Grade 6 Math Practice Test

1. \((5 + 4) + 1 = 5 + (4 + 1)\) : Associativity of addition
2. \(2(4 + 7) = 2 \times 4 + 2 \times 7\) : Distributivity
3. \(11 + 9 = 9 + 11\) : Commutativity of addition
4. \(33 + 0 = 33\) : Identity of addition
5. \(5 \times 1 = 5\) : Identity of multiplication
6. \(9 \times 6 = 6 \times 9\) : Commutativity of multiplication
7. \((7 - 2) \times 6 = 7 \times 6 - 2 \times 6\) : Distributivity
8. \(3 \times 6 - 3 \times 2 = 3(6 - 2)\) : Distributivity in reverse (factoring)

A prime number has exactly two distinct positive divisors: 1 and itself. Note that 0 and 1 are not prime.

The prime numbers are: 2, 3, 5, 7, 11.

The 2 is in the Hundreds place.

1. \((9 - 3) + 2 = 6 + 2 = \mathbf{8}\)
2. \(7 - (5 - 2) = 7 - 3 = \mathbf{4}\)
3. \((3 + 7) \times 3 = 10 \times 3 = \mathbf{30}\)
4. \((8 - 2) \times 3 = 6 \times 3 = \mathbf{18}\)

The first digit after the decimal point is the tenths place. The digit is 8.

1. \(0.41\) rounded is 0 (the tenths digit is 4, which is less than 5).
2. \(1.2999\) rounded is 1 (the tenths digit is 2, which is less than 5).
3. \(123.5\) rounded is 124 (the tenths digit is 5, which rounds up).

1. \(0.1 > 0.3\) : False
2. \(1.2 < 1.3\) : True
3. \(0.5 < 0.05\) : False (0.50 is greater than 0.05)

1. \(0.4 \times 3 = \mathbf{1.2}\)
2. \(8 - 3 \times 0.2 = 8 - 0.6 = \mathbf{7.4}\)
3. \(0.5 \div 5 = \mathbf{0.1}\)

• Factors of 8: 1, 2, 4, 8
• Factors of 12: 1, 2, 3, 4, 6, 12
• Common factors: 1, 2, 4

The greatest common factor is 4.

• Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24...
• Multiples of 7: 7, 14, 21, 28...

The least common multiple is 21.

Rule: Any number ending in 0 or 5 is divisible by 5.

Therefore, 125 and 200 are divisible by 5.

Rule: Any number ending in an even digit (0, 2, 4, 6, 8) is divisible by 2.

Therefore, 280 and 476 are divisible by 2.

Rule: A number is divisible by 3 if the sum of its digits is divisible by 3.

1. \(105 \rightarrow 1+0+5 = 6\) (Divisible by 3)
2. \(101 \rightarrow 1+0+1 = 2\) (Not divisible by 3)
3. \(234 \rightarrow 2+3+4 = 9\) (Divisible by 3)

Therefore, 105 and 234 are divisible by 3.

Definition: An improper fraction is a fraction whose numerator is greater than or equal to its denominator.

Therefore, \(\dfrac{10}{3}\) and \(\dfrac{3}{3}\) are improper fractions.

1. \(\dfrac{7}{5} = \dfrac{5+2}{5} = \dfrac{5}{5} + \dfrac{2}{5} = \mathbf{1 \dfrac{2}{5}}\)
2. \(\dfrac{8}{3} = \dfrac{6+2}{3} = \dfrac{6}{3} + \dfrac{2}{3} = \mathbf{2 \dfrac{2}{3}}\)
3. \(\dfrac{9}{2} = \dfrac{8+1}{2} = \dfrac{8}{2} + \dfrac{1}{2} = \mathbf{4 \dfrac{1}{2}}\)

Rule: Multiply or divide the numerator and denominator by the same number to find equivalent fractions.

