Below are answer explanations to the full-length Math test of the previously released ACT from the current 2015-2016 “Preparing for the ACT Test” (form 1572CPRE) free study guide available here for free PDF download.

The ACT Math test explained below begins on page 24 of the guide. Please note that the 2015-2016 ACT practice test is the same one used in the 2016-2017 and 2017-2018 “Preparing for the ACT Test” guides. Other answer explanations in this series of articles:

When you’re finished reviewing this official practice ACT test, start practicing with our own 10 full-length practice ACT tests—absolutely free during the pandemic.

## ACT Practice Test Math Answer Explanations 2015-2018

Question 1 “Blood types” the answer is D

Probability is calculated by $\frac{\text{Amount&space;of&space;things&space;that&space;fit&space;our&space;criteria}}{\text{total&space;number&space;of&space;things}}$.

1. 67 people have type A blood. 6 people have type AB blood.
2. 67 + 6 = 73 people have type A or type AB blood.
3. There are 150 total people.
4. The answer is $\frac{73}{150}$, which is choice D

Question 2 “Monthly fees” the answer is H

Mean (or average) is calculated by $\frac{\text{sum&space;of&space;the&space;things}}{\text{total&space;number&space;of&space;things}}$

1. The sum of these monthly fees is 370 + 310 + 380 + 340 + 310 = 1710
2. There are 5 total rooms
3. The answer is $\frac{1710}{5}$ = 342, which is choice H

Scale factor problems should be set up as proportions.

1. $\frac{\frac{1}{2}\text{&space;inches}}{18\text{&space;miles}}&space;=&space;\frac{2\frac{1}{2}\text{&space;inches}}{x\text{&space;miles}}$
2. We put inches on the top and miles on the bottom. It’s fine to set it up the other way, as long as you are consistent.
3. Cross multiply to get the equation $\frac{1}{2}&space;\cdot&space;x&space;=&space;2\frac{1}{2}\cdot&space;18$
4. Solve this equation (divide both sides by $\frac{1}{2}$, use a calculator) to obtain x = 90, which is choice E

Question 4 “$f&space;=&space;cd^3$” the answer is F

1. First we plug in our numbers from the problem. $f&space;=&space;cd^3$ becomes $450&space;=&space;c&space;\cdot&space;10^3$
2. Then simplify: $450&space;=&space;c&space;\cdot&space;1000$
3. Finally divide both sides by 1000 to obtain$\frac{450}{1000}&space;=&space;c$, and the decimal for $\frac{450}{1000}$ is .45, which is choice F

Question 5 $f(x)&space;=&space;(3x&space;+&space;7)^2$” the answer is E

This question is testing that we understand function notation.

1. To calculate f(1) plug x = 1 into the equation above to obtain $f(1)&space;=&space;(3&space;\cdot&space;1&space;+&space;7)^2$
2. Simplify to get $f(1)&space;=&space;10^2&space;=&space;100$, which is answer choice E

Question 6 “Jorge’s hourly wage” the answer is H

In order to take a percent of something first convert it to either a fraction or a decimal.

1. 6% =$\frac{6}{100}$ = .06
2. 6% of 12.00 = .06 $\cdot$ 12.00 = 0.72
3. Since his wage increases by 6%, we add that to his current wage we get 12.00 + 0.72 = 12.72 which is choice H

A quicker way to do this is to note that if his wage increases by 6% his new wage will be 100% + 6% = 106% of his old wage. 106% of 12.00 = 1.06 $\cdot$ 12.00 = 12.72

Question 7 “Geometric sequence” the answer is E

Geometric sequences are multiplying sequences.

1. Divide any two consecutive terms to find that the common ratio for this sequence is -3. For example $\frac{-27}{9}$ = -3
2. Now keep multiplying each term by -3 to find the next terms.
3. The 5th term is $\inline&space;-27&space;\cdot&space;-3&space;=&space;81$
4. The 6th term is $\inline&space;81&space;\cdot&space;-3&space;=&space;-243$
5. The 7th term is $-243&space;\cdot&space;-3&space;=&space;729$ which is answer choice E

If you know the formula for geometric series, you may use it to find the answer but it isn’t necessary.

Question 8 “Ship Quick” the answer is H

This question hinges on whether we can read and interpret this chart.

