Use our free Praxis Core Math practice test to review the most important skills and concepts that you must know for this section of your certification exam. The Praxis Core Math test includes a total of 56 questions that must be answered within 85 minutes. An on-screen calculator will be available. The test covers 4 major topics: Numbers & Quantities, Algebra & Functions, Geometry, Statistics & Probability. You may use a calculator.

Question 1 |

### Solve for $x$:

$3(x + 1) = 5(x − 2) + 7$$−2$ | |

$2$ | |

$\dfrac{1}{2}$ | |

$3$ | |

$−3$ |

$3(x + 1) = 5(x − 2) + 7$

$3x + 3 = 5x − 10 + 7$

$3x + 3 = 5x − 3$

Add $3$ to both sides to maintain the equality:

$3x + 6 = 5x$

Subtract $3x$ from both sides and then divide the resulting equation by $2$ to solve for $x$ as follows:

$6 = 2x$

$3 = x$

Question 2 |

### Which of the following numbers is the greatest?

$\dfrac{2}{3} ~~ 0.6 ~~ \dfrac{13}{22} ~~ 0.08$$\dfrac{2}{3}$ | |

$0.6$ | |

$\dfrac{13}{22}$ | |

$0.08$ |

Question 3 |

### Liam is driving to Utah. He travels at $70$ kilometers per hour for $2$ hours, and $63$ kilometers per hour for $5$ hours. Over the $7$ hour time period what was Liam's average speed?

64 km/h | |

65 km/h | |

66 km/h | |

67 km/h | |

68 km/h |

$\text{Average Speed} = \text{Total Distance} ÷ \text{Total Time}$

Total Distance $= 2 \text{hours} × 70 \text{km/h} + 5 \text{hours} × 63 \text{km/h}$

$= 140 \text{km} + 315 \text{km}$

$= 455 \text{km}$

Total Time $= 7$ hours

Average Speed $= 455 \text{km} ÷ 7 \text{hours}$

$= 65 \text{km/h}$

Question 4 |

### Sofía's Restaurant offers the following choices:

If the only dinner you can order*must*include one salad, one main course, and one dessert, how many different combinations of this dinner are possible?

12 | |

60 | |

120 | |

144 | |

360 |

To solve this problem multiply:

$4 \; (\text{the number of salads})$ $× 5 \; (\text{the number of main courses})$ $× 3 \; (\text{the number of desserts})$:

$4 × 5 × 3 = 60$ different combinations

Question 5 |

### If $4x + 3x − 2(x + 5) = −9$, then $x =$ ?

$−\dfrac{1}{2}$ | |

$−\dfrac{14}{5}$ | |

$−\dfrac{19}{5}$ | |

$\dfrac{1}{5}$ | |

$\dfrac{2}{3}$ |

$4x + 3x − 2x − 10 = −9$

$7x − 2x − 10 = −9$

$5x = 1$

$x = \dfrac{1}{5}$

Question 6 |

### Li wants to buy as many bags of mulch as possible with his $\$305$, and he would like them to be delivered to his house. The cost is $\$7.50$ per bag and there is a $\$35.75$ delivery charge. The mulch is only sold in full bags. How many bags can Li buy?

35 | |

36 | |

40 | |

46 | |

41 |

$\$305 − \$35.75 = \$269.25$

We can then divide this amount by the cost per bag to find the total number of bags that Li can buy:

$\$269.25 ÷ \$7.50$ $ = 35.9 \text{ bags}$

However, the question states that the mulch can only be sold in full bags, so we must round our answer down to ensure that Li does not exceed his budget.

Question 7 |

### Mason earns \$8.10 per hour and worked 40 hours. Noah earns \$10.80 per hour. How many hours would Noah need to work to equal Mason’s earnings over 40 hours?

15 | |

25 | |

27 | |

28 | |

30 |

$40 \text{ hours} × \$8.10 \text{ per hour} = \$324$

Next, divide this total by Noah’s hourly rate to find the number of hours Noah would need to work:

$\$324 ÷ \$10.80 \text{ per hour} $ $ = 30 \text{ hours}$

Question 8 |

### Diego’s current age is five times Martina’s age ten years ago. If Martina is currently $m$ years old, what is Diego’s current age in terms of $m$?

5$m$ | |

5$m$ − 10 | |

5$m$ − 50 | |

5$m$ + ($m$ − 10) | |

10($m$ − 5) |

$m − 10$

So Diego’s age is:

$5(m − 10)$

$= 5m − 50$

Question 9 |

### In a coordinate plane, triangle $ABC$ has coordinates: $(−2,7)$, $(−3,6)$, and $(4,5)$. If triangle $ABC$ is reflected over the $y$-axis, what are the coordinates of the new image?

