PSAT Math: No Calculator

This practice test covers the first section of PSAT Math. For these questions there is no calculator allowed.

Congratulations - you have completed .

You scored %%SCORE%% out of %%TOTAL%%.

Your performance has been rated as %%RATING%%


Your answers are highlighted below.
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?

A
$3$
B
$6$
C
$9$
D
$12$
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
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?

A
$f(t) = 8 − 2t$
B
$f(t) = 8 + 2t$
C
$f(t) = 16 − t$
D
$f(t) = 16 − 2t$
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 $f(t) = 8 − 2t$.
Question 3
$3x + 2y = 6$

$7x + 4y = 9$

Based on the system of equations above, what is the value of the product $xy$?

A
$−\dfrac{45}{2}$
B
$−\dfrac{9}{2}$
C
$−3$
D
$\dfrac{15}{2}$
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: $−3 \cdot \dfrac{15}{2} = −\dfrac{45}{2}$
Question 4

If the two roots of $6x^2 + 6x = 12$ are $r_1$ and $r_2$, what is the product $r_1 \cdot r_2$?

A
$−2$
B
$−1$
C
$1$
D
$2$
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:

$1 \cdot −2 = −2$
Question 5

The function $f(x) = ax + 2 + c\,$ passes through the points (0, 4) and (−1, 2). What is the value of $a + c\,$?

A
$2$
B
$4$
C
$6$
D
$8$
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 $a + c$:

$a + c = 2 + 2 = 4$
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 $x$?

A
$\dfrac{\sqrt{6}}{3}$
B
$\dfrac{3\sqrt{6}}{2}$
C
$\dfrac{3}{\sqrt{6}}$
D
$\dfrac{\sqrt{2}}{3}$
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.

$x = \sqrt{\dfrac{2}{3}}$ $= \dfrac{\sqrt{2}}{\sqrt{3}} \dfrac{\sqrt{3}}{\sqrt{3}}$ $= \dfrac{\sqrt{6}}{3}$
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?

A
$|x + y|$
B
$|xy|$
C
$|y − x|$
D
$\text{None}$
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 $b$?

A
$5$
B
$7$
C
$28$
D
$56$
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, $b = 7$.
Question 9

If $b$ is a fraction such that $\frac{1}{4} ≤ b ≤ \frac{3}{4}$, which of the following is a true statement?

A
$b ≥ b^2 ≥ b^{−1}$
B
$b ≤ b^2 ≥ b^{−8}$
C
$b ≥ b^2 ≥ b^{−8}$
D
$b ≥ b^2 ≥ b^8$
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, $b ≥ b^2 ≥ b^8$.
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?

A
$(15, 15)$
B
$(15, 16)$
C
$(15, 17)$
D
$(15, 18)$
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 $10 \cdot 15 + 20 \cdot 18 = 510 ≥ 500$.
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?

A
$5$
B
$5\sqrt{5}$
C
$5\sqrt{6}$
D
$6\sqrt{5}$
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:

$\sqrt{5^2 + 5^2 + 10^2}$

$= \sqrt{150} = \sqrt{25 \cdot 6}$ $= 5\sqrt{6}$
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?

A
$3w^2 + 11w + 10$
B
$3w^2 − 11w + 10$
C
$11w^2 + 3w + 10$
D
$11w^2 − 3w − 10$
Question 12 Explanation: 
The correct answer is (A). The relationship between length and width is given:

$l = 3w$

New area $= (l + 5)( w + 2)$

$= (3w + 5)(w + 2)$

$= 3w^2 + 6w + 5w + 10$

$= 3w^2 + 11w + 10$
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

A
$R(t) = 32t^3 + 60t^2 − 8t + 52$
B
$R(t) = 32t^4 + 60t^3 − 8t^2 + 26t + 52$
C
$R(t) = 32t^4 + 60t^3 − 8t^2 + 26t$
D
$R(t) = 32t^4 + 8t^2 + 26t + 52$
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 $= N(t) \cdot P(t)$

$= (16t^3 − 2t^2 + 13) \cdot (2t + 4)$

$= 32t^4 + 64t^3 − 4t^3 − 8t^2 + 26t + 52$

$= 32t^4 + 60t^3 − 8t^2 + 26t + 52$
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.)

A
$\text{Click}$ $\text{here}$ $\text{to}$ $\text{see}$ $\text{the}$ $\text{correct}$ $\text{answer.}$
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.)

A
$\text{Click}$ $\text{here}$ $\text{to}$ $\text{see}$ $\text{the}$ $\text{correct}$ $\text{answer.}$
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.)

A
$\text{Click}$ $\text{here}$ $\text{to}$ $\text{see}$ $\text{the}$ $\text{correct}$ $\text{answer.}$
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.)

A
$\text{Click}$ $\text{here}$ $\text{to}$ $\text{see}$ $\text{the}$ $\text{correct}$ $\text{answer.}$
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 $x = 5$.

$|−3 + 5| = |2| = 2$
Once you are finished, click the button below. Any items you have not completed will be marked incorrect. Get Results
There are 17 questions to complete.
List
Return
Shaded items are complete.
12345
678910
1112131415
1617End
Return