Add and Subtract Mixed Numbers - 4th Grade Math

Emma walked $2\frac{3}{7}$ miles to school and $1\frac{5}{7}$ miles to the library. Then she walked home, which was $2\frac{1}{7}$ miles farther than her total walking distance to school and the library. How far did Emma walk to get home?
Answer: $6\frac{2}{7}$ miles to walk home Emma walked 2 3/7 miles to school and 1 5/7 miles to the library, which totals 4 1/7 miles, and her walk home was 2 1/7 miles farther than that total, so 4 1/7 plus 2 1/7 equals 6 2/7 miles, making Choice D correct. Choice A, 6 3/7 miles, comes from a small addition slip when combining the fractions. Choice B, 2 miles, is only the extra distance her walk home was longer by, not the full walk home distance. Choice C, 4 1/7 miles, is only her combined distance to school and the library, not the distance home.
Add: $2\ \tfrac{2}{5} + 3\ \tfrac{1}{5}$
Answer: $5\ \tfrac{3}{5}$ Adding the whole numbers (2 + 3 = 5) and the fractions (2/5 + 1/5 = 3/5) gives 5 3/5. Choice B reflects an error in adding the whole-number parts. Choice C incorrectly changes the denominator when adding the fractions. Choice D omits the fractional part of the sum entirely.
A fence has three sections with lengths $3\frac{1}{10}$ feet, $3\frac{4}{10}$ feet, and $2\frac{8}{10}$ feet. If the middle section is replaced with a section that is $1\frac{1}{10}$ feet shorter, what will be the total length of the fence?
Answer: $8\frac{2}{10}$ feet for the complete fence The correct answer is A because the original fence is $3\frac{1}{10} + 3\frac{4}{10} + 2\frac{8}{10} = 9\frac{3}{10}$ feet, and shortening the middle section by $1\frac{1}{10}$ feet gives a new total of $9\frac{3}{10} - 1\frac{1}{10} = 8\frac{2}{10}$ feet. Choice B adds the shortened amount instead of subtracting it. Choice C subtracts too much. Choice D is the original total before the middle section was replaced.
What is $2 \tfrac{2}{5} + 3 \tfrac{1}{5}$ ?
Answer: $5 \tfrac{3}{5}$ The correct answer is A, 5 3/5. Adding the whole numbers gives 2 plus 3 equals 5, and adding the fractions gives 2/5 plus 1/5 equals 3/5, for a total of 5 3/5. Choice B, 6 3/5, comes from an addition error in the whole numbers. Choice C, 5 3/10, comes from also adding the denominators instead of keeping them the same. Choice D, 5, omits the fractional part of the sum.
A rope was $7\frac{2}{9}$ feet long. After cutting off $2\frac{8}{9}$ feet, another piece measuring $1\frac{4}{9}$ feet was tied to the remaining rope. What is the length of the rope now?
Answer: $5\frac{7}{9}$ feet Subtracting $7\frac{2}{9} - 2\frac{8}{9}$ requires regrouping: $7\frac{2}{9} = 6\frac{11}{9}$ , so $6\frac{11}{9} - 2\frac{8}{9} = 4\frac{3}{9}$ . Adding the new piece: $4\frac{3}{9} + 1\frac{4}{9} = 5\frac{7}{9}$ feet. Choice A stops after the subtraction step without adding the new piece. Choice C adds instead of subtracting in the first step. Choice D miscalculates the regrouping in the subtraction step.
Chen had $5\ \tfrac{3}{4}$ liters of water. He used $2\ \tfrac{1}{4}$ liters. How many liters are left?
Answer: $3\ \tfrac{1}{2}$ liters The correct answer is B, $3\frac{1}{2}$ liters, because $5\frac{3}{4} - 2\frac{1}{4} = 3\frac{2}{4} = 3\frac{1}{2}$ . Choice A subtracts too much overall. Choice C comes from subtracting only the whole numbers (5 - 2 = 3) and mistakenly keeping the original $\frac{3}{4}$ unchanged. Choice D comes from a regrouping error that overshoots the true difference.