1. Since 15 is in the range “10-25” we will use that row of the chart.
2. The fee for “10-25” is $10.00 3. The price per pound is$0.65. For 15 pounds this will make 15 $\cdot$ $0.65 =$9.75
4. The total is therefore $10.00 +$9.75 = $19.75 which is answer choice H Question 9 “computer chip” the answer is A 1. The top and bottom layers are a total of 0.03 + 0.03 = 0.06 cm thick. 2. The inner layers use the remaining 0.32 – 0.06 = 0.26 cm 3. Since they are 0.02 cm thick, the question is “how many groups of 0.02 are there inside of 0.26” which is a division problem. $\frac{0.26}{0.02}$ = 13, which is choice A. Note: this problem could also have been solved by working from answer choices. For example, looking at answer choice C, if there are 16 layers of 0.02 each and 2 layers (top and bottom) of 0.03 each, then there is a total of 16 $\cdot$ 0.02 + 2 $\cdot$ 0.03 = 0.38 cm. Since this is too big, we know the correct answer is smaller than 16, so we can repeat this process with either answer choice A or B. Question 10 the answer is K Since the median is the “middle” of the data, our numbers need to be in order. 1. In order, our numbers are 13, 15, 16, 19, 19, 22, 25, 25, 26, 27, 28, 29. 2. The middle of this data set is between 22 and 25. When this happens we take an average of those two numbers. 3. $\frac{22&space;+&space;25}{2}&space;=&space;\frac{47}{2}&space;=&space;23.5$ which is answer choice K Question 11 “rolling cart” the answer is C This problem can be quickly solved by plugging into the answer choices. 1. If we plug in t = 0 into the 5 answer choices, only answer choices A and C give the correct value from the table: d = 14. The others can be eliminated. 2. If we plug in t = 1 answer choice A says d = 1 + 14 = 15, which is not the correct value from the table, so we eliminate choice A. 3. Choice C is the only remaining answer, and therefore must be correct. You may also do this problem by recognizing that a constant rate means a linear equation, and comparing the answer choices to y = mx + b. If you do, the intercept will be 14. Since d increases by 6 when t increases by 1 the slope will be $\frac{6}{1}&space;=&space;6$. This shows that choice C is the correct answer. Question 12 “rectangle” the answer is K It’s always a good idea to draw a picture for geometry questions that don’t have one. The area formula for a rectangle is A = lw, and the perimeter formula is P = 2l + 2w. 1. Working from the area formula we have 54 = 9w. Dividing both sides by 9, we find that w = $\frac{54}{9}$ = 6. The width is 6. 2. Plugging into our perimeter formula we have P = 2 $\cdot$ 9 + 2 $\cdot$ 6 = 18 + 12 = 30. It’s not necessary to remember the perimeter formula for all of your shapes, as long as you know that perimeter means to add up all the sides. Question 13 “angles” the answer is B A good place to start on these “angle puzzle” questions is to start filling out any angles we can find. We can use the angles we find to figure out even more angles until we find what we were looking for. 1. Since $\inline&space;\angle&space;DCE$ and $\inline&space;\angle&space;BCA$ are vertical angles, they are congruent. So, $\inline&space;\angle&space;BCA&space;=&space;\angle&space;DCE&space;=&space;45&space;^{\circ}$ 2. Now that we know that $\inline&space;\angle&space;BCA&space;=&space;45&space;^{\circ}$, we have two out of three angles in triangle BCA. Since a triangle has $\inline&space;180&space;^{\circ}$, we can subtract the other two angles to find $\angle&space;BAC$. 3. $\inline&space;35&space;^{\circ}&space;+&space;45&space;^{\circ}&space;=&space;80&space;^{\circ}$$\inline&space;180&space;^{\circ}&space;-&space;80&space;^{\circ}&space;=&space;100&space;^{\circ}$ which is choice B. Question 14 “circle graph” the answer is H To solve this problem, we need to know that the fraction of hours spent on Core Subjects should be equal to the fraction of the circle it takes up in the graph. 1. $\inline&space;\frac{\text{core&space;subject&space;hours}}{\text{total&space;hours}}&space;=&space;\frac{\text{core&space;subject&space;angle&space;measure}}{\text{total&space;angle&space;measures}}$ 2. There are $\inline&space;360&space;^{\circ}$ in a circle. He spent 4 hours on core subjects. He spent 4 + 3 + 1 + 1 = 9 total hours on all of the subjects. We don’t know the angle measure for the core subjects sector, so we’ll call it “x”. 3. So, our equation becomes: $\inline&space;\frac{4}{9}&space;=&space;\frac{x}{360}$ 4. Cross multiplying we obtain $\inline&space;4&space;\cdot&space;360&space;=&space;9&space;\cdot&space;x$ 5. Dividing both sides by 9, we get x = 160, which is answer choice H. Since there were 9 total hours, we could have also solved this by seeing that 1 hour must be worth $\inline&space;\frac{360}{9}&space;=&space;40&space;^{\circ}$ and since we spent 4 hours on core subjects that makes $\inline&space;4&space;\cdot&space;40&space;^{\circ}&space;=&space;160&space;^{\circ}$. Question 15 “Kami sold 70 figurines…” the answer is B This problem can be solved by working from the answer choices. Try out the math for each answer choice and see which one works. 1. Choice B: If he sold 28 large figurines, he sold $\inline&space;70&space;-&space;28&space;=&space;42$ small figurines, since we know he sold 70 figurines total. 2. His profit for the large figurines is $\inline&space;28&space;\cdot&space;12&space;=&space;336$ 3. His profit for the small figurines is $\inline&space;42&space;\cdot&space;8&space;=&space;336$ 4. Since the problem stated the sales of the large and small figures were “equal”, this lets us know that choice B is correct. 5. If we had tried a different answer choice first, we would have gotten two numbers that were not equal, so we could eliminate that answer choice. If we had wanted to set up an algebra equation for this problem instead, we would let x be the number of large figures, so 70 – x would be the number of small figures. Multiplying these by the cost of the figures we would get the equation $\inline&space;12&space;\cdot&space;x&space;=&space;8&space;\cdot&space;(70-x)$. Question 16 “a car accelerated” the answer is H Don’t get overwhelmed by “feet per second per second” – they’re asking us to find a unit rate. 1. An acceleration is a change in velocity over time, so first we calculate how much the velocity changed: 220 – 88 = 132. 