Adding & Subtracting Fractions - Accuplacer Arithmetic

A number is increased by $2\frac{1}{5}$ , and the result is then decreased by $\frac{3}{4}$ . If the final result is $4\frac{1}{2}$ , what was the original number?
Answer: $3\frac{1}{20}$ To find the original number, reverse the operations. Start with the final result, $4\frac{1}{2}$ , add back the $\frac{3}{4}$ that was subtracted, and then subtract the $2\frac{1}{5}$ that was added. The calculation is $4\frac{1}{2} + \frac{3}{4} - 2\frac{1}{5}$ . Find a common denominator, which is 20. Convert all numbers: $\frac{9}{2} + \frac{3}{4} - \frac{11}{5} = \frac{90}{20} + \frac{15}{20} - \frac{44}{20}$ . Perform the operations: $\frac{105}{20} - \frac{44}{20} = \frac{61}{20}$ . Convert back to a mixed number: $3\frac{1}{20}$ .
A package contains three items. Item A weighs $3\frac{1}{2}$ pounds, Item B weighs $2\frac{1}{3}$ pounds, and the total weight of the package is $7\frac{1}{4}$ pounds. What is the weight of Item C?
Answer: $1\frac{5}{12}$ pounds First, find the combined weight of Item A and Item B: $3\frac{1}{2} + 2\frac{1}{3}$ . Find a common denominator of 6: $3\frac{3}{6} + 2\frac{2}{6} = 5\frac{5}{6}$ pounds. To find the weight of Item C, subtract the combined weight of A and B from the total package weight: $7\frac{1}{4} - 5\frac{5}{6}$ . Find a common denominator of 12: $7\frac{3}{12} - 5\frac{10}{12}$ . To subtract, regroup from the 7: $6\frac{15}{12} - 5\frac{10}{12} = 1\frac{5}{12}$ pounds.
A recipe calls for $2\frac{3}{4}$ cups of flour, but Janet only has $1\frac{5}{6}$ cups available. After borrowing some flour from her neighbor, she now has $3\frac{1}{8}$ cups total. How much flour did Janet borrow?
Answer: $1\frac{7}{24}$ cups of flour To find how much Janet borrowed, subtract what she had from what she has now: $3\frac{1}{8} - 1\frac{5}{6}$ . Converting to improper fractions: $\frac{25}{8} - \frac{11}{6}$ . Using common denominator 24: $\frac{75}{24} - \frac{44}{24} = \frac{31}{24} = 1\frac{7}{24}$ . Choice B results from incorrectly converting $1\frac{5}{6}$ to twenty-fourths. Choice C uses the wrong common denominator of 6. Choice D comes from adding instead of subtracting the fractions.
Which of the following results in the smallest value?
Answer: $\frac{7}{8} - \frac{1}{2}$ Calculate the value of each expression. A) $2\frac{1}{2} - 1\frac{3}{4} = 2\frac{2}{4} - 1\frac{3}{4} = 1\frac{6}{4} - 1\frac{3}{4} = \frac{3}{4}$ . B) $1\frac{1}{3} - \frac{5}{6} = 1\frac{2}{6} - \frac{5}{6} = \frac{8}{6} - \frac{5}{6} = \frac{3}{6} = \frac{1}{2}$ . C) $\frac{7}{8} - \frac{1}{2} = \frac{7}{8} - \frac{4}{8} = \frac{3}{8}$ . D) $3 - 2\frac{1}{5} = 2\frac{5}{5} - 2\frac{1}{5} = \frac{4}{5}$ . Now compare the results: $\frac{3}{4}$ (or 0.75), $\frac{1}{2}$ (or 0.5), $\frac{3}{8}$ (or 0.375), and $\frac{4}{5}$ (or 0.8). The smallest value is $\frac{3}{8}$ .
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?
Answer: $\frac{3}{8}$ First, find the total fraction of students who play a sport by adding the fractions for soccer and basketball: $\frac{3}{8} + \frac{1}{4}$ . Find a common denominator of 8: $\frac{3}{8} + \frac{2}{8} = \frac{5}{8}$ . 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): $1 - \frac{5}{8}$ . Rewrite 1 as $\frac{8}{8}$ . So, $\frac{8}{8} - \frac{5}{8} = \frac{3}{8}$ .
