As you know, using practice questions before the 6th Grade MAP Math test day will help your students learn how the test is structured. Also, by using such questions, your students will understand what mathematical concepts they should master. In this article, we have tried to help your 6th Grade student find their weak area, learn from their mistakes and increase their 6th Grade MAP Math test score by providing 10 common 6th Grade MAP Math test questions and their exact solutions.

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 your students need to practice.

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## 10 Sample **6th Grade MAP** Math Practice Questions

1- What is the equation of a line that passes through points \((0, 4)\) and \((2, 8)\)?

A. \(y = x\)

B. \(y = x + 4\)

C. \(y = 2x + 4\)

D. \(y = 2x – 4\)

2- What is the volume of a box with the following dimensions?

Height \(= 6\) cm , Width \(= 7\) cm, Length \(= 9\) cm

A. \(63 \space cm^3\)

B. \(126\space cm^3\)

C. \(189\space cm^3\)

D. \(378\space cm^3\)

3- Anita’s trick–or–treat bag contains 14 pieces of chocolate, 15 suckers, 16 pieces of gum, 20 pieces of licorice. If she randomly pulls a piece of candy from her bag, what is the probability of her pulling out a piece of sucker?

A. \(\frac{1}{13}\)

B. \(\frac{3}{13}\)

C. \(\frac{14}{65}\)

D. \(\frac{16}{65}\)

4-

In the following rectangle, which statement is false?

A. AD is parallel to BC

B. The measure of the sum of all the angles equals \(360^\circ\).

C. Length of AB equal to length DC.

D. AB is perpendicular to AC.

5- The area of a rectangular yard is 90 square meters. What is its width if its length is 15 meters?

A. 10 meters

B. 8 meters

C. 6 meters

D. 4 meters

6- Which statement about 4 multiplied by \(\frac{3}{5}\) must be true?

A. The product is between 1 and 2

B. The product is greater than 3

C. The product is equal to \(\frac{75}{31}\)

D. The product is between 2 and 2.5

7- Which of the following lists shows the fractions in order from least to greatest?

\(\frac{3}{4}, \frac{2}{7}, \frac{3}{8} , \frac{5}{11}\)

A. \(\frac{3}{8}, \frac{2}{7}, \frac{3}{4}, \frac{5}{11}\)

B. \(\frac{2}{7}, \frac{5}{11}, \frac{3}{8}, \frac{3}{4}\)

C. \(\frac{2}{7}, \frac{3}{8}, \frac{5}{11}, \frac{3}{4}\)

D. \(\frac{3}{8}, \frac{2}{7}, \frac{5}{11}, \frac{3}{4}\)

8- A car costing $300 is discounted \(10\%\). Which of the following expressions can be used to find the selling price of the car?

A. \((300)(0.4)\)

B. \(300-(300×0.1)\)

C. \((300)(0.1)\)

D. \( 300-(300×0.9)\)

9- What is the missing price factor of number 420?

\(420=2^2×3^1×…\)

A. \(2^2×3^1×5^1×7^1\)

B. \(2^2×3^1×7^1×9^1\)

C. \(1^2×2^3×2^1×3^1\)

D. \(3^2×5^1×7^1×9^1\)

10- If the area of the following trapezoid is equal to A, which equation represent \(x\)?

A. \(x = \frac{13}{A}\)

B. \(x = \frac{A}{13}\)

C. \( x=A+13\)

D. \( x=A-13\)

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## Answers:

1- **C **

The slope of the line is:

\(\frac{y_2-y_1}{x_2-x_1 }= \frac{8 – 4}{2 – 0}=\frac{4}{2}=2\)

The equation of a line can be written as:

\( y-y_0=m(x – x_0 )→y-4 = 2(x – 0)→y-4 = 2x→y = 2x + 4\)

2- **D**

Volume of a \(box = length × width × height = 6 × 7 × 9 = 378\)

3- **B**

\(Probability = \frac{number \space of \space desired \space outcomes}{number \space of \space total \space outcomes}=\frac{15}{14+15+16+20}= \frac{15}{65}=\frac{3}{13}\)

4- **D**

In any rectangle, sides are not perpendicular to diagonals.

5- **C**

Let y be the width of the rectangle. Then; \(15×y=90→y=\frac{90}{15}=6\)

6- **D**

\( 4×\frac{3}{5}=\frac{12}{5}=2.4 \)

\( 2.4>2\)

\( 2.4<3\)

\( \frac{75}{31}=2.419≠2.4\)

\(2<2.4<2.5 \) This is the answer!

7- **C**

Let’s compare each fraction:

\(\frac{2}{7}< \frac{3}{8}< \frac{5}{11} < \frac{3}{4}\)

Only choice C provides the right order.

8- **B **

To find the discount, multiply the number (\(100\%\)- rate of discount)

Therefore; \(300(100\%-10\%)=300(1-0.1)=300-(300×0.1)\)

9- **A**

\(420=2^2×3^1×5^1×7^1\)

10- **B**

The area of the trapezoid is: area= \(\frac{(base 1+base 2)}{2}×height= (\frac{10 + 16}{2})x = A\)

\( →13x = A→x = \frac{A}{13}\)

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