that was dropped?
33 | |
32 | |
30 | |
35 |
Question 26 Explanation:
The correct answer is (B). Recall that the formula for mean is:
$\dfrac{\text{sum} \ \text{of} \ \text{data}}{\text{number} \ \text{of} \ \text{data}}= \text{mean}$
Rewriting this formula:
$\text{sum} = $ $ \text{mean} \ast \text{number of data}$
Find the difference between the sums of the 2 sets:
$17 * 26 = 442$
$16 * 25.625 = 410$
$→ 442 − 410 = 32$
$\dfrac{\text{sum} \ \text{of} \ \text{data}}{\text{number} \ \text{of} \ \text{data}}= \text{mean}$
Rewriting this formula:
$\text{sum} = $ $ \text{mean} \ast \text{number of data}$
Find the difference between the sums of the 2 sets:
$17 * 26 = 442$
$16 * 25.625 = 410$
$→ 442 − 410 = 32$
Question 27 |
What is: $\dfrac {\frac{1}{4}+4}{4}$
$\dfrac{17}{16}$ | |
$\dfrac{17}{8}$ | |
$\dfrac{16}{17}$ | |
$1$ |
Question 27 Explanation:
The correct answer is (A). Begin by simplifying the expression in the numerator:
$\dfrac{1}{4} + 4 = \dfrac{1}{4} + \dfrac{16}{4} = \dfrac{17}{4}$
Dividing the numerator by 4:
$\dfrac{\frac{17}{4}}{4} = \dfrac{17}{4} * \dfrac{1}{4} = \dfrac{17}{16}$
$\dfrac{1}{4} + 4 = \dfrac{1}{4} + \dfrac{16}{4} = \dfrac{17}{4}$
Dividing the numerator by 4:
$\dfrac{\frac{17}{4}}{4} = \dfrac{17}{4} * \dfrac{1}{4} = \dfrac{17}{16}$
Question 28 |
Which set contains the solutions for $6x^2 = −5x − 1?$
$\{ \frac{1}{2},1,3\}$ | |
$\{ -\frac{1}{2},-\frac{1}{3}\}$ | |
$\{ \frac{1}{2}, 0 \}$ | |
$\{ -\frac{1}{3}, 0 \}$ |
Question 28 Explanation:
The correct answer is (B). Rearrange the equation and solve for x by factoring:
$6x^2 + 5x + 1 = 0$
$ \to (3x + 1) (2x+1) = 0$
$ \to x = -\frac{1}{3}, x = - \frac{1}{2}$
$6x^2 + 5x + 1 = 0$
$ \to (3x + 1) (2x+1) = 0$
$ \to x = -\frac{1}{3}, x = - \frac{1}{2}$
Question 29 |
What is the radius of a wheel that covers 100 m in 50 rotations?
$π \text{ m}$ | |
$\dfrac{1}{π}\text{ m}$ | |
$2 \text{ m}$ | |
$\dfrac{2}{π} \text{ m}$ |
Question 29 Explanation:
The correct answer is (B). Divide the total distance covered by the number of rotations to find the distance covered in 1 rotation:
$\dfrac{100 \text{ m}}{50 \text{ rotations}} $ $ = 2 \text{ m per rotation}$
This indicates that the circumference of the wheel is 2 m, so:
$C = 2 π r$
$ \to 2 = 2 πr$
$ \to r = \dfrac{1}{π} \text{ m}$
$\dfrac{100 \text{ m}}{50 \text{ rotations}} $ $ = 2 \text{ m per rotation}$
This indicates that the circumference of the wheel is 2 m, so:
$C = 2 π r$
$ \to 2 = 2 πr$
$ \to r = \dfrac{1}{π} \text{ m}$
Question 30 |
What is the $\lim_{x\to -3^-}\dfrac{17}{x + 3}?$
(Limit as $x$ approaches −3 from the left.)
$\infty $ | |
$−3$ | |
$0$ | |
$-\infty $ |
Question 30 Explanation:
The correct answer is (D). Notice that the function is undefined at $x$=−3, because of a division by 0.
