ACCUPLACER Arithmetic Quiz: Adding And Subtracting Fractions
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Adding And Subtracting FractionsQuestion 1 of 20

A container holds 12 gallons of liquid. First, 4234\frac{2}{3} gallons are removed, and then 2122\frac{1}{2} gallons are removed. How much liquid is left in the container?

4564\frac{5}{6} gallons
5165\frac{1}{6} gallons
6166\frac{1}{6} gallons
7167\frac{1}{6} gallons
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ACCUPLACER Arithmetic Quiz

ACCUPLACER Arithmetic Quiz: Adding And Subtracting Fractions

Practice Adding And Subtracting Fractions in ACCUPLACER Arithmetic 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 Adding And Subtracting Fractions, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Arithmetic.

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

A container holds 12 gallons of liquid. First, 4234\frac{2}{3} gallons are removed, and then 2122\frac{1}{2} gallons are removed. How much liquid is left in the container?

  1. 4564\frac{5}{6} gallons (correct answer)
  2. 5165\frac{1}{6} gallons
  3. 6166\frac{1}{6} gallons
  4. 7167\frac{1}{6} gallons
Explanation: First, find the total amount of liquid removed by adding the two quantities: 423+2124\frac{2}{3} + 2\frac{1}{2}. Find a common denominator of 6: 446+236=6764\frac{4}{6} + 2\frac{3}{6} = 6\frac{7}{6}. Since 76=116\frac{7}{6} = 1\frac{1}{6}, the total amount removed is 6+116=7166 + 1\frac{1}{6} = 7\frac{1}{6} gallons. Next, subtract this total from the initial amount: 12−71612 - 7\frac{1}{6}. To subtract, rewrite 12 as 116611\frac{6}{6}. The subtraction becomes 1166−716=45611\frac{6}{6} - 7\frac{1}{6} = 4\frac{5}{6} gallons.

Question 2

Three-eighths of students in a class play soccer, and one-fourth play basketball. If no student plays both sports, what fraction of the class does NOT play either soccer or basketball?

  1. 18\frac{1}{8}
  2. 78\frac{7}{8}
  3. 58\frac{5}{8}
  4. 38\frac{3}{8} (correct answer)
Explanation: First, find the total fraction of students who play a sport by adding the fractions for soccer and basketball: 38+14\frac{3}{8} + \frac{1}{4}. Find a common denominator of 8: 38+28=58\frac{3}{8} + \frac{2}{8} = \frac{5}{8}. This represents the fraction of students who play one of the sports. To find the fraction that does not play either sport, subtract this from the whole class (1): 1−581 - \frac{5}{8}. Rewrite 1 as 88\frac{8}{8}. So, 88−58=38\frac{8}{8} - \frac{5}{8} = \frac{3}{8}.

Question 3

What is the difference between the sum of 34\frac{3}{4} and 23\frac{2}{3}, and the sum of 12\frac{1}{2} and 16\frac{1}{6}?

  1. 14\frac{1}{4}
  2. 12\frac{1}{2}
  3. 23\frac{2}{3}
  4. 34\frac{3}{4} (correct answer)
Explanation: This is a multi-step problem. First, find the first sum: 34+23\frac{3}{4} + \frac{2}{3}. The common denominator is 12. 912+812=1712\frac{9}{12} + \frac{8}{12} = \frac{17}{12}. Second, find the second sum: 12+16\frac{1}{2} + \frac{1}{6}. The common denominator is 6. 36+16=46\frac{3}{6} + \frac{1}{6} = \frac{4}{6}, which simplifies to 23\frac{2}{3}. Finally, find the difference between the two sums: 1712−23\frac{17}{12} - \frac{2}{3}. The common denominator is 12. 1712−812=912\frac{17}{12} - \frac{8}{12} = \frac{9}{12}, which simplifies to 34\frac{3}{4}.

Question 4

From a rope 151215\frac{1}{2} feet long, a piece of 4134\frac{1}{3} feet and another piece of 5145\frac{1}{4} feet are cut off. What is the length of the remaining rope?

  1. 101610\frac{1}{6} feet
  2. 6126\frac{1}{2} feet
  3. 97129\frac{7}{12} feet
  4. 511125\frac{11}{12} feet (correct answer)
Explanation: First, calculate the total length of the rope that was cut off: 413+5144\frac{1}{3} + 5\frac{1}{4}. Find a common denominator of 12: 4412+5312=97124\frac{4}{12} + 5\frac{3}{12} = 9\frac{7}{12} feet. Next, subtract this total from the original length of the rope: 1512−971215\frac{1}{2} - 9\frac{7}{12}. Convert to a common denominator of 12: 15612−971215\frac{6}{12} - 9\frac{7}{12}. Regroup from the 15: 141812−9712=5111214\frac{18}{12} - 9\frac{7}{12} = 5\frac{11}{12} feet.

Question 5

A baker starts with 10 cups of sugar. He uses 2132\frac{1}{3} cups for cookies and 3343\frac{3}{4} cups for a cake. He then adds 1121\frac{1}{2} cups of sugar back to his supply from another bag. How much sugar does he have now?

  1. 311123\frac{11}{12} cups
  2. 55125\frac{5}{12} cups (correct answer)
  3. 61126\frac{1}{12} cups
  4. 75127\frac{5}{12} cups
Explanation: This problem can be represented by the expression 10−213−334+11210 - 2\frac{1}{3} - 3\frac{3}{4} + 1\frac{1}{2}. First, calculate the total sugar used: 213+3342\frac{1}{3} + 3\frac{3}{4}. The common denominator is 12: 2412+3912=51312=61122\frac{4}{12} + 3\frac{9}{12} = 5\frac{13}{12} = 6\frac{1}{12} cups. Now, subtract this from the initial 10 cups: 10−6112=3111210 - 6\frac{1}{12} = 3\frac{11}{12} cups. Finally, add the sugar that was put back: 31112+1123\frac{11}{12} + 1\frac{1}{2}. The common denominator is 12: 31112+1612=41712=55123\frac{11}{12} + 1\frac{6}{12} = 4\frac{17}{12} = 5\frac{5}{12} cups.

Question 6

On a map, the distance between two cities is 5145\frac{1}{4} inches. If a traveler has already covered a distance of 2122\frac{1}{2} inches on the map and then travels another 1131\frac{1}{3} inches, what is the remaining distance in inches on the map?

  1. 3563\frac{5}{6} inches
  2. 17121\frac{7}{12} inches
  3. 15121\frac{5}{12} inches (correct answer)
  4. 41124\frac{1}{12} inches
Explanation: First, find the total distance traveled on the map by adding the two segments: 212+1132\frac{1}{2} + 1\frac{1}{3}. Find a common denominator of 6: 236+126=3562\frac{3}{6} + 1\frac{2}{6} = 3\frac{5}{6} inches. Next, subtract the total distance traveled from the total distance between the cities: 514−3565\frac{1}{4} - 3\frac{5}{6}. Find a common denominator of 12: 5312−310125\frac{3}{12} - 3\frac{10}{12}. Regroup from the 5: 41512−31012=15124\frac{15}{12} - 3\frac{10}{12} = 1\frac{5}{12} inches.

Question 7

What is the result of 10−(314+212)10 - (3\frac{1}{4} + 2\frac{1}{2})?

  1. 4144\frac{1}{4} (correct answer)
  2. 4344\frac{3}{4}
  3. 5345\frac{3}{4}
  4. 9149\frac{1}{4}
Explanation: According to the order of operations, first calculate the sum inside the parentheses: 314+2123\frac{1}{4} + 2\frac{1}{2}. Convert to a common denominator of 4: 314+224=5343\frac{1}{4} + 2\frac{2}{4} = 5\frac{3}{4}. Now, subtract this result from 10: 10−53410 - 5\frac{3}{4}. To subtract, you can rewrite 10 as 9449\frac{4}{4}. Then, 944−534=4149\frac{4}{4} - 5\frac{3}{4} = 4\frac{1}{4}.

Question 8

A rectangular garden has a length of 4124\frac{1}{2} meters and a width of 3133\frac{1}{3} meters. A fence is built around the garden, but a gate with a width of 34\frac{3}{4} meter is left open. What is the total length of the fence used?

  1. 152315\frac{2}{3} meters
  2. 1515 meters
  3. 14111214\frac{11}{12} meters (correct answer)
  4. 1651216\frac{5}{12} meters
Explanation: First, calculate the perimeter of the garden using the formula P=2(L+W)P = 2(L+W). Add the length and width: 412+3134\frac{1}{2} + 3\frac{1}{3}. The common denominator is 6. 436+326=7564\frac{3}{6} + 3\frac{2}{6} = 7\frac{5}{6} meters. The perimeter is 2×756=2×476=946=473=15232 \times 7\frac{5}{6} = 2 \times \frac{47}{6} = \frac{94}{6} = \frac{47}{3} = 15\frac{2}{3} meters. This is the total length around the garden. To find the length of the fence used, subtract the width of the gate: 1523−3415\frac{2}{3} - \frac{3}{4}. The common denominator is 12. 15812−91215\frac{8}{12} - \frac{9}{12}. Regroup from 15: 142012−912=14111214\frac{20}{12} - \frac{9}{12} = 14\frac{11}{12} meters.

Question 9

A package contains three items. Item A weighs 3123\frac{1}{2} pounds, Item B weighs 2132\frac{1}{3} pounds, and the total weight of the package is 7147\frac{1}{4} pounds. What is the weight of Item C?

  1. 15121\frac{5}{12} pounds (correct answer)
  2. 17121\frac{7}{12} pounds
  3. 5565\frac{5}{6} pounds
  4. 1311213\frac{1}{12} pounds
Explanation: First, find the combined weight of Item A and Item B: 312+2133\frac{1}{2} + 2\frac{1}{3}. Find a common denominator of 6: 336+226=5563\frac{3}{6} + 2\frac{2}{6} = 5\frac{5}{6} pounds. To find the weight of Item C, subtract the combined weight of A and B from the total package weight: 714−5567\frac{1}{4} - 5\frac{5}{6}. Find a common denominator of 12: 7312−510127\frac{3}{12} - 5\frac{10}{12}. To subtract, regroup from the 7: 61512−51012=15126\frac{15}{12} - 5\frac{10}{12} = 1\frac{5}{12} pounds.

Question 10

A carpenter is joining two boards together. One board is 1341\frac{3}{4} inches thick and the other is 78\frac{7}{8} of an inch thick. To make them fit, the carpenter shaves 116\frac{1}{16} of an inch off the thicker board. What is the combined thickness of the two boards after the adjustment?

  1. 2122\frac{1}{2} inches
  2. 211162\frac{11}{16} inches
  3. 2582\frac{5}{8} inches
  4. 29162\frac{9}{16} inches (correct answer)
Explanation: First, identify the thicker board. Convert 1341\frac{3}{4} to eighths: 1681\frac{6}{8}. Since 168>781\frac{6}{8} > \frac{7}{8}, the first board is thicker. Shave 116\frac{1}{16} inch off this board: 134−1161\frac{3}{4} - \frac{1}{16}. Convert to a common denominator of 16: 11216−116=111161\frac{12}{16} - \frac{1}{16} = 1\frac{11}{16} inches. Now, find the combined thickness by adding the adjusted thickness of the first board to the thickness of the second board: 11116+781\frac{11}{16} + \frac{7}{8}. Convert to a common denominator of 16: 11116+1416=125161\frac{11}{16} + \frac{14}{16} = 1\frac{25}{16}. Simplify the improper fraction: 1+1916=29161 + 1\frac{9}{16} = 2\frac{9}{16} inches.