Simplify the expression: $(4\frac{1}{3} - 1\frac{1}{2}) - (5\frac{1}{6} - 3\frac{3}{4})$
Answer: $1\frac{5}{12}$ First, evaluate the expression in the first set of parentheses: $4\frac{1}{3} - 1\frac{1}{2}$ . Common denominator is 6. $\frac{13}{3} - \frac{3}{2} = \frac{26}{6} - \frac{9}{6} = \frac{17}{6}$ . Next, evaluate the second set of parentheses: $5\frac{1}{6} - 3\frac{3}{4}$ . Common denominator is 12. $\frac{31}{6} - \frac{15}{4} = \frac{62}{12} - \frac{45}{12} = \frac{17}{12}$ . Finally, subtract the second result from the first: $\frac{17}{6} - \frac{17}{12}$ . Common denominator is 12. $\frac{34}{12} - \frac{17}{12} = \frac{17}{12}$ , which is $1\frac{5}{12}$ .
A board is $8\frac{1}{4}$ feet long. A piece that is $2\frac{2}{3}$ feet long is cut from it. Then, another piece that is $3\frac{1}{2}$ feet long is cut from the remaining part. What is the length of the final piece of the board?
Answer: $2\frac{1}{12}$ feet First, find the total length of the two pieces cut from the board: $2\frac{2}{3} + 3\frac{1}{2}$ . The common denominator is 6. So, $2\frac{4}{6} + 3\frac{3}{6} = 5\frac{7}{6}$ , which simplifies to $6\frac{1}{6}$ feet. Next, subtract this total length from the original length of the board: $8\frac{1}{4} - 6\frac{1}{6}$ . The common denominator is 12. So, $8\frac{3}{12} - 6\frac{2}{12} = 2\frac{1}{12}$ feet. The remaining piece is $2\frac{1}{12}$ feet long.
What is the result of $10 - (3\frac{1}{4} + 2\frac{1}{2})$ ?
Answer: $4\frac{1}{4}$ According to the order of operations, first calculate the sum inside the parentheses: $3\frac{1}{4} + 2\frac{1}{2}$ . Convert to a common denominator of 4: $3\frac{1}{4} + 2\frac{2}{4} = 5\frac{3}{4}$ . Now, subtract this result from 10: $10 - 5\frac{3}{4}$ . To subtract, you can rewrite 10 as $9\frac{4}{4}$ . Then, $9\frac{4}{4} - 5\frac{3}{4} = 4\frac{1}{4}$ .
At a bakery, $\frac{5}{8}$ of the morning's bread production was sold by noon, and $\frac{1}{6}$ of the total production was sold between noon and 3 PM. If $\frac{1}{12}$ of the original production remains unsold, how much of the total production was sold after 3 PM?
Answer: $\frac{1}{8}$ of the total production Total sold = 1 - $\frac{1}{12}$ (remaining) = $\frac{11}{12}$ . Sold by 3 PM = $\frac{5}{8} + \frac{1}{6}$ . Converting to common denominator 24: $\frac{15}{24} + \frac{4}{24} = \frac{19}{24}$ . Sold after 3 PM = $\frac{11}{12} - \frac{19}{24} = \frac{22}{24} - \frac{19}{24} = \frac{3}{24} = \frac{1}{8}$ . Choice B incorrectly uses $\frac{1}{6}$ as the answer. Choice C results from calculation errors with denominators. Choice D comes from using $\frac{6}{24}$ instead of $\frac{3}{24}$ .
What is the difference between the sum of $\frac{3}{4}$ and $\frac{2}{3}$ , and the sum of $\frac{1}{2}$ and $\frac{1}{6}$ ?
Answer: $\frac{3}{4}$ This is a multi-step problem. First, find the first sum: $\frac{3}{4} + \frac{2}{3}$ . The common denominator is 12. $\frac{9}{12} + \frac{8}{12} = \frac{17}{12}$ . Second, find the second sum: $\frac{1}{2} + \frac{1}{6}$ . The common denominator is 6. $\frac{3}{6} + \frac{1}{6} = \frac{4}{6}$ , which simplifies to $\frac{2}{3}$ . Finally, find the difference between the two sums: $\frac{17}{12} - \frac{2}{3}$ . The common denominator is 12. $\frac{17}{12} - \frac{8}{12} = \frac{9}{12}$ , which simplifies to $\frac{3}{4}$ .
A construction project requires $4\frac{5}{6}$ tons of gravel for the foundation and $2\frac{3}{8}$ tons for the driveway. If the supplier delivers $8\frac{1}{4}$ tons total, how much excess gravel will be available for other uses?
