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Read the problem. Sofia walked 842 steps. Chen walked 569 steps. How many more steps did Sofia walk than Chen?
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3rd Grade Math
Practice Test 11 for 3rd Grade Math: real questions and explanations from the Varsity Tutors practice-test pool.
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Question 1 of 25
Read the problem. Sofia walked 842 steps. Chen walked 569 steps. How many more steps did Sofia walk than Chen?
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Read the problem. Sofia walked 842 steps. Chen walked 569 steps. How many more steps did Sofia walk than Chen?
Explanation: This question tests fluent addition and subtraction within 1000 (CCSS.3.NBT.2), specifically using strategies and algorithms based on place value and regrouping. To add 3-digit numbers, line up place values (ones, tens, hundreds). For addition: Add ones (if sum ≥10, regroup to tens), add tens (if sum ≥10, regroup to hundreds), add hundreds. For subtraction: Subtract ones (if can't, borrow from tens), subtract tens (if can't, borrow from hundreds), subtract hundreds. Example: 347+286: Start with ones (7+6=13, write 3 carry 1), then tens (4+8+1=13, write 3 carry 1), then hundreds (3+2+1=6), answer 633. For subtraction across zero like 500-237: Need to borrow: 500 becomes 4 hundreds, 10 tens, 0 ones. Then 10-7=3, 9-3=6 (after borrowing 1 from tens to make 10 ones), 4-2=2, answer 263. In this problem, Sofia walked 842 steps and Chen walked 569, asking how many more Sofia walked. This requires subtraction with borrowing across tens and ones places. Choice A is correct because 842-569=273. This demonstrates proper place value alignment, borrowing correctly. Choice C represents the error of using addition instead of subtraction, calculating 842+569=1,411. This happens when students confuse the operation in comparison problems. To help students with addition/subtraction within 1000: Use place value charts or base-10 blocks to visualize regrouping. Practice expanded form (347 = 300+40+7). Teach borrowing across zero explicitly (500 = 4 hundreds, 10 tens, 0 ones). Emphasize lining up place values vertically. Check answers with estimation (347+286 is about 350+300=650, so 633 reasonable). Practice both with and without regrouping. Watch for students who don't carry/borrow correctly or who apply operations inconsistently across place values.
Divide: 56÷7.
Explanation: This question tests fluent multiplication and division within 100 (CCSS.3.OA.7), specifically computing basic facts to divide efficiently. Fluency means calculating quickly and accurately using efficient strategies or memory. By end of Grade 3, students should know from memory all products of two one-digit numbers (0-10) and related division facts. Strategies include: (1) Using the relationship between multiplication and division (if 8×7=56, then 56÷8=7 and 56÷7=8), (2) Using properties (commutative: 7×8=8×7; distributive: 7×8=7×5+7×3=35+21=56), (3) Using known facts (know 7×7=49, so 7×8=49+7=56), (4) Direct recall from memory. In this problem, we need to find 56÷7. This uses the inverse relationship between multiplication and division. Choice C is correct because 56÷7=8 (since 7×8=56). This demonstrates understanding of operation relationships. Choice A is incorrect because this is 49÷7=7, from the wrong fact family. This error occurs when students confuse adjacent facts. To build fluency with multiplication and division: Practice facts systematically (2s, 5s, 10s first, then 3s, 4s, 6s, then harder 7s, 8s, 9s). Use relationships: teach fact families so learning one fact means knowing four equations. Apply properties: if know 8×5=40, then 8×6=40+8=48 (distributive). Practice with games, flashcards, timed exercises (but low-stress). Emphasize strategies for facts not yet memorized. Connect multiplication to division constantly: every multiplication fact is also two division facts. By end of Grade 3, goal is automatic recall of all single-digit products and related divisions.
Sofia's room is a rectangle 9 feet long and 4 feet wide; what is the area?
