Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

3rd Grade Math

3rd Grade Math Practice Test: Practice Test 4

Practice Test 4 for 3rd Grade Math: real questions and explanations from the Varsity Tutors practice-test pool.

0%

0 / 25 answered

Question 1 of 25

On the number line, 1 and 44\frac{4}{4}44​ are at the same point. Which fraction equals 1?

Question Navigator

All questions

Question 1

On the number line, 1 and 44\frac{4}{4}44​ are at the same point. Which fraction equals 1?

  1. 34\frac{3}{4}43​
  2. 41\frac{4}{1}14​
  3. 14\frac{1}{4}41​
  4. 44\frac{4}{4}44​ (correct answer)

Explanation: This question tests expressing whole numbers as fractions and recognizing fractions that equal whole numbers (CCSS.3.NF.3.c), specifically understanding that n = n/1 and that fractions like 4/4 equal 1. Any whole number can be written as a fraction by putting it over 1. For example, 3 = 3/1 (three ones). Also, when a fraction has the same numerator and denominator (like 4/4, 6/6), all parts are shaded and it equals 1 whole. Multiple wholes work too: 8/4 means eight fourths, which is 2 wholes (because 4 fourths make 1 whole, so 8 fourths make 2 wholes). The number line shows 1 and 4/4 at the same point, demonstrating that 4/4 equals the whole number 1. Choice C is correct because when all 4 fourths are present, it equals 1 whole; this shows understanding that whole numbers can be expressed as fractions and vice versa. Choice A is incorrect because 4/1 is 4, not 1; this error occurs when students don't recognize full fractions equal wholes. To help students understand whole numbers as fractions: Use number lines showing whole numbers and fractions at same point (1 and 2/2, 3 and 6/2). Show physical models: one whole circle = 2/2 = 3/3 = 4/4 (all parts shaded). Teach pattern: any whole number n = n/1 ("n ones"). Emphasize 4/4 = 1, 8/4 = 2 by counting fourths. Practice locating equivalent wholes and fractions on number lines. Watch for students who reverse numerator/denominator or don't recognize full fractions equal wholes.

Question 2

Look at the circle models below. Emma says 28\frac{2}{8}82​ and 14\frac{1}{4}41​ are equivalent because they show the same amount shaded. Which explanation best describes why Emma is correct?

  1. Both fractions have small numerators, so they must represent the same amount of shading
  2. When you divide both parts of 28\frac{2}{8}82​ by 2, you get 14\frac{1}{4}41​, and the shaded areas are equal (correct answer)
  3. Both fractions have even numbers in them, which makes them equivalent to each other
  4. When you multiply both parts of 28\frac{2}{8}82​ by 4, you get 14\frac{1}{4}41​, and the models show this

Explanation: Emma is correct because 2/8 and 1/4 represent the same amount. To show they are equivalent algebraically, we divide both numerator and denominator of 2/8 by 2: (2÷2)/(8÷2) = 1/4. The visual models confirm this by showing identical shaded areas. Choice A gives an irrelevant reason about numerator size. Choice C incorrectly focuses on even numbers. Choice D has the wrong operation (multiplying gives 8/32, not 1/4).

Question 3

Jake wants to place 58\frac{5}{8}85​ on a number line that goes from 0 to 1. He knows that 12=48\frac{1}{2} = \frac{4}{8}21​=84​. Where should Jake place 58\frac{5}{8}85​ compared to 12\frac{1}{2}21​?

  1. 58\frac{5}{8}85​ should be placed exactly at 12\frac{1}{2}21​
  2. 58\frac{5}{8}85​ should be placed slightly to the right of 12\frac{1}{2}21​ (correct answer)
  3. 58\frac{5}{8}85​ should be placed slightly to the left of 12\frac{1}{2}21​
  4. 58\frac{5}{8}85​ should be placed exactly at 34\frac{3}{4}43​

Explanation: Since 12=48\frac{1}{2} = \frac{4}{8}21​=84​ and 58>48\frac{5}{8} > \frac{4}{8}85​>84​, the fraction 58\frac{5}{8}85​ is greater than 12\frac{1}{2}21​ and should be placed to the right of it on the number line. Choice A is wrong because 58≠48\frac{5}{8} \neq \frac{4}{8}85​=84​. Choice C incorrectly suggests 58<12\frac{5}{8} < \frac{1}{2}85​<21​. Choice D is incorrect because 34=68≠58\frac{3}{4} = \frac{6}{8} \neq \frac{5}{8}43​=86​=85​.