Subtract: $4\ \tfrac{5}{6} - 2\ \tfrac{1}{6}$ . Give the difference as a mixed number.
Answer: $2\ \tfrac{4}{6}$ Choice B is correct because subtracting the whole numbers, 4 minus 2 equals 2, and subtracting the fractions, 5/6 minus 1/6 equals 4/6, gives a difference of 2 4/6. Choice A is incorrect because it changes the denominator to 12 instead of keeping it as sixths. Choice C is incorrect because it adds instead of subtracting the whole numbers. Choice D is incorrect because it subtracts the fractions incorrectly, leaving the wrong amount.
Keisha had $4\ \tfrac{1}{2}$ yards of ribbon. She used $2\ \tfrac{1}{2}$ yards. How many yards of ribbon are left?
Answer: $2$ yards Subtracting $4\tfrac{1}{2} - 2\tfrac{1}{2}$ : the whole numbers give $4-2=2$ , and the fractional parts are equal, so $\tfrac{1}{2} - \tfrac{1}{2} = 0$ , leaving exactly $2$ yards. Choice A is one yard too high. Choice B is one yard too low. Choice D comes from forgetting to subtract the fractional parts, leaving the $\tfrac{1}{2}$ in the answer.
Marcus has three containers of paint. The first contains $1\frac{3}{12}$ gallons, the second contains $2\frac{7}{12}$ gallons, and the third contains $1\frac{8}{12}$ gallons. He uses $2\frac{6}{12}$ gallons for a project. How much paint does Marcus have left?
Answer: $3\frac{0}{12}$ gallons of paint remaining When you see a word problem involving adding and subtracting fractions, break it into two clear steps: first find the total amount Marcus started with, then subtract what he used. To find how much paint Marcus had originally, add all three containers: $1\frac{3}{12} + 2\frac{7}{12} + 1\frac{8}{12}$ . Since all fractions have the same denominator (12), you can add the whole numbers together and the fractions together separately. The whole numbers: $1 + 2 + 1 = 4$ . The fractions: $\frac{3}{12} + \frac{7}{12} + \frac{8}{12} = \frac{18}{12}$ . Since $\frac{18}{12} = 1\frac{6}{12}$ , Marcus started with $4 + 1\frac{6}{12} = 5\frac{6}{12}$ gallons total. Next, subtract what he used: $5\frac{6}{12} - 2\frac{6}{12} = 3\frac{0}{12}$ gallons remaining. This confirms answer B is correct. Looking at the wrong answers: Choice A ( $8\frac{0}{12}$ ) likely comes from adding all four amounts instead of subtracting the used paint. Choice C ( $5\frac{6}{12}$ ) is the total amount Marcus started with before using any paint. Choice D ( $2\frac{6}{12}$ ) is simply the amount he used, not what remains. Study tip: In multi-step fraction problems, always identify what you're looking for first. Here, "remaining" tells you it's a subtraction problem, but you need the total first. Write out each step clearly to avoid mixing up intermediate answers with your final answer.
Sofia walked $3\ \tfrac{2}{5}$ miles in the morning and $1\ \tfrac{4}{5}$ miles after lunch. How many miles did she walk in all?
Answer: $5\ \tfrac{1}{5}$ miles The correct answer is 5 and 1/5 miles, because 3 and 2/5 plus 1 and 4/5 gives 4 wholes plus 6/5, and 6/5 equals 1 and 1/5 more, for a total of 5 and 1/5. Choice A, 4 and 1/5 miles, forgets to carry the extra whole from combining the fifths. Choice B, 4 and 6/10 miles, mistakenly changes the denominator instead of keeping fifths. Choice C, 5 and 6/5 miles, correctly finds 6/5 but fails to convert it into a whole number plus a proper fraction.
Maya walked $1\frac{3}{4}$ miles in the morning and $2\frac{1}{4}$ miles after school. How many miles did she walk in all?