2. Since this happened over 3 seconds, we divide by 3 to find the unit rate. 3. $\inline&space;\frac{132}{3}&space;=&space;44$, which is answer choice H. Question 17 “lines in a plane” the answer is D This problem is very difficult to do without a picture! Draw a picture for any geometry problems that don’t already have one. 1. If we draw a good picture, we can see that $\inline&space;\angle&space;BAC$ and $\inline&space;\angle&space;BAD$ are a linear pair since they form a straight line together. 2. Since a straight line has $\inline&space;180&space;^{\circ}$, these two lines should make $\inline&space;180&space;^{\circ}$ together. 3. We can subtract to find our missing angle. $\inline&space;180&space;-&space;47&space;=&space;133$, which is answer choice D. Question 18 “arranging fractions” the answer is F Ordering fractions can sometimes be tricky; try using your calculator to turn them into decimals first. 1. $\inline&space;\frac{1}{2}&space;=&space;0.5$ 2. $\inline&space;\frac{5}{6}&space;\approx&space;0.833$ 3. $\inline&space;\frac{5}{8}&space;\approx&space;0.625$ 4. Now we can put the decimals in order: $\inline&space;0.5&space;<&space;0.625&space;<&space;0.833$ 5. If we turn those back into fractions we get: $\inline&space;\frac{1}{2}&space;<&space;\frac{5}{8}&space;<&space;\frac{5}{6}$, which is choice F. Question 19 “scientific notation” the answer is D Don’t try to go to scientific notation right away, do the addition first. 1. 670,000,000 + 700,000,000 = 1,370,000,000 2. To turn this into scientific notation we move the decimal place 9 times, so we get $\inline&space;1.37&space;\times&space;10^9$. If you’re not sure about how to put the numbers into scientific notation you can use your calculator to check the answer choices and see if they make 1,370,000,000. Question 20 “trapezoid angles” the answer is F This question is testing a geometry fact: these angles in a trapezoid are always supplementary, so the answer is F. If you don’t know this fact, there are still some ways of approaching this problem. The steps below will assume that we don’t know this fact. 1. If we draw a vertical line through the center of this figure, it will create a quadrilateral with two right angles, an angle of x, and an unknown angle. 2. Since there are 360 degrees in a quadrilateral, the missing angle must be 360 – 90 – 90 – x = 180 – x Another approach is to think of this as a two parallel lines cut by a transversal question, but you will need to know those rules. If you’re completely unsure how to begin this problem, you can make a very educated guess. The problem says the angles are “distinct” so choice K is incorrect. Choices H and J are angles that are greater than 180, which is impossible in a trapezoid. You now have a 50/50 chance between choices F and G. Question 21 “driver’s license” the answer is B Don’t get ahead of yourself on this problem. Work each step one at a time. 1. 80% of the applicants will pass the written test. $\inline&space;80%&space;\text{&space;of&space;}&space;1000&space;=&space;.80&space;\cdot&space;1000&space;=&space;800$. 2. 60% of these applicants will now pass the driving test. $\inline&space;60%&space;\text{&space;of&space;}&space;800&space;=&space;.60&space;\cdot&space;800&space;=&space;480$, which is choice B Question 22 “a^b = x” the answer is H. 1. Since $\inline&space;x&space;=&space;a^b$ and $\inline&space;y&space;=&space;c^b$, we can substitute into the equation for xy. 2. $xy&space;=&space;(a^b)(c^b)&space;=&space;(ac)^b$, which is choice H. If you’re uncomfortable with the exponent rules here, you can solve this by plugging in some numbers. For instance, take a = 3, b = 2, c = 5. This means $\inline&space;x&space;=&space;3^2&space;=&space;9$ and $\inline&space;y&space;=&space;5^2&space;=&space;25$, so $\inline&space;xy&space;=&space;225$. If we substitute these values for a, b, and c into our answer choices (use your calculator!) only choice H gives a matching value of 225. Question 23 “$\mathbf{\frac{1}{2}y^2(6x&space;+&space;2y&space;+&space;12x&space;-&space;2y)}$” the answer is A This problem is easier if we combine like terms before we distribute. 1. Working in the parentheses: $\inline&space;6x&space;+&space;12x&space;=&space;18x$ and $\inline&space;2y&space;-2y&space;=&space;0$ 2. $\inline&space;\frac{1}{2}y^2(6x&space;+&space;2y&space;+&space;12x&space;-&space;2y)&space;=&space;\frac{1}{2}y^2(18x)$ 3. Simplifying, we obtain $\inline&space;\frac{18}{2}xy^2&space;=&space;9xy^2$ which is choice A If you find this algebra too difficult, you can solve this one by plugging in some numbers. If we take x = 1 and y = 1, our equation becomes$\inline&space;\frac{1}{2}y^2(6x&space;+&space;2y&space;+&space;12x&space;-&space;2y)&space;=&space;\frac{1}{2}(18)&space;=&space;9$. If we substitute these values for x and y into our answer choices (use your calculator!) only choice A gives a matching value of 9. Question 24 “An artist…” the answer is H. 1. If our profit is “at least” 60,000 then it is greater than or equal to 60,000. 2. $\inline&space;(500p&space;-&space;p^2)&space;\geq&space;60000$. Subtracting 60000 from both sides we obtain: 3. $\inline&space;-p^2&space;+&space;500p&space;-&space;60000&space;\geq&space;0$. 4. Rather than solving this by hand, we can graph this equation on our graphing calculator. The answer choices tell us our x-window should include 100 and 600, so 90 to 610 would be a good choice. 5. Our picture will show our function crossing the x-axis and becoming positive at x = 200, which is choice H. If you wanted to do the algebra, you would need to solve the equation in 3) either by factoring or using the quadratic formula. If you find this algebra too difficult, you can solve this one by plugging in the answer choices. Since it asks for “the fewest number”, start with the smallest answer choice and plug it into the formula. Work your way up until you find the first number that yields a profit greater than or equal to 60000. Question 25 “Pie chart” the answer is B First we’ll find the fraction of her total expenditures, and then we’ll convert that fraction to a percent. 1. Clothes are her greatest expense at$254 dollars.