Question 4

Carlos has a backpack that weighs 5 kg and a bicycle that weighs 8 kg. What is the total mass of both?

  1. 3 kg
  2. 13 kg (correct answer)
  3. 40 kg
  4. 13 g

Explanation: This question tests 3rd grade measurement: measuring and calculating with mass (grams, kilograms) and volume (liters), and solving one-step word problems (CCSS.3.MD.2). Mass measures how heavy something is. We use grams for lighter objects and kilograms for heavier objects (1 kg = 1000 g). Volume measures how much liquid something holds, using liters. To solve measurement word problems, identify the operation (add, subtract, multiply, divide) and make sure units are the same. The problem gives a backpack with 5 kg and a bicycle with 8 kg, and asks us to find the total mass of both. Choice B is correct because 5 kg + 8 kg = 13 kg. This shows understanding of adding masses in kilograms. Choice D represents unit confusion: 13 g, which typically happens because students might confuse heavy items like bicycles with light units. To help students: Provide hands-on measurement experiences using scales (balance and digital) and measuring cups/beakers. Have students hold objects and estimate mass before measuring. Create reference points ('A pencil is about 10 grams, a textbook is about 500 grams, I weigh about 35 kilograms'). Use real containers to understand liters (water bottle = 1 liter, juice box = smaller). Practice with manipulatives and real measurements. Watch for: Students who don't include units in answers, students who use unrealistic measurements (person weighing 5 g), students who confuse grams and kilograms, and students who don't convert units when needed (adding 2 kg + 500 g without converting to same unit).

Question 5

Ava has 7 bags with 10 marbles each. How many marbles?​

  1. 70 marbles (correct answer)
  2. 17 marbles
  3. 700 marbles
  4. 7 marbles

Explanation: This question tests multiplying one-digit numbers by multiples of 10 in the range 10-90 (CCSS.3.NBT.3), specifically using place value strategies and properties of operations. To multiply a digit by a multiple of 10, use the pattern: first multiply the digit by the unit digit, then multiply the result by 10. For example, 7×10: This is a special case where we multiply directly by 10, so 7×10=70. Another way: 7×10 means 7 groups of 10, which equals 70. In this problem, Ava has 7 bags with 10 marbles each. This represents the multiplication 7×10. Choice B is correct because 7×10=70, which is the basic pattern for multiplying by 10. This demonstrates understanding of multiplying by multiples of 10. Choice C is incorrect because it shows only 7 without multiplying by 10. This error occurs when students forget to perform the multiplication operation. To help students multiply by multiples of 10: Connect to basic facts (7×1=7, so 7×10=70). Use place value language (7×10 = 7 tens = 70). Model with base-10 blocks (7 tens rods). Practice skip counting by 10s: 10, 20, 30, 40, 50, 60, 70.

Question 6

Sarah measured three objects with a ruler. Look at her measurement data in the table. She wants to arrange the objects from shortest to longest. What is the correct order?

  1. Marker, Scissors, Stapler
  2. Stapler, Marker, Scissors (correct answer)
  3. Scissors, Stapler, Marker
  4. Stapler, Scissors, Marker

Explanation: From the table: Stapler = 12 cm, Marker = 14 cm, Scissors = 16 cm. Arranging from shortest to longest: 12 cm < 14 cm < 16 cm, so Stapler, Marker, Scissors. Choice A uses incorrect measurements. Choice C puts 14 cm as the largest number, which is wrong. Choice D incorrectly orders 16 cm before 14 cm.

Question 7

A school orders pencils in packs of 888. They need 969696 pencils total. After getting the packs, they want to distribute all pencils equally to 666 classrooms. How many pencils will each classroom receive?

