For the Classic ACT exam:
The ACT Mathematics test is a timed exam...60 questions in 60 minutes
This implies that you have to solve each question in one minute.
Each of the first 20 questions (less challenging) will typically take less than a minute a solve.
Each of the next 20 questions (medium challenging) may take about a minute to solve.
Each of the last 20 questions (more challenging) may take more than a minute to solve.
The goal is to maximize your time.
You use the time saved on the questions you solve in less than a minute to solve questions that will take more
than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.
Also: please note that unless specified otherwise, geometric figures are drawn to scale. So, you can figure out
the correct answer by eliminating the incorrect options.
Other suggestions are listed in the solutions/explanations as applicable.
These are the solutions to the ACT past questions on the topics: Numbers, Fractions, Decimals, and Percents.
When applicable, the TI-84 Plus CE calculator (also applicable to TI-84 Plus calculator) solutions are provided
for some questions.
The link to the video solutions will be provided for you. Please
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(2.) While her mother drives their car along the highway, Mia is noticing some of the mile marker signs.
She sees mile marker 117 at noon, and exactly 20 minutes later she sees mile marker 97.
What is the average speed, in miles per hour, of their car over these 20 minutes.
At noon, Mia sees mile marker 117
Then, after 20 minutes, she sees mile marker 97
This implies that her mother drove 117 − 97 = 20 miles in 20 minutes
$
distance = 20\;miles \\[3ex]
time = 20\;minutes \\[3ex]
speed = \dfrac{distance}{time} \\[5ex]
= \dfrac{20}{20} \\[5ex]
= 1\;mile/minute \\[3ex]
$
On average, Mia's mother drove 1 mile in 1 minute.
In other words, her average speed is 1 mile per minute.
But the question wants us to calculate the speed in miles per hour
$
60\;minutes = 1\;hour \\[3ex]
1\;mile\;\;per\;\;minute = ?\;mile\;\;per\;\;hour \\[3ex]
\underline{Unity\;\;Fraction\;\;Method} \\[3ex]
\dfrac{1\;mile}{minute} * \dfrac{...\;minute}{...\;hour} \\[5ex]
\dfrac{1\;mile}{minute} * \dfrac{60\;minute}{1\;hour} \\[5ex]
60\;miles/hour
$
On average, Mia's mother drove 60 miles in 1 hour.
Her average speed is 60 miles per hour. (60 mph)
(3.) Which of the following lists gives the numbers below arranged in order from least to greatest?
$$
0.6, \;\; 0.08, \;\; \dfrac{5}{8}
$$
$
A.\;\; \dfrac{5}{8}, \;\; 0.6, \;\; 0.08 \\[5ex]
B.\;\; 0.6, \;\; \dfrac{5}{8}, \;\; 0.08 \\[5ex]
C.\;\; 0.6, \;\; 0.08, \;\; \dfrac{5}{8} \\[5ex]
D.\;\; 0.08, \;\; \dfrac{5}{8}, \;\; 0.6 \\[5ex]
E.\;\; 0.08, \;\; 0.6, \;\; \dfrac{5}{8} \\[5ex]
$
(4.) The Student Council is preparing a budget for an upcoming fund-raising dance.
They have decided to spend $150 for a light and sound show, $400 for refreshments,
and $50 for decorations.
These are the only expenses.
Given that the Student Council estimates 500 students will attend the dance, what should be the price, per
student, for admission to the
dance if the Student Council wants to raise as close as possible to $300 after paying expenses?
Total Expenses = $150 + $400 + $50 = $600
Expected Profit = $300
Expected Income = $600 + $300 = $900
Expected Number of students = 500
Price per student to generate expected income = 900 ÷ 500 = $1.80
(5.) JoAnna drives a route that is exactly 450 miles from Little Rock, Arkansas, to Mobile, Alabama.
JoAnna has already driven 180 miles at an average speed of 60 miles per hour.
What is the minimum average speed, in miles per hour, that JoAnna can drive for the remainder of the
route and have a driving time of 9 hours for the entire trip?
$
A.\;\; 45 \\[3ex]
B.\;\; 50 \\[3ex]
C.\;\; 55 \\[3ex]
D.\;\; 60 \\[3ex]
E.\;\;\text{Cannot be determined from the given information} \\[3ex]
$
$
s...t...d \implies speed \cdot time = distance \\[3ex]
\underline{Entire\;\;Journey} \\[3ex]
Distance = 450\;miles \\[3ex]
Time = 9\;hours \\[5ex]
\underline{Already\;\;Driven} \\[3ex]
Distance = 180\;miles \\[3ex]
Speed = 60\;miles\;per\;hour \\[3ex]
Time = \dfrac{Distance}{Speed} = \dfrac{180}{60} = 3\;hours \\[5ex]
\underline{Remaining\;\;Journey} \\[3ex]
Distance = 450 - 180 = 270\;miles \\[3ex]
Time = 9 - 3 = 6\;hours \\[3ex]
Speed = \dfrac{Distance}{Speed} \\[5ex]
Speed = \dfrac{270}{6} \\[5ex]
Speed = 45\;miles\;per\;hour \\[3ex]
$
The minimum average speed that JoAnna can drive for the remainder of the
route and have a driving time of 9 hours for the entire trip is 45 miles per hour.