Explanation: This question tests 3rd grade perimeter and area: finding perimeter given sides, finding unknown side lengths, and understanding same perimeter can have different areas or same area can have different perimeters (CCSS.3.MD.8). Perimeter is the distance around a shape, calculated by adding all side lengths, measured in linear units (feet, meters). Area is the space inside a shape, calculated for rectangles by multiplying length × width, measured in square units (square feet, square meters). For rectangles, perimeter = 2×length + 2×width or add all four sides. The room is a rectangle 9 feet long and 4 feet wide, and the question asks for the area. Choice C is correct because area = 9×4 = 36 square feet, showing understanding of area calculation for the space inside the room. Choice A represents a common error of adding instead of multiplying, like confusing perimeter (9+4=13) with area, which happens because students mix up the operations for border distance versus inside space. To help students: Distinguish perimeter and area with context—fence/border/frame = perimeter (around), carpet/tile/paint = area (inside). Practice both: 'This room is 9 by 4. Area for carpet: 9×4=36 square feet. Perimeter for baseboard: 9+4+9+4=26 feet.' Watch for: Students who add for area, forget square units, or confuse area with perimeter.
Use the figure shown to answer this question. Anna wants to carpet this T-shaped room. What is the total area she needs to cover?
Explanation: The T-shaped room can be split into two rectangles. The horizontal top section is 12 ft × 4 ft = 48 square feet. The vertical bottom section is 5 ft × 8 ft = 40 square feet. Total area = 48 + 40 = 88 square feet.
The figure below shows a shape made of unit squares arranged in rows. Row 1 has 2 unit squares, Row 2 has 4 unit squares, and Row 3 has 3 unit squares. What is the area of the entire shape?
Explanation: Add the unit squares in each row: Row 1 + Row 2 + Row 3 = 2 + 4 + 3 = 9 square units total.
Which number completes the equation 42÷6=□?
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 42 ÷ 6 = □. The unknown is the quotient. To solve, we need to divide 42 by 6. Choice C is correct because 42÷6=7, which can be written as 6×7=42. This value makes the equation true. Choice A is incorrect because selecting 6 (the divisor) instead of solving for the unknown doesn't make the equation true: 42÷6 is not 6. 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 8×6=48, then 48÷8=6, 48÷6=8, and 6×8=48 (all related). Model thinking aloud: "42 divided by 6 equals what? I know 6×7=42, so □ is 7." Cover up the unknown with your finger, say the known information, and think what number fits. Practice with manipulatives or arrays (42 objects divided into 6 groups, count 7 per group). Always check by substituting back: Does 42÷6 really equal 7? 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.
Look at Shapes A–E. Why are all these shapes quadrilaterals?
Explanation: This question tests 3rd grade geometry: classifying quadrilaterals by shared attributes and recognizing that shapes in different categories may share properties (CCSS.3.G.1). A quadrilateral is defined as any closed shape with exactly 4 straight sides, regardless of their lengths, angles, or other properties. The correct answer is A: 'They all have 4 sides,' which is the defining characteristic of quadrilaterals. The distractors represent more specific properties that only some quadrilaterals have: 4 equal sides (only rhombuses and squares), 4 right angles (only rectangles and squares), or being the same size (not a geometric property). To help students: Start with the prefix 'quad' meaning four (like quadruplets or quad bike). Have students count sides on various shapes, emphasizing that 4 sides = quadrilateral, regardless of how the shape looks. Create a shape sort with triangles (3 sides), quadrilaterals (4 sides), and pentagons (5 sides) to reinforce that the number of sides is the key attribute. Use the motto: 'If it has 4 sides, it's in the quadrilateral family!'
Sophie is filling bags with marbles for a party. She puts the same number of marbles in each bag. After filling 6 bags completely, she has 8 marbles left over. If she started with 74 marbles total, how many marbles did she put in each bag?
Explanation: When you see a problem about dividing items equally with leftovers, you're working with division and remainders. This type of problem asks you to figure out how many items go in each group when you can't divide evenly. Let's work backwards from what we know. Sophie started with 74 marbles total, filled 6 bags with the same amount in each, and had 8 marbles left over. This means: (marbles per bag × 6 bags) + 8 leftover marbles = 74 total marbles. First, subtract the leftover marbles: 74−8=66 marbles were used to fill the 6 bags completely. Now divide those 66 marbles by 6 bags: 66÷6=11 marbles per bag. Let's check: 11×6+8=66+8=74 ✓ Looking at the wrong answers: Choice A gives us 12×6+8=80 marbles, which is too many. Choice C gives us 10×6+8=68 marbles, which is too few. Choice D gives us 13×6+8=86 marbles, which is way too many. The answer is B) 11 marbles were put in each bag. Strategy tip: For division with remainder problems, always work backwards by subtracting the remainder first, then dividing. Double-check your answer by multiplying back and adding the remainder to make sure you get the original total.