  1. 121212 pencils per classroom
  2. 141414 pencils per classroom
  3. 161616 pencils per classroom (correct answer)
  4. 181818 pencils per classroom

Explanation: The school has 96 pencils total to distribute equally to 6 classrooms: 96÷6=1696 ÷ 6 = 1696÷6=16 pencils per classroom. The information about packs of 8 is extra information that doesn't affect the answer. Choice A results from 96÷896 ÷ 896÷8. Choice B and D result from calculation errors or confusion with the pack size.

Question 8

A rectangular mat is 888 units long and 555 units wide. Kevin covers it with 1×11 \times 11×1 unit squares, but he overlaps some squares and leaves some gaps. He counts 383838 squares total. What can you conclude about his work?

  1. He covered the mat perfectly since 383838 is close to 404040
  2. He used too few squares since the area is 8×5=408 \times 5 = 408×5=40 square units
  3. He made mistakes since he should use exactly 8×5=408 \times 5 = 408×5=40 squares with no overlaps or gaps (correct answer)
  4. He worked efficiently since 38<4038 < 4038<40 means he saved 222 squares

Explanation: The mat's area is 8×5=408 \times 5 = 408×5=40 square units. To cover it completely without overlaps or gaps, Kevin needs exactly 40 unit squares. Since he used 38 squares but had overlaps and gaps, his coverage is incorrect. Choice A incorrectly accepts approximation. Choice B ignores the overlaps mentioned. Choice D misunderstands that he didn't actually save squares—he made errors.

Question 9

A garden is 555 meters by 444 meters in 111-meter squares; what area?

  1. 18 square meters
  2. 16 square meters
  3. 9 square meters
  4. 20 square meters (correct answer)

Explanation: This question tests 3rd grade area: finding the area of a rectangle by tiling it with unit squares, and showing that the area equals the product of the side lengths (CCSS.3.MD.7.a). When we tile a rectangle with unit squares, we create rows and columns. For example, a 3-by-5 rectangle has 3 rows with 5 squares in each row, giving 3×5=15 total squares. Multiplying the side lengths (length × width) gives the same answer as counting all the tiles because multiplication counts equal groups efficiently. The rectangle has dimensions 5 meters by 4 meters. When tiled with unit squares, it has 5 rows of 4 squares each (or 4 columns of 5 squares each). Choice C is correct because 5 rows of 4 squares = 5×4 = 20 square meters, which can be verified by counting all tiles OR multiplying length times width: 5×4=20. This shows understanding that tiling and multiplication give the same area. Choice B represents adding instead of multiplying, wrong calculation, perimeter confusion, missing units. This typically happens because students confuse addition with multiplication, miscalculate or miscount, confuse area (inside space) with perimeter (distance around), or forget area is measured in square units. To help students: Use physical tiles or graph paper to build rectangles, then count AND multiply to see they match. Show that '3 rows of 5' means 5+5+5 (repeated addition) which equals 3×5 (multiplication). Practice with various rectangle sizes: 2×4, 3×5, 4×6. Help students see the connection: rows × squares per row = total squares. Watch for: Students who add dimensions instead of multiply (3+5 instead of 3×5), students who confuse area with perimeter, students who multiply but forget to say 'square units,' and students who don't connect the visual tiling to the multiplication. Use the language 'rows of' to bridge to multiplication: 'I see 4 rows of 7 squares, so 4 times 7 equals 28.' This develops fluency with multiplication as counting equal groups while building area understanding.

Question 10

Maya has 5 bags with 7 cookies each. Which expression shows the total?

  1. 353535
  2. 5×75 \times 75×7 (correct answer)
  3. 5+75 + 75+7
  4. 7×57 \times 57×5