(6.) Because of rising expenses, a motel manager raises the room rate of $80.00 by 20% to get the
new room rate for the motel.
What is the new room rate?
Due to the fact that one minute is allocated to each question on the ACT (60 minutes for 60 questions), it
is better to check the solution to this question by their answer options.
So, let us check and eliminate until we get the answer.
$
\underline{Option\;F} \\[3ex]
18 \div 5 = 3 \;R\; 3 \; \checkmark \\[3ex]
18 \div 6 = 3 \;R\;0 \;\text{remainder is 0, not 4} \\[3ex]
NEXT \\[5ex]
\underline{Option\;G} \\[3ex]
23 \div 5 = 4 \;R\; 3 \; \checkmark \\[3ex]
23 \div 6 = 3 \;R\; 5 \;\text{remainder is 3, not 4} \\[3ex]
NEXT \\[5ex]
\underline{Option\;H} \\[3ex]
28 \div 5 = 5 \;R\; 3 \; \checkmark \\[3ex]
28 \div 6 = 4 \;R\; 4 \; \checkmark \\[3ex]
STOP \\[3ex]
$
Option H is the correct answer.
This is the answer because the answer choices are in ascending order.
Student: Is there another way to do this question without checking by the answer options? Teacher: Yes, we can do it: Modular Arithmetic
However, I think it takes more than a minute to do.
$
Let: \\[3ex]
dividend = d \\[3ex]
quotient = q \\[3ex]
1st:\;\; \text{Remainder of 3 when divided by 5} \\[3ex]
d \equiv 3 \mod 5...cong.(1) \\[5ex]
2nd:\;\; \text{Remainder of 4 when divided by 6} \\[3ex]
d \equiv 4 \mod 6 ...cong.(2) \\[5ex]
\implies \\[3ex]
d = 6q + 4 ...eqn.(1) \\[3ex]
Substitute\;\;eqn.(1) \;\;for\;\;d\;\;in\;\;cong.(1) \\[3ex]
6q + 4 \equiv 3 \mod 5 \\[3ex]
\text{Test positive integers for q beginning from the first positive integer} \\[3ex]
6(1) + 4 = 10 \equiv 0 \mod 5...Not\;\;3 \\[3ex]
6(2) + 4 = 16 \equiv 1 \mod 5...Not\;\;3 \\[3ex]
6(3) + 4 = 22 \equiv 2 \mod 5...Not\;\;3 \\[3ex]
6(4) + 4 = 28 \equiv 3 \mod 5 \;\checkmark \\[3ex]
\implies \\[3ex]
d = 28
$
(8.) The number a is positive and even.
The number b is negative and odd.
The number a − b is:
F. positive and even. G. positive and odd. H. negative and even. J. negative and odd. K. zero.
Let us test this statement with some values of a and b
$
Let: \\[3ex]
\underline{Example\;1} \\[3ex]
a = 6 ...\text{positive and even} \\[3ex]
b = -3 ...\text{negative and odd} \\[3ex]
a - b \\[3ex]
6 - (-3)...\text{subtraction operation} \\[3ex]
6 + 3 \\[3ex]
9...\text{positve and odd} \\[5ex]
\underline{Example\;2} \\[3ex]
a = 2 ...\text{positive and even} \\[3ex]
b = -5 ...\text{negative and odd} \\[3ex]
a - b \\[3ex]
2 - (-5)...\text{subtraction operation} \\[3ex]
2 + 5 \\[3ex]
7...\text{positve and odd}
$
(9.) For how many integers x is the value of the expression $(x - 1)(x - 4)$ a positive prime number?
Let us test some integers for x to see what we can get.
We shall skip all positive numbers less than or equal to 4 because we need to get a positive prime
number.
$
(x - 1)(x - 4) \\[3ex]
Test\;\;x = 5 \\[3ex]
(5 - 1)(5 - 4) = 4(1) = 4 ...Not\;\;prime \\[5ex]
Test\;\;x = 7 \\[3ex]
(7 - 1)(7 - 4) = 6(3) = 18 ...Not\;\;prime \\[5ex]
Test\;\;x = -1 \\[3ex]
(-1 - 1)(-1 - 4) = (-2)(-5) = 10 ...Not\;\;prime \\[5ex]
Test\;\;x = -3 \\[3ex]
(-3 - 1)(-3 - 4) = (-4)(-7) = 28 ...Not\;\;prime \\[5ex]
\text{This applies to all odd numbers: 9, 11, 13, ... and −1, −3, −5, ...} \\[3ex]
\text{The odd numbers will give a positive composite (not positive prime)} \\[5ex]
Test\;\;x = 6 \\[3ex]
(6 - 1)(6 - 4) = 5(2) = 10 ...Not\;\;prime \\[5ex]
Test\;\;x = 8 \\[3ex]
(8 - 1)(8 - 4) = 7(4) = 28 ...Not\;\;prime \\[5ex]
Test\;\;x = -2 \\[3ex]
(-2 - 1)(-2 - 4) = (-3)(-6) = 18 ...Not\;\;prime \\[5ex]
Test\;\;x = -4 \\[3ex]
(-4 - 1)(-4 - 4) = (-5)(-8) = 40 ...Not\;\;prime \\[5ex]
\text{This applies to all even numbers: 10, 12, 14, ... and −2, −4, −6, ...} \\[3ex]
\text{The even numbers will give a positive composite (not positive prime)} \\[3ex]
$
As you can see, there is no number that we substutute for x which will give a positive prime number.