$(−2,−7), (−3,−6), (−4,−5)$ | |

$(−2,−7), (−3,−6), (4,−5)$ | |

$(2,7), (3,6), (−4,5)$ | |

$(2,7), (3,6), (4,5)$ | |

$(2,−7), (3,−6), (−4,−5)$ |

Reflecting $(−2,7), (−3,6), (4,5)$ across the $y$-axis produces the points $(2,7), (3,6), (−4,5)$.

Question 10 |

### Arrange the following fractions in order from least to greatest.

$\dfrac{7}{5}, \dfrac{15}{4}, \dfrac{3}{2}, \dfrac{11}{4}, \dfrac{13}{3}$$\dfrac{7}{5}, \dfrac{15}{4}, \dfrac{3}{2}, \dfrac{11}{4}, \dfrac{13}{3}$ | |

$\dfrac{7}{5}, \dfrac{3}{2}, \dfrac{15}{4}, \dfrac{11}{4}, \dfrac{13}{3}$ | |

$\dfrac{7}{5}, \dfrac{3}{2}, \dfrac{11}{4}, \dfrac{15}{4}, \dfrac{13}{3}$ | |

$\dfrac{7}{5}, \dfrac{15}{4}, \dfrac{11}{4}, \dfrac{3}{2}, \dfrac{13}{3}$ | |

$\dfrac{3}{2}, \dfrac{7}{5}, \dfrac{11}{4}, \dfrac{15}{4}, \dfrac{13}{3}$ |

$\frac{7}{5} = 1.4$, $\frac{15}{4} = 3.75$, $\frac{3}{2} = 1.5$, $\frac{11}{4} = 2.75$, $\frac{13}{3} = 1.33$

Now, order the decimals from least to greatest, and match the resulting list with its corresponding list of fractions: $1.4, 1.5, 2.75, 3.75$, and $4.33$ which corresponds to:

$\dfrac{7}{5}, \dfrac{3}{2}, \dfrac{11}{4}, \dfrac{15}{4}, \dfrac{13}{3}$

Question 11 |

### This pie chart shows Al’s monthly expenses. If Al spent a total of \$1,550 in one month, how much did he spend on clothes in that month?

\$77.50 | |

\$232.50 | |

\$775 | |

\$7.75 | |

\$1,627.50 |

$.05 × \$1,550 = \$77.50$

Question 12 |

### Calculate the value of $x$ for the right triangle shown below.

$66$ | |

$89$ | |

$56$ | |

$65$ | |

$75$ |

Substitute the known values into their appropriate places and solve for the unknown side length:

$33^2 + 56^2 = c^2$

Evaluate the squares:

$1,089 + 3,136 = c^2$

$4,225 = c^2$

Evaluate the square root of both sides:

$\sqrt{c^2} = \sqrt{4,225}$

$c = 65$

Question 13 |

### Oscar purchased a new hat that was on sale for $\$5.06$. The original price was $\$9.20$. What percentage discount was the sale price?

4.5% | |

41.4% | |

45% | |

55% | |

5.5% |

$\$9.20 − \$5.06 = \$4.14$

The percentage discount is this difference divided by the original price:

$\$4.14 ÷ \$9.20 = 0.45$

Convert the decimal to a percentage by multiplying by 100%:

$0.45 × 100\% = 45\%$

Question 14 |

### Anastasia needs to order liquid fertilizer for her landscaping company. She plans to keep the fertilizer in a large cylindrical storage tank, but isn’t sure how much it will hold. The tank is 10 feet tall and the circular base has a diameter of 10 feet. What is the volume of her storage tank?

78.5 ft³ | |

157 ft³ | |

3,140 ft³ | |

1,570 ft³ | |

785 ft³ |

*Volume = (area of the base) (height)*

The base of a cylinder is a circle, so you will also need the formula for the area of a circle, which is also provided on the formula sheet:

$\text{Area} = π(\text{radius})^2$

Combining these formulas, we get:

$V = πr^2h$

$V$ is volume, $r$ is the radius, and $h$ is the height of the cylinder. (Remember that the radius of a circle is half of the diameter.) Substitute the given values into the equation to solve for the volume:

$V = π \cdot (5)^2 \cdot 10$

$V = 250π$

Recall that $π$ (pi) is approximately equal to 3.14:

$250 \cdot 3.14 = 785$

Question 15 |

### Kayla owns a house cleaning company and must give price quotes to potential customers. She determines her price by assuming a $\$25$ base charge and then adding $\$8$ for each bathroom and $\$4$ for every other room. If she uses $P$ to represent the price, $B$ to represent the bathrooms, and $R$ to represent the other rooms, which of the following defines her price quote formula?