Answer: $1\frac{1}{24}$ tons of excess gravel available 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: $4\frac{5}{6} + 2\frac{3}{8}$ . To add these mixed numbers, convert to improper fractions or work with the parts separately. Finding a common denominator of 24: $4\frac{20}{24} + 2\frac{9}{24} = 6\frac{29}{24} = 7\frac{5}{24}$ tons needed. Next, subtract the required amount from what was delivered: $8\frac{1}{4} - 7\frac{5}{24}$ . Convert $8\frac{1}{4}$ to twenty-fourths: $8\frac{6}{24}$ . Since $\frac{6}{24} < \frac{5}{24}$ , you need to borrow from the whole number: $7\frac{30}{24} - 7\frac{5}{24} = 0\frac{25}{24} = 1\frac{1}{24}$ tons excess. Answer A ( $1\frac{7}{24}$ ) likely results from calculation errors when finding common denominators. Answer B ( $1\frac{5}{24}$ ) might come from incorrectly subtracting fractions without proper borrowing. Answer C ( $1\frac{1}{12}$ ) probably stems from using 12 instead of 24 as the common denominator and making computational mistakes. The correct answer is D: $1\frac{1}{24}$ 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.
Three friends are sharing a pizza. The first friend eats $\frac{1}{3}$ of the pizza. The second friend eats $\frac{1}{4}$ of the pizza. What fraction of the pizza is left for the third friend?
Answer: $\frac{5}{12}$ First, find the total fraction of the pizza eaten by the first two friends by adding their shares: $\frac{1}{3} + \frac{1}{4}$ . The common denominator is 12. So, $\frac{4}{12} + \frac{3}{12} = \frac{7}{12}$ . To find the fraction remaining, subtract the eaten portion from the whole pizza (1): $1 - \frac{7}{12}$ . Rewrite 1 as $\frac{12}{12}$ . The subtraction is $\frac{12}{12} - \frac{7}{12} = \frac{5}{12}$ .
From a rope $15\frac{1}{2}$ feet long, a piece of $4\frac{1}{3}$ feet and another piece of $5\frac{1}{4}$ feet are cut off. What is the length of the remaining rope?
Answer: $5\frac{11}{12}$ feet First, calculate the total length of the rope that was cut off: $4\frac{1}{3} + 5\frac{1}{4}$ . Find a common denominator of 12: $4\frac{4}{12} + 5\frac{3}{12} = 9\frac{7}{12}$ feet. Next, subtract this total from the original length of the rope: $15\frac{1}{2} - 9\frac{7}{12}$ . Convert to a common denominator of 12: $15\frac{6}{12} - 9\frac{7}{12}$ . Regroup from the 15: $14\frac{18}{12} - 9\frac{7}{12} = 5\frac{11}{12}$ feet.
A carpenter is joining two boards together. One board is $1\frac{3}{4}$ inches thick and the other is $\frac{7}{8}$ of an inch thick. To make them fit, the carpenter shaves $\frac{1}{16}$ of an inch off the thicker board. What is the combined thickness of the two boards after the adjustment?
Answer: $2\frac{9}{16}$ inches First, identify the thicker board. Convert $1\frac{3}{4}$ to eighths: $1\frac{6}{8}$ . Since $1\frac{6}{8} > \frac{7}{8}$ , the first board is thicker. Shave $\frac{1}{16}$ inch off this board: $1\frac{3}{4} - \frac{1}{16}$ . Convert to a common denominator of 16: $1\frac{12}{16} - \frac{1}{16} = 1\frac{11}{16}$ inches. Now, find the combined thickness by adding the adjusted thickness of the first board to the thickness of the second board: $1\frac{11}{16} + \frac{7}{8}$ . Convert to a common denominator of 16: $1\frac{11}{16} + \frac{14}{16} = 1\frac{25}{16}$ . Simplify the improper fraction: $1 + 1\frac{9}{16} = 2\frac{9}{16}$ inches.
A recipe for fruit punch calls for $1\frac{2}{3}$ cups of orange juice and $2\frac{1}{4}$ cups of pineapple juice. If Lisa wants to make $\frac{3}{4}$ of the original recipe, how much total juice will she need?
Answer: $2\frac{15}{16}$ cups of total juice needed Original total juice: $1\frac{2}{3} + 2\frac{1}{4} = 1\frac{8}{12} + 2\frac{3}{12} = 3\frac{11}{12}$ . Amount needed for $\frac{3}{4}$ recipe: $3\frac{11}{12} \times \frac{3}{4} = \frac{47}{12} \times \frac{3}{4} = \frac{141}{48} = 2\frac{45}{48} = 2\frac{15}{16}$ . Choice A shows an incorrect mixed number conversion. Choice B results from computational errors in the multiplication. Choice D adds instead of finding $\frac{3}{4}$ of the total.
