Prealgebra Quiz: Adding And Subtracting Fractions
4 questions · exam conditions
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Adding And Subtracting FractionsQuestion 1 of 4

A carpenter cuts a board into three pieces. The first piece is 25\frac{2}{5} of the original length, and the second piece is 13\frac{1}{3} of the original length. If 115\frac{1}{15} of the original board length was lost to sawdust during cutting, what fraction of the original length is the third piece?

15\frac{1}{5} of the original length
415\frac{4}{15} of the original length
115\frac{1}{15} of the original length
215\frac{2}{15} of the original length
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Prealgebra Quiz

Prealgebra Quiz: Adding And Subtracting Fractions

Practice Adding And Subtracting Fractions in Prealgebra 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 Prealgebra.

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 carpenter cuts a board into three pieces. The first piece is 25\frac{2}{5} of the original length, and the second piece is 13\frac{1}{3} of the original length. If 115\frac{1}{15} of the original board length was lost to sawdust during cutting, what fraction of the original length is the third piece?

  1. 15\frac{1}{5} of the original length (correct answer)
  2. 415\frac{4}{15} of the original length
  3. 115\frac{1}{15} of the original length
  4. 215\frac{2}{15} of the original length
Explanation: The total must equal 1 (the whole board). We have: first piece + second piece + third piece + sawdust = 1. So: 25+13+third piece+115=1\frac{2}{5} + \frac{1}{3} + \text{third piece} + \frac{1}{15} = 1. Converting to common denominator 15: 615+515+third piece+115=1\frac{6}{15} + \frac{5}{15} + \text{third piece} + \frac{1}{15} = 1. This gives us: 1215+third piece=1\frac{12}{15} + \text{third piece} = 1. Therefore: third piece = 1−1215=1515−1215=315=151 - \frac{12}{15} = \frac{15}{15} - \frac{12}{15} = \frac{3}{15} = \frac{1}{5}. Choice A is correct. Choice B (415\frac{4}{15}) would result if students forgot to account for sawdust. Choice C (115\frac{1}{15}) might result from calculation errors. Choice D (215\frac{2}{15}) could result from misreading the sawdust amount.

Question 2

A garden sprinkler system has three zones. Zone A uses 38\frac{3}{8} of the total daily water allocation, Zone B uses 16\frac{1}{6} of the total allocation, and Zone C uses the remaining water. If the system malfunctions and Zone A uses an additional 112\frac{1}{12} of the total daily allocation, what fraction of the total daily allocation does Zone C now receive?

  1. 38\frac{3}{8} of the allocation (correct answer)
  2. 13\frac{1}{3} of the allocation
  3. 512\frac{5}{12} of the allocation
  4. 14\frac{1}{4} of the allocation
Explanation: Originally, Zone C gets: 1−38−161 - \frac{3}{8} - \frac{1}{6}. Converting to common denominator 24: 1−924−424=24−9−424=11241 - \frac{9}{24} - \frac{4}{24} = \frac{24-9-4}{24} = \frac{11}{24}. After malfunction, Zone A uses 38+112=924+224=1124\frac{3}{8} + \frac{1}{12} = \frac{9}{24} + \frac{2}{24} = \frac{11}{24}. Zone B still uses 16=424\frac{1}{6} = \frac{4}{24}. Zone C now gets: 1−1124−424=924=381 - \frac{11}{24} - \frac{4}{24} = \frac{9}{24} = \frac{3}{8}. Choice B (13=824\frac{1}{3} = \frac{8}{24}) might result from minor calculation errors. Choice C (512=1024\frac{5}{12} = \frac{10}{24}) could result from not accounting for the malfunction properly. Choice D (14=624\frac{1}{4} = \frac{6}{24}) might result from various computational mistakes.

Question 3

A tank is 58\frac{5}{8} full of water. After using 16\frac{1}{6} of the tank's total capacity for irrigation, rain adds water equal to 112\frac{1}{12} of the tank's total capacity. What fraction of the tank is now filled with water?

  1. 712\frac{7}{12} of the tank
  2. 1324\frac{13}{24} of the tank (correct answer)
  3. 1124\frac{11}{24} of the tank
  4. 512\frac{5}{12} of the tank
Explanation: Starting with 58\frac{5}{8} of the tank full, then subtracting 16\frac{1}{6} (used for irrigation) and adding 112\frac{1}{12} (from rain): 58−16+112\frac{5}{8} - \frac{1}{6} + \frac{1}{12}. Converting to common denominator 24: 1524−424+224=15−4+224=1324\frac{15}{24} - \frac{4}{24} + \frac{2}{24} = \frac{15-4+2}{24} = \frac{13}{24}. Choice A (712=1424\frac{7}{12} = \frac{14}{24}) might result from computational errors. Choice C (1124\frac{11}{24}) could result from adding instead of subtracting the irrigation amount. Choice D (512=1024\frac{5}{12} = \frac{10}{24}) might result from multiple calculation errors.

Question 4

A recipe for trail mix calls for 35\frac{3}{5} cup of nuts and 14\frac{1}{4} cup of dried fruit. Lisa wants to make 23\frac{2}{3} of this recipe, but she accidentally adds 13\frac{1}{3} cup of nuts instead of the correct amount. How much more nuts does she need to add to have the right amount for 23\frac{2}{3} of the recipe?

  1. 115\frac{1}{15} cup more (correct answer)
  2. 215\frac{2}{15} cup more
  3. 15\frac{1}{5} cup more
  4. 415\frac{4}{15} cup more
Explanation: For 23\frac{2}{3} of the recipe, Lisa needs 23×35=615=25\frac{2}{3} \times \frac{3}{5} = \frac{6}{15} = \frac{2}{5} cup of nuts. She added 13=515\frac{1}{3} = \frac{5}{15} cup. The correct amount needed is 615\frac{6}{15} cup. Additional nuts needed: 615−515=115\frac{6}{15} - \frac{5}{15} = \frac{1}{15} cup. Choice B (215\frac{2}{15}) might result from calculation errors. Choice C (15=315\frac{1}{5} = \frac{3}{15}) could result from finding the difference between the full recipe amount and what she added. Choice D (415\frac{4}{15}) might result from various computational mistakes.