2. The problem already tells us the total expenditures are $900, we don’t need to add all these numbers to figure that out. 3. $\inline&space;\frac{254}{900}&space;=&space;.282$ 4. $\inline&space;.282&space;\cdot&space;100%&space;=&space;28.2%$ which rounds to 28%, which is choice B. Question 26 “Angles” the answer is G 1. Since our picture shows that $\inline&space;\angle&space;BAD&space;=&space;\angle&space;BAC&space;+&space;\angle&space;CAD$, we can subtract to find the value of $\inline&space;\angle&space;CAD$ 2. $\inline&space;\angle&space;CAD&space;=&space;90&space;-&space;(x&space;+&space;20)&space;=&space;70&space;-&space;x$, which is answer choice G. If you weren’t sure how to start this problem, you can still make an educated guess. From the picture, $\inline&space;\angle&space;CAD$ is less than 90, so eliminate choices J and K. Question 27 “isosceles right triangle” the answer is E If you know your special right triangles, this is a 45-45-90 and you can use that knowledge to get the answer very quickly. If you don’t know that fact, you can still get the answer. Since they asked for the perimeter, we should begin by trying to find the missing sides. 1. Since the triangle is isosceles, the two legs of this right triangle will be equal. We’ll call them both “x”. 2. Now, we can use the pythagorean theorem to set up and solve this problem. Be careful, this triangle is facing a weird direction – the hypotenuse is the bottom side. 3. The pythagorean theorem tells us $\inline&space;a^2&space;+&space;b^2&space;=&space;c^2$, which for us means $\inline&space;x^2&space;+&space;x^2&space;=&space;(8\sqrt{2})^2$. 4. Simplifying, we obtain $\inline&space;2x^2&space;=&space;128$. Dividing by two we obtain $\inline&space;x^2&space;=&space;64$, so x = 8. Both of the legs of this triangle are 8. 5. Don’t be fooled by choice A! They asked for the perimeter, we need to add up all of the sides. $\inline&space;8&space;+&space;8&space;+&space;8\sqrt{2}&space;=&space;16&space;+&space;8\sqrt{2}$, which is choice E. Question 28 “$\mathbf{y&space;=&space;ax^2&space;+&space;bx&space;+&space;c}$” the answer is H This question asks us to understand what the “solutions” of an equation look like on a graph. Solutions will occur where the graph crosses the x-axis. 1. Since the graph crosses the x-axis twice, there will be two real solutions. 2. It crosses once on the positive side and once on the negative side, so we will have 1 positive and 1 negative solution, which is choice H. Tip: The terms “solutions”, “roots”, “zeroes”, and “x-intercepts” all mean the same thing and can be used interchangeably! Question 29 “product of complex numbers” the answer is C This question tests our understanding of complex numbers. In particular, we need to know that i^2 = -1. 1. First we should FOIL this expression. $\inline&space;(-3i&space;+&space;4)(3i&space;+&space;4)&space;=&space;-9i^2&space;-&space;12i&space;+&space;12i&space;+16&space;=&space;-9i^2&space;+&space;16$. 2. But since $\inline&space;i^2&space;=&space;-1$, $\inline&space;-9i^2&space;+&space;16&space;=&space;-9\cdot-1&space;+&space;16&space;=&space;9&space;+&space;16&space;=&space;25$, which is choice C. Question 30 “right circular cone” the answer is G This question tests whether we know the basic definitions of trig functions. 1. In the triangle we know the side opposite of $\theta$ and the side adjacent to $\theta$. 2. The trig function that relates opposite to adjacent is tangent. 3. $\inline&space;tan(\theta)&space;=&space;\frac{\text{opposite}}{\text{adjacent}}&space;=&space;\frac{7}{5}$, which is choice G. Tip: Remember “SOH” “CAH” “TOA”: $\inline&space;sin&space;=&space;\frac{\text{opposite}}{\text{hypotenuse}}$, $\inline&space;cos&space;=&space;\frac{\text{adjacent}}{\text{hypotenuse}}$, $\inline&space;tan&space;=&space;\frac{\text{opposite}}{\text{adjacent}}$ Question 31 “jigsaw puzzle” the answer is D Probability is calculated by $\inline&space;\frac{\text{Amount&space;of&space;things&space;that&space;fit&space;our&space;criteria}}{\text{total&space;number&space;of&space;things}}$. 1. There are 5 extra pieces 2. There are 750 + 5 = 755 total pieces 3. The answer is $\inline&space;\frac{5}{755}$, which is choice D. Question 32halfway between fractions” the answer is K. To find what is halfway between two numbers, we average the two numbers. 1. We average these two numbers to get $\inline&space;\frac{\frac{2}{3}&space;+&space;\frac{3}{4}}{2}$ 2. Rather than do the fraction simplification by hand, we should use our calculators. The calculator will tell us the answer is $\inline&space;\frac{17}{24}$ which is choice K. Depending on what kind of calculator you have, you may need to enter the expression as $\inline&space;(\frac{1}{2})((\frac{2}{3})&space;+&space;(\frac{3}{4}))$. If your calculator only gives decimal answers and can’t give fraction answers, you can convert each answer choice into a decimal to find the one that matches. Question 33 “scale factor” the answer is B Scale factor problems should always be set up as proportions. 1. The information we need to work this problem is “.25 inch represents 2 feet” 2. $\inline&space;\frac{.25\text{&space;inches}}{2\text{feet}}&space;=&space;\frac{x\text{&space;inches}}{15\text{&space;feet}}$ 3. We put inches on the top and feet on the bottom. It’s fine to set it up the other way, as long as you are consistent. 4. Cross multiply to get the equation $\inline&space;.25&space;\cdot&space;15&space;=&space;2\cdot&space;x$ 5. Solve this equation (divide both sides by 2, use a calculator) to obtain x = 1.875, which is choice B Question 34 “installing tile” the answer is H This question is asking us to calculate the area of rectangles. We just need to be sure that we find all of the information we need before we get started. 1. We want to calculate the area of the “floor that will NOT be covered by cabinets”. 2. From the picture, we can tell that one portion of the cabinets is $\inline&space;12&space;\times&space;2$. We are also told that in the middle of the room there will be “4 cabinets” that are “2 feet wide and 2 feet deep”. The total area of the cabinets will therefore be $\inline&space;12&space;\cdot&space;2&space;+&space;4&space;\cdot&space;(2&space;\cdot&space;2)&space;=&space;40$. 3. The room is $\inline&space;12&space;\times&space;15$, so the area NOT covered by cabinets will be $\inline&space;12&space;\cdot&space;15&space;-&space;40&space;=&space;140$, which is choice H. Question 35 “installation cost” the answer is D 1. The labor cost is$650 and it will “remain the same”. Since the estimate was for \$2150, the charge for the cabinets must have been 2150 – 650 = 1500.
2. We are told “the charge PER cabinet remains the same” so twice as many cabinets should cost twice as much. $\inline&space;1500&space;\cdot&space;2&space;=&space;3000$.
3. Therefore, her total charge is 3000 + 650 = 3650, which is choice D.

