# Top 10 7th Grade OST Math Practice Questions

One way to increase the speed of 7th Grade students’ action in the 7th Grade OST Math test is to use practice tests. After studying the material, 7th Grade students should continue to use the 7th Grade OST Math sample practice questions until the questions seem repetitive to them. Of course, not all questions are going to be solved in one step, but in the first step, it is enough for students to get acquainted with the most common 7th Grade OST Math test questions. This is what we have done in this article! In other words, we have prepared the top 10 common questions in the 7th Grade OST Math test in this article and explained their answers.

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.

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**7th Grade OST Math Practice Questions**

1- What is the area of the shaded region?

A. \(9 π\) cm\(^2\)

B. \(25 π\) cm\(^2\)

C. \(39 π\) cm\(^2\)

D. \(64 π\) cm\(^2\)

2- A pizza cut into 8 parts. William and his sister Ella ordered two pizzas. William ate \(\frac{1}{4}\) of his pizza and Ella ate \(\frac{1}{2}\) of her pizza. What part of the two pizzas was left?

A. \(\frac{1}{2}\)

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

C. \(\frac{3}{8}\)

D. \(\frac{5}{8}\)

3- Simplify \(6x^2 y^3 (2x^2 y)^3\).

A. \( 12x^4 y^6\)

B. \( 12x^8 y^6\)

C. \( 48x^4 y^6\)

D. \( 48x^8 y^6\)

4- Right triangle ABC has two legs of lengths 6 cm (AB) and 8 cm (AC). What is the length of the third side (BC)?

A. 4 cm

B. 6 cm

C. 8 cm

D. 10 cm

5- The marked price of a computer is D dollar. Its price decreased by \(20\%\) in January and later increased by \(10\%\) in February. What is the final price of the computer in D dollars?

A. 0.80 D

B. 0.88 D

C. 0.90 D

D. 1.20 D

6- \([6 × (–24) + 8] – (–4) + [4 × 5] ÷ 2 =\) ?

A. \(-122\)

B. \(-112\)

C. \(-102\)

D. \(-92\)

7- The area of a circle is \(64 π\). What is the circumference of the circle?

A. \(8 π\)

B. \(16 π\)

C. \(32 π\)

D. \(64 π\)

8- A $40 shirt now selling for $28 is discounted by what percent?

A. \(20 \%\)

B. \(30 \%\)

C. \(40 \%\)

D. \(60 \%\)

9- From last year, the price of gasoline has increased from $1.25 per gallon to $1.75 per gallon. The new price is what percent of the original price?

A. \(72 \%\)

B. \(120 \%\)

C. \(140 \%\)

D. \(160 \%\)

10- If \(40 \%\) of a class are girls, and \(25 \%\) of girls play tennis, what fraction of the class play tennis?

A. \(10 \%\)

B. \(15 \%\)

C. \(20 \%\)

D. \(40 \%\)

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## Answers:

1- **C**

In this case, to find the area of the shaded region, subtract the area of two circles. (S1: the area of the big circle. S2: the area of a little circle)

Use the area of a circle formula.

S = \(π\)r\(^2\)

\(S_{1} – S_{2}= π( 5 + 3 cm)^2 – π(5 cm)^2\) ⇒ \(S_{1} – S_{2} = π64 cm^2 – π25 cm^2\) ⇒ \(S_{1} – S_{2} = 39π cm^2\)

2- **D**

William ate \(\frac{1}{4}\) of 8 parts of his pizza which means 2 parts out of 8 parts ( \(\frac{1}{4}\) of 8 parts = x ⇒ x = 2) and left 6 parts.

Ella ate \(\frac{1}{2}\) of 8 parts of her pizza which means 4 parts out of 8 parts ( \(\frac{1}{2}\) of 8 parts = x ⇒ x = 4) and left 4 parts.

Therefore, they ate \((4 + 2)\) parts out of\( (8 + 8)\) parts of their pizza and left \((6 + 4)\) parts out of \((8 + 8)\) parts of their pizza that it means: \(\frac{10}{16}\)

After simplification we have: \(\frac{5}{8}\)

3- **D**

Simplify.

\(6x^2 y^3 (2x^2 y)^3= 6x^2 y^3 (8x^6 y^3 ) = 48x^8 y^6\)

4- **D**

Use Pythagorean Theorem: \(a^2 + b^2 = c^2\)

\(62 + 82 = c^2 ⇒ 100 = c^2 ⇒ c = 10\)

5- **B**

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

Therefore, for the first discount, we get:

\((D) (100\% – 20\%) = (D) (0.80) = 0.80 D\)

For increase of \(10 \%\):

\((0.80 D) (100\% + 10\%) = (0.85 D) (1.10) = 0.88 D = 88\% D\)

6- **A**

Use PEMDAS (order of operation):

\([6 × (– 24) + 8] – (– 4) + [4 × 5] ÷ 2 = [– 144 + 8] – (– 4) + [20] ÷ 2 = [– 144 + 8] – (– 4) + 10 =\)

\([– 136] – (– 4) + 10 = [– 136] + 4 + 10 = – 122\)

7- **B**

Use the formula of areas of circles.

Area =\( πr^2 ⇒ 64 π = πr^2 ⇒ 64 = r^2 ⇒ r = 8\)

The radius of the circle is 8. Now, use the circumference formula:

Circumference = \(2πr = 2π (8) = 16 π\)

8- **B**

Use the formula for Percent of Change

\(\frac{New Value-Old Value}{Old Value}× 100 \%\)

\(\frac{28-40}{40}× 100 \% = – 30 \% \)

(negative sign here means that the new price is less than the old price).

9- **C**

The question is this: 1.75 is what percent of 1.25?

Use the percent formula:

part =\( \frac{percent}{100}× whole\)

\(\frac{percent}{100}× 1.25\) ⇒ \(1.75 = \frac{percent ×1.25}{100}⇒175 =\) percent \(×1.25\) ⇒ percent \(= \frac{175}{1.25}= 140\)

10- **A**

Let x be the number of students in the class.

\(40 \%\) of x = girls

\(25 \%\) of girls = tennis player

Input \(40\%\) of a class instead of girls in the second formula. Therefore, \(25\%\) of \(40\%\) of a class = tennis player

tennis player = \(10\%\)

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