A stock's price started the day at $\$45\frac{1}{8}$ . During the day, it rose by $\$2\frac{3}{4}$ and then fell by $\$3\frac{1}{2}$ . What was the final price of the stock at the end of the day?
Answer: $\$44\frac{3}{8}$ Start with the initial price and perform the two changes. The calculation is $45\frac{1}{8} + 2\frac{3}{4} - 3\frac{1}{2}$ . Find a common denominator of 8. The expression becomes $45\frac{1}{8} + 2\frac{6}{8} - 3\frac{4}{8}$ . First, add the rise in price: $45\frac{1}{8} + 2\frac{6}{8} = 47\frac{7}{8}$ . Then, subtract the fall in price: $47\frac{7}{8} - 3\frac{4}{8} = 44\frac{3}{8}$ . The final price is $\$44\frac{3}{8}$ .
A carpenter cut $2\frac{5}{12}$ feet from a board, then cut another $1\frac{3}{8}$ feet from the same board. If $6\frac{3}{4}$ feet of the original board remains, what was the length of the original board?
Answer: $10\frac{13}{24}$ feet was the original length Add all the pieces together: cuts of $2\frac{5}{12} + 1\frac{3}{8}$ plus remaining $6\frac{3}{4}$ . Converting to common denominator 24: $2\frac{10}{24} + 1\frac{9}{24} + 6\frac{18}{24} = 9\frac{37}{24} = 10\frac{13}{24}$ . Choice A results from miscalculating $\frac{37}{24}$ as $\frac{31}{24}$ . Choice C uses the wrong common denominator and gets $9\frac{5}{6}$ . Choice D rounds incorrectly to $10\frac{1}{2}$ .
On a map, the distance between two cities is $5\frac{1}{4}$ inches. If a traveler has already covered a distance of $2\frac{1}{2}$ inches on the map and then travels another $1\frac{1}{3}$ inches, what is the remaining distance in inches on the map?
Answer: $1\frac{5}{12}$ inches First, find the total distance traveled on the map by adding the two segments: $2\frac{1}{2} + 1\frac{1}{3}$ . Find a common denominator of 6: $2\frac{3}{6} + 1\frac{2}{6} = 3\frac{5}{6}$ inches. Next, subtract the total distance traveled from the total distance between the cities: $5\frac{1}{4} - 3\frac{5}{6}$ . Find a common denominator of 12: $5\frac{3}{12} - 3\frac{10}{12}$ . Regroup from the 5: $4\frac{15}{12} - 3\frac{10}{12} = 1\frac{5}{12}$ inches.
A baker starts with 10 cups of sugar. He uses $2\frac{1}{3}$ cups for cookies and $3\frac{3}{4}$ cups for a cake. He then adds $1\frac{1}{2}$ cups of sugar back to his supply from another bag. How much sugar does he have now?
Answer: $5\frac{5}{12}$ cups This problem can be represented by the expression $10 - 2\frac{1}{3} - 3\frac{3}{4} + 1\frac{1}{2}$ . First, calculate the total sugar used: $2\frac{1}{3} + 3\frac{3}{4}$ . The common denominator is 12: $2\frac{4}{12} + 3\frac{9}{12} = 5\frac{13}{12} = 6\frac{1}{12}$ cups. Now, subtract this from the initial 10 cups: $10 - 6\frac{1}{12} = 3\frac{11}{12}$ cups. Finally, add the sugar that was put back: $3\frac{11}{12} + 1\frac{1}{2}$ . The common denominator is 12: $3\frac{11}{12} + 1\frac{6}{12} = 4\frac{17}{12} = 5\frac{5}{12}$ cups.
Calculate: $5\frac{1}{6} - 2\frac{3}{4} + 1\frac{1}{3}$
Answer: $3\frac{3}{4}$ To solve this, convert all mixed numbers to improper fractions and find a common denominator, which is 12. The expression is $\frac{31}{6} - \frac{11}{4} + \frac{4}{3}$ . Convert the fractions: $\frac{62}{12} - \frac{33}{12} + \frac{16}{12}$ . Perform the operations from left to right: $\frac{62 - 33}{12} + \frac{16}{12} = \frac{29}{12} + \frac{16}{12} = \frac{45}{12}$ . