A student writes: "When you multiply two numbers, the product is always greater than both factors." The teacher wants one example, with an explanation, that shows the product of two positive numbers can be less than one of the factors. Which example best meets this goal?

0.75 × 12 = 9; multiplying 12 by a factor between 0 and 1 takes only part of 12, so the product is less than 12

1.25 × 12 = 15; multiplying 12 by any decimal factor shrinks it, so the product is less than one factor

1 × 12 = 12; multiplying 12 by the factor 1 leaves the size of 12 unchanged, so the product is not greater

−3 × 12 = −36; multiplying 12 by a negative factor gives a result that is less than both of the factors

A student states a rule: "If a whole number is divisible by both 4 and 6, then it is divisible by 24." Which response best addresses the student's rule?

The rule is false: 12 is divisible by 4 and by 6 but not by 24; since 4 and 6 share a factor of 2, the number need only be divisible by LCM(4, 6) = 12

The rule is true: 4 × 6 = 24, and whenever a number is divisible by two numbers, it must also be divisible by the product of those two numbers

The rule is false: 18 is divisible by 6 but not by 24, which shows that a number divisible by 4 and by 6 does not need to be divisible by 24

The rule is true: 48 is divisible by 4, by 6, and by 24, and checking several more multiples of 24, such as 72 and 96, confirms that it always holds

A student solves 7 − 3x > 19 as follows. Step 1: −3x > 12 Step 2: x > −4 Which response best identifies the student's error and gives the correct solution set?

Dividing both sides by −3 reverses the inequality, so the solution is x < −4.

Subtracting 7 from both sides reverses the inequality, so the solution is x < −4.

The student should have added 3x first instead, so the solution is x > 4.

Dividing both sides by −3 reverses the inequality, so the solution is x < 4.

Given f(x) = x² + 1, a student writes f(x + 2) = x² + 3. Which statement correctly diagnoses the student’s work?

The student added 2 to the output instead of the input; f(x + 2) = (x + 2)² + 1 = x² + 4x + 5.

The work is correct, since adding 2 to an input of f always adds exactly 2 to the output of f.

The student squared x but not 2; the correct value is f(x + 2) = x² + 2² + 1, which is x² + 5.

The student should multiply: f(x + 2) = f(x) · f(2) = 5(x² + 1), which simplifies to 5x² + 5.

A student writes 0.3 × 0.2 = 0.6. Which model would best help the student see why the product is 0.06?

On a 10-by-10 grid representing 1, shade 3 columns (0.3) and 2 rows (0.2); the overlap is 6 of 100 squares, or 0.06

On a number line, draw 3 jumps of 0.2 starting at 0; the jumps end at 0.6, so the student should check 0.2 + 0.2 + 0.2

Use base-ten rods: 3 tenths rods and 2 tenths rods are combined to make 5 tenths rods, so the product should be 0.5

Count decimal places: 0.3 has one and 0.2 has one, so the product has one decimal place, which shows that 0.6 is correct

To decide whether 221 is prime, a student plans to divide 221 by every prime number up to 110, "because no factor can be more than half of 221." Which response would best improve the student's method?

Only primes up to √221 ≈ 14.9 need testing, because a factor pair always contains a factor at most √221; testing 13 shows 221 = 13 × 17, so 221 is composite

Only the primes 2, 3, and 5 need testing, because every composite number is divisible by at least one of them; none divides 221, so 221 is prime

Every whole number from 2 to 220 must be tested, including composites, because a factor of 221 might be composite; doing so shows that 221 is prime

Only primes up to 221 ÷ 3 ≈ 73 need testing, because 3 is the smallest odd prime that could divide 221; testing these primes shows that 221 is prime

A student solves 4(x − 2) + 3 = 4x − 5 and gets 4x − 5 = 4x − 5, then 0 = 0. The student writes "x = 0." Which response best addresses the student's conclusion?

The equation is an identity, so every real number is a solution, not only x = 0.

The student is correct, because the variable was eliminated and 0 = 0 is true.

The student made an error distributing the 4; the correct solution is x = 5/4.

