Star Math Practice Questions (2026) — Free Sample Questions with Answers

Answer: A

1. 1. $7^2=49$ and $8^2=64$, so $\sqrt{52}$ is between 7 and 8, closer to 7.
2. 2. $7.2^2=51.84$ and $7.3^2=53.29$
3. 3. $52-51.84=0.16$ and $53.29-52=1.29$, so 52 is closer to $7.2^2$.
4. 4. $\sqrt{52}\approx 7.2$

7.2: Correct. $7.2^2=51.84$ and $7.3^2=53.29$; 52 is much closer to 51.84, so $\sqrt{52}\approx 7.2$.

7.3: Sees that 52 is between $7.2^2$ and $7.3^2$ and rounds up, but 52 is closer to $7.2^2=51.84$.

7.5: Assumes $\sqrt{52}$ is halfway between 7 and 8; 52 is much closer to 49 than to 64.

26: Divides 52 by 2 instead of finding the square root.

Estimate a square root by finding the perfect squares on either side, then squaring decimals to narrow it down.

Answer: D

1. 1. Exponents: $2^3=8$ and $3^2=9$
2. 2. Parentheses: $8+4=12$
3. 3. Divide by the fraction: $12÷\frac{1}{2}=12\times 2=24$
4. 4. Subtract: $24-9=15$

$-3$: Multiplies by $\frac{1}{2}$ instead of dividing: $12\times \frac{1}{2}-9=-3$.

11: Treats $2^3$ as $2\times 3=6$: $(6+4)÷\frac{1}{2}-9=11$.

14: Treats both powers as base × exponent: $(6+4)÷\frac{1}{2}-6=14$.

15: Correct. $(8+4)÷\frac{1}{2}-9=12\times 2-9=24-9=15$.

Follow the order of operations: parentheses, exponents, multiplication and division, then addition and subtraction. Dividing by ½ is the same as multiplying by 2.

Answer: D

1. 1. One batch is 4 + 6 = 10 cups of paint.
2. 2. 30 ÷ 10 = 3, so every amount is 3 times the batch.
3. 3. White paint = 3 × 6 = 18 cups
4. 4. Check: blue = 3 × 4 = 12, 12 + 18 = 30, and 12 : 18 = 4 : 6

6: Gives the white paint for one 10-cup batch without scaling up to 30 cups.

12: Finds the amount of blue paint (3 × 4) instead of white paint.

15: Splits the 30 cups equally between the two colors, ignoring the 4 : 6 ratio.

18: Correct. 30 ÷ 10 = 3, so he makes 3 times the batch: white paint = 3 × 6 = 18 cups.

Use the totals to find the scale factor, then multiply each part of the ratio by that factor.

Answer: D

1. Rate of change: (62 − 46) ÷ (4 − 2) = 16 ÷ 2 = 8 dollars per week.
2. Work back 2 weeks from (2, 46): 46 − 2 × 8 = 30.
3. The y-intercept is (0, 30): she had $30 in the account at week 0.

She had $46 in the account at the start.: Uses the first value in the table, which is the amount after 2 weeks, not at 0 weeks.

She adds $8 to the account each week.: Describes the slope (the rate of change), not the y-intercept.

She had $0 in the account at the start.: Assumes the line passes through the origin; working back from the table gives $30 at 0 weeks.

She had $30 in the account at the start.: Correct. The amount grows by $8 each week, and working back 2 weeks from $46 gives $46 − $16 = $30 at week 0.

The y-intercept is the value of y when x = 0. In context it is the starting amount, found by working back from the table using the rate of change.

Answer: D

1. 1. $m=\frac{9-1}{6-2}=\frac{8}{4}=2$
2. 2. Substitute (2, 1): $1=2(2)+b$, so $b=-3$
3. 3. $y=2x-3$
4. 4. Check (6, 9): $2(6)-3=9$

$y=\frac{1}{2}x$: Uses run over rise, $\frac{4}{8}$, for the slope.

$y=2x+1$: Uses the $y$-coordinate of (2, 1) as the $y$-intercept; (2, 1) is not on the $y$-axis.

$y=2x-1$: Finds $b$ as $1-2$, forgetting to multiply the slope by $x=2$.

$y=2x-3$: Correct. $m=\frac{9-1}{6-2}=2$; $1=2(2)+b$, so $b=-3$.

Find the slope from the two points, then substitute one point into y = mx + b to solve for b.

Answer: B

1. 1. Compare the rows: 3, 6, 12, 24. Each number of tiles is 2 times the one before.
2. 2. Continue the pattern: Figure 5 has 48, Figure 6 has 96, and Figure 7 has 192.
3. 3. 96 is not more than 100, but 192 is, so Figure 7 is the first.

Figure 6: Figure 6 has 96 tiles, which is close to 100 but not more than 100.

Figure 7: Correct. Each figure has twice as many tiles as the one before: 48, 96, 192. Figure 7, with 192 tiles, is the first over 100.

Figure 11: Keeps adding 12, the last difference in the table (36, 48, 60, …), instead of doubling.

Figure 34: Assumes each figure adds 3 tiles, as from Figure 1 to Figure 2, so finds 3 × 34 = 102.

Check how each term relates to the one before it. When the differences keep doubling, the pattern multiplies by 2 instead of adding a fixed amount.

Answer: B

1. Cone: ⅓ × π × 3² × 8 = ⅓ × 72π = 24π.
2. Whole sphere: 4/3 × π × 3³ = 4/3 × 27π = 36π, so the hemisphere is 18π.
3. Total: 24π + 18π = 42π cubic centimeters.

$24\pi$ cubic centimeters: Finds only the volume of the cone.

$42\pi$ cubic centimeters: Correct. Cone: ⅓ × 9 × 8π = 24π. Hemisphere: ½ × 4/3 × 27π = 18π. Total: 24π + 18π = 42π.

