This practice test covers the first section of PSAT Math. For these questions there is no calculator allowed.
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Question 1 |
George takes 3 hours to mow a lawn. Jacob takes 6 hours to mow the same lawn. How long will it take if both of them mow 3 lawns together?
$3$ | |
$6$ | |
$9$ | |
$12$ |
Question 1 Explanation:
The correct answer is (B).
George takes 3 hrs/lawn, so his rate of work is $\frac{1}{3}$ lawn/hr.
Jacob takes 6 hrs/lawn, so his rate of work is $\frac{1}{6}$ lawn/hr.
If they work together, they will mow $\frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ lawn/hr.
We can now set up equivalent proportions:
$\dfrac{\frac{1}{2} \text{ lawn}}{1 \text{ hour}}$ $= \dfrac{3 \text{ lawns}}{x \text{ hours}}$
Cross multiplying, we get:
$(1)(3) = \left(\dfrac{1}{2}\right)(x)$
$3 = \dfrac{1}{2}x$
$x = 6$ hours
George takes 3 hrs/lawn, so his rate of work is $\frac{1}{3}$ lawn/hr.
Jacob takes 6 hrs/lawn, so his rate of work is $\frac{1}{6}$ lawn/hr.
If they work together, they will mow $\frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ lawn/hr.
We can now set up equivalent proportions:
$\dfrac{\frac{1}{2} \text{ lawn}}{1 \text{ hour}}$ $= \dfrac{3 \text{ lawns}}{x \text{ hours}}$
Cross multiplying, we get:
$(1)(3) = \left(\dfrac{1}{2}\right)(x)$
$3 = \dfrac{1}{2}x$
$x = 6$ hours
Question 2 |
A new electric hybrid car has a gas mileage of 45 miles per gallon. The car has a gas tank that holds 16 gallons of gas. The car with a half-full tank is driven at 90 miles an hour.
Which of the following functions $f$ models the amount of gas remaining in the tank after $t$ hours?$f(t) = 8 − 2t$ | |
$f(t) = 8 + 2t$ | |
$f(t) = 16 − t$ | |
$f(t) = 16 − 2t$ |
Question 2 Explanation:
The correct answer is (A). The car starts off with $8$ gallons of gas. We need to find "gallons per hour".
$\dfrac{45 \text{ miles}}{1 \text{ gallon}}$ or $\dfrac{1 \text{ gallon}}{45 \text{ miles}}$
$\require{cancel} \dfrac{1 \text{ gallon}}{45 \cancel{\text{ miles}}} \cdot \dfrac{90 \cancel{\text{ miles}}}{1 \text{ hour}}$ $= \dfrac{(1)(90) \text{ gallons}}{(45)(1) \text{ hours}}$ $= \dfrac{90 \text{ gallons}}{45 \text{ hours}}$ $= \dfrac{2 \text{ gallons}}{1 \text{ hour}}$
Each hour ($t$), it loses $2$ gallons of gas. This is represented by $f(t) = 8 − 2t$.
$\dfrac{45 \text{ miles}}{1 \text{ gallon}}$ or $\dfrac{1 \text{ gallon}}{45 \text{ miles}}$
$\require{cancel} \dfrac{1 \text{ gallon}}{45 \cancel{\text{ miles}}} \cdot \dfrac{90 \cancel{\text{ miles}}}{1 \text{ hour}}$ $= \dfrac{(1)(90) \text{ gallons}}{(45)(1) \text{ hours}}$ $= \dfrac{90 \text{ gallons}}{45 \text{ hours}}$ $= \dfrac{2 \text{ gallons}}{1 \text{ hour}}$
Each hour ($t$), it loses $2$ gallons of gas. This is represented by $f(t) = 8 − 2t$.
Question 3 |
$3x + 2y = 6$ $7x + 4y = 9$
Based on the system of equations above, what is the value of the product $xy$?