$P = 25 + 12(BR)$ | |

$P = 25(4R + 8B)$ | |

$P = 25 + 8B + 4R$ | |

$P = (4)(8)(R + B) + 25$ | |

$P = (25 + 8B) + (25 + 4R)$ |

Question 16 |

### Use the information below to answer the question that follows.

A business recorded the number of customers who visited the store throughout the week. How many days had a number of visitors greater than the average number of visitors for the entire week?$6$ days | |

$4$ days | |

$5$ days | |

$2$ days | |

$3$ days |

$\dfrac{13 + 12 + 16 + 19 + 25 + 33 + 22}{7}$ $= \dfrac{140}{7} = 20$

How many days had more than $20$ visitors?

Fri $= 25$

Sat $= 33$

Sun $= 22$

$3$ days exceeded the $20$-visitor average.

Question 17 |

### In a factory there are two separate containers of stress balls. In the first container are $80$ balls, and in the second are $90$ balls. $40\%$ of the balls in the first container are defective, and $20\%$ in the second container are defective. In total, how many balls in the two containers are defective?

$102$ | |

$50$ | |

$60$ | |

$51$ |

$40\%$ of $80 = 0.40 \cdot 80 = 32$

$20\%$ of $90 = 0.20 \cdot 90 = 18$

$32 + 18 = 50$

Question 18 |

### Which of the following fractions is greater than $0.4$ and less than $0.5$?

$\dfrac{6}{11}$ | |

$\dfrac{3}{10}$ | |

$\dfrac{12}{23}$ | |

$\dfrac{9}{20}$ |

Since $5.5$ is half of $11$, $\frac{6}{11}$ is greater than $\frac{1}{2}$. No good.

Since $11.5$ is half of $23$, $\frac{12}{23}$ is greater than $\frac{1}{2}$. No good.

This means our answer is either $\frac{3}{10}$ or ${9}{20}$.

$\frac{3}{10}$ is equivalent to $0.3$, which is less than $0.4$. No good.

$\frac{9}{20}$ must be correct. You could also solve by recognizing that $0.4 = \frac{4}{10} = \frac{8}{20}$ and $0.5 = \frac{5}{10} = \frac{10}{20}$, so $\frac{9}{20}$ works.

Question 19 |

### Use the menu below to answer the question that follows.

Neil ordered the following for his family: two cheeseburgers, one hamburger, two large fries, one small fries, and three small sodas. If the total calories for the order is $3370$, what is the missing calorie information on the menu?$1000$ | |

$400$ | |

$450$ | |

$480$ | |

$900$ |

Two cheeseburgers $= 2 × 530 = 1,060$ calories

One hamburger $= 430$ calories

One small fries $= 230$ calories

Three small sodas $= 3 × 250 = 750$ calories

If we subtract each of the known calories from the total calories for the order we will know how many calories came from two large fries:

$3,370 − 1,060 − 430 − 230 − 750 = 900$

$900$ calories came from two orders of large fries. Each order of fries is $450$ calories.

Question 20 |

### Aisha wants to paint the walls of a room. She knows that each can of paint contains one gallon. A half gallon will completely cover a $55$ square feet of wall. Each of the four walls of the room is $10$ feet high. Two of the walls are $10$ feet wide and two of the walls are $15$ feet wide. How many $1$-gallon buckets of paint does Aisha need to buy in order to fully paint the room?

$4$ | |

$5$ | |

$9$ | |

$10$ | |

$15$ |

$2 (10 × 10) = 200$ sq. ft. $2 (10 × 15) = 300$ sq. ft. So the total square footage of the walls is $500$. If a half gallon of paint will cover $55$ square feet, then each gallon will cover $2 × 55 = 110$ square feet. Four gallons can only cover $440$ square feet. Five gallons will cover $550$ square feet, which will be enough for the entire area of the walls.

Question 21 |

### The data in the table below shows the results of Tracy trying to train her dog Snowy.

In what percentage of these trials did Snowy obey the command given?25% | |

37.5% | |

50% | |

62.5% | |

75% |

$\dfrac{4}{8}$ $= \dfrac{1}{2} = 50\%$

Question 22 |

### Max struggled with his math class early in the year, but he has been working hard to improve his scores. There is one test left, and he is hoping that his final average test score will be 75. What score will he need to get on Test 6 to finish the year with an average score of 75?

75 | |

85 | |

92 | |

98 | |

100 |

Average = sum of data points ÷ number of data points

$75 = (50 + 52 + 77 + 88 + 91 + x) ÷ 6$

Now solve for $x$. Start with the addition:

$75 = (358 + x) ÷ 6$

Next, eliminate the denominator by multiplying both sides by 6:

$450 = 358 + x$

The last step is to subtract 358 from both sides:

$x = 92$

Question 23 |

### A student is calculating the average of a list of seven numbers. After adding the seven numbers together, he accidentally multiplied by 7 instead of dividing and got 392. What could the student do to get the correct average without having to clear the calculator and start over?