Hence, the correct answer is zero: Option F.
(10.) Each student in a particular classroom was given a string that was 72 inches long.
Each student cut his or her string into pieces of equal length.
Which of the following CANNOT be the length, in inches, of any student’s pieces?
72 inches would be cut into equal pieces (piecese of equal length)
Let us analyze each option to determine the correct answer.
Option A
1 ÷ 16 = 0.0625 inches
Can we have equal lengths of 0.0625 inch from a string length of 72 inches?
72 ÷ 0.0625 = 1152
Yes, we can have 1152 pieces of equal lengths of 0.0625 inch each
Option B
1 ÷ 2 = 0.5 inches
Can we have equal lengths of 0.5 inch from a string length of 72 inches?
72 ÷ 0.5 = 144
Yes, we can have 144 pieces of equal lengths of 0.5 inch each
Option C
1 inch
Can we have equal lengths of 1 inch from a string length of 72 inches?
72 ÷ 1 = 72
Yes, we can have 72 pieces of equal lengths of 1 inch each
Option D
16 inches
Can we have equal lengths of 16 inches from a string length of 72 inches?
72 ÷ 16 = 4.5
This is not an integer. There is a remainder.
We can have 4 equal lengths of 16 inches each; however the remainder will not be an equal length of 16
inches.
So, this option is the correct answer but let us go ahead and analyze the final option.
Option E
36 inches
Can we have equal lengths of 36 inches from a string length of 72 inches?
72 ÷ 36 = 22
Yes, we can have 2 pieces of equal lengths of 36 inches each
Option D is the correct answer.
(11.) Joe, Tom, and Alexis are planning to participate in a 3-person relay for charity.
The plan is for Joe to jog his 3 miles at 4 miles per hour (mph), Tom to jog his 3 miles at 5 mph, and then
Alexis to jog her 3 miles at 6 mph.
In how many hours and minutes does this 3-person team plan to complete the 9-mile relay?
$
s...t...d \\[3ex]
speed * time = distance \\[3ex]
time = \dfrac{distance}{speed} \\[5ex]
\underline{Joe} \\[3ex]
d = 3\;miles \\[3ex]
s = 4\;mph \\[3ex]
t = \dfrac{3}{4}\;hour \\[5ex]
\underline{Tom} \\[3ex]
d = 3\;miles \\[3ex]
s = 5\;mph \\[3ex]
t = \dfrac{3}{5}\;hour \\[5ex]
\underline{Alexis} \\[3ex]
d = 3\;miles \\[3ex]
s = 6\;mph \\[3ex]
t = \dfrac{3}{6} = \dfrac{1}{2}\;hour \\[5ex]
\underline{3-Person\;\;Team} \\[3ex]
time = \dfrac{3}{4} + \dfrac{3}{5} + \dfrac{1}{2} \\[5ex]
= \dfrac{15 + 12 + 10}{20} \\[5ex]
= \dfrac{37}{20} \\[5ex]
= 1.85\;hours \\[3ex]
= 1\;hour + 0.85\;hour \\[3ex]
= 1\;hour + \left(0.85\;hour * \dfrac{60\;minutes}{1\;hour}\right) \\[5ex]
= 1\;hour + 51\;minutes
$
(12.) The number line shown below is marked in equal intervals.
Two fractions are indicated on the number line.
One of the following fractions corresponds to the point marked P.
Which one?
$
\text{Distance between the fractions} = \dfrac{5}{4} - \dfrac{1}{4} \\[5ex]
= \dfrac{4}{4} \\[5ex]
= 1...\text{represented by 8 lines} \\[5ex]
8\;lines\;\;from\;\; \dfrac{1}{4} \;\;to\;\; \dfrac{5}{4}...\text{not including the line on}\;\;\dfrac{1}{4}
\\[5ex]
\dfrac{1}{4}\;\;\text{will be included when finding the value} \\[5ex]
P\;\;\text{is on the 7th line} \\[3ex]
8\;\;lines \rightarrow 1 \\[3ex]
7th\;\;line \rightarrow \dfrac{7(1)}{8} = \dfrac{7}{8} \\[5ex]
\text{Value on the 7th line} = \dfrac{1}{4} + \dfrac{7}{8} \\[5ex]
= \dfrac{2}{8} + \dfrac{7}{8} \\[5ex]
= \dfrac{9}{8}
$
(13.) Only juniors and seniors are enrolled in Algebra III.
