Elementary School Math Quiz: Add And Subtract Mixed Numbers
20 questions · exam conditions
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Add And Subtract Mixed NumbersQuestion 1 of 20

Subtract: 4 5/6 - 2 1/6. Give the difference as a mixed number.

2 4/6
2 1/6
2 6/12
3 4/6
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Elementary School Math Quiz

Elementary School Math Quiz: Add And Subtract Mixed Numbers

Practice Add And Subtract Mixed Numbers in Elementary School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Add And Subtract Mixed Numbers, giving you a quick way to practice the rules, question types, and explanations that matter most for Elementary School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Subtract: 4 5/6 - 2 1/6. Give the difference as a mixed number.

  1. 2 4/6 (correct answer)
  2. 2 1/6
  3. 2 6/12
  4. 3 4/6
Explanation: Subtract the whole numbers (4 - 2 = 2) and the fractions (5/6 - 1/6 = 4/6) separately, giving 2 4/6. Choice D (3 4/6) comes from subtracting the whole numbers incorrectly. Choice B (2 1/6) mistakenly keeps the second fraction instead of subtracting. Choice C (2 6/12) changes the denominator instead of keeping it the same.

Question 2

Maya has 2382\frac{3}{8} yards of ribbon. She uses 1581\frac{5}{8} yards to wrap a gift and then buys 3283\frac{2}{8} more yards. How much ribbon does Maya have now?

  1. 4084\frac{0}{8} yards (correct answer)
  2. 4284\frac{2}{8} yards
  3. 7287\frac{2}{8} yards
  4. 5285\frac{2}{8} yards
Explanation: Starting with 2382\frac{3}{8} yards, subtracting the 1581\frac{5}{8} yards used leaves 68\frac{6}{8} yard, and adding the 3283\frac{2}{8} yards purchased gives exactly 4 yards. Choice B comes from a small error in regrouping when subtracting the amount used for the gift. Choice C results from adding all three amounts together instead of subtracting the amount used. Choice D comes from a similar addition error that overstates the final total. Choice A correctly combines the subtraction and addition steps to reach exactly 4 yards.

Question 3

Carmen ran 3163\frac{1}{6} miles on Monday and 2562\frac{5}{6} miles on Tuesday. On Wednesday, she ran 1261\frac{2}{6} miles less than her total for Monday and Tuesday combined. How far did Carmen run on Wednesday?

  1. 4464\frac{4}{6} miles on Wednesday (correct answer)
  2. 5065\frac{0}{6} miles on Wednesday
  3. 7267\frac{2}{6} miles on Wednesday
  4. 4564\frac{5}{6} miles on Wednesday
Explanation: Monday and Tuesday total: 316+256=606=63\frac{1}{6} + 2\frac{5}{6} = 6\frac{0}{6} = 6 miles. Wednesday distance: 6−126=4466 - 1\frac{2}{6} = 4\frac{4}{6} miles. Choice B makes an error in the subtraction. Choice C incorrectly adds 1261\frac{2}{6} instead of subtracting. Choice D represents a calculation error in the final subtraction step.

Question 4

Chen walked 2 342\ \tfrac{3}{4} miles in the morning and 1 241\ \tfrac{2}{4} miles after lunch. How many miles did Chen walk total?

  1. 3 143\ \tfrac{1}{4} miles
  2. 3 543\ \tfrac{5}{4} miles
  3. 4 144\ \tfrac{1}{4} miles (correct answer)
  4. 3 583\ \tfrac{5}{8} miles
Explanation: Adding 2 342\ \tfrac{3}{4} and 1 241\ \tfrac{2}{4} gives 3 543\ \tfrac{5}{4}, and since 54\tfrac{5}{4} equals 1 141\ \tfrac{1}{4}, the total simplifies to 4 144\ \tfrac{1}{4} miles. Choice A adds the whole numbers correctly but drops the extra whole mile when the fraction part carries over. Choice B stops right before simplifying the improper fraction, leaving the answer unsimplified. Choice D miscombines the fraction parts using the wrong denominator. Choice C is the only option that adds both distances correctly and simplifies fully.

Question 5

Add: 2 25+3 152\ \tfrac{2}{5} + 3\ \tfrac{1}{5}

  1. 5 355\ \tfrac{3}{5} (correct answer)
  2. 6 356\ \tfrac{3}{5}
  3. 5 3105\ \tfrac{3}{10}
  4. 55
Explanation: 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.

Question 6

Maya walked 1341\frac{3}{4} miles in the morning and 2142\frac{1}{4} miles after school. How many miles did she walk in all?

  1. 4 miles (correct answer)
  2. 3343\frac{3}{4} miles
  3. 3123\frac{1}{2} miles
  4. 3 miles
Explanation: 4 miles is correct because 134+214=41\frac{3}{4} + 2\frac{1}{4} = 4. 3343\frac{3}{4} miles is incorrect because it does not carry the whole number correctly when the fractional parts sum to 1. 3123\frac{1}{2} miles is incorrect because it undercounts the fractional sum. 3 miles is incorrect because it drops the fractional parts entirely.

Question 7

A rope was 7297\frac{2}{9} feet long. After cutting off 2892\frac{8}{9} feet, another piece measuring 1491\frac{4}{9} feet was tied to the remaining rope. What is the length of the rope now?

  1. 4394\frac{3}{9} feet
  2. 5795\frac{7}{9} feet (correct answer)
  3. 115911\frac{5}{9} feet
  4. 6196\frac{1}{9} feet
Explanation: Subtracting 729−2897\frac{2}{9} - 2\frac{8}{9} requires regrouping: 729=61197\frac{2}{9} = 6\frac{11}{9}, so 6119−289=4396\frac{11}{9} - 2\frac{8}{9} = 4\frac{3}{9}. Adding the new piece: 439+149=5794\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.

Question 8

What is 225+3152 \tfrac{2}{5} + 3 \tfrac{1}{5}?

  1. 5355 \tfrac{3}{5} (correct answer)
  2. 6356 \tfrac{3}{5}
  3. 53105 \tfrac{3}{10}
  4. 55
Explanation: 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.

Question 9

What is 1 56+2 561\ \tfrac{5}{6} + 2\ \tfrac{5}{6}?

  1. 4 464\ \tfrac{4}{6} (correct answer)
  2. 3 10123\ \tfrac{10}{12}
  3. 3 263\ \tfrac{2}{6}
  4. 3 5123\ \tfrac{5}{12}
Explanation: 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.

Question 10

A fence has three sections with lengths 31103\frac{1}{10} feet, 34103\frac{4}{10} feet, and 28102\frac{8}{10} feet. If the middle section is replaced with a section that is 11101\frac{1}{10} feet shorter, what will be the total length of the fence?

  1. 82108\frac{2}{10} feet for the complete fence (correct answer)
  2. 1041010\frac{4}{10} feet for the complete fence
  3. 71107\frac{1}{10} feet for the complete fence
  4. 93109\frac{3}{10} feet for the complete fence
Explanation: The correct answer is A because the original fence is 3110+3410+2810=93103\frac{1}{10} + 3\frac{4}{10} + 2\frac{8}{10} = 9\frac{3}{10} feet, and shortening the middle section by 11101\frac{1}{10} feet gives a new total of 9310−1110=82109\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.

Question 11

Chen had 5 345\ \tfrac{3}{4} liters of water. He used 2 142\ \tfrac{1}{4} liters. How many liters are left?

  1. 2 122\ \tfrac{1}{2} liters
  2. 3 123\ \tfrac{1}{2} liters (correct answer)
  3. 3 343\ \tfrac{3}{4} liters
  4. 4 124\ \tfrac{1}{2} liters
Explanation: The correct answer is B, 3123\frac{1}{2} liters, because 534−214=324=3125\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 34\frac{3}{4} unchanged. Choice D comes from a regrouping error that overshoots the true difference.

Question 12

Chen had 5345\frac{3}{4} liters of water. He used 2142\frac{1}{4} liters. How many liters are left?

  1. 3123\frac{1}{2} liters (correct answer)
  2. 2122\frac{1}{2} liters
  3. 3 liters
  4. 8 liters
Explanation: 3123\frac{1}{2} liters is correct because 534−214=3125\frac{3}{4} - 2\frac{1}{4} = 3\frac{1}{2}. 2122\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.

Question 13

Chen had 4384\frac{3}{8} feet of ribbon. He used 1781\frac{7}{8} feet. How much ribbon is left?

  1. 3123\frac{1}{2} feet
  2. 3483\frac{4}{8} feet
  3. 2122\frac{1}{2} feet (correct answer)
  4. 2482\frac{4}{8} feet
Explanation: 2122\frac{1}{2} feet is correct because 438−178=2484\frac{3}{8} - 1\frac{7}{8} = 2\frac{4}{8}, which simplifies to 2122\frac{1}{2}. 3123\frac{1}{2} feet is incorrect because the whole-number part is too large by one. 3483\frac{4}{8} feet is incorrect for the same reason, without simplifying. 2482\frac{4}{8} feet is incorrect only because it is left unsimplified, not because the value is wrong.