Explanation: This question tests interpreting multiplication as equal groups (CCSS.3.OA.1), specifically understanding that a product like 5 × 7 represents the total number of objects when there are 5 equal groups with 7 objects in each group. Multiplication describes situations with equal groups: [number of groups] × [objects per group] = [total]. For example, 5 × 7 means "5 groups of 7 objects each" or "5 times 7." The first factor (5) tells how many groups; the second factor (7) tells how many in each group. In this problem, Maya has 5 bags with 7 cookies in each bag. This represents 5 × 7. Choice C is correct because it accurately represents 5 groups × 7 objects per group shown in the scenario. The first factor (5) is the number of bags, and the second factor (7) is the number of cookies in each bag, giving the correct total of 35 cookies. Choice A is incorrect because it reverses the factors (shows 7 × 5 instead of 5 × 7). This error occurs when students don't understand that the first factor represents the number of groups and the second factor represents the size of each group. To help students interpret multiplication as equal groups: Use concrete materials (blocks, counters) to build equal groups physically. Draw arrays or circles with objects to visualize groups. Practice translating: "5 bags with 7 cookies each" → 5 × 7. Emphasize language: "[#] groups OF [#] objects each." Connect to repeated addition: 7+7+7+7+7 is the same as 5×7.

Question 11

Jake knows these two facts: 8×9=728 × 9 = 728×9=72 and 72÷8=972 ÷ 8 = 972÷8=9. Using the same three numbers, what is another division fact Jake can write?

  1. 72÷6=1272 ÷ 6 = 1272÷6=12
  2. 81÷9=981 ÷ 9 = 981÷9=9
  3. 72÷9=872 ÷ 9 = 872÷9=8 (correct answer)
  4. 64÷8=864 ÷ 8 = 864÷8=8

Explanation: The three numbers in Jake's facts are 8, 9, and 72. From 8×9=728 × 9 = 728×9=72, we can write two division facts: 72÷8=972 ÷ 8 = 972÷8=9 (given) and 72÷9=872 ÷ 9 = 872÷9=8. Choice A uses different numbers (6 and 12). Choice B uses different numbers (81). Choice D uses different numbers (64).

Question 12

A bakery opens at 6:00 AM. The first batch of cookies takes 35 minutes to bake, and the second batch takes 40 minutes to bake. If both batches bake one after the other, what time are all the cookies finished?

  1. 6:75 AM
  2. 7:05 AM
  3. 7:15 AM (correct answer)
  4. 6:45 AM

Explanation: When you see a time problem involving adding minutes, you need to carefully track both hours and minutes, especially when the minutes add up to more than 60. Let's work through this step by step. The bakery opens at 6:00 AM. The first batch takes 35 minutes, so after the first batch finishes, it's 6:35 AM. Then the second batch takes 40 minutes, starting at 6:35 AM. To find when the second batch finishes, add 40 minutes to 6:35 AM. Break this down: 6:35 + 40 minutes = 6:75. But wait - there are only 60 minutes in an hour! When you get 75 minutes, you need to convert: 75 minutes = 1 hour and 15 minutes. So 6:75 becomes 7:15 AM. Looking at the wrong answers: Choice A (6:75 AM) shows the math before converting minutes properly - this isn't a real time since clocks don't show 75 minutes. Choice B (7:05 AM) might come from miscounting the minutes or adding incorrectly. Choice D (6:45 AM) only accounts for adding one of the baking times, not both batches. The correct answer is C) 7:15 AM because 6:00+35+40=6:75=7:156:00 + 35 + 40 = 6:75 = 7:156:00+35+40=6:75=7:15 AM. Study tip: Whenever you're adding time and get more than 60 minutes, remember to convert: subtract 60 from the minutes and add 1 to the hour. Practice converting "impossible" times like 4:73 (which becomes 5:13) or 9:68 (which becomes 10:08).

Question 13

Maya studies the pattern when multiplying by 5: 5 × 1 = 5, 5 × 2 = 10, 5 × 3 = 15, 5 × 4 = 20, 5 × 5 = 25. She observes that the ones digit follows a specific pattern. What explains why this ones digit pattern occurs?

  1. The ones digits alternate 5, 0, 5, 0 because 5 × odd = ends in 5, and 5 × even = ends in 0 (correct answer)
  2. The ones digits are always 5 because you're multiplying by 5, so the answer must end in 5
  3. The ones digits increase by 5 each time because you add 5 to get the next multiple
  4. The ones digits alternate 0, 5, 0, 5 because even multiples end in 0 and odd multiples end in 5

Explanation: When multiplying by 5, the ones digit alternates between 5 and 0 because 5 × any odd number ends in 5, while 5 × any even number ends in 0. This creates the pattern 5, 0, 5, 0, 5... Choice B is wrong because not all end in 5. Choice C misunderstands how ones digits work in multiplication. Choice D has the alternating pattern backwards.

