Imagine that a test taker at the beginning of a 7th Grade NYSE Math test session is faced with a question that seems simple but at that particular moment is not able to identify the correct option or the method of solving that question. One of the reasons for this problem is lack of enough practice. Fortunately, this problem can be solved by using 7th Grade NYSE Math sample practice questions in the weeks leading up to the 7th Grade NYSE Math test. In this article, we provide 7th Grade students with 10 examples of the best 7th Grade NYSE Math test practice questions. These questions come with answers that are explained in detail so there is no need to worry if students are not able to solve a problem.

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 7th Grade NYSE Math** Test

## 7th Grade NYSE 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 dollar?

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. \(40 \%\)

## Best **7th Grade NYSE **Math Prep Resource for 2021

## Answers:

1- **C**

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

Use the area of 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 that it 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 that it 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 =\( πr^2 ⇒ 64 π = πr^2 ⇒ 64 = r^2 ⇒ r = 8\)

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 old price).

9- **C**

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

Use 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 amount of students in the class.

\(40 \%\) of x = girls

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

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

tennis player = \(10\%\)

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