Question 36 “1 < x + y < 2” the answer is J

1. If we subtract x from all three parts of this equation, we get $\inline&space;1&space;-&space;x&space;<&space;y&space;<&space;2&space;-&space;x$.
2. This tells us our region is above the line y = 1 – x, and below the line y = 2 – x. These lines both have a slope of -1, and y-intercepts of 1 and 2 respectively. Choice J is the only choice that has these lines graphed correctly.

If you find this algebra too difficult, you can solve this one by plugging in some numbers from the shaded areas in the answer choices. If we choose the point (0,0) our equation says $\inline&space;1&space;<&space;0&space;<&space;2$, which isn’t true. We can eliminate choices F and K because they have (0,0) in their shaded region. Similarly, if we plug in the points (1.5, 0) and (0, 1.5) we see that they work in our equation, so they should be in the shaded region. Choice J is the only remaining choice that contains both of these points.

Question 37 “mean and median” the answer is A

Mean (or average) is calculated by $\inline&space;\frac{\text{sum&space;of&space;the&space;things}}{\text{total&space;number&space;of&space;things}}$. The median is the “middle” of the data.

1. Our mean will be $\inline&space;\frac{3+8+10+15}{4}&space;=&space;9$.
2. For our median, the “middle” of our data falls between 8 and 10, so we average those two numbers. $\inline&space;\frac{8&space;+&space;10}{2}&space;=&space;9$.
3. The difference between the two is $\inline&space;9&space;-&space;9&space;=&space;0$.

Question 38 “graphs of functions” the answer is F

This question is testing our understanding of the relationship between functions and their graphs. If the graphs of two functions intersect, those functions are equal at those points.

1. Since these two graphs intersect at two points, the functions are equal at those two points. The correct answer is A.

Question 39 “slope” the answer is B

The formula to calculate slope given two points is $\inline&space;\frac{y_2&space;-&space;y_1}{x_2&space;-&space;x_1}$

1. Are two points are $\inline&space;(x_1,y_1)&space;=&space;(9,4)$ and $\inline&space;(x_2,y_2)&space;=&space;(12,1)$
2. The slope will be $\inline&space;\frac{y_2&space;-&space;y_1}{x_2&space;-&space;x_1}&space;=&space;\frac{12-9}{1-4}&space;=&space;\frac{3}{-3}=-1$, which is choice B.

Question 40 “reflection” the answer is F

The formula for reflection over the y-axis is $\inline&space;(x,y)&space;\mapsto&space;(-x,y)$.

1. Using the formula $\inline&space;(12,1)&space;\mapsto&space;(-12,1)$, which is answer choice F

If you didn’t know this formula, a good strategy is to draw approximately where D’ will be. Imagine folding your test booklet over along the y-axis. Where does the point D “land”? If your picture is reasonably good, choice F is the only one that will be remotely close.

Question 41 “2 trapezoids of equal area” the answer is E.

The formula for the area of a trapezoid is $\inline&space;\frac{b_1&space;+&space;b_2}{2}\cdot&space;h$. It is possible to solve this problem by working from the answer choices, but this will prove to be very tedious. The algebra provides a faster path to the solution. If the algebra proves too difficult, see the note after this solution for another approach.

1. If we cut this trapezoid along the line x = a, the left trapezoid will have a lower base of $\inline&space;a-2$, an upper base of $\inline&space;a-3$, and a height of 3. The area of this trapezoid will therefore be $\inline&space;\frac{a&space;-&space;3&space;+&space;a&space;-&space;2}{2}\cdot&space;3&space;=&space;3a&space;-&space;7.5$.
2. The right trapezoid will have a lower base of $\inline&space;12-a$, an upper base of $\inline&space;9-a$, and a height of 3. The area of this trapezoid will therefore be $\inline&space;\frac{12&space;-&space;a&space;+&space;9&space;-&space;a}{2}\cdot&space;3&space;=&space;31.5&space;-&space;3a$.
3. If these two shapes have equal area then $\inline&space;3a&space;-&space;7.5&space;=&space;31.5&space;-&space;3a$, so $\inline&space;6a&space;=&space;39$, and $\inline&space;a&space;=&space;\frac{39}{6}=6.5$, which is choice E.