Consider the behavior of the function when gradually nearing $x = -3$ from the left ($x$ = −3.01, $x$ = −3.001, $x$ = −3.0001). The denominator becomes progressively smaller and has a negative sign:
$\dfrac{17}{-3.001 + 3} = \dfrac{17}{-.001} $ $ = -17{,}000$
$\dfrac{17}{-3.0001 + 3} = \dfrac{17}{-.0001} $ $ = -170{,}000$
As this continues, the function approaches infinity but remains negative. Consequently, the limit approaches negative infinity.
$\dfrac{17}{-3.001 + 3} = \dfrac{17}{-.001} $ $ = -17{,}000$
$\dfrac{17}{-3.0001 + 3} = \dfrac{17}{-.0001} $ $ = -170{,}000$
As this continues, the function approaches infinity but remains negative. Consequently, the limit approaches negative infinity.
Question 31 |
Simplify $\left(\cot(x)\sin(x) \right)^2$
$1 - \cos x$ | |
$1 - \sin x$ | |
$1 - \cos^2 x$ | |
$1 - \sin^2 x$ |
Question 31 Explanation:
The correct answer is (D).
Recall that $\cot(x) = \dfrac{\cos(x)}{\sin(x)}$, so the expression becomes:
$\left(\dfrac{\cos(x)}{\sin(x)} \times \sin(x)\right)^2$
$= (\cos(x))^2 = \cos^2(x)$
$= 1 - \sin^2(x)$
By way of the trigonometric identity:
$\cos^2x + \sin^2x=1$
Recall that $\cot(x) = \dfrac{\cos(x)}{\sin(x)}$, so the expression becomes:
$\left(\dfrac{\cos(x)}{\sin(x)} \times \sin(x)\right)^2$
$= (\cos(x))^2 = \cos^2(x)$
$= 1 - \sin^2(x)$
By way of the trigonometric identity:
$\cos^2x + \sin^2x=1$
Question 32 |
If Tim is riding a Ferris wheel which is traveling at 3 rpm, how long will it take him to reach maximum height from the ground?
20 seconds | |
10 seconds | |
5 seconds | |
Unsolvable without radius |
Question 32 Explanation:
The correct answer is (B). If it travels at 3 rpm, then the wheel makes a complete rotation every 20 seconds:
$\dfrac{3 \text{ rotations}}{1 \text{ minute}} * \dfrac{1 \text{ minute}}{60 \text{ seconds}} $ $ = \dfrac{1 \text{ rotation}}{20 \text{ seconds}}$
If it takes 20 seconds for the bottom of the wheel to fully rotate and reach the bottom again, it will take half this time, or 10 seconds, to rotate from the minimum height to the maximum height.
$\dfrac{3 \text{ rotations}}{1 \text{ minute}} * \dfrac{1 \text{ minute}}{60 \text{ seconds}} $ $ = \dfrac{1 \text{ rotation}}{20 \text{ seconds}}$
If it takes 20 seconds for the bottom of the wheel to fully rotate and reach the bottom again, it will take half this time, or 10 seconds, to rotate from the minimum height to the maximum height.
Question 33 |
$f(x) = -\cos x$
$g(x) = 1 - x^2$
What is $\{g [f'(x) ] \}^{\frac{1}{2}} ?$
$ \cos (x)$ | |
$\sin (x)$ | |
$1- \sin^2 (x)$ | |
$\tan (x)$ |
Question 33 Explanation:
The correct answer is (A). Begin by evaluating the derivative of $f(x):f'(x)=\sin(x)$.
Evaluate $g(\sin(x)):1 - (\sin(x))^2 $ $ = 1 - \sin^2(x)$.
Use the trig identity $\sin^2(x) + \cos^2(x) = 1$, to show $1 - \sin^2(x) = \cos^2(x)$
$g [f'(x) ] = \cos^2(x)$
$\{g[f'(x)]\}^{\frac{1}{2}} = \sqrt{g[f'(x)]}$
$= \sqrt{\cos^2(x)} = \cos(x)$
Evaluate $g(\sin(x)):1 - (\sin(x))^2 $ $ = 1 - \sin^2(x)$.