1. Multiply numerator and denominator by 2: \(\dfrac{1 \times 2}{2 \times 2} = \dfrac{\mathbf{2}}{4}\)
2. Multiply numerator and denominator by 3: \(\dfrac{2 \times 3}{5 \times 3} = \dfrac{6}{\mathbf{15}}\)
3. Multiply numerator and denominator by 3: \(\dfrac{1 \times 3}{3 \times 3} = \dfrac{\mathbf{3}}{9}\)

1. \(\dfrac{4-1}{10} = \mathbf{\dfrac{3}{10}}\)
2. \(\dfrac{2}{4} + \dfrac{3}{4} = \mathbf{\dfrac{5}{4}}\)
3. \(\dfrac{1 \times 2}{2 \times 3} = \dfrac{2}{6} = \mathbf{\dfrac{1}{3}}\)
4. \(\dfrac{3}{4} \times \dfrac{2}{1} = \dfrac{6}{4} = \mathbf{\dfrac{3}{2}}\)
5. \(\dfrac{3}{4} \times \dfrac{1}{3} = \dfrac{3}{12} = \mathbf{\dfrac{1}{4}}\)
6. \(\dfrac{2}{1} \times \dfrac{2}{6} = \dfrac{4}{6} = \mathbf{\dfrac{2}{3}}\)
7. \((1+2) + \left(\dfrac{1}{4} + \dfrac{1}{4}\right) = 3\dfrac{2}{4} = \mathbf{3\dfrac{1}{2}}\)
8. \((3-1) + \left(\dfrac{2}{5} - \dfrac{1}{5}\right) = \mathbf{2\dfrac{1}{5}}\)

1. \(7 \div 10 = \mathbf{0.7}\)
2. \(17 \div 100 = \mathbf{0.17}\)

1. There is one full circle and one-fourth of another. Answer: \(1 \dfrac{1}{4}\)
2. There is one-fourth of the square shaded. Answer: \(\dfrac{1}{4}\)
3. There are three full squares and three-eighths of another. Answer: \(3 \dfrac{3}{8}\)

1. \( 2 \times 2 \times 2 \times 2 \times 2 = \mathbf{2^5} \)
2. Five squared = \(\mathbf{5^2}\)
3. 4 cubed = \(\mathbf{4^3}\)
4. 6 to the seventh power = \(\mathbf{6^7}\)

1. \( 2^3 = 2 \times 2 \times 2 = \mathbf{8} \)
2. \( 1^5 = 1 \times 1 \times 1 \times 1 \times 1 = \mathbf{1} \)
3. \( 4^2 = 4 \times 4 = \mathbf{16} \)
4. \( 1000^0 = \mathbf{1} \) (Any non-zero number to the power of 0 is 1)

1. Red to blue = 2:3
2. Blue to red = 3:2
3. Blue to total (which is 2+3=5) = 3:5

1. Girls to boys = 11:8
2. Boys to total (which is 11+8=19) = 8:19

\(\$15 \div 5 \text{ kg} = \) \(\$3 / \text{kg}\)

\(120 \text{ km} \div 2 \text{ hours} = \) \(60 \text{ km/hr}\)

The ratio of boys to girls is \(1:3\). This means for every 1 boy, there are 3 girls, making a total of 4 "parts" in the student body.

If \(x\) is the number of boys, then \(3x\) is the number of girls.

$$x + 3x = 600$$

$$4x = 600$$

$$x = 600 \div 4 = \mathbf{150 \text{ boys}}$$

Check: If there are 150 boys, there are \(3 \times 150 = 450\) girls. \(150 + 450 = 600\).

\(\dfrac{60}{100} \times 20 = \dfrac{1200}{100} = \mathbf{12}\)

\(35\% = \dfrac{35}{100} = \mathbf{0.35}\)

\(15\% = \dfrac{15}{100}\). Divide numerator and denominator by 5: \(\mathbf{\dfrac{3}{20}}\)

\(\dfrac{50}{100} \times \dfrac{1}{4} = \dfrac{1}{2} \times \dfrac{1}{4} = \mathbf{\dfrac{1}{8}}\)

\(\dfrac{3}{5} = 0.6 = \dfrac{60}{100} = \mathbf{60\%}\)

\(\dfrac{600}{3000} = \dfrac{6}{30} = \dfrac{1}{5} = 0.20 = \mathbf{20\%}\)

Change in percent = \(\dfrac{\text{new price - original price}}{\text{original price}}\)

$$= \dfrac{100 - 125}{125} = \dfrac{-25}{125} = -0.2 = \mathbf{-20\%}$$

Discount = \(40\% \times 40 = \dfrac{40}{100} \times 40 = \$16\)