Divide the incorrect result by 7 | |

Divide the incorrect result by 49 | |

Multiply the incorrect result by 7 | |

Divide the incorrect result by 14 | |

Divide the incorrect result by 77 |

To get the correct answer, the student needs to divide by 7 twice, which is the same as dividing by 49:

$\dfrac{7x}{49} = \dfrac{x}{7}$

Question 24 |

### List A consists of the numbers ${2, 9, 5, 1, 13}$, and list B consists of the numbers ${7, 4, 12, 15, 18}$.

If the two lists are combined, what is the median of the combined list?5 | |

6 | |

7 | |

8 | |

9 |

${1, 2, 4, 5, 7, 9, 12, 13, 15, 18}$

Since there are an even number of items in the resulting list, the median is the average of the two middle numbers.

Median $= (7 + 9) ÷ 2 = 8$

Question 25 |

### Tavon's flight is 270 minutes long. How many hours does the flight last?

$4$ hours | |

$4 \frac{1}{2}$ hours | |

$5$ hours | |

$5 \frac{1}{2}$ hours |

$\require{cancel} 270 \cancel{\text{minutes}} \cdot \dfrac{1 \text{ hour}}{60 \cancel{\text{minutes}}}$ $= \dfrac{270}{60} \text{ hours}$ $= \dfrac{27}{6} \text{hours}$

$\dfrac{27}{6} = \dfrac{9}{2} = 4 \dfrac{1}{2}$

Question 26 |

### $6.6 × 10^{−4}$

0.000066 | |

0.00066 | |

0.0066 | |

0.066 | |

0.66 |

*In the case of a negative exponent, the decimal is moved to the left*(this is the same as dividing by 10 a number of times).

*In the case of a positive exponent, the decimal is moved to the right*(this is the same as multiplying by 10 a number of times).

The negative exponent here, −4, indicates that the decimal point is to be moved to the left 4 places:

$6.6 × 10^{−4} = 0.66 × 10^{−3}$

$= 0.066 × 10^{−2}$

$= 0.0066 × 10^{−1}$

$= 0.00066 × 10^{0}$

$= 0.00066$

Question 27 |

### $ABC$ is a triangle with coordinates $(3, 1),$ $(6, 1),$ and $(1, 3)$. It is translated to points $A′(−3, 3),$ $B′( 0, 3),$ and $C′(−5, 5)$. Find the rule for the translation.

6 units right and 2 units down | |

2 units left and 6 units up | |

6 units down and 2 units right | |

6 units left and 2 units up | |

2 units left and 6 units right |

$(x_2 − x_1) = (−3 − 3) $ $ = (0 − 6) = (−5 − 1) = −6$

This corresponds to a shift of 6 units to the left.

Likewise, for the $y$ values:

$(y_2 − y_1) = (3 − 1) $ $ = (3 − 1) = (5 − 3) = 2$

This corresponds to a vertical shift of 2 units.

Question 28 |

### There are 5 blue marbles, 4 red marbles, and 3 yellow marbles in a box. If Isabella randomly selects a marble from the box, what is the probability of her selecting a red or yellow marble?

$\dfrac{1}{4}$ | |

$\dfrac{1}{3}$ | |

$\dfrac{7}{12}$ | |

$\dfrac{3}{4}$ | |

$\dfrac{2}{3}$ |

Number of red and yellow marbles: $4 + 3 = 7$

Total number of marbles: $5 + 4 + 3 = 12$

Probability: $\dfrac{7}{12}$

Question 29 |

### A river rafting guide offers group trips. She charges \$250 for 2 people and \$40 more for each additional person. If 5 friends share the cost of a trip equally, how much will each person pay?

\$40 | |

\$58 | |

\$74 | |

\$90 | |

\$370 |

$250 + 40 + 40 + 40 = \$370$

Then divide this amount by 5 to determine the amount that each person will pay:

$\$370 ÷ 5 = \$74$

Question 30 |

### Consider the list:

$2, 2, 3, 5, 9, 11, 17, 21$ If the number $23$ is added to the list, which measurement will NOT change?Mean | |

Median | |

Mode | |

Range | |

Average |

The mean is the average, and in this case, it will increase because a number larger than the current average is added to the list.

The median is the number in the middle, and it will change from $7$ to $9$.

The range is the difference between the highest and lowest numbers and will change from $19$ to $21$ because a new maximum value is added to the list ($21 − 2 = 19$ and $23 − 2 = 21$).