$−\dfrac{45}{2}$ | |
$−\dfrac{9}{2}$ | |
$−3$ | |
$\dfrac{15}{2}$ |
Question 3 Explanation:
Solution: The correct answer is (A). Solve the system of linear equations:
$3x + 2y = 6$
$7x + 4y = 9$
Multiply the first equation by 2 throughout:
$6x + 4y = 12$
$7x + 4y = 9$
Subtract the first equation from the second:
$~~~7x + 4y = 9$
$−(6x + 4y = 12)$
––––––––––––––––––
$~~~x = −3$
Substitute $x = −3$ into either equation:
$7x + 4y = 9$
$7(−3) + 4y = 9$
$−21 + 4y = 9$
$4y = 30$
$y = \dfrac{30}{4}$
$y = \dfrac{15}{2}$
The product is: $−3 \cdot \dfrac{15}{2} = −\dfrac{45}{2}$
$3x + 2y = 6$
$7x + 4y = 9$
Multiply the first equation by 2 throughout:
$6x + 4y = 12$
$7x + 4y = 9$
Subtract the first equation from the second:
$~~~7x + 4y = 9$
$−(6x + 4y = 12)$
––––––––––––––––––
$~~~x = −3$
Substitute $x = −3$ into either equation:
$7x + 4y = 9$
$7(−3) + 4y = 9$
$−21 + 4y = 9$
$4y = 30$
$y = \dfrac{30}{4}$
$y = \dfrac{15}{2}$
The product is: $−3 \cdot \dfrac{15}{2} = −\dfrac{45}{2}$
Question 4 |
If the two roots of $6x^2 + 6x = 12$ are $r_1$ and $r_2$, what is the product $r_1 \cdot r_2$?
$−2$ | |
$−1$ | |
$1$ | |
$2$ |
Question 4 Explanation:
The correct answer is (A). Dividing the equation by 6 gives:
$x^2 + x − 2 = 0$
Factoring gives:
$(x - 1) (x + 2) = 0$
The roots are 1 and −2. The product of the roots is:
$1 \cdot −2 = −2$
$x^2 + x − 2 = 0$
Factoring gives:
$(x - 1) (x + 2) = 0$
The roots are 1 and −2. The product of the roots is:
$1 \cdot −2 = −2$
Question 5 |
The function $f(x) = ax + 2 + c\,$ passes through the points (0, 4) and (−1, 2). What is the value of $a + c\,$?
$2$ | |
$4$ | |
$6$ | |
$8$ |
Question 5 Explanation:
The correct answer is (B).
$f(x) = ax + 2 + c$
Since the slope-intercept form of a line is $y = mx + b$, $m = a$ (our slope is "$a$") and $b = 2 + c$ (our $y$-intercept is "$2 + c$").
Since the point $(0, 4)$ lies on the line, we know the $y$-intercept is $4$.
$2 + c = 4$
$c = 2$
Slope $m = \dfrac{y_2 − y_1}{x_2 − x_1}$ $= \dfrac{2 − 4}{−1 − 0}$ $= \dfrac{−2}{−1} = 2$
$a = 2$
We are asked to find $a + c$:
$a + c = 2 + 2 = 4$
$f(x) = ax + 2 + c$
Since the slope-intercept form of a line is $y = mx + b$, $m = a$ (our slope is "$a$") and $b = 2 + c$ (our $y$-intercept is "$2 + c$").
Since the point $(0, 4)$ lies on the line, we know the $y$-intercept is $4$.
$2 + c = 4$
$c = 2$
Slope $m = \dfrac{y_2 − y_1}{x_2 − x_1}$ $= \dfrac{2 − 4}{−1 − 0}$ $= \dfrac{−2}{−1} = 2$
$a = 2$
We are asked to find $a + c$:
$a + c = 2 + 2 = 4$
Question 6 |
If the solution to $\frac{\sqrt{2}}{x + \sqrt{2}} − \frac{\sqrt{2}}{x − \sqrt{2}} = 3$ is positive, what is the value of $x$?
$\dfrac{\sqrt{6}}{3}$ | |
$\dfrac{3\sqrt{6}}{2}$ | |
$\dfrac{3}{\sqrt{6}}$ | |
$\dfrac{\sqrt{2}}{3}$ |
Question 6 Explanation:
The correct answer is (A). Find the common denominator and simplify:
$\dfrac{\sqrt{2} \cdot (x − \sqrt{2})}{(x + \sqrt{2})(x − \sqrt{2})}$ $− \dfrac{\sqrt{2} \cdot (x + \sqrt{2})}{(x − \sqrt{2})(x + \sqrt{2})} = 3$
"FOIL"ing the denominator, we get:
$(x + \sqrt{2})(x − \sqrt{2})$ $= (x)(x) + (x)(−\sqrt{2}) + (\sqrt{2})(x) + (\sqrt{2})(\sqrt{−2})$
$= x^2 − \sqrt{2}x + \sqrt{2}x − 2$
$= x^2 − 2$
Rewriting our fractions:
$\dfrac{\sqrt{2} \cdot (x − \sqrt{2})}{x^2 − 2}$ $− \dfrac{\sqrt{2} \cdot (x + \sqrt{2})}{x^2 − 2} = 3$
$\dfrac{x\sqrt{2} − 2}{x^2 − 2}$ $− \dfrac{x\sqrt{2} + 2}{x^2 − 2} = 3$
$\dfrac{x\sqrt{2} − 2 − x\sqrt{2} − 2}{x^2 − 2} = 3$
$ \dfrac{−4}{x^2 − 2} = 3$
$\dfrac{−4}{x^2 − 2}$ $= \dfrac{3}{1}$
$(−4)(1) = (3)(x^2 − 2)$
$−4 = 3x^2 − 6$
$2 = 3x^2$
$x^2 = \dfrac{2}{3}$
$x = ± \sqrt{\dfrac{2}{3}}$, but we want the positive solution.