Use the trig identity $\sin^2(x) + \cos^2(x) = 1$, to show $1 - \sin^2(x) = \cos^2(x)$
$g [f'(x) ] = \cos^2(x)$
$\{g[f'(x)]\}^{\frac{1}{2}} = \sqrt{g[f'(x)]}$
$= \sqrt{\cos^2(x)} = \cos(x)$
Question 34 |
What is $i^{100}?$
$i$ | |
$-i$ | |
$1$ | |
$-1$ |
Question 34 Explanation:
The correct answer is (C). Recall that $i=\sqrt{-1}$, so $i^2=-1$.
Rewrite the expression in terms of $i^2$:
$i^{100}=(i^2)^{50} \to (-1)^{50}=1$
(Because 50 is an even power.)
Rewrite the expression in terms of $i^2$:
$i^{100}=(i^2)^{50} \to (-1)^{50}=1$
(Because 50 is an even power.)
Question 35 |
What is the sum of the numbers from 1 to 100?
5050 | |
5049
| |
5051
| |
4950 |
Question 35 Explanation:
The correct answer is (A). Recall that the sum of an arithmetic series is:
$s= \dfrac{n(a_1+a_n)}{2}$
Substituting values into the formula:
$s=\dfrac{100*(1+100)}{2} $ $ =50*101=5050$
$s= \dfrac{n(a_1+a_n)}{2}$
Substituting values into the formula:
$s=\dfrac{100*(1+100)}{2} $ $ =50*101=5050$
Question 36 |
What is the arc length of a circle which has a radius of 2 and an angle of 30 degrees?
$π$ | |
$\dfrac{π}{3}$ | |
$\dfrac{π}{4}$ | |
$\dfrac{π}{2}$ |
Question 36 Explanation:
The correct answer is (B). The arc length, $s$, is equal to the product of the radius, $r$, and the angle, $\theta$, measured in radians. Convert 30° to radians and multiply by the radius to find the arc length:
$θ = 30^\circ * \dfrac{2π}{360^\circ} = \; \dfrac{π}{6}$
$s = r * θ = 2* \dfrac{π}{6}= \; \dfrac{π}{3}$
$θ = 30^\circ * \dfrac{2π}{360^\circ} = \; \dfrac{π}{6}$
$s = r * θ = 2* \dfrac{π}{6}= \; \dfrac{π}{3}$
Question 37 |
If a circular wheel makes 5 revolutions in one minute, how many seconds does it take to rotate 30°?
1 second
| |
2 seconds
| |
4 seconds
| |
12 seconds
|
Question 37 Explanation:
The correct answer is (A). Begin by converting the revolutions per minute to revolutions per second:
$\dfrac{5 \text{ rev}}{1 \text{ min}} * \dfrac{1 \text{ min}}{60 \text { seconds}} $ $ = \dfrac{1 \text{ rev}}{12 \text{ seconds}}$
$\dfrac{1 \text{ rev}}{12 \text{ seconds}} * \dfrac{360°}{1 \text { rev}} $ $ = \dfrac{360°}{12 \text { seconds}} = \dfrac{30°}{1 \text{ second}}$
The wheel will take 1 second to rotate 30°.
$\dfrac{5 \text{ rev}}{1 \text{ min}} * \dfrac{1 \text{ min}}{60 \text { seconds}} $ $ = \dfrac{1 \text{ rev}}{12 \text{ seconds}}$
$\dfrac{1 \text{ rev}}{12 \text{ seconds}} * \dfrac{360°}{1 \text { rev}} $ $ = \dfrac{360°}{12 \text { seconds}} = \dfrac{30°}{1 \text{ second}}$
The wheel will take 1 second to rotate 30°.
Question 38 |
Evaluate: $\lim_{x\to \infty} \dfrac{x^2 + 1000 x}{x^3}$
$1$ | |
$+ \infty$ | |
$- \infty$ | |
$0$ |
Question 38 Explanation:
The correct answer is (D). Factor out $x^3$ from both numerator and denominator:
$\dfrac{x^3(\frac{1}{x} + \frac{1000}{x^2})}{x^3(1)} \to \dfrac{1}{x} + \dfrac{1000}{x^2}$
The limit as $x$ approaches $\infty$ of this sum is 0.
$\dfrac{x^3(\frac{1}{x} + \frac{1000}{x^2})}{x^3(1)} \to \dfrac{1}{x} + \dfrac{1000}{x^2}$
The limit as $x$ approaches $\infty$ of this sum is 0.