There are 3 juniors for each senior.
On an Algebra III test, 80% of the juniors and 70% of the seniors passed.
What percent of the students enrolled in Algebra III did NOT pass the test?
$
\underline{Algebra\;III\;\;Test} \\[3ex]
Let: \\[3ex]
number\;\;of\;\;juniors = x \\[3ex]
number\;\;of\;\;seniors = y \\[3ex]
Enrolled = x + y \\[3ex]
\text{3 juniors for each senior} \implies x = 3y \\[3ex]
\implies \\[3ex]
Enrolled = 3y + y = 4y \\[5ex]
80\% = \dfrac{80}{100} = 0.8 \\[5ex]
70\% = \dfrac{70}{100} = 0.7 \\[5ex]
Passed:\;\;0.8x \\[3ex]
Did\;\;not\;\;pass:\;\;x - 0.8x = 0.2x \\[5ex]
Passed:\;\;0.7y \\[3ex]
Did\;\;not\;\;pass:\;\;y - 0.7y = 0.3y \\[5ex]
\implies \\[3ex]
Did\;\;not\;\;pass = 0.2x + 0.3y \\[3ex]
= 0.2(3y) + 0.3y \\[3ex]
= 0.6y + 0.3y \\[3ex]
= 0.9y \\[5ex]
\text{Percent who did not pass} = \dfrac{\text{Number who did not pass}}{\text{Number of Enrolled Students}}
* 100 \\[5ex]
=\dfrac{0.9y}{4y} * 100 \\[5ex]
= 22.5\% \\[3ex]
= 22\dfrac{1}{2}\%
$
(14.) Ms. Clark is scoring her class’s geography test.
The test had 30 questions, each worth 1 point.
Ms. Clark is currently scoring Tomás's test paper.
So far, she has marked 24 of his answers correct and 3 incorrect.
What is the maximum percent correct, to the nearest percent, that Tomás can earn on the test?
30 questions on the test
Correct ones so far = 24
Incorrect ones so far = 3
Remaining to be marked = 30 − (24 + 3)
= 30 − 27
= 3
To find the maximum percent that Tomás can earn, we shall assume that he will get the
remaining 3 questions correct
Hence, the number of correct ones (marked and assumed) = 24 + 3 = 27
$
\text{Percent of correct questions} \\[3ex]
= \dfrac{\text{Number of correct questions}}{\text{Number of questions}} \cdot 100 \\[5ex]
= \dfrac{27}{30} \cdot 100 \\[5ex]
= 9 \cdot 10 \\[3ex]
= 90\%
$
(15.) In the ordered pairs given below, the first number is time, in hours, and the second number is distance,
in miles, traveled in that time.
Which of these ordered pairs represent speeds less than 4 miles per hour? F.A and B only G.B and C only H.A, B, and C only J.C, D, and E only K.A, B, C, D, and E
$
s.......t.......d \\[3ex]
s \cdot t = d \\[3ex]
s = \dfrac{d}{t} \\[5ex]
$
(time, distance)
A
B
C
D
E
$\dfrac{1}{0.25} = 4$
$\dfrac{0.5}{0.2} = \color{darkblue}{2.5}$
$\dfrac{7}{2} = \color{darkblue}{3.5}$
$\dfrac{8}{0.5} = 16$
$\dfrac{9}{0.75} = 12$
The ordered pairs that represents speeds less than 4 miles per hour (in darkblue color) are: B and
C
(16.) Let A be the greatest whole number that is less than $\sqrt{420}$.
Let B be the least whole number that is greater than $\sqrt{56}$.
What is A − B?
Due to the fact that one minute is allocated to each question on the ACT (60 minutes for 60 questions), it
is better to check the solution to this question by their answer options.
So, let us check and eliminate until we get the answer.
$
\underline{Option\;F} \\[3ex]
22 \div 6 = 3 \;R\; 4 \; \checkmark \\[3ex]
22 \div 7 = 3 \;R\; 1 \;\text{remainder is 1, not 5} \\[3ex]
NEXT \\[5ex]
\underline{Option\;G} \\[3ex]
33 \div 6 = 5 \;R\; 3 \;\;\text{remainder is 3, not 4} \\[3ex]
NEXT \\[5ex]
\underline{Option\;H} \\[3ex]
40 \div 6 = 6 \;R\; 4 \; \checkmark \\[3ex]
40 \div 7 = 5 \;R\; 5 \; \checkmark \\[3ex]
STOP \\[3ex]
$
Option H is the correct answer.
The answer choices in the options are arranged is ascending order.
So, 40 is the least positive number among the remaining choices that we did not test.
Student: Is there another way to do this question without checking by the answer options? Teacher: Yes, we can do it: Modular Arithmetic
However, I think it takes more than a minute to do.
$
Let: \\[3ex]
dividend = d \\[3ex]
quotient = q \\[3ex]
1st:\;\; \text{Remainder of 4 when divided by 6} \\[3ex]
d \equiv 4 \mod 6...cong.(1) \\[5ex]
2nd:\;\; \text{Remainder of 5 when divided by 7} \\[3ex]
d \equiv 5 \mod 7 ...cong.(2) \\[5ex]
\implies \\[3ex]
d = 7q + 5 ...eqn.(1) \\[3ex]
Substitute\;\;eqn.(1) \;\;for\;\;d\;\;in\;\;cong.(1) \\[3ex]
7q + 5 \equiv 4 \mod 6 \\[3ex]
\text{Test positive integers for q beginning from the first positive integer} \\[3ex]
7(1) + 5 = 12 \equiv 0 \mod 6...Not\;\;4 \\[3ex]
7(2) + 5 = 19 \equiv 1 \mod 6...Not\;\;4 \\[3ex]
7(3) + 5 = 26 \equiv 2 \mod 6...Not\;\;4 \\[3ex]
7(4) + 5 = 33 \equiv 3 \mod 6...Not\;\;4 \\[3ex]
7(5) + 5 = 40 \equiv 4 \mod 6 \;\checkmark \\[3ex]
\implies \\[3ex]
d = 40
$
(18.) The product of 2 real numbers is a nonzero rational number.
Which of the following statements CANNOT be true?
A. Both numbers are irrational. B. Both numbers are rational. C. Both numbers are integers. D. One number is positive, and the other is negative. E. One number is rational, and the other is irrational.
Let us test each option.
In other words, let us determine any two real numbers whose product would not be a nonzero rational
number.
(19.) Five different stores have different prices for a particular brand and style of T-shirts.
Those prices are shown in the table below.
Which of the 5 stores offers the lowest price for 6 of these T-shirts?
Store
Price
1
2
3
4
5
$7.00 each
2 for $14.99
3 for $20.95
6 for $41.95
Buy 2 for $21.10, get 1 free
Store 3 offers the lowest price for 6 of these T-shirts.
Student: Mr. C Teacher: Yes, my dear Student Student: Can we just determine the unit price of each shirt?
I mean the actual unit price of each shirt, because the free shirt in Store 5 is not really free
The lowest unit price would give the lowest price for 6 shirts
Can we do that? Teacher: Yes, we can.
That's a nice suggestion.
Let's do it.
Store
Price
Unit Price
1
$7.00 each
$\$7.00$
2
2 for $14.99
$\dfrac{\$14.99}{2} = \$7.495$
3
3 for $20.95
$\dfrac{\$20.95}{3} = \$6.983333333$
4
6 for $41.95
$\dfrac{\$41.95}{6} = \$6.991666667$
5
Buy 2 for $21.10, get 1 free
1 free is not free
You must buy two to get that extra 1
If you do not buy two, you do not get 1
So, getting an extra 1 is dependent on the condition that you must buy 2
This is the same as:
Buy 3 for $21.10
$\dfrac{\$21.10}{3} = \$7.033333333$
This approach also confirms Store 3
(20.) For certain positive integers a and b, the greatest common divisor of a and
b is 1, and 9a = 4b.
If it can be determined, which of the following statements must be true for a and b?
F. 2 is a prime factor of a, and 3 is a prime factor of b. G. 2 is a prime factor of a, and 3 is not a prime factor of b. H. 2 is not a prime factor of a, and 3 is a prime factor of b. J. 2 is not a prime factor of a, and 3 is not a prime factor of b. K. Cannot be determined from the given information.
$
9a = 4b \\[3ex]
\implies \\[3ex]
\dfrac{a}{b} = \dfrac{4}{9} \\[5ex]
\implies \\[3ex]
a = 4 \\[3ex]
b = 9 \\[3ex]
$
2 is a prime factor of a (because 2 is a factor of 4 and 2 is a prime number)
3 is a prime factor of b (because 3 is a factor of 9 and 3 is a prime number)
The correct answer is Option F.
(22.) For what values of x, if any, is $-|-x| \lt 0$ true?
F. No real values of x G. Only negative values of x H. Only positive values of x J. All real values of x except 0 K. All real values of x
Let us test values of real numbers to determine the correct option.
$
\text{Left-Hand Side of the Inequality} \\[3ex]
-|-x| \\[3ex]
Test:\;\;x = -3 \\[3ex]
-|-(-3)| \\[3ex]
= -|3| \\[3ex]
= -3 \\[3ex]
-3 \lt 0 ...works \\[5ex]
Test:\;\;x = 0 \\[3ex]
-|-0| \\[3ex]
= -0 \\[3ex]
= 0 \\[3ex]
0 = 0 \\[3ex]
0 \text{is not less than} 0 ...\text{does not work} \\[5ex]
Test:\;\;x = 3 \\[3ex]
-|-3| \\[3ex]
= -3 \\[3ex]
-3 \lt 0 ...works \\[5ex]
$
All real values of x except 0 does not work.
The correct option is Option J.
(23.) An automobile gasoline tank is $\dfrac{3}{8}$ full.
After 6 gallons of gasoline are added to the tank, it is $\dfrac{3}{4}$ full.
Gasoline sells for $1.50 per gallon.
If the tank is empty, what would be the cost to fill it $\dfrac{3}{4}$ full?
$
\text{Fraction of the Initial Volume of Gasoline Tank} = \dfrac{3}{8} \\[5ex]
\text{6 Gallons of Gasoline Added} \\[3ex]
\text{Fraction of New Volume of Gasoline Tank} = \dfrac{3}{4} \\[5ex]
\implies \\[3ex]
\text{Fraction of the Volume of Gasoline Tank Due to the 6 Gallons of Gasoline} \\[3ex]
= \dfrac{3}{4} - \dfrac{3}{8} \\[5ex]
= \dfrac{6}{8} - \dfrac{3}{8} \\[5ex]
= \dfrac{6 - 3}{8} \\[5ex]
= \dfrac{3}{8} \\[5ex]
\text{6 Gallons account for } \dfrac{3}{8} \\[5ex]
\text{How many gallons accounts for } \dfrac{3}{4}? \\[5ex]
$
(24.) An investment doubles in worth every 7 years.
The worth of this investment was $24,000 exactly 21 years after the investment was made.
The worth of the investment exactly 8 years after the investment was made was between:
F.. $0 and $3,000 G.$3,000 and $6,000 H.$6,000 and $9,000 J.$9,000 and $12,000 K.$12,000 and $24,000
Investment doubles in worth every 7 years.
In 21 years:
21 ÷ 7 = 3
This means that it will double in worth 3 times ...over 21 years.
Investment amount is $24,000 in 21 years
So, we have to work backwards to find the investment amount after 8 years.
In working backwards:
We shall divide the investment amount by 2 (rather than multiplying by 2 because we
are working backwards).
We shall subtract 7 years from the years (rather than adding 7 years to the years because we are working
backwards).
Let us represent this information using a table.
It might make more sense that way.
Investment Worth ($)
Years
24000
21
24000 ÷ 2 = 12000
21 − 7 = 14
12000 ÷ 2 = 6000
14 − 7 = 7
The investment amount was $6000 in exactly 7 years.
Ceteris paribus, this implies that the investment amount is between $6000 and
$9000 in exactly 8 years after the investment was made.
(25.) Five points (P, Q, R, S, and T) are on a line in the order given.
The length of $\overline{PR}$ is 12 inches, the length of $\overline{QT}$ is 15 inches, $\overline{QR}$ is the
same length as $\overline{ST}$, and $\overline{PQ}$ is the same length as $\overline{RS}$.
How many inches long is $\overline{PT}$?
(26.) You want to list all the positive two-digit numbers for which the units digit is larger than the tens
digit and for which the sum of the digits is 12.
How many two-digit numbers should be on your list?
A. 1 B. 3 C. 5 D. 6 E. 7
For a two-digit number say: xy:
the tens-digit = x
the units digit = y
Positive two-digit numbers are: 10 – 99
For which:
The units digit is larger than the tens digit are:
12 – 19; 23 – 29; 34 – 39; 45 – 49; 56 – 59; 67 – 69; 78, 79, 89
For which:
The sum of the digits is 12 are:
39, 48, 57
There are only 3 numbers in the list.
(27.) For $\overleftrightarrow{RT}$ shown below, point S is on $\overline{RT}$, the length of
$\overline{RS}$ is 8 cm, and
the length of $\overline{ST}$ is 20 cm.
What is the distance, in centimeters, between T and the midpoint of $\overline{RS}$?
$
p = 40 \\[3ex]
q = -12 \\[3ex]
p + q \\[3ex]
= 40 + (-12) \\[3ex]
= 40 - 12 \\[3ex]
= 28 \\[5ex]
-4 \cdot what = 28 \\[3ex]
what = \dfrac{28}{-4} \\[5ex]
what = -7
$
(29.) Ben and Shawnee are painting a room in the library.
They started with 7 gallons of paint.
On the first day, Ben used $\dfrac{3}{4}$ gallon of paint and Shawnee used $3\dfrac{1}{2}$ gallons of paint.
How many gallons of paint were left when they completed their first day of painting?
(30.) Ken is paid a regular hour wage of $15 per hour, before taxes and benefits are deducted,
for working up to and including 40 hours in 1 week.
For each additional hour he works in a week, Ken is paid 2 times his regular hourly wage.
Ken worked 44 hours this week.
What was his pay for this week before taxes and benefits were deducted?
(31.) Consider the 4 expressions below, where m and n are distinct integers greater than 2.