Question 11

A board is 8148\frac{1}{4} feet long. A piece that is 2232\frac{2}{3} feet long is cut from it. Then, another piece that is 3123\frac{1}{2} feet long is cut from the remaining part. What is the length of the final piece of the board?

  1. 1561\frac{5}{6} feet
  2. 21122\frac{1}{12} feet (correct answer)
  3. 2132\frac{1}{3} feet
  4. 6166\frac{1}{6} feet
Explanation: First, find the total length of the two pieces cut from the board: 223+3122\frac{2}{3} + 3\frac{1}{2}. The common denominator is 6. So, 246+336=5762\frac{4}{6} + 3\frac{3}{6} = 5\frac{7}{6}, which simplifies to 6166\frac{1}{6} feet. Next, subtract this total length from the original length of the board: 814−6168\frac{1}{4} - 6\frac{1}{6}. The common denominator is 12. So, 8312−6212=21128\frac{3}{12} - 6\frac{2}{12} = 2\frac{1}{12} feet. The remaining piece is 21122\frac{1}{12} feet long.

Question 12

Which of the following results in the smallest value?

  1. 212−1342\frac{1}{2} - 1\frac{3}{4}
  2. 113−561\frac{1}{3} - \frac{5}{6}
  3. 78−12\frac{7}{8} - \frac{1}{2} (correct answer)
  4. 3−2153 - 2\frac{1}{5}
Explanation: Calculate the value of each expression. A) 212−134=224−134=164−134=342\frac{1}{2} - 1\frac{3}{4} = 2\frac{2}{4} - 1\frac{3}{4} = 1\frac{6}{4} - 1\frac{3}{4} = \frac{3}{4}. B) 113−56=126−56=86−56=36=121\frac{1}{3} - \frac{5}{6} = 1\frac{2}{6} - \frac{5}{6} = \frac{8}{6} - \frac{5}{6} = \frac{3}{6} = \frac{1}{2}. C) 78−12=78−48=38\frac{7}{8} - \frac{1}{2} = \frac{7}{8} - \frac{4}{8} = \frac{3}{8}. D) 3−215=255−215=453 - 2\frac{1}{5} = 2\frac{5}{5} - 2\frac{1}{5} = \frac{4}{5}. Now compare the results: 34\frac{3}{4} (or 0.75), 12\frac{1}{2} (or 0.5), 38\frac{3}{8} (or 0.375), and 45\frac{4}{5} (or 0.8). The smallest value is 38\frac{3}{8}.

Question 13

A carpenter cut 25122\frac{5}{12} feet from a board, then cut another 1381\frac{3}{8} feet from the same board. If 6346\frac{3}{4} feet of the original board remains, what was the length of the original board?

  1. 1072410\frac{7}{24} feet was the original length
  2. 10132410\frac{13}{24} feet was the original length (correct answer)
  3. 9569\frac{5}{6} feet was the original length
  4. 101210\frac{1}{2} feet was the original length
Explanation: Add all the pieces together: cuts of 2512+1382\frac{5}{12} + 1\frac{3}{8} plus remaining 6346\frac{3}{4}. Converting to common denominator 24: 21024+1924+61824=93724=1013242\frac{10}{24} + 1\frac{9}{24} + 6\frac{18}{24} = 9\frac{37}{24} = 10\frac{13}{24}. Choice A results from miscalculating 3724\frac{37}{24} as 3124\frac{31}{24}. Choice C uses the wrong common denominator and gets 9569\frac{5}{6}. Choice D rounds incorrectly to 101210\frac{1}{2}.

Question 14

A recipe for fruit punch calls for 1231\frac{2}{3} cups of orange juice and 2142\frac{1}{4} cups of pineapple juice. If Lisa wants to make 34\frac{3}{4} of the original recipe, how much total juice will she need?