Question 14

A recipe calls for 4254\frac{2}{5} cups of flour, but Jin only has 2452\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?

  1. 2152\frac{1}{5} cups of flour
  2. 2252\frac{2}{5} cups of flour
  3. 1351\frac{3}{5} cups of flour (correct answer)
  4. 6656\frac{6}{5} cups of flour
Explanation: To find how much flour Jin bought, subtract what he already had from the total the recipe needs: 425−245=1354\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.

Question 15

Marcus has three containers of paint. The first contains 13121\frac{3}{12} gallons, the second contains 27122\frac{7}{12} gallons, and the third contains 18121\frac{8}{12} gallons. He uses 26122\frac{6}{12} gallons for a project. How much paint does Marcus have left?

  1. 80128\frac{0}{12} gallons of paint remaining
  2. 30123\frac{0}{12} gallons of paint remaining (correct answer)
  3. 56125\frac{6}{12} gallons of paint remaining
  4. 26122\frac{6}{12} gallons of paint remaining
Explanation: 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: 1312+2712+18121\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=41 + 2 + 1 = 4. The fractions: 312+712+812=1812\frac{3}{12} + \frac{7}{12} + \frac{8}{12} = \frac{18}{12}. Since 1812=1612\frac{18}{12} = 1\frac{6}{12}, Marcus started with 4+1612=56124 + 1\frac{6}{12} = 5\frac{6}{12} gallons total. Next, subtract what he used: 5612−2612=30125\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 (80128\frac{0}{12}) likely comes from adding all four amounts instead of subtracting the used paint. Choice C (56125\frac{6}{12}) is the total amount Marcus started with before using any paint. Choice D (26122\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.

Question 16

Emma walked 2372\frac{3}{7} miles to school and 1571\frac{5}{7} miles to the library. Then she walked home, which was 2172\frac{1}{7} miles farther than her total walking distance to school and the library. How far did Emma walk to get home?

  1. 6376\frac{3}{7} miles to walk home
  2. 22 miles to walk home
  3. 4174\frac{1}{7} miles to walk home
  4. 6276\frac{2}{7} miles to walk home (correct answer)
Explanation: 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.

Question 17

Add 2 13+3 232\ \tfrac{1}{3} + 3\ \tfrac{2}{3}. What is the sum?

  1. 6 136\ \tfrac{1}{3}
  2. 66 (correct answer)
  3. 5 135\ \tfrac{1}{3}
  4. 55
Explanation: 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.

Question 18

Add 2 15+3 452\ \tfrac{1}{5} + 3\ \tfrac{4}{5}. Write the sum as a mixed number or whole number.

  1. 5 355\ \tfrac{3}{5}
  2. 6 156\ \tfrac{1}{5}
  3. 5 455\ \tfrac{4}{5}
  4. 66 (correct answer)
Explanation: Add whole numbers and fractions separately: wholes: 2 + 3 = 5; fractions: 1/5 + 4/5 = 5/5 = 1; combined: 5 + 1 = 6. The correct answer is D. Choice A (5 3/5) uses the wrong fractional part — 1/5 + 4/5 = 5/5 = 1 whole, not 3/5. Choice B (6 1/5) overcounts, adding an extra 1/5 not present in the problem. Choice C (5 4/5) forgets to carry the 1 whole from the fraction sum.

Question 19

Subtract: 4 56−2 164\ \tfrac{5}{6} - 2\ \tfrac{1}{6}. Give the difference as a mixed number.

  1. 2 6122\ \tfrac{6}{12}
  2. 2 462\ \tfrac{4}{6} (correct answer)
  3. 3 463\ \tfrac{4}{6}
  4. 2 162\ \tfrac{1}{6}
Explanation: 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.

Question 20

Luis ate 1231\frac{2}{3} slices of pizza for lunch and 2132\frac{1}{3} slices for dinner. His sister ate 1131\frac{1}{3} slices less than Luis's total. How many slices did his sister eat?

  1. 5135\frac{1}{3} slices
  2. 4234\frac{2}{3} slices
  3. 2232\frac{2}{3} slices (correct answer)
  4. 33 slices
Explanation: Luis ate 123+213=41\frac{2}{3} + 2\frac{1}{3} = 4 slices total, and his sister ate 1131\frac{1}{3} less than that, which is 4−113=2234 - 1\frac{1}{3} = 2\frac{2}{3} slices. Choice A comes from adding Luis's amount and the difference instead of subtracting. Choice B results from a regrouping error when subtracting the fraction parts. Choice D rounds the result to a whole number instead of keeping the exact fraction. Choice C correctly finds Luis's total and subtracts the difference to get his sister's amount.