Alex needs to find the area of a rectangular carpet that is 7 feet by 5 feet. The carpet has a complex pattern that makes it hard to visualize. Which tool would help most?
Explanation: When you need to find the area of a rectangle, you're looking for how many square units fit inside it. Area is always calculated by multiplying length times width, regardless of what the rectangle looks like or what pattern it has. Since Alex already knows the carpet measures 7 feet by 5 feet, he has all the information needed to find the area. The calculation is straightforward: 7×5=35 square feet. The complex pattern on the carpet doesn't change its dimensions or affect the area calculation at all. Let's look at why the other options aren't the best choice. Option A suggests using grid paper to draw a scaled rectangle. While this could work, it's unnecessary extra work when you already have the measurements. Option B recommends using measuring tape on the carpet, but Alex already knows the carpet is 7 feet by 5 feet - no additional measuring is needed. Option D suggests using estimation, but when you have exact measurements, there's no reason to estimate and lose accuracy. The key insight is that area depends only on the measurements, not on visual appearance. Whether a carpet has stripes, polka dots, or swirls doesn't change how you calculate its area. Remember: when you have exact measurements for length and width, mental math using the area formula (length × width) is the most direct and accurate approach. Don't let decorative details distract you from the mathematical facts you need.
Sarah found some coins under her couch cushions. She found 2 quarters, 5 dimes, and 9 pennies. She already had $2.18 in her piggy bank. What is the total amount of money Sarah has now?
Explanation: The coins Sarah found are worth 2 quarters (50¢) + 5 dimes (50¢) + 9 pennies (9¢) = 109¢ = 1.09.Addingthistoherexisting2.18: 2.18+1.09 = $3.27. Choice A represents forgetting to add some coins. Choice C would be if she had found 119¢. Choice D would be if she had found 129¢.
Based on the coordinate plane shown, what tools and strategies would best help find the distance between points P and Q using the grid system?
Explanation: Choice A correctly uses the coordinate grid for counting and calculation tools for addition to find grid distance. Choice B measures the graph paper rather than using coordinate values. Choice C lacks a systematic approach. Choice D doesn't utilize the coordinate system effectively.
Use the data table shown to find the total volume of liquid that Marcus drank during the week.
Explanation: Adding all daily amounts: 650 + 500 + 750 + 600 + 800 + 700 + 350 = 4350 mL. Convert to liters: 4350 mL = 4.35 L. Choice B gives the correct number but in wrong units. Choice C omits one day's consumption. Choice D moves decimal point incorrectly.
Leo has 4 boxes with 90 crayons each. How many crayons total?
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, 4×90: Think of 90 as 9 tens. Multiply 4×9=36, then multiply by 10 to get 360 (or think: 36 tens = 360). In this problem, Leo has 4 boxes with 90 crayons each. This represents the multiplication 4×90. Choice B is correct because 4×90=360 using the pattern (4×9=36, then ×10=360) or place value (4×9 tens = 36 tens = 360). This demonstrates understanding of multiplying by multiples of 10. Choice C is incorrect because it shows only 4×9=36 and forgot to multiply by 10. This error occurs when students don't complete the pattern. To help students multiply by multiples of 10: Connect to basic facts (if you know 4×9=36, then 4×90=360). Use place value language (4×90 = 4×9 tens = 36 tens = 360). Model with base-10 blocks (4 groups of 9 tens rods).
Look at the figure. Tommy marks three points: P, Q, and R. He draws line segment PQ and ray QR. How many different line segments can be formed using these three points?