  1. 259482\frac{59}{48} cups of total juice needed
  2. 211162\frac{11}{16} cups of total juice needed
  3. 215162\frac{15}{16} cups of total juice needed (correct answer)
  4. 31163\frac{1}{16} cups of total juice needed
Explanation: Original total juice: 123+214=1812+2312=311121\frac{2}{3} + 2\frac{1}{4} = 1\frac{8}{12} + 2\frac{3}{12} = 3\frac{11}{12}. Amount needed for 34\frac{3}{4} recipe: 31112×34=4712×34=14148=24548=215163\frac{11}{12} \times \frac{3}{4} = \frac{47}{12} \times \frac{3}{4} = \frac{141}{48} = 2\frac{45}{48} = 2\frac{15}{16}. Choice A shows an incorrect mixed number conversion. Choice B results from computational errors in the multiplication. Choice D adds instead of finding 34\frac{3}{4} of the total.

Question 15

A construction project requires 4564\frac{5}{6} tons of gravel for the foundation and 2382\frac{3}{8} tons for the driveway. If the supplier delivers 8148\frac{1}{4} tons total, how much excess gravel will be available for other uses?

  1. 17241\frac{7}{24} tons of excess gravel available
  2. 15241\frac{5}{24} tons of excess gravel available
  3. 11121\frac{1}{12} tons of excess gravel available
  4. 11241\frac{1}{24} tons of excess gravel available (correct answer)
Explanation: When you encounter word problems involving mixed numbers, you need to identify what operation to perform and then work systematically with fractions that have different denominators. This problem asks for excess gravel, which means you need to find the difference between what was delivered and what's needed. First, calculate the total gravel required: 456+2384\frac{5}{6} + 2\frac{3}{8}. To add these mixed numbers, convert to improper fractions or work with the parts separately. Finding a common denominator of 24: 42024+2924=62924=75244\frac{20}{24} + 2\frac{9}{24} = 6\frac{29}{24} = 7\frac{5}{24} tons needed. Next, subtract the required amount from what was delivered: 814−75248\frac{1}{4} - 7\frac{5}{24}. Convert 8148\frac{1}{4} to twenty-fourths: 86248\frac{6}{24}. Since 624<524\frac{6}{24} < \frac{5}{24}, you need to borrow from the whole number: 73024−7524=02524=11247\frac{30}{24} - 7\frac{5}{24} = 0\frac{25}{24} = 1\frac{1}{24} tons excess. Answer A (17241\frac{7}{24}) likely results from calculation errors when finding common denominators. Answer B (15241\frac{5}{24}) might come from incorrectly subtracting fractions without proper borrowing. Answer C (11121\frac{1}{12}) probably stems from using 12 instead of 24 as the common denominator and making computational mistakes. The correct answer is D: 11241\frac{1}{24} tons of excess gravel available. Study tip: With mixed number problems, always find the least common denominator first, then double-check your borrowing when subtracting fractions where the minuend's fraction is smaller than the subtrahend's.

Question 16

A recipe calls for 2342\frac{3}{4} cups of flour, but Janet only has 1561\frac{5}{6} cups available. After borrowing some flour from her neighbor, she now has 3183\frac{1}{8} cups total. How much flour did Janet borrow?

  1. 17241\frac{7}{24} cups of flour (correct answer)
  2. 15241\frac{5}{24} cups of flour
  3. 1131\frac{1}{3} cups of flour
  4. 111241\frac{11}{24} cups of flour
Explanation: To find how much Janet borrowed, subtract what she had from what she has now: 318−1563\frac{1}{8} - 1\frac{5}{6}. Converting to improper fractions: 258−116\frac{25}{8} - \frac{11}{6}. Using common denominator 24: 7524−4424=3124=1724\frac{75}{24} - \frac{44}{24} = \frac{31}{24} = 1\frac{7}{24}. Choice B results from incorrectly converting 1561\frac{5}{6} to twenty-fourths. Choice C uses the wrong common denominator of 6. Choice D comes from adding instead of subtracting the fractions.