Question 14

Solve: 30÷5=?30 \div 5 = ?30÷5=? What is ???

  1. 5
  2. 6 (correct answer)
  3. 35
  4. 7

Explanation: This question tests determining the unknown whole number in multiplication or division equations (CCSS.3.OA.4), specifically finding the missing value that makes an equation true. To find the unknown in an equation, identify what's missing (factor, product, dividend, divisor, or quotient), then use the relationship between multiplication and division. For multiplication: If one factor and the product are known, divide to find the other factor (8×?=48, so ?=48÷8=6). If both factors are known, multiply to find product (8×6=?, so ?=8×6=48). For division: If dividend and divisor are known, divide to find quotient (48÷8=?, so ?=48÷8=6). If dividend and quotient are known, multiply to find divisor (48÷?=6, so ?=48÷6=8). If divisor and quotient are known, multiply to find dividend (?÷8=6, so ?=6×8=48). In this problem, the equation is 30 ÷ 5 = ?. The unknown is the quotient. To solve, we need to divide 30 by 5. Choice A is correct because 30÷5=6, which can be written as 5×6=30. This value makes the equation true. Choice C is incorrect because using 30+5=35 instead of 30÷5=6 shows wrong operation. This error occurs when students use wrong operation. To help students find unknowns in equations: Teach the inverse relationship (multiplication ↔ division). Use fact families: If 8×6=48, then 48÷8=6, 48÷6=8, and 6×8=48 (all related). Model thinking aloud: "8 times what number equals 48? I know 8×6=48, so ? is 6." Cover up the unknown with your finger, say the known information, and think what number fits. Practice with manipulatives or arrays (8 rows of ? objects = 48 total, count 6 per row). Always check by substituting back: Does 8×6 really equal 48? Yes! Watch for students who add/subtract instead of multiply/divide, or who don't understand the three numbers are related through multiplication and division.

Question 15

A school orders notebooks for students. They order 777 boxes, with each box containing 888 notebooks. After distributing the notebooks equally among 444 classrooms, the principal finds that each classroom received the same number of notebooks with none left over. Later, one classroom gives half of their notebooks to the library. How many notebooks does that classroom have left?

  1. 777 notebooks remaining in the classroom (correct answer)
  2. 121212 notebooks remaining in the classroom
  3. 141414 notebooks remaining in the classroom
  4. 282828 notebooks remaining in the classroom

Explanation: Total notebooks: 7 × 8 = 56 notebooks. Per classroom: 56 ÷ 4 = 14 notebooks. After giving half to library: 14 ÷ 2 = 7 notebooks remaining. Choice B miscalculates the division by 2. Choice C shows the original amount before giving any away. Choice D shows an incorrect calculation in the distribution step.

Question 16

Lina has 42 beads and makes groups of 6. How many groups?

  1. 48 groups
  2. 7 groups (correct answer)
  3. 8 groups
  4. 6 groups

Explanation: This question tests interpreting division as equal shares or equal groups (CCSS.3.OA.2), specifically understanding that a÷b can mean (1) a objects divided into b equal shares (partition), or (2) a objects with b per group, how many groups (measurement). Division has two interpretations. Partition (equal shares): When you have a total and need to divide it into a specific number of shares, asking "how many in each share?" For example, 24÷6 can mean "24 cookies divided equally among 6 children—how many does each child get?" Answer: 4 cookies per child. Measurement (equal groups): When you have a total and put a specific amount in each group, asking "how many groups?" For example, 24÷6 can also mean "24 cookies, put 6 in each bag—how many bags needed?" Answer: 4 bags. Both use 24÷6=4 but ask different questions. In this problem, Lina has 42 beads and makes groups of 6, asking how many groups she can make. This represents measurement division, asking for the number of groups. Choice B is correct because 42÷6=7, meaning 7 groups when putting 6 beads per group. This accurately interprets the division as measurement: number of groups. Choice A is incorrect because it gives the divisor (6) instead of the quotient (7), perhaps confusing the group size with the number of groups. This error occurs when students confuse the two interpretations of division. To help students interpret division: Teach both meanings explicitly using the same numbers (24÷6 as partition: 6 shares of 4 each; as measurement: 4 groups of 6 each). Use concrete materials (counters, cubes) to physically divide and group. Draw pictures showing both interpretations. Connect to real contexts: sharing food (partition), packaging items (measurement). Language cues: "divided among" or "each person gets" suggests partition; "put X in each" or "per group" suggests measurement. Practice writing story problems for division expressions. Connect to multiplication: If 8×7=56, then 56÷8=7 and 56÷7=8.

Question 17

Maya cut a pizza into 8 equal slices. Each slice is what fraction?

  1. 1/41/41/4
  2. 8/18/18/1
  3. 1/61/61/6
  4. 1/81/81/8 (correct answer)

Explanation: This question tests 3rd grade fractions: partitioning shapes into equal parts and expressing the area of each part as a unit fraction (CCSS.3.G.2). When a shape is divided into equal parts, each part has the same area. A unit fraction describes one part: 1/8 means 1 out of 8 equal parts. The denominator (bottom number) tells how many equal parts in the whole; the numerator (top number) tells how many parts you're describing (for unit fractions, it's always 1). The pizza is divided into 8 equal slices, like cutting a circle into eight identical wedges. Each slice has the same area. Choice A is correct because there are 8 equal parts, so each slice is 1/8. This shows understanding that equal partitioning creates unit fractions. Choice C represents a reversal error, where students write 8/1 instead of 1/8. This typically happens because they confuse the roles of numerator and denominator or miscount the parts. To help students: Use real objects like play pizzas or fraction circles to demonstrate equal slicing. Have students cut paper circles into eight parts and label each with 1/8. Practice counting: '1 through 8 equal slices, each is 1/8.' Watch for reversal mistakes or thinking larger denominators mean bigger pieces, and reinforce with visuals showing more slices mean smaller portions.

Question 18

Tim knows that 6×8=486 \times 8 = 486×8=48. He uses this fact to solve 48÷648 \div 648÷6. What other division problem can he solve using the same multiplication fact?

  1. 48÷8=648 \div 8 = 648÷8=6 (correct answer)
  2. 48÷12=448 \div 12 = 448÷12=4
  3. 54÷6=954 \div 6 = 954÷6=9
  4. 42÷6=742 \div 6 = 742÷6=7

Explanation: From 6×8=486 \times 8 = 486×8=48, we can solve both 48÷6=848 \div 6 = 848÷6=8 and 48÷8=648 \div 8 = 648÷8=6. Choice B uses 48 but with wrong factors, while choices C and D use different products that don't come from 6×8=486 \times 8 = 486×8=48.

Question 19

Mrs. Johnson's class collected 4 boxes of 125 pencils each. They gave away 167 pencils to another class. How many pencils do they have left?

  1. 667667667 pencils
  2. 500500500 pencils
  3. 292292292 pencils
  4. 333333333 pencils (correct answer)

Explanation: This is a two-step word problem that combines multiplication and subtraction. When you see problems involving "boxes of" items followed by giving some away, you'll need to find the total first, then subtract what was removed. Start by finding how many pencils Mrs. Johnson's class collected in total. They had 4 boxes with 125 pencils in each box, so multiply: 4×125=5004 \times 125 = 5004×125=500 pencils total. Next, subtract the pencils they gave away: 500−167=333500 - 167 = 333500−167=333 pencils remaining. The correct answer is D) 333333333 pencils. Let's see why the other answers are incorrect. Choice A) 667667667 pencils happens if you accidentally add instead of subtract: 500+167=667500 + 167 = 667500+167=667. This is a common mistake when students mix up the operation. Choice B) 500500500 pencils is the total number of pencils before giving any away—this means you forgot to do the subtraction step entirely. Choice C) 292292292 pencils likely comes from calculation errors, possibly in either the multiplication or subtraction steps. For multi-step word problems like this, always identify what you need to find first (the total), then what happens to that amount (pencils given away). Write out each step clearly: multiply to find the total, then subtract what was removed. Double-check your arithmetic, especially when subtracting larger numbers, and make sure you're performing the right operation at each step.