If you’re not sure how to work the algebra, this problem can be solved by estimation. Since we’re cutting this into two equal sized pieces, the line should go roughly down the middle of the figure. If you sketch that in, the x-coordinate should be about halfway between the bottom two points. The average of 2 and 12 is 7, making 6.5 a very strong guess.

Question 42 “$\mathbf{f(g(\frac{1}{2}))}$” the answer is K

Don’t try to calculate the formula for f(g(x)), just work in pieces.

1. To calculate $\inline&space;f(g(\frac{1}{2}))$ we work from the inside out. We’ll start with $\inline&space;g(\frac{1}{2})$ and then plug our answer into f(x).
2. Since $\inline&space;g(x)&space;=&space;\frac{1}{x}$, $\inline&space;g(\frac{1}{2})&space;=&space;\frac{1}{\frac{1}{2}}&space;=&space;2$ (use your calculator!).
3. $\inline&space;f(2)&space;=&space;2&space;-&space;\frac{1}{2}&space;=&space;\frac{3}{2}$, which is choice K.

Question 43 “monthly payment” the answer is D

Tracing how changes in variables affect the answer can be simple or very complicated depending on the formula. One approach that always works is to plug in some numbers and see what happens.

1. If we let a = 1, r = 1, and y = 1 then the expression $\inline&space;\frac{\frac{1}{2}ary&space;+&space;a}{12y}$ becomes $\inline&space;\frac{\frac{1}{2}&space;\cdot&space;1&space;\cdot&space;1&space;\cdot&space;1&space;+&space;1}{12&space;\cdot&space;1}&space;=&space;\frac{1.5}{12}&space;=&space;.125$. So $\inline&space;p&space;=&space;.125$.
2. Now, if we multiply a by 2, we have a = 2 and the expression $\inline&space;\frac{\frac{1}{2}ary&space;+&space;a}{12y}$ becomes $\inline&space;\frac{\frac{1}{2}&space;\cdot&space;2&space;\cdot&space;1&space;\cdot&space;1&space;+&space;2}{12&space;\cdot&space;1}&space;=&space;\frac{3}{12}&space;=&space;.25$. So, P = .25. Since this is twice as much, the answer is D.

If you wanted to do this with algebra, substitute 2a into the formula instead of a. You should get $\inline&space;\frac{\frac{1}{2}(2a)ry&space;+&space;2a}{12y}&space;=&space;2&space;\cdot&space;\frac{\frac{1}{2}ary&space;+&space;a}{12y}$. Since the coefficient of 2 can be factored out, the answer is D.

Question 44 “coordinates of D” the answer is G.

It might be tempting to try to apply the distance formula, but that will make this problem very difficult. A better way to approach it is similar to the midpoint formula. We will calculate the values for each coordinate separately.

1. The x-coordinates are 14 – 6 = 8 apart. Since this distance is 4 times the length of DE, we will divide this by 4. $\inline&space;\frac{8}{4}&space;=&space;2$. Since DE contains the point E, we will count our distance from the coordinates of E. $\inline&space;6&space;+&space;2&space;=&space;8$. The x-coordinate of D is 8.
2. The y-coordinates are 12 – 4 = 8 apart. Since this distance is 4 times the length of DE, we will divide this by 4. $\inline&space;\frac{8}{4}&space;=&space;2$. Since DE contains the point E, we will count our distance from the coordinates of E.$\inline&space;4&space;+&space;2&space;=&space;6$. The y-coordinate of D is 6.
3. So, the coordinates of D are (8,6) which is choice G.

If you’re unsure of how to start this problem, you can make a very educated guess. If you place a dot for point D $\inline&space;\frac{1}{4}$th of the way along EF, you’ll see that point D is very close to point E. F, and G would both be good guesses.

Question 45 “matrices” the answer is D

Multiplying a matrix by a constant is as simple as multiplying each entry by that number.

1. $a\begin{bmatrix}&space;2&space;&&space;6\\&space;1&space;&&space;4&space;\end{bmatrix}&space;=&space;\begin{bmatrix}&space;2a&space;&&space;6a&space;\\&space;a&space;&&space;4a&space;\end{bmatrix}$
2. So,$\begin{bmatrix}&space;2a&space;&&space;6a&space;\\&space;a&space;&&space;4a&space;\end{bmatrix}&space;=\begin{bmatrix}&space;x&space;&&space;27\\&space;y&space;&&space;z&space;\end{bmatrix}$
3. Now we can work from corresponding pieces. We should have $6a=27$, so $a&space;=&space;4.5$
4. This tells us that $\inline&space;x&space;=&space;2&space;\cdot&space;4.5&space;=&space;9$ and $\inline&space;z&space;=&space;4&space;\cdot&space;4.5&space;=&space;18$, so $\inline&space;x+z=9+8=27$, which is choice D.

Question 46 “a container full of water” the answer is J

1. The volume of water in the tank increased from $\inline&space;\frac{1}{8}$ to $\inline&space;\frac{3}{4}$. This is a difference of $\inline&space;\frac{3}{4}&space;-&space;\frac{1}{8}&space;=&space;\frac{5}{8}$. So, $\inline&space;\frac{5}{8}$ of the total capacity must equal the amount of water that was added: 10 cups.
2. So, if x represents the total capacity we have $\frac{5}{8}&space;x&space;=&space;10$. Dividing both sides by $\inline&space;\frac{5}{8}$ (use your calculator!) we find that $\inline&space;x&space;=&space;16$.