Question 17

At a bakery, 58\frac{5}{8} of the morning's bread production was sold by noon, and 16\frac{1}{6} of the total production was sold between noon and 3 PM. If 112\frac{1}{12} of the original production remains unsold, how much of the total production was sold after 3 PM?

  1. 18\frac{1}{8} of the total production (correct answer)
  2. 16\frac{1}{6} of the total production
  3. 124\frac{1}{24} of the total production
  4. 14\frac{1}{4} of the total production
Explanation: Total sold = 1 - 112\frac{1}{12} (remaining) = 1112\frac{11}{12}. Sold by 3 PM = 58+16\frac{5}{8} + \frac{1}{6}. Converting to common denominator 24: 1524+424=1924\frac{15}{24} + \frac{4}{24} = \frac{19}{24}. Sold after 3 PM = 1112−1924=2224−1924=324=18\frac{11}{12} - \frac{19}{24} = \frac{22}{24} - \frac{19}{24} = \frac{3}{24} = \frac{1}{8}. Choice B incorrectly uses 16\frac{1}{6} as the answer. Choice C results from calculation errors with denominators. Choice D comes from using 624\frac{6}{24} instead of 324\frac{3}{24}.

Question 18

A stock's price started the day at \45\frac{1}{8}.Duringtheday,itroseby. During the day, it rose by $2\frac{3}{4}andthenfellbyand then fell by$3\frac{1}{2}$. What was the final price of the stock at the end of the day?

  1. \44\frac{3}{8}$ (correct answer)
  2. \46\frac{1}{8}$
  3. \47\frac{7}{8}$
  4. \51\frac{3}{8}$
Explanation: Start with the initial price and perform the two changes. The calculation is 4518+234−31245\frac{1}{8} + 2\frac{3}{4} - 3\frac{1}{2}. Find a common denominator of 8. The expression becomes 4518+268−34845\frac{1}{8} + 2\frac{6}{8} - 3\frac{4}{8}. First, add the rise in price: 4518+268=477845\frac{1}{8} + 2\frac{6}{8} = 47\frac{7}{8}. Then, subtract the fall in price: 4778−348=443847\frac{7}{8} - 3\frac{4}{8} = 44\frac{3}{8}. The final price is \44\frac{3}{8}$.

Question 19

What number must be added to 3583\frac{5}{8} to get a sum of 5145\frac{1}{4}?

  1. 1381\frac{3}{8}
  2. 1581\frac{5}{8} (correct answer)
  3. 2382\frac{3}{8}
  4. 8788\frac{7}{8}
Explanation: This problem requires finding the difference between the two numbers. The calculation is 514−3585\frac{1}{4} - 3\frac{5}{8}. First, find a common denominator, which is 8. The expression becomes 528−3585\frac{2}{8} - 3\frac{5}{8}. Since 28\frac{2}{8} is smaller than 58\frac{5}{8}, you need to regroup from the whole number 5. Rewrite 5285\frac{2}{8} as 41084\frac{10}{8}. Now subtract: 4108−358=1584\frac{10}{8} - 3\frac{5}{8} = 1\frac{5}{8}.

Question 20

What is the sum of 1781\frac{7}{8} and 2132\frac{1}{3} subtracted from 66?

  1. 45244\frac{5}{24}
  2. 21242\frac{1}{24}
  3. 119241\frac{19}{24} (correct answer)
  4. 419244\frac{19}{24}
Explanation: First, find the sum of 1781\frac{7}{8} and 2132\frac{1}{3}. The common denominator is 24. So, 12124+2824=329241\frac{21}{24} + 2\frac{8}{24} = 3\frac{29}{24}. This simplifies to 45244\frac{5}{24}. Next, subtract this sum from 6: 6−45246 - 4\frac{5}{24}. To subtract, rewrite 6 as 524245\frac{24}{24}. The problem becomes 52424−4524=119245\frac{24}{24} - 4\frac{5}{24} = 1\frac{19}{24}.