Question 39 |
A ladder leans against a wall. The top of the ladder is the same distance from the ground as the base of the ladder is from the wall. What is the length of the ladder?
$(\text{Distance from wall}) \ast \sqrt{2}$ | |
$(\text{Distance from ground}) \ast \sqrt{2}$ | |
$\text {Both (A) and (B)}$
| |
$\text{None of the above}$ |
Question 39 Explanation:
The correct answer is (C). Draw a picture representing this situation, using the variable $x$ for the equal side lengths. You should have a 45°, 45°, 90° right triangle, which has corresponding side lengths of $x$, $x$, $x \sqrt{2}$.
Because the distance of the base from the wall is the same as the distance of the top from the ground, and the length of the ladder is the hypotenuse of the resulting right triangle, answer choice (C) must be correct.
Because the distance of the base from the wall is the same as the distance of the top from the ground, and the length of the ladder is the hypotenuse of the resulting right triangle, answer choice (C) must be correct.
Question 40 |
A solution is composed of 11 ions of $\text{Na}^+$ and 8 ions of $\text{Cl}^−$. What will be the sum of the number of molecules and the number of ions after the solution settles?
14 | |
11 | |
8 | |
15 |
Question 40 Explanation:
The correct answer is (B). The 8 chlorine (Cl−) ions will bond with the 8 sodium (Na+) ions to create 8 total NaCl molecules. Combining this with the number of sodium ions remaining, 3, gives a total of 11 molecules and ions.
Question 41 |
In which portion of a cone-shaped water tank will the rate of change of the height of the water be greatest, if the water is drained at a constant volumetric rate (e.g. 3 liters per second)?
The top (base)
| |
The middle
| |
The bottom (pointed side)
| |
The rate of change in height will remain constant
|
Question 41 Explanation:
The correct answer is (C). Since the cross-sectional area becomes smaller as the water level decreases in the cone, the rate of change of the height of the water will increase to maintain a constant volumetric rate.
Question 42 |
List the numbers in increasing order:
I. $e^2$
II. $π ^2$
III. $i^2$
IV. |-20|
III, I, IV, II
| |
III, I, II, IV
| |
IV, III, I, II
| |
II, III, IV, I
|
Question 42 Explanation:
The correct answer is (B). Convert each value to a numeric expression:
$e^2 \approx 2.7^2 \approx 7$
$π ^2 \approx 9$
$i^2=-1$
$|-20| = 20$
In increasing order:
$i^2, \; e^2, \; π ^2, \; |-20|$
$e^2 \approx 2.7^2 \approx 7$
$π ^2 \approx 9$
$i^2=-1$
$|-20| = 20$
In increasing order:
$i^2, \; e^2, \; π ^2, \; |-20|$
Question 43 |
What is the first derivative of:
$y= \dfrac{x^4 +3x}{7x^6}$
$\dfrac{7x^6}{-2x^3 +15}$ | |
$x^7$ | |
$\dfrac{-2x^3 - 15}{7x^6}$ | |
$\left(\dfrac{x^5}{5} + \dfrac{3x^2}{2}\right)$ |
Question 43 Explanation:
The correct answer is (C). Begin by factoring out an $x$ from the numerator and denominator:
$y= \dfrac{x(x^3+3)}{x(7x^5)} \to y= \dfrac{x^3+3}{7x^5}$
Use the quotient rule to evaluate the derivative:
$y'= $ $ \dfrac{(7x^5*3x^2)-(x^3+3)(35x^4)}{(7x^5)^2}$
$= \dfrac{21x^7 - 35x^7 - 105x^4}{49x^{10}}$
$= \dfrac{-14x^7 - 105x^4}{49x^{10}} $
$= \dfrac{7x^4(-2x^3-15)}{7x^4(7x^6)}$
$= \dfrac{-2x^3-15}{7x^6}$
$y= \dfrac{x(x^3+3)}{x(7x^5)} \to y= \dfrac{x^3+3}{7x^5}$
Use the quotient rule to evaluate the derivative:
$y'= $ $ \dfrac{(7x^5*3x^2)-(x^3+3)(35x^4)}{(7x^5)^2}$
$= \dfrac{21x^7 - 35x^7 - 105x^4}{49x^{10}}$
$= \dfrac{-14x^7 - 105x^4}{49x^{10}} $
$= \dfrac{7x^4(-2x^3-15)}{7x^4(7x^6)}$
$= \dfrac{-2x^3-15}{7x^6}$
Question 44 |
What is the y value of the absolute maximum of:
$y= \dfrac{x^4+3x}{7x^6}$
$0$ | |
$\text{There is no}$ $\text{absolute}$ $\text{maximum}$
| |
$\sqrt[3]{\frac {15}{2}}$ | |
$- \sqrt[3]{\frac {15}{2}}$ |
Question 44 Explanation:
The correct answer is (B). Notice that the function has a vertical asymptote at $x$ = 0. As $x$ approaches 0 from the right, the function approaches infinity. Because the function approaches infinity, there is no absolute maximum.