Question 20

Emma is sorting shapes for an art project. She has a cube, a cylinder, a triangular prism, and a sphere. She wants to put all shapes that have at least one flat face in one group and shapes with no flat faces in another group. Which shape will be alone in its group?

  1. The cube will be alone because it has the most flat faces
  2. The sphere will be alone because it has no flat faces (correct answer)
  3. The cylinder will be alone because it has curved and flat faces
  4. The triangular prism will be alone because it has triangular faces

Explanation: The sphere is the only shape with no flat faces - it is completely curved. The cube has 6 flat faces, the cylinder has 2 flat circular faces, and the triangular prism has 5 flat faces (2 triangular and 3 rectangular). So the sphere will be alone in the 'no flat faces' group while the other three shapes will be together in the 'at least one flat face' group.

Question 21

Jamal’s pencil case is 150 g and his notebook is 300 g; what is the total mass?

  1. 450 kg
  2. 450 g (correct answer)
  3. 350 g
  4. 150 g

Explanation: This question tests 3rd grade measurement: measuring and calculating with mass (grams, kilograms) and volume (liters), and solving one-step word problems (CCSS.3.MD.2). Mass measures how heavy something is. We use grams for lighter objects and kilograms for heavier objects (1 kg = 1000 g). To solve measurement word problems, identify the operation (add, subtract, multiply, divide) and make sure units are the same. The problem gives a pencil case with mass 150 g and a notebook with mass 300 g, and asks us to find the total mass. Choice B (450 g) is correct because 150 g + 300 g = 450 g. This shows understanding of adding masses with the same unit. Choice D (450 kg) represents a unit confusion error where students got the right number but wrong unit. This typically happens because students don't pay attention to whether the problem uses grams or kilograms. To help students: Provide hands-on measurement experiences using scales (balance and digital) and measuring cups/beakers. Have students hold objects and estimate mass before measuring. Create reference points ('A pencil is about 10 grams, a textbook is about 500 grams, I weigh about 35 kilograms'). Practice with manipulatives and real measurements. Watch for: Students who don't include units in answers, students who use unrealistic measurements (pencil case weighing 150 kg), and students who confuse grams and kilograms.

Question 22

Maya has 74\frac{7}{4}47​ pizzas. She wants to write this amount as a mixed number. But first, she needs to figure out how many unit fractions of 14\frac{1}{4}41​ make up 74\frac{7}{4}47​. How many unit fractions of 14\frac{1}{4}41​ does Maya need?

  1. 4 unit fractions because there are 4 parts in each whole
  2. 6 unit fractions because 7−1=67 - 1 = 67−1=6 and we don't count the whole
  3. 7 unit fractions because the numerator tells us how many fourths we have (correct answer)
  4. 8 unit fractions because we need to round up to the next whole number

Explanation: To find how many unit fractions of 14\frac{1}{4}41​ make up 74\frac{7}{4}47​, we look at the numerator. The fraction 74\frac{7}{4}47​ means 7 pieces, where each piece is 14\frac{1}{4}41​. So Maya needs exactly 7 unit fractions of 14\frac{1}{4}41​. Choice A confuses the number of fourths in one whole with the total needed. Choice B incorrectly subtracts 1 from the numerator. Choice D adds an extra unit fraction unnecessarily.

Question 23

What is the value of nnn in 9×n=819 \times n = 819×n=81?