$
\dfrac{m}{n - 1}, \;\;\; \dfrac{m}{n}, \;\;\; \dfrac{m}{n + 1}, \;\;\; \dfrac{m - 1}{n} \\[5ex]
$
If it can be determined, which of the 4 expressions must have the greatest value?
$
A.\;\; \dfrac{m}{n - 1} \\[5ex]
B.\;\; \dfrac{m}{n} \\[5ex]
C.\;\; \dfrac{m}{n + 1} \\[5ex]
D.\;\; \dfrac{m - 1}{n} \\[5ex]
E.\;\;\text{Cannot be determined from the given information} \\[3ex]
$
Let us test values to determine the correct option
$m, n > 2$
Test:
$m = 3, n = 3$
$m = 3, n = 4$
$\dfrac{m}{n - 1}$
$
\dfrac{3}{3 - 1} \\[5ex]
1.5
$
$
\dfrac{3}{4 - 1} \\[5ex]
1
$
$\dfrac{m}{n}$
$
\dfrac{3}{3} \\[5ex]
1
$
$
\dfrac{3}{4} \\[5ex]
0.75
$
$\dfrac{m}{n + 1}$
$
\dfrac{3}{3 + 1} \\[5ex]
0.75
$
$
\dfrac{3}{4 + 1} \\[5ex]
0.6
$
$\dfrac{m - 1}{n}$
$
\dfrac{3 - 1}{3} \\[5ex]
0.6\bar{6}
$
$
\dfrac{3 - 1}{4} \\[5ex]
0.5
$
Compare:
$1.5 \gt 1 \gt 0.75 \gt 0.6\bar{6}$
$1 \gt 0.75 \gt 0.6 \gt 0.5$
The expression that has the greatest value is $\dfrac{m}{n - 1}$
(33.) The original price of an item was decreased by 20%.
The 1st reduced price was decreased by 20% and then that 2nd reduced price was decreased by 50%.
The price that resulted from these 3 decreases was what percent less than the original price?
Original price: reduced by 20%
20% of p
= 0.2(p)
= 0.2p p decreased by 20%
= 20% off p
= p − 0.2p
= 0.8p
1st reduced price: reduced by 20%
20% of 0.8p
= 0.2(0.8p)
= 0.16p 0.8p decreased by 20%
= 20% off 0.8p
= 0.8p − 0.16p
= 0.64p
2nd reduced price: reduced by 50%
50% of 0.64p
= 0.5(0.64p)
= 0.32p 0.64p decreased by 50%
= 50% off 0.64p
= 0.64p − 0.32p
= 0.32p
The price that resulted from these 3 decreases was what percent less than the original price?
0.32p is what percent less than p?
Let's do it this way:
0.32p is what number less than p? p − 0.32p = 0.68p
0.68p is what percent of p?
Student: Mr. C, is there a way we can attempt this question without using a variable?
The question did not ask us to find the original price.
Do you not think it's much better to just assume that the original price is something like a dollar?
Then, work from there. Teacher: Good observation. We can do it that way.
Let's do it.
Let the original price of the item = $1.00
Original price: reduced by 20%
20% of 1
= 0.2(1)
= 0.2
1 decreased by 20%
= 20% off 1
= 1 − 0.2
= 0.8
1st reduced price: reduced by 20%
20% of 0.8
= 0.2(0.8)
= 0.16
0.8 decreased by 20%
= 20% off 0.8
= 0.8 − 0.16
= 0.64
2nd reduced price: reduced by 50%
50% of 0.64
= 0.5(0.64)
= 0.32
0.64 decreased by 50%
= 50% off 0.64
= 0.64 − 0.32
= 0.32
The price that resulted from these 3 decreases was what percent less than the original price?
0.32 is what percent less than 1?
Let's do it this way:
0.32 is what number less than 1?
1 − 0.32 = 0.68
0.68 is what percent of 1?
$
\dfrac{is}{of} = \dfrac{what}{100} ...Percent-Proportion \\[5ex]
\dfrac{0.68}{1} = \dfrac{what}{100} \\[5ex]
what = 0.68(100) \\[3ex]
what = 68\% \\[3ex]
$
The answer is 68%
0.32p is 68% less than p
(34.) Which of the following values is equal to $4\sqrt{24} - 5\sqrt{54}$
(35.) The Student Council is preparing a budget for an upcoming fund-raising dance.
They have decided to spend $225 for a light and sound show, $600 for refreshments,
and $75 for decorations.
These are the only expenses.
Given that the Student Council estimates 400 students will attend the dance, what should be the price, per
student, for admission to the
dance if the Student Council wants to raise as close as possible to $500 after paying expenses?
In the context of the question, the only even numbers are the multiples of 2.
This is $2n$
For $\sqrt{n}$, the square root of several even numbers are decimals (not even numbers.)
Example: $\sqrt{6}$ is not an even number.
For $\dfrac{n}{2}$, some even-number dividends give odd-number quotients when divided by 2.
Examples are 10, 14, etc.