Question 45 |
An object’s position is given by the equation:
$x(t)=t^4+4t^2$
What is the object’s acceleration at $t=3 ?$
117 | |
132 | |
116 | |
0 |
Question 45 Explanation:
The correct answer is (C). Recall that acceleration is the second derivative of position. Evaluate the first and second derivative and plug in $t$ = 3:
$v(t)=x'(t)=4t^3+8t$
$\to a(t)=v'(t)=x''(t) $ $ =12t^2 + 8$
$\to a(3)=12(3)^2+8=116$
$v(t)=x'(t)=4t^3+8t$
$\to a(t)=v'(t)=x''(t) $ $ =12t^2 + 8$
$\to a(3)=12(3)^2+8=116$
Question 46 |
Bill and Joe start from the same point before running in opposite directions. Bill runs at 20 m/s west and Joe runs at 10 m/s east. In how many seconds will they be 120 m apart?
2 | |
3 | |
3.5 | |
4 |
Question 46 Explanation:
The correct answer is (D). Every second, the distance between Bill and Joe increases by 30 meters:
$\dfrac{120 \text{ m}}{30 \text{ m/s}}=4$ seconds
$\dfrac{120 \text{ m}}{30 \text{ m/s}}=4$ seconds
Question 47 |
Bill and Joe start from the same point before running in opposite directions. Bill runs at 20 m/s west and Joe runs at 10 m/s east. Assume that Bill turns around after 2 seconds and runs in the same direction as Joe. How far apart will they be 3 seconds after Bill turns around if he continues to run 20 m/s?
50 m
| |
40 m
| |
30 m
| |
20 m
|
Question 47 Explanation:
The correct Answer is (C). After the initial 2 seconds, the distance between Bill and Joe will be 60 meters, since the distance between them increases by 30 meters per second. After Bill turns around and runs in Joe’s direction, the distance between them decreases 10 meters for every second (because Bill runs 10 m/s faster than Joe). Three seconds after turning around, when a total of five seconds has elapsed, the original distance between them, 60 meters, will be reduced by:
$3 \text{ seconds} * 10 \frac{\text{meters}}{\text{second}} = 30 \text{ m}$
$60\text{ m} - 30 \text{ m}= 30 \text{ m}$
$3 \text{ seconds} * 10 \frac{\text{meters}}{\text{second}} = 30 \text{ m}$
$60\text{ m} - 30 \text{ m}= 30 \text{ m}$
Question 48 |
What is the reciprocal of $\dfrac{\frac43}{\frac23}?$
$2$ | |
$1$ | |
$\dfrac32$ | |
$\dfrac12$ |
Question 48 Explanation:
The correct Answer is (D). Divide the fractions and simplify:
$\dfrac{\frac43}{\frac23}=\dfrac{4}{3}*\dfrac{3}{2} = \dfrac{12}{6} = \dfrac{2}{1}$
Find the reciprocal:
$\dfrac{2}{1} \to \dfrac{1}{2}$
$\dfrac{\frac43}{\frac23}=\dfrac{4}{3}*\dfrac{3}{2} = \dfrac{12}{6} = \dfrac{2}{1}$
Find the reciprocal:
$\dfrac{2}{1} \to \dfrac{1}{2}$
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