New Price = \(\$40 - \$16 = \mathbf{\$24}\)

Divide by 100: \(320 \div 100 = \mathbf{3.2 \text{ hL}}\)

Divide by 1000: \(234500 \div 1000 = \mathbf{234.5 \text{ m}}\)

Divide by 1000: \(2300 \div 1000 = \mathbf{2.3 \text{ km}}\)

Multiply by 1000: \(1.2 \times 1000 = \mathbf{1200 \text{ mL}}\)

Divide by 3600: \(7200 \div 3600 = \mathbf{2 \text{ hrs}}\)

Divide by 1760: \(2640 \div 1760 = \mathbf{1.5 \text{ mi}}\)

Multiply by 39.37: \(3 \times 39.37 = \mathbf{118.11 \text{ in}}\)

\( 0.2 + 2 = \mathbf{2.2} \)

\( 2 (3 + 2) = 2(5) = \mathbf{10} \)

\( 3 - 2 = \mathbf{1} \)

\( \dfrac{2(6)}{3} = \dfrac{12}{3} = \mathbf{4} \)

1. \(\mathbf{x - 3}\)
2. \(\mathbf{5(x + 2)}\)
3. \(\mathbf{(2x + 1)^2}\)
4. \(\mathbf{3(x + 1) = 2}\)

1. \(\mathbf{3 < x}\)
2. \(\mathbf{5 \ge x}\)
3. \(\mathbf{y \le 9}\)
4. \(\mathbf{x - 2 \le -2}\)

1. \( 2x + 8 \)
2. \( 3a + 3b + 6 \)
3. \( \dfrac{8}{4}x + \dfrac{4}{4} = \mathbf{2x + 1} \)
4. \( 0.2x + 0.4 \)

1. Factors of 9 are 1, 3, 9. Factors of 6 are 1, 2, 3, 6. The GCF is 3.
2. \(9 = \mathbf{3 \times 3}\), and \(6 = \mathbf{3 \times 2}\).
3. \( 9x + 6 = 3(3x) + 3(2) = \mathbf{3(3x + 2)} \)

1. \(x + 2 - 2 = 8 - 2 \implies \mathbf{x = 6}\)
2. \(2x \div 2 = 6 \div 2 \implies \mathbf{x = 3}\)
3. \(x - 3 + 3 = 7 + 3 \implies \mathbf{x = 10}\)

1. Perimeter = \(2 \times L + 2 \times W\). So, \(10 = 2(3) + 2x \implies \mathbf{2x + 6 = 10}\).
2. \(2x = 10 - 6 \implies 2x = 4 \implies \mathbf{x = 2}\) meters.
3. Check: \(2(3) + 2(2) = 6 + 4 = 10\). This matches the given perimeter.

1. \((0,1)\) is on the positive y-axis.
2. \((-2,-3)\) is in Quadrant III.
3. \((2,9)\) is in Quadrant I.
4. \((-4,6)\) is in Quadrant II.
5. \((3,-4)\) is in Quadrant IV.
6. \((-3,0)\) is on the negative x-axis.

• \(A = \mathbf{(-1, 1)}\)
• \(B = \mathbf{(0, -1)}\)
• \(C = \mathbf{(2, 0)}\)
• \(D = \mathbf{(3, 1)}\)
• \(E = \mathbf{(-2, -1)}\)
• \(F = \mathbf{(3, -2)}\)
• \(G = \mathbf{(0, 3)}\)
• \(H = \mathbf{(0, 0)}\)

a) Distance \(d = 5 \times t\):

Time \(t\) (hrs)	0	1	2	3	4	5
Distance \(d\) (km)	0	5	10	15	20	25

b) Graph of the line:

Numbers further left are smaller.

1. \(-4 < 0\) : True ($-4$ is to the left of 0)
2. \(0 > 4\) : False
3. \(-6 < -4\) : True
4. \(-6 > 5\) : False
5. \(0 < 6\) : True (assuming 6 is plotted to the right of 5)

The correct answer is b) \( 180^{\circ} \).

Because they form a straight line, they sum to \(180^{\circ}\). The correct answer is b) Supplementary.