The statement 0 = 0 means the equation has no solution, so x = 0 must be rejected.

A student claims Mapping P {1 → 4, 2 → 4, 3 → 7} is not a function but Mapping Q {4 → 1, 4 → 2, 7 → 3} is a function, “because a function can’t repeat a number on the output side.” Which response is correct?

P is a function and Q is not, since only an input paired with two outputs, like 4 in Q, breaks the rule.

Neither is a function, because each mapping contains a number that appears twice on one of its sides.

Both are functions, because every number in the domain of each mapping is paired with some output.

The student is right: P fails since outputs repeat, while Q passes since each of its outputs is distinct.

A student claims that 1 square yard equals 3 square feet, because 1 yard equals 3 feet. Which activity would best help the student correct this error?

Cover a 1-yard-by-1-yard square with 1-foot-by-1-foot tiles and count the 9 tiles that are needed.

Lay three 1-foot rulers end to end along a yardstick and confirm that they match its full length.

Measure the perimeter of a 1-yard square in feet and confirm that the total distance is 12 feet.

Convert 1 yard to 36 inches and then divide 36 by 12 to confirm the 3-to-1 ratio of the two units.

Praxis Middle School Mathematics (5164)

Free Praxis Middle Scho 5164 Practice Questions

10 sample questions for the Praxis Middle School Mathematics (5164), with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.

Sections covered on the exam: Numbers and Operations, Algebra, Functions, Geometry and Measurement, Statistics and Probability.