Question 31 |

### The points represented by the $(x, y)$ coordinate pairs in the table below all lie on line $k$.

What is the slope of line $k$?−5 | |

−3 | |

−2 | |

2 | |

5 |

Slope $= \dfrac{y_2 − y_1}{x_2 − x_1}$

Slope $= \dfrac{−5 − 3}{5 − 1}$

Slope $= \dfrac{−8}{4}$

Slope $= −2$

Question 32 |

### This pie chart shows Al’s monthly expenses. If Al spent $\$210$ on transportation in one month, how much did he spend on food in that month?

$\$126$ | |

$\$250$ | |

$\$325$ | |

$\$350$ | |

$\$325$ |

$\dfrac{\text{Amount spent on food}}{\text{Amount spent on transportation}}$

$= \dfrac{\text{Percentage spent on food}}{\text{Percentage spent on transportation}}$

Substitute the known values to then solve for the unknown amount spent on food:

$\dfrac{\text{Amount spent on food}}{\$210} = \dfrac{25}{15}$

Cross multiply and simplify to solve:

Amount spent on food $= \dfrac{\$210 \cdot 25}{15} = \$350$

Question 33 |

### Which single piece of information is necessary to solve the problem above?

The number of multiple choice questions | |

The time the teacher expects each short-answer question to take | |

The percentage of questions that are multiple choice | |

The number of students taking the exam | |

The average time the teacher expects each question on the exam to take |

Let $a$ = average time required for each multiple choice question.

Let $b$ = average time required for each short-answer question.

Since the exam is expected to take a total of 90 minutes:

$a(56) + b(19) = 90$

If we are given $b$, the time required for each short-answer question, we can use the above equation to solve for $a$, the time required for each multiple choice question.

Question 34 |

### Susan, Gus, Harold, and Jeff were left an inheritance by their grandfather. If Susan receives $\frac{5}{16}$ of the inheritance, Gus receives $\frac{1}{8}$ of the inheritance, Harold receives $\frac{3}{16}$ of the inheritance, and Jeff receives the remainder, what fraction of the inheritance does Jeff receive?

$\dfrac{5}{16}$ | |

$\dfrac{7}{16}$ | |

$\dfrac{3}{8}$ | |

$\dfrac{1}{8}$ | |

$\dfrac{5}{8}$ |

Susan’s Fraction + Gus’ Fraction + Harold’s Fraction + Jeff’s Fraction = 1

$s+g+h+j=1$

$\dfrac{5}{16}+\dfrac{1}{8}+\dfrac{3}{16}+j=1$

$j=1-\dfrac{5}{16}-\dfrac{1}{8}-\dfrac{3}{16}$

When adding or subtracting fractions we need a common denominator (least common multiple) which is 16 in this case.

$j = \dfrac{16}{16}-\dfrac{5}{16}-\dfrac{2}{16}-\dfrac{3}{16}$

$j = \dfrac{16}{16}-\dfrac{10}{16}$

$j=\dfrac{6}{16}$

$j=\dfrac{3}{8}$

Question 35 |

### A manufacturer packages their high-quality pencils in packs that hold 15 pencils. There are 281 pencils left to be packed. How many *complete* packs can be filled.

18 | |

20 | |

16 | |

17 | |

19 |

The simplest way to tackle this is to use the calculator provided on the test.

$\dfrac{281}{15}= 18.73$

We are able to complete 18 complete packs and 0.73 of the 19th pack.

The question specifically asks for the number of complete packs, so the correct answer is 18 packs.

Question 36 |

### Tim is in the process of painting a large mural on the side of a building. The mural was 28% complete before Tim began working on it today. Today, after spending 4 hours on the mural, it is now 40% complete. If he continues progressing at the same rate, how many more hours will be needed to complete the mural?

16 hours | |

12 hours | |

10 hours | |

24 hours | |

20 hours |

It took him 4 hours to paint that 12% portion.

60% of the mural remains to be painted. How many hours will this require?

To find out we can set up an equivalent proportion:

$\dfrac{4 \text{ hours}}{12\%}=\dfrac{x \text{ hours}}{60\%}$

$(4)(60) = (12)(x)$

$240 = 12x$

$x=\dfrac{240}{12}$

$x = 20$

It will take 20 hours to complete the mural.

(Note: You might also solve by noticing that 12% goes into 60% five times, and 5 × 4 hours = 20 hours.)

Question 37 |

### A university gave 2 Bachelor of Science (BS) degrees for every 7 Bachelor of Arts (BA) degrees. If the combined total number of BS and BA degrees was 6,327, how many Bachelor of Science degrees were given?

1406 | |

1808 | |

1346 | |

1039 | |

703 |

This means that 2 out of 9 of the degrees was a BS degree. We need to determine what $\frac{2}{9}$ of 6,327 is.

$\dfrac{2}{9}*6372=1406$

1406 of the Bachelor degrees given were a Bachelor of Science degree.