Explanation: With three points P, Q, and R, three different line segments can be formed: PQ, QR, and PR. Even though Tommy only drew line segment PQ and ray QR, the question asks how many line segments can be formed using these points. Choice A is wrong because it doesn't count line segment PR. Choice C is wrong because a ray is not a line segment. Choice D is wrong because line segment QR can be formed even though ray QR was drawn.
Observe: 6+8=14, 12+4=16, 20+10=30. Explain why even + even is even.
Explanation: This question tests identifying and explaining arithmetic patterns (CCSS.3.OA.9), specifically observing patterns in addition sequences and explaining them using properties of operations. When adding two even numbers, the sum is even because each even can be paired completely, and combining them allows all to be paired with none left over. This occurs because even numbers are multiples of 2, so even + even = 2a + 2b = 2(a+b), still a multiple of 2. For example, 6+8=14: 6=3 pairs, 8=4 pairs, total 7 pairs (14). In this problem, the pattern shown is even + even equaling even, like 6+8=14, 12+4=16, 20+10=30. This pattern continues because the pairing property of evens preserves evenness in the sum. Choice A is correct because it accurately identifies the pattern and explains it using the pairing property with no leftovers, demonstrating understanding both the pattern and its mathematical structure. Choice B is incorrect because it cites the wrong property (associative) and falsely claims it changes the sum to odd. This error occurs when students confuse properties or don't connect observations to mathematical structure. To help students identify and explain patterns: Show patterns in tables, sequences, and visuals. Ask 'What do you notice?' then 'Why does this happen?' Use color-coding to highlight patterns in tables. Teach properties explicitly and then USE them to explain patterns (not just memorize). Connect to structure: '4×n is even because 4 equals 2+2, and anything with two equal parts is even.' Have students create their own examples of patterns. Compare patterns: why are 2s patterns different from 3s patterns? Practice explaining, not just identifying—the explanation using properties is key. Watch for students who see patterns but can't explain why they occur structurally.
A library has 463 books on Monday. On Tuesday, they receive 7 boxes of new books with 25 books in each box. On Wednesday, 149 books are checked out. How many books are in the library now?
Explanation: This is a multi-step word problem that requires you to track changes to a starting amount. When you see problems like this, work through each day's changes in order and keep a running total. Start with Monday's 463 books. On Tuesday, the library receives 7 boxes with 25 books each. First, calculate how many new books arrive: 7×25=175 books. Add these to your running total: 463+175=638 books after Tuesday. On Wednesday, 149 books are checked out, which means they leave the library. Subtract this from your total: 638−149=489 books remaining. Answer choice A (175 books) is just the number of new books that arrived on Tuesday - this ignores both the original books and the Wednesday checkout. Answer choice B (314 books) comes from subtracting the new books from the original amount (463−149=314), but this incorrectly treats Tuesday's delivery as books leaving rather than arriving. Answer choice C (638 books) is the total after Tuesday but before Wednesday's checkout - this misses the final step of the problem. The correct answer is D (489 books). Remember for multi-step word problems: read carefully to identify whether each change adds to or subtracts from your total, then work through the days in chronological order. Don't skip steps, and make sure your final answer accounts for all the changes mentioned in the problem.
Read the problem. Maya had 458 stickers. She got 276 more stickers and gave away 139 stickers. How many stickers does she have now?
Explanation: This question tests fluent addition and subtraction within 1000 (CCSS.3.NBT.2), specifically using strategies and algorithms based on place value and regrouping. To add 3-digit numbers, line up place values (ones, tens, hundreds). For addition: Add ones (if sum ≥10, regroup to tens), add tens (if sum ≥10, regroup to hundreds), add hundreds. For subtraction: Subtract ones (if can't, borrow from tens), subtract tens (if can't, borrow from hundreds), subtract hundreds. Example: 347+286: Start with ones (7+6=13, write 3 carry 1), then tens (4+8+1=13, write 3 carry 1), then hundreds (3+2+1=6), answer 633. For subtraction across zero like 500-237: Need to borrow: 500 becomes 4 hundreds, 10 tens, 0 ones. Then 10-7=3, 9-3=6 (after borrowing 1 from tens to make 10 ones), 4-2=2, answer 263. In this problem, Maya had 458 stickers, got 276 more, and gave away 139, asking how many she has now. This requires multi-step operations: addition with regrouping followed by subtraction with borrowing. Choice A is correct because 458+276=734, then 734-139=595. This demonstrates proper place value alignment, regrouping when needed, borrowing correctly, multi-step execution. Choice B represents the error of forgetting to subtract or misordering operations, leaving the sum as 734. This happens when students confuse multi-step sequences. To help students with addition/subtraction within 1000: Use place value charts or base-10 blocks to visualize regrouping. Practice expanded form (347 = 300+40+7). Teach borrowing across zero explicitly (500 = 4 hundreds, 10 tens, 0 ones). Emphasize lining up place values vertically. Check answers with estimation (347+286 is about 350+300=650, so 633 reasonable). Practice both with and without regrouping. Watch for students who don't carry/borrow correctly or who apply operations inconsistently across place values.