$10 \div 2 = 5$
5 is not an even number.
(37.) What is the least positive number that has a remainder of 4 when divided by 7 and a remainder of 3 when
divided by 6?
Due to the fact that one minute is allocated to each question on the ACT (60 minutes for 60 questions), it
is better to check the solution to this question by their answer options.
So, let us check and eliminate until we get the answer.
$
\underline{Option\;F} \\[3ex]
20 \div 7 = 2 \;R\; 6 \; \\[3ex]
\text{remainder is 6, not 4} \\[3ex]
NEXT \\[5ex]
\underline{Option\;G} \\[3ex]
35 \div 7 = 5 \;R\; 0 \\[3ex]
\text{remainder is 0, not 4} \\[3ex]
NEXT \\[5ex]
\underline{Option\;H} \\[3ex]
39 \div 7 = 5 \;R\; 4 \checkmark \\[3ex]
39 \div 6 = 6 \;R\; 3 \; \checkmark \\[3ex]
STOP \\[3ex]
$
Option H is the correct answer.
It is less than the remaining choices even if those choices work.
Student: I do not want to make any assumptions, however, I am beginning to notice a pattern with
Option H.
Similar questions in Numbers (7.) and (17.) has option H. as the answer.
Is there another way to do this question without checking by the answer options? Teacher: Yes, we can do it: Modular Arithmetic
However, I think it takes more than a minute to do.
$
Let: \\[3ex]
dividend = d \\[3ex]
quotient = q \\[3ex]
1st:\;\; \text{Remainder of 4 when divided by 7} \\[3ex]
d \equiv 4 \mod 7...cong.(1) \\[5ex]
2nd:\;\; \text{Remainder of 3 when divided by 6} \\[3ex]
d \equiv 3 \mod 6 ...cong.(2) \\[5ex]
\implies \\[3ex]
d = 6q + 3 ...eqn.(1) \\[3ex]
Substitute\;\;eqn.(1) \;\;for\;\;d\;\;in\;\;cong.(1) \\[3ex]
6q + 3 \equiv 4 \mod 7 \\[3ex]
\text{Test positive integers for q beginning from the first positive integer} \\[3ex]
6(1) + 3 = 9 \equiv 2 \mod 7...Not\;\;4 \\[3ex]
6(2) + 3 = 15 \equiv 1 \mod 7...Not\;\;4 \\[3ex]
6(3) + 3 = 21 \equiv 0 \mod 7...Not\;\;4 \\[3ex]
6(4) + 3 = 27 \equiv 6 \mod 7...Not\;\;4 \\[3ex]
6(5) + 3 = 33 \equiv 5 \mod 7 \;\checkmark \\[3ex]
6(6) + 3 = 39 \equiv 4 \mod 7 \;\checkmark \\[3ex]
\implies \\[3ex]
d = 39
$
(38.) Consider the number $630.6 \times 10^a$, where a is an integer.
What is scientific notation for this number?
$
x = 5 \\[3ex]
y = 3 \\[3ex]
z = -6 \\[5ex]
x + y - z \\[3ex]
= 5 + 3 - (-6) \\[3ex]
= 5 + 3 + 6 \\[3ex]
= 14 \\[5ex]
y + z \\[3ex]
= 3 + (-6) \\[3ex]
= 3 - 6 \\[3ex]
= -3 \\[5ex]
(x + y - z)(y + z) \\[3ex]
= (14)(-3) \\[3ex]
= -42
$
(40.) Coach Shannon is buying packages of granola bars, juice boxes, and apples as snacks for her soccer team.
The table below gives the number of snacks per package and the price per package.
Snack type
Snacks per package
Price per package
Granola bars
Juice boxes
Apples
3
4
5
$2.50 $3.00 $4.50
What is the minimum total price of the snacks, all bought in whole packages, Coach Shannon buys so that each
of the 15 girls on the team gets at least 1 snack of each type?
15 girls on the soccer team
Each girl should get at least 1 snack of each type
Granola: 3 snacks per package; $2.50 per package
15 ÷ 3 = 5
5 packages of Granola are needed
This costs: $5(\$2.50) = \$12.50$
Juice boxes: 4 snacks per package; $3.00 per package
15 ÷ 4 = 3.75
This is not a whole number...because each girl will get at least 1 snack
Round it up
4 packages of Juice boxes are needed
This costs: $5(\$3.00) = \$12.00$
Apples: 5 snacks per package; $4.50 per package
15 ÷ 5 = 3
3 packages of Apples are needed
This costs: $3(\$4.50) = \$13.50$
The minimum total price of the snacks
= $12.50 + $12.00 + $13.50
= $38.00
(41.) A cake recipe requires $\dfrac{5}{8}$ cup of flour.
Mary and Haloa decide to make the cake together.
Mary has $\dfrac{1}{3}$ cup of flour and Haloa has $\dfrac{1}{4}$ cup of flour.
How many more cups of flour do they need to make the cake?