  1. 8
  2. 81
  3. 18
  4. 9 (correct answer)

Explanation: This question tests determining the unknown whole number in multiplication or division equations (CCSS.3.OA.4), specifically finding the missing value that makes an equation true. To find the unknown in an equation, identify what's missing (factor, product, dividend, divisor, or quotient), then use the relationship between multiplication and division. For multiplication: If one factor and the product are known, divide to find the other factor (8×?=488 \times ? = 488×?=48, so ?=48÷8=6? = 48 \div 8 = 6?=48÷8=6). If both factors are known, multiply to find product (8×6=?8 \times 6 = ?8×6=?, so ?=8×6=48? = 8 \times 6 = 48?=8×6=48). For division: If dividend and divisor are known, divide to find quotient (48÷8=?48 \div 8 = ?48÷8=?, so ?=48÷8=6? = 48 \div 8 = 6?=48÷8=6). If dividend and quotient are known, multiply to find divisor (48÷?=648 \div ? = 648÷?=6, so ?=48÷6=8? = 48 \div 6 = 8?=48÷6=8). If divisor and quotient are known, multiply to find dividend (?÷8=6? \div 8 = 6?÷8=6, so ?=6×8=48? = 6 \times 8 = 48?=6×8=48). In this problem, the equation is 9 \times n = 81. The unknown is a factor. To solve, we need to divide 81 by 9. Choice A is correct because 9\times9=81, so 9 makes the equation 9\times n=81 true. This value makes the equation true. Choice D is incorrect because selecting 81 (the product) instead of solving for the unknown doesn't make the equation true: 9\times81=729, not 81. This error occurs when students don't solve for the unknown. To help students find unknowns in equations: Teach the inverse relationship (multiplication ↔ division). Use fact families: If 8\times6=48, then 48\div8=6, 48\div6=8, and 6\times8=48 (all related). Model thinking aloud: "9 times what number equals 81? I know 9\times9=81, so n is 9." Cover up the unknown with your finger, say the known information, and think what number fits. Practice with manipulatives or arrays (9 rows of ? objects = 81 total, count 9 per row). Always check by substituting back: Does 9\times9 really equal 81? Yes! Watch for students who add/subtract instead of multiply/divide, or who don't understand the three numbers are related through multiplication and division.

Question 24

Anna's dance class is 1 hour and 15 minutes long. The class has a 10-minute break in the middle. How many minutes does Anna actually spend dancing?

  1. 75 minutes
  2. 65 minutes (correct answer)
  3. 85 minutes
  4. 55 minutes

Explanation: When you see a time problem with breaks or interruptions, you need to subtract the non-activity time from the total time to find the actual activity time. Let's work through this step by step. Anna's dance class is 1 hour and 15 minutes long. First, convert everything to minutes to make the math easier: 1 hour = 60 minutes, so the total class time is 60 + 15 = 75 minutes. However, there's a 10-minute break in the middle where Anna isn't dancing. To find how much time she actually spends dancing, subtract the break time from the total class time: 75 minutes - 10 minutes = 65 minutes of actual dancing. Looking at the wrong answers: Choice A (75 minutes) is the trap of forgetting to subtract the break time - this is just the total class length. Choice C (85 minutes) incorrectly adds the break time to the total instead of subtracting it. Choice D (55 minutes) makes an error in the initial time conversion, possibly calculating 45 minutes + 10 minutes instead of properly converting 1 hour 15 minutes. The correct answer is B (65 minutes). Study tip: In time problems involving breaks or pauses, always remember to subtract the inactive time from the total time. Convert everything to the same units (usually minutes) first to avoid confusion, then do your subtraction. Watch out for answer choices that represent the total time without accounting for breaks - these are common distractors.

Question 25

Maria has 23\frac{2}{3}32​ of a pizza. She wants to trade it for an equal amount from Jake's pizza, which is cut into 6 equal pieces. How many pieces of Jake's pizza should Maria get to make a fair trade?

  1. 2 pieces, because the numerators are the same
  2. 3 pieces, because that matches the denominator
  3. 4 pieces, because 46=23\frac{4}{6} = \frac{2}{3}64​=32​ (correct answer)
  4. 5 pieces, because she needs more than half

Explanation: To find equivalent fractions, we need 23=?6\frac{2}{3} = \frac{?}{6}32​=6?​. Since 3×2=63 \times 2 = 63×2=6, we multiply both numerator and denominator by 2: 2×23×2=46\frac{2 \times 2}{3 \times 2} = \frac{4}{6}3×22×2​=64​. Choice A incorrectly focuses only on numerators. Choice B uses the original denominator. Choice D assumes more than half without calculating the exact equivalent.