Question 38 |

### Doug and Heidi work at the same clinic. They are both scheduled to work on January 1st. If Doug is scheduled to work every 3 days (1st, 4th, 7th, ...) and Heidi is scheduled to work every 5 days (1st, 6th, 11th, ...), including Jan 1st, how many times will they work together over the course of 56 days?

3 | |

6 | |

4 | |

5 | |

2 |

Day 1, 16, 31, 46 are the

__four__days when they will work on the same day.

Another approach would be to list out the days each will be working and look for the days that the two lists have in common (underlined):

Doug:

__1__, 4, 7, 10, 13,

__16__, 19, 22, 25, 28,

__31__, 34, 37, 40, 43,

__46__, 49, 52, 55

Heidi:

__1__, 6, 11,

__16__, 21, 26,

__31__, 36, 41,

__46__, 51, 56

They will work 4 days together.

Question 39 |

### A bicycle manufacturer makes two models of bikes: “mountain” and “beach.” In their warehouse this morning they had 2,112 total bicycles, and 7 out of every 12 of those bikes was a “mountain” model. After shipping 75 “beach” models from the warehouse, how many “beach” model bikes will remain?

978 | |

1,157 | |

880 | |

671 | |

805 |

Number of “beach” bikes initially $= \dfrac{5}{12}*2{,}112=880$

After shipping 75 bikes how many will remain?

Number of “beach” bikes remaining $= 880 − 75 = 805$

Question 40 |

### Aiden’s new toy car is a 1/64th model of the actual car, meaning a feature measuring 1 mm on his toy would be 64 mm on the actual car. He measures the wheel base on his toy and gets 44 mm. What would be the expected wheelbase of the actual car in *meters*?

2.8m | |

4.1m | |

6.9m | |

4.8m | |

7.5m |

$ \dfrac{1\text{ mm}}{64\text{ mm}}=\dfrac{44\text{ mm}}{x\text{ mm}}$

Cross multiplying:

$(1)(x) = (64)(44) $

$x = 2816 \text{ mm}$

To convert millimeters to meters, we divide by 1000:

$\dfrac{2816}{1000} = 2.816 \approx 2.8 \text{ m}$

Question 41 |

### What is the least common multiple of 105 and 231?

4,179 | |

2,310 | |

1,155 | |

24,255 | |

495 |

Next, list the prime factors of each number:

Prime factors of 105: 3, 5, 7

Prime factors of 231: 3, 7, 11

Combine the lists, but do not duplicate any factor that appears in both lists: Prime factors without duplicate factors: 3, 5, 7, 11

(Notice that 3 and 7 are only listed once since they appear in the factor trees of both numbers.)

Find the product of the numbers:

3 ∗ 5 ∗ 7 ∗ 11 = 1, 155

The least common multiple is 1,155.

Because this question is multiple choice, we could instead work backwards from the answers.

We check each answer choice to see whether or not it can be divided by 105 and by 231:

4,179 – NO

2,310 – YES

1,155 – YES

24,255 – YES

495 – NO

We then choose the multiple that is the least. In this case 1,155 is the least multiple of each number (it is the smallest number that can be divided by both 105 and 231).

Question 42 |

### A bag contains only green and red balls. A player draws a single ball from the bag. If the ball is green they win, and if it is red they lose. Initially, the bag contained 20 green balls and an unknown (non-zero) number of red balls.

When 24 additional green balls are added to the bag with no change to the number of red balls, it causes the probability of losing to be halved (i.e. a player is now half as likely to lose).

How many red balls are in the bag?

5 | |

6 | |

4 | |

8 | |

12 |

*total*be the total number of balls originally in the bag (= $r$ + $g$).

The probability of losing can be expressed as:

$\text{P(Losing)} = \dfrac {r}{r+g} = \dfrac{r}{\text{total}}$

**“Logical” Approach:**

We are told that when the number of green balls was increased by 24 without changing the number of red balls, the probability of losing was halved. Put another way, when

*total*was increased by 24, the probability of losing was halved.

But, for the probability of losing to be halved, the

*total*must have been doubled (since we know that the number of red balls did not change).

Adding 24 balls doubled the

*total*, therefore there must have been 24 balls in the bag originally.

And if 20 of the 24 original balls was green, the balance of 4 must have been red.