$x = \sqrt{\dfrac{2}{3}}$ $= \dfrac{\sqrt{2}}{\sqrt{3}} \dfrac{\sqrt{3}}{\sqrt{3}}$ $= \dfrac{\sqrt{6}}{3}$
$\dfrac{\sqrt{2} \cdot (x − \sqrt{2})}{(x + \sqrt{2})(x − \sqrt{2})}$ $− \dfrac{\sqrt{2} \cdot (x + \sqrt{2})}{(x − \sqrt{2})(x + \sqrt{2})} = 3$
"FOIL"ing the denominator, we get:
$(x + \sqrt{2})(x − \sqrt{2})$ $= (x)(x) + (x)(−\sqrt{2}) + (\sqrt{2})(x) + (\sqrt{2})(\sqrt{−2})$
$= x^2 − \sqrt{2}x + \sqrt{2}x − 2$
$= x^2 − 2$
Rewriting our fractions:
$\dfrac{\sqrt{2} \cdot (x − \sqrt{2})}{x^2 − 2}$ $− \dfrac{\sqrt{2} \cdot (x + \sqrt{2})}{x^2 − 2} = 3$
$\dfrac{x\sqrt{2} − 2}{x^2 − 2}$ $− \dfrac{x\sqrt{2} + 2}{x^2 − 2} = 3$
$\dfrac{x\sqrt{2} − 2 − x\sqrt{2} − 2}{x^2 − 2} = 3$
$ \dfrac{−4}{x^2 − 2} = 3$
$\dfrac{−4}{x^2 − 2}$ $= \dfrac{3}{1}$
$(−4)(1) = (3)(x^2 − 2)$
$−4 = 3x^2 − 6$
$2 = 3x^2$
$x^2 = \dfrac{2}{3}$
$x = ± \sqrt{\dfrac{2}{3}}$, but we want the positive solution.
$x = \sqrt{\dfrac{2}{3}}$ $= \dfrac{\sqrt{2}}{\sqrt{3}} \dfrac{\sqrt{3}}{\sqrt{3}}$ $= \dfrac{\sqrt{6}}{3}$
Question 7 |
Monday's high temperature was $x$°. Tuesday’s high temperature was $y$°. The high temperature on Wednesday was the absolute difference between Monday’s and Tuesday’s high temperatures.
Which of the following represents the high temperature on Wednesday?$|x + y|$ | |
$|xy|$ | |
$|y − x|$ | |
$\text{None}$ |
Question 7 Explanation:
The correct answer is (C). The difference in temperatures between Monday and Tuesday is $|x − y|$. This is not an answer choice. Note that $|x − y| = |y − x|$, since the difference is absolute.
Question 8 |
$\dfrac{1}{2}x − \dfrac{1}{4}y = 5$ $14x − by = 140$
In the system of linear equations above, $b$ is a constant. If the system has infinite solutions, what is the value of $b$?
$5$ | |
$7$ | |
$28$ | |
$56$ |
Question 8 Explanation:
The correct answer is (B). Multiplying the first equation by 28 gives:
$(28)(\frac{1}{2}x) + (28)(−\frac{1}{4}y) = (28)(5)$
$14x − 7y = 140$
$14x − by = 140$
Since the system has infinite solutions, the equations/lines are the same. So, $b = 7$.
$(28)(\frac{1}{2}x) + (28)(−\frac{1}{4}y) = (28)(5)$
$14x − 7y = 140$
$14x − by = 140$
Since the system has infinite solutions, the equations/lines are the same. So, $b = 7$.
Question 9 |
If $b$ is a fraction such that $\frac{1}{4} ≤ b ≤ \frac{3}{4}$, which of the following is a true statement?
$b ≥ b^2 ≥ b^{−1}$ | |
$b ≤ b^2 ≥ b^{−8}$ | |
$b ≥ b^2 ≥ b^{−8}$ | |
$b ≥ b^2 ≥ b^8$ |
Question 9 Explanation:
The correct answer is (D). Use the fact that multiplying a fraction by itself will make the result smaller each time. Thus $b$ will be greater than $b^n$ where $n > 1$. Using this fact, you can conclude that the correct choice is choice (D).
