Welcome to our FREE 7th Grade OST Math practice test, with answer key and answer explanations. This practice test’s realistic format and high-quality practice questions can help your student succeed on the 7th Grade OST Math test. Not only does the test closely match what students will see on the real OST, but it also comes with detailed answer explanations.

For this practice test, we’ve selected 20 real questions from past exams for your student’s OST Practice test. Your student will have the chance to try out the most common 7th Grade OST Math questions. For every question, there is an in-depth explanation of how to solve the question and how to avoid mistakes next time.

Use our free 7th Grade OST Math practice tests and study resources (updated for 2021) to help your students ace the 7th Grade OST Math test! Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice.

## The Absolute Best Book** to Ace the 7th Grade OST** **Math** Test

## 10 Sample **7th Grade OST** Math Practice Questions

1- What is the slope of a line that is perpendicular to the line \(4x-2y=12\)?

A. \(-2\)

B. 2

C. 4

D. 12

2- Simplify: \(5(x-2y)+(2-x)^2 = (5x-10y)+(4-4x+x^2) = x -10y +4 +x^2\)

When \(x=3\) and \(y=-2\) ,therefore:

\(x -10y +4 +x^2 =3+20+4+9 =36\)

A. \(-4\)

B. 20

C. 36

D. 50

3- The mean of 50 test scores was calculated as 88. But, it turned out that one of the scores was misread as 94 but it was 69. What is the mean?

A. 85

B. 87

C. 87.5

D. 88.5

4- The width of a box is one third of its length. The height of the box is one third of its width. If the length of the box is 27 cm, what is the volume of the box?

A. 81 cm\(^3\)

B. 162 cm\(^3\)

C. 243 cm\(^3\)

D. 729 cm\(^3\)

5- In five successive hours, a car travels 40 km, 45 km, 50 km, 35 km and 55 km. In the next five hours, it travels with an average speed of 50 km per hour. Find the total distance the car traveled in 10 hours.

A. 425 km

B. 450 km

C. 475 km

D. 500 km

6- The ratio of boys to girls in a school is 2:3. If there are 600 students in a school, how many boys are in the school.____________

7- The perimeter of the trapezoid below is 54 cm. What is its area? _________

8- In 1999, the average worker’s income increased $2,000 per year starting from $24,000 annual salary. Which equation represents income greater than average? (\(I =\) income, \(x = \)number of years after 1999)

A. \(I > 2000 x + 24000\)

B. \(I > -2000 x + 24000\)

C. \(I < -2000 x + 24000\)

D. \(I < 2000 x – 24000\)

9- Which of the following graphs represents the compound inequality?

A. graph 1

B. graph 2

C. graph 3

D. graph 4

10- A football team had $20,000 to spend on supplies. The team spent $14,000 on new balls. New sport shoes cost $120 each. Which of the following inequalities represent how many new shoes the team can purchase.

A. \( 120x+14,000 ≤20,000 \)

B. \(120x+14,000 ≥20,000\)

C. \(14,000x+12,0 ≤20,000 \)

D. \( 14,000x+12,0 ≥20,000 \)

11- Two dice are thrown simultaneously, what is the probability of getting a sum of 6 or 9?

A. \(\frac{1}{3}\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{6}\)

D. \(\frac{11}{36}\)

12- A swimming pool holds 2,000 cubic feet of water. The swimming pool is 25 feet long and 10 feet wide. How deep is the swimming pool? __________

13- Which graph corresponds to the following inequalities?

\(y≤ x + 4\)

\(2x + y ≤ – 4\)

A.

B.

C.

D.

14- A bank is offering \(4.5\%\) simple interest on a savings account. If you deposit $8,000, how much interest will you earn in five years?

A. $360

B. $720

C. $1800

D. $3600

15- A card is drawn at random from a standard 52–card deck, what is the probability that the card is of Hearts? (The deck includes 13 of each suit clubs, diamonds, hearts, and spades)

A. \(\frac{1}{3}\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{6}\)

D. \(\frac{1}{52}\)

16- How long does a 420–miles trip take moving at 50 miles per hour (mph)?

A. 4 hours

B. 6 hours and 24 minutes

C. 8 hours and 24 minutes

D. 8 hours and 30 minutes

17- 11 yards 6 feet and 4 inches equals to how many inches?

A. 388

B. 468

C. 472

D. 476

18- A shirt costing $200 is discounted \(15\%\). After a month, the shirt is discounted another \(15\%\). Which of the following expressions can be used to find the selling price of the shirt?

A. \((200) (0.70)\)

B. \((200) – 200 (0.30)\)

C. \((200) (0.15) – (200) (0.15)\)

D. \((200) (0.85) (0.85)\)

19- Which of the following points lies on the line \(2x + 4y = 10\)

A. \((2, 1)\)

B. \((–1, 3)\)

C. \((–2, 2)\)

D. \((2, 2)\)

20- \(5 + 8 × (–2) – [4 + 22 ×5] ÷ 6 = \)?

A. \(-30\)

B. \(-20\)

C. \(-10\)

D. 0

## Best **7th Grade OST** Math Workbook Resource for 2021

## Answers:

1- **A**

The equation of a line in slope intercept form is: \(y = mx + b\)

Solve for \(y\).

\(4x-2y=12 ⇒ -2y=12-4x ⇒ y=(12-4x)÷(-2) ⇒ y=2x-6\)

The slope of this line is 2.

The product of the slopes of two perpendicular lines is\( -1\).