Question 1 · Numbers and Operations
A student writes: "When you multiply two numbers, the product is always greater than both factors." The teacher wants one example, with an explanation, that shows the product of two positive numbers can be less than one of the factors. Which example best meets this goal?
- A.1.25 × 12 = 15; multiplying 12 by any decimal factor shrinks it, so the product is less than one factor
- B.1 × 12 = 12; multiplying 12 by the factor 1 leaves the size of 12 unchanged, so the product is not greater
- C.−3 × 12 = −36; multiplying 12 by a negative factor gives a result that is less than both of the factors
- D.0.75 × 12 = 9; multiplying 12 by a factor between 0 and 1 takes only part of 12, so the product is less than 12
Show answer and explanation
Answer: D. 0.75 × 12 = 9; multiplying 12 by a factor between 0 and 1 takes only part of 12, so the product is less than 12
0.75 × 12 = 9, and 9 < 12, because multiplying by a number between 0 and 1 produces a part of the other factor. The example 1 × 12 = 12 shows a product equal to a factor, not less than it. The example 1.25 × 12 = 15 actually supports the student's claim, and its explanation is false. The example with −3 uses a negative factor, which the teacher's goal excludes.
Question 2 · Numbers and Operations
A student states a rule: "If a whole number is divisible by both 4 and 6, then it is divisible by 24." Which response best addresses the student's rule?
- A.The rule is true: 4 × 6 = 24, and whenever a number is divisible by two numbers, it must also be divisible by the product of those two numbers
- B.The rule is false: 18 is divisible by 6 but not by 24, which shows that a number divisible by 4 and by 6 does not need to be divisible by 24
- C.The rule is false: 12 is divisible by 4 and by 6 but not by 24; since 4 and 6 share a factor of 2, the number need only be divisible by LCM(4, 6) = 12
- D.The rule is true: 48 is divisible by 4, by 6, and by 24, and checking several more multiples of 24, such as 72 and 96, confirms that it always holds
Show answer and explanation
Answer: C. The rule is false: 12 is divisible by 4 and by 6 but not by 24; since 4 and 6 share a factor of 2, the number need only be divisible by LCM(4, 6) = 12
The number 12 is divisible by 4 and by 6 but not by 24, so the rule fails. Divisibility by both a and b guarantees divisibility by LCM(a, b), which equals a · b only when a and b are relatively prime; here LCM(4, 6) = 12. The example 18 is not a counterexample because 18 is not divisible by 4, and checking examples like 48 cannot prove a general rule.
Question 3 · Algebra
A student solves 7 − 3x > 19 as follows. Step 1: −3x > 12 Step 2: x > −4 Which response best identifies the student's error and gives the correct solution set?
- A.Subtracting 7 from both sides reverses the inequality, so the solution is x < −4.
- B.Dividing both sides by −3 reverses the inequality, so the solution is x < −4.
- C.The student should have added 3x first instead, so the solution is x > 4.
- D.Dividing both sides by −3 reverses the inequality, so the solution is x < 4.
Show answer and explanation
Answer: B. Dividing both sides by −3 reverses the inequality, so the solution is x < −4.
Step 1 is correct: subtracting 7 does not change the direction of the inequality. In Step 2, dividing by a negative number must reverse the inequality, so 12 ÷ (−3) gives x < −4. Check: x = −5 gives 7 + 15 = 22 > 19, while x = 0 gives 7 > 19, which is false. Adding 3x first gives −12 > 3x, which also leads to x < −4.
Question 4 · Functions
Given f(x) = x² + 1, a student writes f(x + 2) = x² + 3. Which statement correctly diagnoses the student’s work?
- A.The work is correct, since adding 2 to an input of f always adds exactly 2 to the output of f.
- B.The student added 2 to the output instead of the input; f(x + 2) = (x + 2)² + 1 = x² + 4x + 5.
- C.The student squared x but not 2; the correct value is f(x + 2) = x² + 2² + 1, which is x² + 5.
- D.The student should multiply: f(x + 2) = f(x) · f(2) = 5(x² + 1), which simplifies to 5x² + 5.
Show answer and explanation
Answer: B. The student added 2 to the output instead of the input; f(x + 2) = (x + 2)² + 1 = x² + 4x + 5.
The student computed f(x) + 2, which shifts the graph up, rather than f(x + 2), which replaces every x with x + 2. Correctly, f(x + 2) = (x + 2)² + 1 = x² + 4x + 4 + 1 = x² + 4x + 5. Writing x² + 2² ignores the middle term 4x of the binomial square, and f(x) · f(2) is not a valid rule for function inputs.
Question 5 · Geometry and Measurement
A rectangular patio slab will measure 18 feet by 12 feet and be 4 inches thick. Concrete is sold only in whole cubic yards. What is the least number of cubic yards that must be ordered?
- A.24
- B.3
- C.2
- D.8
Show answer and explanation
Answer: B. 3
Volume = 18 · 12 · (4/12) = 72 ft³. Since 1 yd³ = 3³ = 27 ft³, 72 ÷ 27 ≈ 2.67 yd³, which must be rounded up to 3. Dividing by 9 (the square-yard factor) gives 8, dividing by 3 gives 24, and rounding down to 2 would leave the slab short.
Question 6 · Numbers and Operations
A student writes 0.3 × 0.2 = 0.6. Which model would best help the student see why the product is 0.06?
- A.On a 10-by-10 grid representing 1, shade 3 columns (0.3) and 2 rows (0.2); the overlap is 6 of 100 squares, or 0.06