Answer: 4 miles 4 miles is correct because $1\frac{3}{4} + 2\frac{1}{4} = 4$ . 3 miles is incorrect because it drops the fractional parts entirely. $3\frac{1}{2}$ miles is incorrect because it undercounts the fractional sum. $3\frac{3}{4}$ miles is incorrect because it does not carry the whole number correctly when the fractional parts sum to 1.
Add $2\ \tfrac{1}{3} + 3\ \tfrac{2}{3}$ . What is the sum?
Answer: $6$ Add whole numbers and fractions separately: wholes: 2 + 3 = 5; fractions: 1/3 + 2/3 = 3/3 = 1; combined: 5 + 1 = 6. The correct answer is B. Choice A (6 1/3) overcounts — the fractions sum to exactly 1 whole (not 1 1/3), so the total is 6, not 6 1/3. Choice C (5 1/3) uses the wrong fractional part and misses the carry. Choice D (5) adds the wholes correctly but forgets to add the 1 whole from the fraction sum.
A recipe calls for $4\frac{2}{5}$ cups of flour, but Jin only has $2\frac{4}{5}$ cups. After going to the store, he has exactly enough flour for the recipe. How much flour did Jin buy at the store?
Answer: $1\frac{3}{5}$ cups of flour To find how much flour Jin bought, subtract what he already had from the total the recipe needs: $4\frac{2}{5} - 2\frac{4}{5} = 1\frac{3}{5}$ cups. Choice A comes from subtracting without regrouping the whole number correctly. Choice B comes from a similar subtraction error that misapplies the fifths. Choice D results from adding the two amounts instead of subtracting them and leaves the fraction unsimplified. Choice C reflects the correct subtraction with proper regrouping.
What is $1\ \tfrac{5}{6} + 2\ \tfrac{5}{6}$ ?
Answer: $4\ \tfrac{4}{6}$ Adding the whole numbers (1 + 2 = 3) and the fractions (5/6 + 5/6 = 10/6, or 1 4/6) gives 3 + 1 4/6 = 4 4/6. Choice B changes the denominator when combining the fractions instead of keeping a common denominator of 6. Choice C reflects an error in adding the fractional parts. Choice D reflects a combination of denominator and numerator errors when adding the fractions.
Chen had $5\frac{3}{4}$ liters of water. He used $2\frac{1}{4}$ liters. How many liters are left?
Answer: $3\frac{1}{2}$ liters $3\frac{1}{2}$ liters is correct because $5\frac{3}{4} - 2\frac{1}{4} = 3\frac{1}{2}$ . $2\frac{1}{2}$ liters is incorrect because the whole-number part is too small by one. 3 liters is incorrect because it drops the fractional part entirely. 8 liters is incorrect because it results from adding instead of subtracting.
Subtract: $4\frac{7}{8} - 2\frac{3}{8}$ . Write the difference as a mixed number.
Answer: $2\frac{4}{8}$ $2\frac{4}{8}$ is correct because $4\frac{7}{8} - 2\frac{3}{8} = 2\frac{4}{8}$ . $1\frac{4}{8}$ is incorrect because the whole-number part is too small by one. $3\frac{4}{8}$ is incorrect because the whole-number part is too large by one. $2\frac{3}{8}$ is incorrect because it repeats the fraction being subtracted instead of finding the difference.
Chen had $4\frac{3}{8}$ feet of ribbon. He used $1\frac{7}{8}$ feet. How much ribbon is left?
Answer: $2\frac{1}{2}$ feet $2\frac{1}{2}$ feet is correct because $4\frac{3}{8} - 1\frac{7}{8} = 2\frac{4}{8}$ , which simplifies to $2\frac{1}{2}$ . $3\frac{1}{2}$ feet is incorrect because the whole-number part is too large by one. $3\frac{4}{8}$ feet is incorrect for the same reason, without simplifying. $2\frac{4}{8}$ feet is incorrect only because it is left unsimplified, not because the value is wrong.
A baker used $3\frac{1}{8}$ cups of sugar in the morning and $2\frac{7}{8}$ cups in the afternoon. If she started with $9\frac{2}{8}$ cups of sugar, how much sugar does she have left?
Answer: $3\frac{2}{8}$ cups of sugar remaining When you see a word problem involving mixed numbers, you need to identify what operation to use and work systematically through the steps. This problem asks how much sugar remains after using some, so you'll subtract the amounts used from the starting amount. Start with $9\frac{2}{8}$ cups and subtract both amounts used. First, add the sugar used: $3\frac{1}{8} + 2\frac{7}{8}$ . Add the whole numbers: $3 + 2 = 5$ . Add the fractions: $\frac{1}{8} + \frac{7}{8} = \frac{8}{8} = 1$ . So the total used is $5 + 1 = 6$ cups. Now subtract: $9\frac{2}{8} - 6 = 3\frac{2}{8}$ cups remaining. This matches answer D. Looking at the wrong answers: Answer A ( $3\frac{4}{8}$ ) likely comes from incorrectly adding the fractions $\frac{2}{8} + \frac{1}{8} + \frac{7}{8} = \frac{10}{8}$ , but forgetting to convert $\frac{10}{8}$ to $1\frac{2}{8}$ . Answer B ( $15\frac{2}{8}$ ) results from adding all three amounts instead of subtracting: $9\frac{2}{8} + 3\frac{1}{8} + 2\frac{7}{8}$ . Answer C ( $6\frac{0}{8}$ ) comes from only calculating how much sugar was used total, not how much remains. When solving mixed number word problems, always read carefully to determine the correct operation, then work step-by-step: handle whole numbers and fractions separately, and remember to convert improper fractions back to mixed numbers when needed.
Carmen ran $3\frac{1}{6}$ miles on Monday and $2\frac{5}{6}$ miles on Tuesday. On Wednesday, she ran $1\frac{2}{6}$ miles less than her total for Monday and Tuesday combined. How far did Carmen run on Wednesday?
Answer: $4\frac{4}{6}$ miles on Wednesday Monday and Tuesday total: $3\frac{1}{6} + 2\frac{5}{6} = 6\frac{0}{6} = 6$ miles. Wednesday distance: $6 - 1\frac{2}{6} = 4\frac{4}{6}$ miles. Choice B makes an error in the subtraction. Choice C incorrectly adds $1\frac{2}{6}$ instead of subtracting. Choice D represents a calculation error in the final subtraction step.
Maya poured $1\ \tfrac{2}{8}$ liters of water into a jug, then poured in $2\ \tfrac{3}{8}$ more liters. How many liters are in the jug now?
Answer: $3\ \tfrac{5}{8}$ liters This question tests 4th grade ability to add and subtract mixed numbers with like denominators, using strategies such as converting to improper fractions or using properties of operations (CCSS.4.NF.3.c). Mixed numbers combine a whole number and a fraction (like 1 2/8). To add or subtract mixed numbers with the same denominator, we can either: (1) convert both to improper fractions, add/subtract, then convert back to mixed number, or (2) add/subtract the whole number parts and fraction parts separately using properties of operations. For subtraction, if the fraction part being subtracted is larger than the fraction part starting with, we must regroup by converting 1 whole into fractions. To add 1 2/8 and 2 3/8, students can convert to improper: 10/8 and 19/8, add: 29/8 = 3 5/8, or add separately: wholes (1+2=3), fractions (2/8+3/8=5/8), combine: 3 5/8. Choice A is correct because adding separately: 1+2=3, 2/8+3/8=5/8, combining: 3 5/8, demonstrating understanding of mixed number operations with like denominators. Choice B represents adding denominators by mistake, which happens when students think denominators change in addition. To help students: Practice both methods. For improper fraction method: convert mixed to improper (a b/c = (a×c+b)/c), add/subtract numerators keeping denominator same, convert result back to mixed (divide numerator by denominator for quotient=whole, remainder=numerator). For separate parts method: add/subtract wholes, add/subtract fractions, combine. Check answers: addition result > either addend. Watch for: adding denominators, conversion errors, and not carrying over if needed.

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