Vertical angles are opposite each other when two lines cross. The valid vertical pairs from the list are:

• b) \(\angle BOC\) and \(\angle EOF\)
• d) \(\angle FOB\) and \(\angle COE\)
• f) \(\angle BOD\) and \(\angle EOA\)

1. Pentagon: 5 sides
2. Trapezoid: 4 sides
3. Triangle: 3 sides
4. Kite: 4 sides

The correct answer is b). To be precise: an isosceles triangle has two sides that are equal in length, and the angles opposite to those equal sides are also equal in measure.

Perimeter \(= 2 \times \text{Length} + 2 \times \text{Width} = 2(10) + 2(5) = 20 + 10 = \mathbf{30 \text{ cm}}\).

Area \(= \pi \times r^2 \approx 3.14 \times (1)^2 = \mathbf{3.14 \text{ square meters}}\).

Area of the entire rectangle ABCD \(= \text{length} \times \text{width} = 10 \times 7 = 70 \text{ cm}^2\).

Area of the unshaded right triangle ADE \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 7 = 7 \text{ cm}^2\).

Area of the shaded region \(= \text{Total Area} - \text{Triangle Area} = 70 - 7 = \mathbf{63 \text{ cm}^2}\).

Number of edges = 4 (top) + 4 (bottom) + 4 (sides) = 12 edges.

Number of faces = 1 (top) + 1 (bottom) + 4 (sides) = 6 faces.

1. Area ABCD (top) \(= 6 \times 4 = \mathbf{24 \text{ sq units}}\)
   Area ADHE (side) \(= 4 \times 10 = \mathbf{40 \text{ sq units}}\)
   Area DCGH (front) \(= 6 \times 10 = \mathbf{60 \text{ sq units}}\)
2. Surface area is twice the sum of the three distinct faces:
   \(SA = 2 \times (24 + 40 + 60) = 2 \times 124 = \mathbf{248 \text{ sq units}}\).
3. Volume \(= \text{length} \times \text{width} \times \text{height} = 6 \times 4 \times 10 = \mathbf{240 \text{ cubic units}}\).

First, find the area of the triangular base. It is a right triangle with legs 3 and 4:
Area of base \(a = \frac{1}{2} \times 3 \times 4 = 6\).

The volume is the area of the base times the length of the prism (\(AF\)):
\(24 = 6 \times AF \implies AF = 4\).

The total surface area is the sum of the two triangular bases and the three rectangular faces:
Two bases: \(6 + 6 = 12\).
Three rectangles: \((4 \times 4) + (3 \times 4) + (5 \times 4) = 16 + 12 + 20 = 48\).

Total Surface Area \(= 12 + 48 = \mathbf{60 \text{ sq units}}\).

Add the hours for each day: 3 (Mon) + 3 (Tue) + 2 (Wed) + 4 (Thu) + 3 (Fri).

Total \(= 3 + 3 + 2 + 4 + 3 = \mathbf{15 \text{ hours}}\).

1. From the histogram, 3 students scored in the 90-99 range.
2. 7 students scored 80-89, and 4 students scored 60-69.
   Difference \(= 7 - 4 = \) 3 more students.

First, order the data: \(1, 2, 2, 2, 3, 4, 7\)

• Range: Largest value - Smallest value \(= 7 - 1 = \mathbf{6}\).
• Mean: \(\frac{1 + 2 + 2 + 2 + 3 + 4 + 7}{7} = \frac{21}{7} = \mathbf{3}\).
• Mode: The most frequent value is 2 (appears three times).
• Median: The middle value in the ordered list is 2.

Probability must always be a number between 0 and 1 (inclusive). Therefore, b) -0.5 and c) 2 cannot be probabilities.

1. Flipping a coin has 2 outcomes (heads, tails).
2. Selecting from 5 cards has 5 outcomes.
3. Doing both creates \(2 \times 5 =\) 10 possible outcomes.

1. The 6 possible outcomes are: {1, 2, 3, 4, 5, 6}.
2. 0 is not on a standard die, so the probability is 0.
3. There is one 5, so the probability is \(\dfrac{1}{6}\).
4. The numbers greater than 4 are 5 and 6 (two outcomes). The probability is \(\dfrac{2}{6} = \mathbf{\dfrac{1}{3}}\).