**Algebraic Approach:**

$P\text{(Losing originally)} = \dfrac {r}{r+g} $ $ = \dfrac {r}{r+20}$

$P\text{(Losing after adding 24 more green balls)}$

$= \dfrac {r}{r+20+24} = \dfrac {r}{r+44}$

We also know that:

$P\text{(Losing after adding 24 more green balls)}$

$= \dfrac {1}{2} \ast P\text{(Losing originally)}$

Which means that:

$2 \ast P\text{(Losing after adding 24 more green balls)}$

$= P\text{(Losing originally)}$

$2*\dfrac {r}{r+44}=\dfrac {r}{r+20}$

$\dfrac {2r}{r+44}=\dfrac {r}{r+20}$

Cross multiplying:

$(2r)(r + 20) = (r + 44)(r)$

Dividing both sides by $r$:

$(2)(r + 20) = (r + 44)$

$2r + 40 = r + 44$

$r = 4$

Question 43 |

### The table below shows the distribution of attendees of a conference by sex and by whether they leased or bought the car they drive.

### Among those attendees who lease their car, what percentage are female?

41.9% | |

15.0% | |

72.0% | |

48.6% | |

32.7% |

Notice that we are NOT being asked the percentage of attendees that are both female and lease their car.

Instead, we are told to only look “among those attendees who lease their car.”

There are a total of 43 (= 25 + 18) attendees who lease their car.

18 of these 43 are female.

Thus, among those attendees who lease their car, the percentage that are female. is:

$ \dfrac{18}{43} \approx 0.419 = 41.9\%$

Question 44 |

### Which of the following scatter plots shows a weak negative correlation between $x$ and $y$?

Answer choice (A) also has a negative slope, but here the points are tightly grouped about the line, making it a “strong negative correlation.”

Question 45 |

### A company is analyzing data which shows the relationship between the hourly wage of employees (*W*) and the number of vacation days taken (*D*). Which statement below best describes the relationship between *W* and *D*?

An increase in W causes an increase in D | |

There is no correlation between W and D | |

There is a positive correlation between W and D | |

There is a negative correlation between W and D | |

An increase in D causes an increase in W |

*D*increases,

*W*also tends to increase.

However, just because there is a positive correlation, we cannot assume that either variable causes a change in the other variable.

Question 46 |

### The scatter plot below shows data points along with a line of best fit. Which of the following is the equation for the graph of the line of best fit?

−10$x$ + 10 | |

10$x$ + 9 | |

2$x$ + 10 | |

5$x$ + 10 | |

−5$x$ − 9 |

To determine the slope look at the change in $y$ relative to the change in $x$. It appears that the $y$-value of the graph increases by 10 hours as the $x$-value increases by 5 days.

$\text{slope} = m = \dfrac{\text{Change in } y}{\text{Change in } x}$

$m = \dfrac{10}{5} = 2$

Recall that the slope-intercept form of a line is: $y = mx + b$

Only answer choice (C) has a line with a slope of $m$ = 2.

Question 47 |

### The pictograph below shows bicycle sales by category in the USA for 2017. According to the pictograph, what was the mean sales amongst the three categories of bicycles?

$3.7 Million | |

$400 Million | |

$37 Million | |

$4.0 Million | |

$370 Million |

This means there were:

6 × \$100 = \$600 Million in Mountain bike sales.

4 × \$100 = \$400 Million in Road bike sales.

1 × \$100 = \$100 Million in Electric bike sales.

The sum of these sales figures for the 3 categories:

\$600 + \$400 + \$100 Million = \$1100 Million

To find the mean we need to divide by the “number of numbers”:

$\text{Mean} = \dfrac{\$1100 \text{ Million}}{3}$

$\text{Mean} = \$366.6 \text{ Million}$

\$370 Million is the best answer.

Question 48 |

### The graph below shows data for two products, Widgets and Gizmos. For each product the graph shows the sales in units sold and the sales as a percentage of total company sales. For the data shown, what was the greatest decrease in Gizmos unit sales experienced from one month to the next (consecutive months)?

7382 | |

1209 | |

8279 | |

3730 | |

9755 |

We can visually deduce that the greatest decrease in Gizmos unit sales in consecutive months occurred between March and April. That is where we see the green bars change most drastically in height.

Subtract these monthly totals to calculate the decrease:

$44{,}482 − 34{,}727 = 9{,}755 \text{ units}$

Question 49 |

### According to the frequency distribution chart shown below, how many students chose a favorite color other than Red, Green, or Blue?

21 | |

57 | |

66 | |

17 | |

34 |

__other than__Red, Green, or Blue”, means that we are looking for the number of students who chose Orange, Yellow, Purple, or Pink.

Orange = 5 students

Yellow = 5 students

Purple = 20 students

Pink = 4 students

TOTAL of these 4 colors = 34 students

Question 50 |

### There are 35 data points shown in the scatter-plot below. What fraction of the data points represent an hourly wage (*W*) less than 50?