You can also plug in a value for $b$, say $\frac{1}{4}$ and test the inequalities. This may be an easy way to solve these kinds of problems. Number lines are very helpful to visualize abstract problems like this. In this problem, $b^8$, $b^2$, $b$ appear in this order in the number line. Mark them on a number line to help you visualize quickly. So, $b ≥ b^2 ≥ b^8$.
You can also plug in a value for $b$, say $\frac{1}{4}$ and test the inequalities. This may be an easy way to solve these kinds of problems. Number lines are very helpful to visualize abstract problems like this. In this problem, $b^8$, $b^2$, $b$ appear in this order in the number line. Mark them on a number line to help you visualize quickly. So, $b ≥ b^2 ≥ b^8$.
Question 10 |
At Sing & Dance, customers can rent singing and dancing rooms in increments of 1 minute with a minimum total of 30 minutes. The cost for renting the singing room is \$10 per minute and the cost for renting the dancing room is \$20 per minute.
If the minimum total must be \$500, then which of the following choices indicate a possible combination of singing and dancing minutes that can be rented?$(15, 15)$ | |
$(15, 16)$ | |
$(15, 17)$ | |
$(15, 18)$ |
Question 10 Explanation:
The correct answer is (D).
Let $S$ be the number of minutes rented from the song room and $D$ be the number of minutes rented from the dance room. Minimum total is $30$ minutes, so $S + D ≥ 30$.
Minimum total cost of dancing $= 10S + 20D ≥ 500$.
Plug in the choices. For $S = 15$ and $D = 18$, both inequalities are satisfied. $S + D = 15 + 18$ $= 33 ≥ 30$ and $10 \cdot 15 + 20 \cdot 18 = 510 ≥ 500$.
Let $S$ be the number of minutes rented from the song room and $D$ be the number of minutes rented from the dance room. Minimum total is $30$ minutes, so $S + D ≥ 30$.
Minimum total cost of dancing $= 10S + 20D ≥ 500$.
Plug in the choices. For $S = 15$ and $D = 18$, both inequalities are satisfied. $S + D = 15 + 18$ $= 33 ≥ 30$ and $10 \cdot 15 + 20 \cdot 18 = 510 ≥ 500$.
Question 11 |
Two cubes with side lengths of 5 are placed adjacently as shown in the diagram below. What is the length of the interior diagonal of the newly formed rectangular prism?
$5$ | |
$5\sqrt{5}$ | |
$5\sqrt{6}$ | |
$6\sqrt{5}$ |
Question 11 Explanation:
The correct answer is (C). The length of the diagonal is the same as a rectangular prism with dimensions of $5 × 5 × 10$, which is:
$\sqrt{5^2 + 5^2 + 10^2}$
$= \sqrt{150} = \sqrt{25 \cdot 6}$ $= 5\sqrt{6}$
$\sqrt{5^2 + 5^2 + 10^2}$
$= \sqrt{150} = \sqrt{25 \cdot 6}$ $= 5\sqrt{6}$
Question 12 |
The length of a rectangle $l$, is three times its width $w$. If the length is increased by 5 units and width is increased by 2 units, the area of a rectangle is given by which of the following expression?
$3w^2 + 11w + 10$ | |
$3w^2 − 11w + 10$ | |
$11w^2 + 3w + 10$ | |
$11w^2 − 3w − 10$ |
Question 12 Explanation:
The correct answer is (A). The relationship between length and width is given:
$l = 3w$
New area $= (l + 5)( w + 2)$
$= (3w + 5)(w + 2)$
$= 3w^2 + 6w + 5w + 10$
$= 3w^2 + 11w + 10$
$l = 3w$
New area $= (l + 5)( w + 2)$
$= (3w + 5)(w + 2)$
$= 3w^2 + 6w + 5w + 10$
$= 3w^2 + 11w + 10$
Question 13 |
The polynomial function that can be used to model the number of items sold at a store is $N(t) = 16t^3 − 2t^2 + 13$, where $t$ is the month (Jan = 1, Dec = 12).
The average price of the items each month and can be modeled as $P(t) = 2t + 4$. The function $R(t)$ that can be used to model the revenue from these items is$R(t) = 32t^3 + 60t^2 − 8t + 52$ | |
$R(t) = 32t^4 + 60t^3 − 8t^2 + 26t + 52$ | |
$R(t) = 32t^4 + 60t^3 − 8t^2 + 26t$ | |
$R(t) = 32t^4 + 8t^2 + 26t + 52$ |
Question 13 Explanation:
The correct answer is (B).