We also could have set this problem up as a proportion: $\inline&space;\frac{\frac{5}{8}\text{&space;tank}}{10\text{&space;cups}}&space;=&space;\frac{1\text{&space;tank}}{x\text{&space;cups}}$.

Question 47 “Washington High School” the answer is B

We can’t actually figure out how many students attend this school, but we don’t need too. All that matters are the ratios. It can help to write these ratios as fractions, rather than using colons.

1. Since $\inline&space;\frac{86}{255}$ of the school are 10th graders and $\inline&space;\frac{18}{51}$ are 11th graders, $\inline&space;\frac{86}{255}&space;+&space;\frac{18}{51}&space;=&space;\frac{176}{255}$ of the school are in either 10th or 11th grade.
2. This means $\inline&space;1&space;-&space;\frac{176}{255}&space;=&space;\frac{79}{255}$ of the school are 12th graders.
3. Of these three ratios the one for 11th graders ($\inline&space;\frac{18}{51}&space;=&space;\frac{90}{255}$) is the greatest, so the answer is B.

A different way to conceptualize this process is to pretend there are 255 students. Since $\inline&space;\frac{18}{51}&space;=&space;\frac{90}{255}$, there would be 90 11th graders, 86 10th graders, and 79 12th graders.

This problem may seem challenging, but the same rules for adding fractions still apply. We need to find a common denominator.

1. A common denominator between $\inline&space;\sqrt{2}$ and $\inline&space;\sqrt{3}$ is $\inline&space;\sqrt{2}&space;\cdot&space;\sqrt{3}&space;=&space;\sqrt{6}$. All we need to do is multiply to take our fractions to the common denominator, and then add.
2. $\inline&space;\frac{4}{\sqrt{2}}&space;\cdot&space;\frac{\sqrt{3}}{\sqrt{3}}&space;+&space;\frac{2}{\sqrt{3}}&space;\cdot&space;\frac{\sqrt{2}}{\sqrt{2}}&space;=&space;\frac{4\sqrt{3}+2\sqrt{2}}{\sqrt{6}}$, which is answer choice G.

If working with the square roots is too difficult, another option is to use your calculator to get a decimal for this expression, and then check it against the decimal values of each answer choice.

1. We are below the line y =-x+2, so our region should have $\inline&space;y<-x+2$.
2. We are inside the circle $\inline&space;(x-1)^2&space;+&space;(y-2)^2&space;=&space;9$, so our region should have $\inline&space;(x-1)^2&space;+&space;(y-2)^2&space;<&space;9$.
3. Only choice A shows both of these

A different approach is to try to work from the graph. The only point in our shaded region that we can identify is (0,0). Since it is in our shaded region it should work if we plug it into our answer choices. If we plug in (0,0) into each answer choice, the only one for which both equations work is choice A.

Question 50 “submerged object” the answer is F

The volume of a rectangular prism is given by $\inline&space;V=lwh$.

1. The volume of the water by itself is $\inline&space;V=40\cdot&space;30\cdot&space;20&space;=&space;24000$.
2. The volume of the water and object together is $\inline&space;V=40\cdot&space;30\cdot&space;20.25&space;=&space;24300$.
3. The volume of the object must be $\inline&space;24300&space;-&space;24000&space;=&space;300$, which is choice F.

Tip: When Archimedes discovered this principle of volume by displacement, he was so happy that he ran through the streets naked shouting “Eureka”. This approach to problem solving is not a recommended strategy on the ACT.

Question 51 “ratios” the answer is E

Doing the algebra here is actually rather tricky, it’s nicer to work with numbers. It’s also often nicer to write ratios as fractions rather than using a colon, and we’ll do that here.

1. If we let x = 5, then the first ratio tells us y = 2.
2. This makes the equation $\inline&space;\frac{y}{z}&space;=&space;\frac{3}{2}$ become $\inline&space;\frac{2}{z}&space;=&space;\frac{3}{2}$. This is an equation we can solve.
3. Cross multiplying, we obtain $\inline&space;2\cdot&space;2&space;=&space;3z$, so $\inline&space;z&space;=&space;\frac{4}{3}$.
4. Finally, we have $\inline&space;\frac{x}{z}&space;=&space;\frac{5}{\frac{4}{3}}&space;=&space;\frac{15}{4}$ (use your calculator!), which is choice E.

Question 52 “three part inequality” the answer is H

We can work three part equations almost the same way we work two part equations. What we do to one side, we need to do to all three sides. We need to be careful to not forget that multiplying or dividing by a negative switches the signs of our inequalities.

1. To solve $\inline&space;-5&space;<&space;1&space;-3x&space;<&space;10$ first subtract one from all three sides to obtain $\inline&space;-6&space;<&space;-3x&space;<&space;9$. Next divide all three sides by -3 to get $\inline&space;2&space;>&space;x&space;>&space;-3$, which is choice H.

If this algebra is tricky, you can solve this by plugging in numbers. Be sure to plug in numbers that will help you distinguish between your answer choices. For example, if we plug in x = 5, the equation –$\inline&space;5&space;<&space;1&space;-&space;3\cdot&space;5$ is not true. We can eliminate any answer choices that contain 5. This means F, G, and K are all incorrect.

Question 53 “surface area” the answer is B

Tracing how changes in variables affect the answer can be simple or very complicated depending on the formula. One approach that always works is to plug in some numbers and see what happens.