Read the problem. Amir needs 650 tickets for a prize. He already has 385 tickets. How many more tickets are needed?
Explanation: This question tests fluent addition and subtraction within 1000 (CCSS.3.NBT.2), specifically using strategies and algorithms based on place value and regrouping. To add 3-digit numbers, line up place values (ones, tens, hundreds). For addition: Add ones (if sum ≥10, regroup to tens), add tens (if sum ≥10, regroup to hundreds), add hundreds. For subtraction: Subtract ones (if can't, borrow from tens), subtract tens (if can't, borrow from hundreds), subtract hundreds. Example: 347+286: Start with ones (7+6=13, write 3 carry 1), then tens (4+8+1=13, write 3 carry 1), then hundreds (3+2+1=6), answer 633. For subtraction across zero like 500-237: Need to borrow: 500 becomes 4 hundreds, 10 tens, 0 ones. Then 10-7=3, 9-3=6 (after borrowing 1 from tens to make 10 ones), 4-2=2, answer 263. In this problem, Amir needs 650 tickets and has 385, asking how many more are needed. This requires subtraction with borrowing across hundreds and tens places. Choice B is correct because 650-385=265. This demonstrates proper place value alignment, borrowing correctly. Choice D represents the error of using addition instead of subtraction, calculating 650+385=1,035. This happens when students misread the problem and add instead. To help students with addition/subtraction within 1000: Use place value charts or base-10 blocks to visualize regrouping. Practice expanded form (347 = 300+40+7). Teach borrowing across zero explicitly (500 = 4 hundreds, 10 tens, 0 ones). Emphasize lining up place values vertically. Check answers with estimation (347+286 is about 350+300=650, so 633 reasonable). Practice both with and without regrouping. Watch for students who don't carry/borrow correctly or who apply operations inconsistently across place values.
Round 300 to the nearest hundred.
Explanation: This question tests 3rd grade place value and rounding: using place value understanding to round whole numbers to the nearest 10 or 100 (CCSS.3.NBT.1). To round to the nearest ten, look at the ones digit: if it's 0-4, round down (keep tens digit); if it's 5-9, round up (increase tens digit by 1). To round to the nearest hundred, look at the tens digit: if it's 0-4, round down (keep hundreds digit); if it's 5-9, round up (increase hundreds digit by 1). The result is always a multiple of 10 or 100. The number 300 is being rounded to the nearest hundred. We look at the tens digit, which is 0. Choice B is correct because the tens digit is 0, which is <5, so we round down to 300 (it stays the same as it's already a multiple of 100). This shows understanding of rounding rules and place value. Choice C represents rounding up unnecessarily. This typically happens because students think exact multiples need adjustment. To help students: Use number lines to show proximity to tens or hundreds. For 300, mark 200 and 400, show 300 is exactly at 300. Teach the rule as '5 or more, round up the floor; 4 or less, let it rest.' Practice identifying the critical digit: 'To round to nearest hundred, circle the TENS digit. If 0-4, stay at current hundred. If 5-9, go to next hundred.' Use place value charts to visualize: Ones|Tens|Hundreds. Watch for: Students who round the wrong direction, students who look at wrong digit (ones digit when rounding to nearest hundred), students who truncate (547→500 by dropping), and students who don't recognize multiples of 100 (100,200,300...). Connect to real contexts: 'If game had 78 people, we could say 'about 80 people' by rounding.'