$60\pi$ cubic centimeters: Adds a whole sphere (36π) instead of half of one.

$90\pi$ cubic centimeters: Forgets the ⅓ for the cone: 72π + 18π.

For a composite solid, find the volume of each part and add. A hemisphere is half of the sphere formula.

Answer: A

1. The labeled angles are vertical angles, so 3x + 10 = 5x − 30.
2. Subtract 3x and add 30: 40 = 2x, so x = 20.
3. Angle measure: 3(20) + 10 = 70°. Check: 5(20) − 30 = 70°.

70°: Correct. The two labeled angles are vertical angles, so they are equal: 3x + 10 = 5x − 30 gives x = 20, and 3(20) + 10 = 70.

85°: Treats the angles as supplementary: 8x − 20 = 180 gives x = 25 and 3(25) + 10 = 85.

110°: Gives the measure of an adjacent angle, 180° − 70°.

140°: Adds the two vertical angles together.

Vertical angles (opposite angles formed by two intersecting lines) are equal. Adjacent angles along a line are supplementary.

Answer: B

1. 1. Rectangle: $10\times 6=60$ m²
2. 2. The half circle has diameter 6 m, so $r=3$ m.
3. 3. Half circle: $\frac{1}{2}\times 3.14\times 3^2=\frac{1}{2}\times 28.26=14.13$ m²
4. 4. Total: $60+14.13=74.13$ m²

69.42 m²: Uses half of the circumference formula, ½ × 3.14 × 6 = 9.42, for the half circle.

74.13 m²: Correct. Rectangle 10 × 6 = 60; half circle ½ × 3.14 × 3² = 14.13; total 74.13 m².

88.26 m²: Adds the area of a whole circle (28.26) instead of half of it.

116.52 m²: Uses the 6 m diameter as the radius: ½ × 3.14 × 36 = 56.52.

For a composite figure, add the parts; a half circle has half the area of the full circle with the same radius.

Answer: C

1. Base JK: from x = −2 to x = 4 is 6 units.
2. Height: from the base at y = −1 up to L at y = 3 is 4 units.
3. Area = ½ × 6 × 4 = 12 square units.

4 square units: Finds the base as 4 − 2 = 2 instead of 4 − (−2) = 6.

9 square units: Uses the y-coordinate of L, 3, as the height instead of the distance from y = −1 to y = 3.

12 square units: Correct. The base JK is 6 units and the height from L down to JK is 4 units, so A = ½ × 6 × 4 = 12.

24 square units: Multiplies base × height and forgets the ½.

The area of a triangle is ½ × base × height. The height is the distance from the base to the opposite vertex, not that vertex's coordinate.

Answer: B

1. 1. Rectangle: $10\times 6=60$ m²
2. 2. Triangle height: $10-6=4$ m
3. 3. Triangle: $\frac{1}{2}\times 10\times 4=20$ m²
4. 4. Total: $60+20=80$ m²

50 m²: Treats the whole front as one triangle with base 10 m and height 10 m.

80 m²: Correct. Rectangle: 10 × 6 = 60; triangle height 10 − 6 = 4, so ½ × 10 × 4 = 20; total 80 m².

100 m²: Forgets the ½ for the triangle: 60 + 10 × 4 = 100.

110 m²: Uses the full 10 m as the triangle's height: 60 + ½ × 10 × 10 = 110.

Split a composite figure into simple shapes, find any missing dimension from the given ones, and add the areas.

Answer: D

1. Pizza or pasta in the sample: 14 + 9 = 23 of 40 students.
2. Set up a proportion: 23/40 = x/600.
3. The school is 600 ÷ 40 = 15 times the sample, so x = 23 × 15 = 345.

23 students: Gives the number in the sample, not the estimate for the whole school.

135 students: Estimates only the students who would choose pasta.

210 students: Estimates only the students who would choose pizza.

345 students: Correct. In the sample, 14 + 9 = 23 of 40 chose pizza or pasta; 23/40 × 600 = 23 × 15 = 345.

A random sample can be used to estimate a population: the part of the sample with a trait times the population size.

Answer: A

1. 1. Total tiles: 4 + 6 + 2 = 12
2. 2. First draw blue: $\frac{6}{12}$
3. 3. Second draw blue (11 left, 5 blue): $\frac{5}{11}$
4. 4. Both: $\frac{6}{12}\times \frac{5}{11}=\frac{30}{132}=\frac{5}{22}$

$\frac{5}{22}$: Correct. $\frac{6}{12}\times \frac{5}{11}=\frac{30}{132}=\frac{5}{22}$.

$\frac{1}{4}$: Puts the first tile back: $\frac{6}{12}\times \frac{6}{12}=\frac{1}{4}$; here the tile is kept out.

$\frac{1}{2}$: Finds the probability of blue on one draw only.

$\frac{21}{22}$: Adds the two probabilities, $\frac{6}{12}+\frac{5}{11}$, instead of multiplying them.

For "and" events, multiply; without replacement, reduce both the count and the total for the second draw.

Answer: D

1. 1. Five-day total: 5 × 17 = 85 cans
2. 2. Monday–Thursday: 12 + 22 + 8 + 16 = 58 cans
3. 3. Friday: 85 − 58 = 27 cans
4. 4. Check: (58 + 27) ÷ 5 = 17

10: Uses 4 × 17 = 68 for the total, as if there were only four days: 68 − 58 = 10.

14.5: Finds the mean of the four days shown (58 ÷ 4) instead of Friday's count.

17: Assumes Friday's count equals the mean.

27: Correct. The five-day total is 5 × 17 = 85, and 85 − 58 = 27.

Mean × number of values gives the total; subtract the known values to find the missing one.