- B.On a number line, draw 3 jumps of 0.2 starting at 0; the jumps end at 0.6, so the student should check 0.2 + 0.2 + 0.2
- C.Use base-ten rods: 3 tenths rods and 2 tenths rods are combined to make 5 tenths rods, so the product should be 0.5
- D.Count decimal places: 0.3 has one and 0.2 has one, so the product has one decimal place, which shows that 0.6 is correct
Show answer and explanation
Answer: A. On a 10-by-10 grid representing 1, shade 3 columns (0.3) and 2 rows (0.2); the overlap is 6 of 100 squares, or 0.06
In the area model, 0.3 of 0.2 is the overlapping region, which is 6 hundredths of the whole grid, so 0.3 × 0.2 = 0.06. Three jumps of 0.2 model 3 × 0.2 = 0.6, not 0.3 × 0.2. The decimal-place rule actually requires two places in the product, and combining rods models addition.
Question 7 · Numbers and Operations
To decide whether 221 is prime, a student plans to divide 221 by every prime number up to 110, "because no factor can be more than half of 221." Which response would best improve the student's method?
- A.Only the primes 2, 3, and 5 need testing, because every composite number is divisible by at least one of them; none divides 221, so 221 is prime
- B.Every whole number from 2 to 220 must be tested, including composites, because a factor of 221 might be composite; doing so shows that 221 is prime
- C.Only primes up to 221 ÷ 3 ≈ 73 need testing, because 3 is the smallest odd prime that could divide 221; testing these primes shows that 221 is prime
- D.Only primes up to √221 ≈ 14.9 need testing, because a factor pair always contains a factor at most √221; testing 13 shows 221 = 13 × 17, so 221 is composite
Show answer and explanation
Answer: D. Only primes up to √221 ≈ 14.9 need testing, because a factor pair always contains a factor at most √221; testing 13 shows 221 = 13 × 17, so 221 is composite
If 221 = a × b with a ≤ b, then a² ≤ 221, so a ≤ √221 ≈ 14.9; only 2, 3, 5, 7, 11, and 13 need testing. Since 221 ÷ 13 = 17, the number is composite. Testing only 2, 3, and 5 misses composites like 13 × 17, and any composite factor has a smaller prime factor that would already be found.
Question 8 · Algebra
A student solves 4(x − 2) + 3 = 4x − 5 and gets 4x − 5 = 4x − 5, then 0 = 0. The student writes "x = 0." Which response best addresses the student's conclusion?
- A.The student is correct, because the variable was eliminated and 0 = 0 is true.
- B.The equation is an identity, so every real number is a solution, not only x = 0.
- C.The student made an error distributing the 4; the correct solution is x = 5/4.
- D.The statement 0 = 0 means the equation has no solution, so x = 0 must be rejected.
Show answer and explanation
Answer: B. The equation is an identity, so every real number is a solution, not only x = 0.
Both sides simplify to 4x − 5, so the equation is true for every value of x; for example, x = 10 gives 35 = 35. The final statement 0 = 0 is a true statement with no variable, which signals an identity, not the value x = 0. A false statement such as 0 = 3 would signal no solution. The distribution 4(x − 2) = 4x − 8 was done correctly.
Question 9 · Functions
A student claims Mapping P {1 → 4, 2 → 4, 3 → 7} is not a function but Mapping Q {4 → 1, 4 → 2, 7 → 3} is a function, “because a function can’t repeat a number on the output side.” Which response is correct?
- A.Neither is a function, because each mapping contains a number that appears twice on one of its sides.
- B.P is a function and Q is not, since only an input paired with two outputs, like 4 in Q, breaks the rule.
- C.Both are functions, because every number in the domain of each mapping is paired with some output.
- D.The student is right: P fails since outputs repeat, while Q passes since each of its outputs is distinct.
Show answer and explanation
Answer: B. P is a function and Q is not, since only an input paired with two outputs, like 4 in Q, breaks the rule.
A function assigns exactly one output to each input; different inputs may share an output. In P, each of 1, 2, and 3 has one output, so P is a function (1 and 2 sharing 4 is allowed). In Q, the input 4 is paired with both 1 and 2, so Q is not a function. The student has reversed the condition, checking outputs instead of inputs.
Question 10 · Geometry and Measurement
A student claims that 1 square yard equals 3 square feet, because 1 yard equals 3 feet. Which activity would best help the student correct this error?
- A.Cover a 1-yard-by-1-yard square with 1-foot-by-1-foot tiles and count the 9 tiles that are needed.
- B.Lay three 1-foot rulers end to end along a yardstick and confirm that they match its full length.
- C.Measure the perimeter of a 1-yard square in feet and confirm that the total distance is 12 feet.
- D.Convert 1 yard to 36 inches and then divide 36 by 12 to confirm the 3-to-1 ratio of the two units.
Show answer and explanation
Answer: A. Cover a 1-yard-by-1-yard square with 1-foot-by-1-foot tiles and count the 9 tiles that are needed.
Tiling the square shows 3 rows of 3 square feet, so 1 yd² = 3² = 9 ft²; area conversions use the square of the length factor. The other activities all reinforce the linear 3-to-1 relationship (or perimeter) and do not address area, so they would not challenge the misconception.

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