Items sold each month $= N(t) = 16t^3 − 2t^2 + 13$
Price of items each month $= P(t) = 2t + 4$.
Revenue $= N(t) \cdot P(t)$
$= (16t^3 − 2t^2 + 13) \cdot (2t + 4)$
$= 32t^4 + 64t^3 − 4t^3 − 8t^2 + 26t + 52$
$= 32t^4 + 60t^3 − 8t^2 + 26t + 52$
Items sold each month $= N(t) = 16t^3 − 2t^2 + 13$
Price of items each month $= P(t) = 2t + 4$.
Revenue $= N(t) \cdot P(t)$
$= (16t^3 − 2t^2 + 13) \cdot (2t + 4)$
$= 32t^4 + 64t^3 − 4t^3 − 8t^2 + 26t + 52$
$= 32t^4 + 60t^3 − 8t^2 + 26t + 52$
Question 14 |
Amanda pays her cell phone company 50 dollars a month, plus 1 dollar for every minute spoken and 50 cents for each text message sent. Betty pays her cell phone company 25 dollars a month, plus 2 dollars for every minute spoken and 75 cents for each text message sent.
If both speak for 100 minutes a month, and send 200 text messages a month, how much more money will Betty spend than Amanda in 1 year?(Write down your answer.)
$\text{Click}$ $\text{here}$ $\text{to}$ $\text{see}$ $\text{the}$ $\text{correct}$ $\text{answer.}$ |
Question 14 Explanation:
The correct answer is $\$1{,}500$.
Both speak for 100 minutes and send 200 texts, in addition to the base monthly payment.
Amanda:
$50 + 1(100) + 0.5(200)$
$= 50 + 100 + 100$
$= \$250 \text{ per month}$
$250 \cdot 12 = \$3{,}000 \text{ per year}$
Betty:
$25 + 2(100) + 0.75(200)$
$= 25 + 200 + 150$
$= \$375 \text { per month}$
$375 \cdot 12 = \$4500 \text{ per year}$
The difference between their yearly costs is:
$\$4{,}500 − \$3{,}000 = \$1{,}500$
Question 15 |
A cab driver charges \$5 for the first 0.75 miles, \$8 for the subsequent 0.25 miles, and \$2 for each additional mile. How many miles can you travel in the cab if you paid \$50?
(Write down your answer.)
$\text{Click}$ $\text{here}$ $\text{to}$ $\text{see}$ $\text{the}$ $\text{correct}$ $\text{answer.}$ |
Question 15 Explanation:
The correct answer is $19.5$.
Let $m$ be the miles traveled for $\$50$. Then $m$ miles can be broken down, based on each fare segment. This must to add up to $m$ miles.
$m = 0.75 + 0.25 + (m − 1)$
The first mile is included in the $m$. Now apply the fares to each segment:
$5 + 8 + (m − 1) \cdot 2 = 50$
$13 + 2m − 2 = 50$
$2m = 39$
$m = 19.5 \text{ miles}$
Question 16 |
In the diagram shown above, $AE$ has a length of $3$. $AD$ has a length of $9$. $AB$ and $CD$ are parallel. What is the ratio of the area of triangle $CED$ to triangle $AEB$?
(Write down your answer.)
$\text{Click}$ $\text{here}$ $\text{to}$ $\text{see}$ $\text{the}$ $\text{correct}$ $\text{answer.}$ |
Question 16 Explanation:
The correct answer is $4$.
$∠ CED ≅ ∠ BEA$ (vertical angles)
Since $AB$ and $CD$ are parallel, $∠EBA ≅ ∠ECD$ and $∠EAB ≅ ∠EDC$ (alternate interior angles).
This means that triangles $ABE$ and $ECD$ are similar by AAA similarity. $AE = 3$ and $ED = 6$. This means that the side lengths of the larger triangle are twice the smaller triangle. Hence, the area of the larger triangle is $2^2 = 4$ times the area of the smaller triangle.
Question 17 |
What is the absolute value of the sum of the solutions to the polynomial shown below?
$x^2 − 2x − 15 = 0$(Write down your answer.)
$\text{Click}$ $\text{here}$ $\text{to}$ $\text{see}$ $\text{the}$ $\text{correct}$ $\text{answer.}$ |
Question 17 Explanation:
The correct answer is $2$.
The polynomial $x^2 − 2x − 15$ can be factored into:
$(x − 5)(x + 3)$
The solutions to this are $x = −3$ and $x = 5$.
$|−3 + 5| = |2| = 2$
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