$\dfrac{33}{35}$ | |

$\dfrac{11}{35}$ | |

$\dfrac{32}{35}$ | |

$\dfrac{26}{35}$ | |

$\dfrac{14}{35}$ |

Rather than count the number of data points with a wage less than 50, let’s save time and count the number of data points with a wage of 50 or greater and then deduct this from the 35 total points.

There are 9 such points (circled):

This means there are 26 (= 35 − 9) points representing a wage less than 50.

Since there are 35 total points, the fraction of data points representing an hourly wage less than 50 is $\frac{26}{35}$

Question 51 |

### Two hundred attendees of a consumer electronics show were surveyed to find their preferred gaming system.

### One-fifth of those who were surveyed preferred system “X.” Twice as many preferred system “N” as system “X.” Thirty fewer preferred system “S” compared to system “N.” The rest preferred system “P.”

### Which bar graph correctly represents the results of the survey?

Because only 1 of the 5 bar graphs has an “X” bar denoting 40, we actually have enough information to answer the question. However, for the sake of practice, we will look at calculating the remaining information:

Twice as many preferred “N” as “X”:

$N = 2 \ast X = 2 \ast 40 = 80$

30 fewer preferred “S” compared to“N”:

$S = X − 30 = 80 − 30 = 50$

The rest preferred “P” :

$P = 200 − X − N − S$

$P = 200 −40 − 80 − 50 = 30$

Question 52 |

### The figure below shows a rectangle with exactly one-quarter of it removed. After removing the upper-left quarter, the remaining shape has an area of 96 cm^{2}. What is the value of $x$?

9 | |

4 | |

8 | |

10 | |

7 |

$A = 2x * x = 2x^2$

With one-quarter of the full rectangle removed, three-quarters of the full rectangle will remain. We are told that the area of the remaining three-quarters is 96 cm

^{2}.

$\dfrac{3}{4}A = 96$

$\dfrac{3}{4}(2x^2) = 96$

$2x^2 = \dfrac{4}{3}(96)$

$2x^2 = 128$

$x^2 = 64$

$x = 8$

(Must be positive since the lengths of the sides are positive.)

Question 53 |

### For the triangle shown below, $\overline{AB} \cong \overline{BC}$. If $∠B = 128^\circ$, what is the measure of $∠C\,?$

33° | |

26° | |

64° | |

52° | |

16° |

We also know that the angles of a triangle sum to 180°:

$m∠A+m∠B+m∠C =180^\circ$

$m∠A + 128^\circ + m∠C = 180^\circ$

Let $x$ be the measure of $∠A$. And, since $∠A \cong ∠C$, $x$ will also be the measure of $∠C$.

$m∠A + 128^\circ + m∠C = 180^\circ$

$x + 128 + x = 180$

$2x+128 = 180$

$2x = 52$

$x = 26$

$∠C = 26^\circ$

Question 54 |

### The length of the equator of the earth (shown in pink in the illustration below) is approximately 40,000 kilometers. Given that the equator represents the circumference of a circle with a diameter equal to the diameter of the earth, what is the approximate diameter of the earth?

15,000 km | |

13,000 km | |

19,000 km | |

17,000 km | |

11,000 km |

We are given that: $C = 40{,}000 \text{ km}$

$40{,}000 = π * D$

$D= \dfrac{40{,}000}{π}$

$D \approx 12{,}732$

$D \approx 13{,}000 \text{ km}$

Question 55 |

### $∆ABC \sim ∆DAC$ (the triangles are similar).

### What is the length of $\overline{DC}$?

35 | |

55 | |

45 | |

50 | |

40 |

Also, for similar triangles the side lengths are proportional.

Based on these facts, we can set up an equivalent proportion:

$\dfrac{BC}{AC}=\dfrac{AC}{DC}$

$\dfrac{315}{105}=\dfrac{105}{DC}$

$(315)(DC) = (105)(105)$

$(315)(DC) = 11{,}025$

$DC = \dfrac{11{,}025}{315}$

$DC = 35$

Question 56 |

### In a survey of 100 people the following information was obtained: 40 like cats, 30 like both dogs and cats, and 20 do not like either dogs or cats. This information is shown in the diagram below.

### How many of those surveyed like dogs but not cats?

35 | |

25 | |

30 | |

40 | |

45 |

Since 40 of those surveyed like cats, and of these 30 like both dogs and cats, then 40 − 30 = 10 must like cats but not like dogs, as shown (in green) below:

There were 100 surveyed in total and of these 20 did not like either dogs or cats, which means that 100 − 20 = 80 liked either dogs or cats or both.

“Liked Dogs (but not cats)” + “Liked Cats (but not dogs)” + “Liked Both” = 80

“Liked Dogs (but not cats)” + 10 + 30 = 80

“Liked Dogs (but not cats)” + 40 = 80

“Liked Dogs (but not cats)” = 40

The final Venn diagram is shown below:

List |