Therefore, the slope of a line that is perpendicular to this line is:

\(m_1 × m_2 = -1 ⇒ 2 × (m_2) = -1 ⇒ (m_2) = -\frac{1}{2}\)

2- **C**

Simplify: \(5(x-2y)+(2-x)^2 = (5x-10y)+(4-4x+x^2) = x -10y +4 +x^2\)

When \(x=3\) and \(y=-2\) ,therefore:

\(x -10y +4 +x^2 =3+20+4+9 =36\)

3-** C**

\(average (mean) = \frac{sum \space of \space terms }{number \space of \space terms}⇒ 88 = \frac{sum \space of \space terms}{50}⇒ sum = 88 × 50 = 4400\)

The difference of 94 and 69 is 25. Therefore, 25 should be subtracted from the sum.

\(4400 – 25 = 4375\)

\(mean =\frac{sum \space of \space terms}{number \space of \space terms}⇒ mean = \frac{4375 }{50}= 87.5\)

4- **D**

If the length of the box is 27, then the width of the box is one third of it, 9, and the height of the box is 3 (one third of the width). The volume of the box is:

\(V = lwh = (27) (9) (3) = 729\)

5- **C**

Add the first 5 numbers. \(40 + 45 + 50 + 35 + 55 = 225\)

To find the distance traveled in the next 5 hours, multiply the average by number of hours.

\(Distance = Average × Rate = 50 × 5 = 250\)

Add both numbers.

\(250 + 225 = 475\)

6- **240**

The ratio of boy to girls is 2:3. Therefore, there are 2 boys out of 5 students. To find the answer, first divide the total number of students by 5, then multiply the result by 2.

\(600 ÷ 5 = 120 ⇒ 120 × 2 = 240\)

7- **130**

The perimeter of the trapezoid is 54 cm.

Therefore, the missing side (high) is \(= 54 – 18 – 12 – 14 = 10\)

Area of a trapezoid:

\( A = \frac{1}{2}h (b_1 + b_2) = \frac{1}{2}(10) (12 + 14) = 130\)

8- **A**

Let \(x\) be the number of years. Therefore, $2,000 per year equals 2000\(x\).

starting from $24,000 annual salary means you should add that amount to 2000\(x\).

Income more than that is:

\(I > 2000x + 24000\)

9- **D**

Solve for \(x\).

\(-2≤2x-4<8 \)⇒ (add 4 all sides)\( -2+4≤2x-4+4<8+4 \)

\(⇒ 2≤2x<12 \)

⇒ (divide all sides by 2)\( 1≤x<6\)

\(x\) is between 1 and 6.

10- **A**

Let \(x\) be the number of new shoes the team can purchase. Therefore, the team can purchase 120 \(x\).

The team had $20,000 and spent $14000. Now the team can spend on new shoes $6000 at most.

Now, write the inequality:

\(120x+14.000 ≤20.000\)

11- **B**

For Sum 6: (1 & 5) and (5 & 1), (2 & 4) and (4 & 2), (3 & 3), so we have 5 options.

For sum 9: (3 & 6) and (6 & 3), (4 & 5) and (5 & 4), we have 4 options.

To get a sum of 6 or 9 for two dice: \(5+4=9\)

Since we have \(6 × 6 = 36\) total options, the probability of getting a sum of 6 and 9 is 9 out of 36 or \(\frac{1}{4}\).

12- **8**

Use formula of rectangle prism volume.

\(V = (length) (width) (height) ⇒ 2000 = (25) (10) (height)\)

\( ⇒ height = 2000 ÷ 250 = 8\)

13- **A**

For each option, choose a point in the solution part and check it on both inequalities.

A. Point \((–4, –4)\) is in the solution section. Let’s check the point in both inequalities.

\(–4 ≤ – 4 + 4, \space It \space works\)

\(2 (–4) + (–4) ≤ –4 ⇒ – 12 ≤ – 4\) it works (this point works in both)

B. Let’s choose this point \((0, 0)\)

\(0 ≤ 0 + 4, \space It \space works\)

\(2 (0) + (0) ≤ –4, \space \space That’s \space not \space true!\)

C. Let’s choose this point \((–5, 0)\)

\(0 ≤ –5 + 4, \space That’s \space not \space true!\)

D. Let’s choose this point \((0, 5)\)

\(5 ≤ 0 + 4, \space That’s \space not \space true!\)

14- **C**

Use simple interest formula:

I=prt

(I = interest, p = principal, r = rate, t = time)

\(I=(8000)(0.045)(5)=1800\)

15- **B**

The probability of choosing a Hearts is \(\frac{13}{52}=\frac{1}{4}\)

16- **C**

Use distance formula:

\(Distance = Rate × time ⇒ 420 = 50 × T\)

divide both sides by 50.

\(\frac{420}{50} = T ⇒ T = 8.4 \space hours\)

Change hours to minutes for the decimal part.

\(0.4 \space hours = 0.4 × 60 = 24 \space minutes\)

17- **C**

\(11 × 36 + 6 × 12 + 4 = 472\)

18- **D**

To find the discount, multiply the number by (\(100\% – \)rate of discount).

Therefore, for the first discount we get:

\((200) (100\% – 15\%) = (200) (0.85) = 170\)

For the next \(15%\) discount: \((200) (0.85) (0.85)\)

19- **B**

Input \((-1, 3)\) in the \(2x + 4y = 10\) formula instead of \(x\) and y. So we have:

\( 2(-1) + 4(3) = 10\)

\(-2 + 12 = 10\)

20- **A**

Use PEMDAS (order of operation):

\(5 + 8 × (–2) – [4 + 22 ×5] ÷ 6 = 5 + 8 × (–2) – [4 + 110] ÷ 6 = 5 + 8 × (–2) – [114] ÷ 6 = 5 + (–16) – 19 = 5 + (–16) – 19 = –11 – 19 = –30\)

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