Emma divides a number line from 0 to 1 into thirds, then marks 32. Next, she wants to mark a fraction that has the same denominator as 32 but is smaller than 32. Which fraction should Emma mark?
Explanation: The fraction must have denominator 3 (same as 32) and be smaller than 32. The only fraction with denominator 3 that is smaller than 32 is 31 (since 30=0). Choice B has denominator 2, not 3. Choice C has denominator 6, and while 62=31, it doesn't have denominator 3. Choice D has denominator 4 and is larger than 32.
Maya wants to create a scaled picture graph for the data shown in the table. If she uses 1 star symbol to represent 4 books, how many star symbols will she need for the Fiction row?
Explanation: Fiction has 20 books. With each star representing 4 books, Maya needs 20 ÷ 4 = 5 stars for Fiction. Choice B incorrectly rounds up 20 ÷ 4. Choice C uses the actual number of books instead of converting to symbols. Choice D uses the scale value instead of calculating the needed symbols.
Look at the multiplication table shown. Marcus observes that all products in the 6 times table (6 × 1, 6 × 2, 6 × 3, etc.) follow a specific pattern. What is the best explanation for this pattern?
Explanation: All products in the 6 times table are even because 6 is even, and when you multiply an even number by any whole number, the result is always even. This is because 6 = 2 × 3, so 6 × n = 2 × 3 × n, which always contains the factor 2. Choice B is wrong because the products are even, not odd. Choice C is wrong because all products are even. Choice D mentions factors but doesn't correctly explain why this makes results even.
The number line shows 1 and which fraction at the same point?
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=1n and that fractions like 44 equal 1. Any whole number can be written as a fraction by putting it over 1. For example, 3=13 (three ones). Also, when a fraction has the same numerator and denominator (like 44, 66), all parts are shaded and it equals 1 whole. Multiple wholes work too: 48 means eight fourths, which is 2 wholes (because 4 fourths make 1 whole, so 8 fourths make 2 wholes). The number line shows the whole number 1 and equivalent fractions at the same point, demonstrating equivalence like 1 and 22 at the same location. Choice C is correct because 22 equals 1, showing understanding that whole numbers can be expressed as fractions and vice versa. Choice A is incorrect because 21 equals 0.5, 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 22, 3 and 26). Show physical models: one whole circle = 22 = 33 = 44 (all parts shaded). Teach pattern: any whole number n = 1n ("n ones"). Emphasize 44=1, 48=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.
Maria wants to cover her rectangular table with decorative paper. She measures the table and finds it is 4 feet long and 3 feet wide. What property of the table does Maria need to know to determine how much paper she needs?
Explanation: Maria needs to know the area (amount of surface space) to determine how much paper is needed to cover the table. Choice A describes perimeter, C describes a dimension not relevant to covering the surface, and D describes weight capacity.
A parking garage has 3 levels. Level 1 has 178 cars, Level 2 has 94 fewer cars than Level 1, and Level 3 has 126 cars. How many cars are in the garage total?
Explanation: This is a multi-step word problem that requires you to find missing information before calculating a total. When you see "fewer than" in a problem, it signals subtraction, and when you need a "total," you'll be adding all the parts together. Let's work through this step by step. You know Level 1 has 178 cars and Level 3 has 126 cars, but you need to figure out how many cars are on Level 2 first. Since Level 2 has "94 fewer cars than Level 1," you subtract: 178−94=84 cars on Level 2. Now you can find the total by adding all three levels: 178+84+126=388 cars total. This matches answer choice B. Let's check why the other answers are wrong. Choice A (398) likely comes from adding 178 + 94 + 126, incorrectly treating "94 fewer" as if Level 2 simply had 94 cars instead of calculating 94 fewer than 178. Choice C (304) probably results from adding 178 + 126 and forgetting to include Level 2 entirely. Choice D (272) might come from subtracting instead of adding somewhere in the final step. When solving multi-step word problems, always identify what information you're missing first, then calculate that missing piece before moving to the final answer. Pay special attention to key phrases like "fewer than" or "more than" – they tell you exactly which operation to use.