1. If we let l = 1, w = 1, and h = 1 then the equation $\inline&space;A&space;=&space;2lw&space;+&space;2lh&space;+&space;2wh$ becomes $\inline&space;A&space;=&space;2\cdot&space;1&space;\cdot&space;1&space;+2\cdot&space;1&space;\cdot&space;1&space;+2\cdot&space;1&space;\cdot&space;1&space;=&space;6$.
2. If we double each of the dimensions we have l = 2, w = 2, and h = 2. Now our equation becomes $\inline&space;A&space;=&space;2\cdot&space;2&space;\cdot&space;2&space;+2\cdot&space;2&space;\cdot&space;2&space;+2\cdot&space;2&space;\cdot&space;2&space;=&space;24$.
3. Since the volume was multiplied by 4 ($\inline&space;6&space;\cdot&space;4&space;=&space;24$) the answer is B.

Question 54 “dog food” the answer is K

Setting this up as a proportion is a great approach.

1. $\inline&space;\frac{7\text{cans}}{3\text{days}}&space;=&space;\frac{x\text{cans}}{3+d\text{days}}$
2. Cross multiply to obtain $\inline&space;7(3+d)&space;=&space;3x$.
3. Next, distribute to obtain $\inline&space;21&space;+&space;7d&space;=&space;3x$.
4. Divide by three to obtain $\inline&space;7&space;+&space;\frac{7}{3}d$, which is choice K.

If you’re not sure how to start this problem, you can make a strong educated guess. Since the dog eats 7 cans in 3 days, in 3 + d days the dog should eat 7 cans + some amount more. Choices J and K both match this situation.

Another strong approach would be to plug in a number for d. If d = 3, 6 days have passed so the dog should have eaten 14 cans. Plug three into your answer choices, and you’ll see that only choice K correctly shows 14 cans.

Question 55 “skiing” the answer is E

The formula that applies here is Total = A + B – Both + Neither

1. The total is 120. 55 people did neither. 28 people ski downhill. 45 people ski cross country. We can plug into the formula.
2. $\inline&space;120&space;=&space;28&space;+&space;45&space;-&space;x&space;+&space;55$. Solving this we obtain x = 8, which is choice E.

Another way to approach this is to notice that 28 + 45 = 73, but only 65 people said that they skied. Where did the 8 extra people come from? The ones that do both got counted twice.

Question 56 “a square in 3 rows” the answer is K

1. Each row, in total, will have $\inline&space;\frac{1}{3}$ of the total area of the square.
2. In the top row, box A will have an area of  $\inline&space;\frac{1}{2}&space;\cdot&space;\frac{1}{3}&space;=&space;\frac{1}{6}$.
3. In the middle row, box A will have an area of  $\inline&space;\frac{1}{3}&space;\cdot&space;\frac{1}{3}&space;=&space;\frac{1}{9}$.
4. In the bottom row, box A will have an area of  $\inline&space;\frac{1}{4}&space;\cdot&space;\frac{1}{3}&space;=&space;\frac{1}{12}$.
5. Putting it all together  $\inline&space;\frac{1}{6}&space;+&space;\frac{1}{9}&space;+&space;\frac{1}{12}&space;=&space;\frac{13}{36}$ (use your calculator!), which is choice K.

If working this problem was too challenging, estimation can give you a strong educated guess. From the picture it appears that the “A” blocks cover approximately $\inline&space;\frac{1}{3}$rd of the square. The answer choices that are close to that are G and K, give you a 50/50 chance at the problem.

Question 57 “translating functions” the answer is A

We need to know that $\inline&space;f(x&space;-&space;a)&space;+&space;b$ results in a vertical shift up b and a horizontal shift right a.

1. Since this function has a left/right shift and not a vertical shift, we must have that b = 0. This means answer choice A is correct.

If you don’t know your rules for function translation, your best approach is to plug in some numbers for a and b, and use your graphing calculator to see if they make a correct picture or not.

Question 58 the answer is K

This is a trick question, don’t be fooled.

1. Since the absolute value of a number is never negative, there is no way to make $|x-5|<-1$. This equation has no solution, so the answer is K.

If you aren’t comfortable with absolute value equations, another approach is to start plugging in numbers from the number line to see if they work. You will find that none of them do.

Question 59 “a probability experiment” the answer is E

For probabilities of sequential events like this, the probabilities multiply.

1. Since one answer out of three is correct, the probability that he gets one question right is $\inline&space;\frac{1}{3}$.
2. The probability that he gets all 4 correct will then be $\inline&space;\frac{1}{3}&space;\cdot&space;\frac{1}{3}&space;\cdot&space;\frac{1}{3}&space;\cdot&space;\frac{1}{3}&space;=&space;\frac{1}{81}$, which is choice E.

If you don’t know how to start this one, you can make an educated guess. The probability that he gets all of the answers right by luck, is certainly very low. You should guess one of the smaller answers, such as choices C, D, or E.

Question 60 “the law of cosines” the answer is J

This problem looks very intimidating, but they give us the formulas we need in the problem. We just need to not panic! It’s always a good idea to draw a picture for geometry questions that don’t have one.

1. In a triangle the smallest angle is opposite the smallest side, so $\theta$ will be opposite the 14cm side.
2. We have all three sides and want to know one angle. The formula that relates these numbers to each other is $\inline&space;c^2&space;=&space;a^2&space;+&space;b^2&space;-&space;2ab&space;\cos(C)$. The other formula has multiple angles in it, so we can’t use it since we don’t know any angles.
3. Since $\theta$ was opposite the 14cm side those will be C and c, respectively.
4. Plugging in, our formula becomes $\inline&space;14^2&space;=&space;18^2&space;+&space;20^2&space;-2\cdot18\cdot20\cdot&space;\cos(\theta)$ which is answer choice J.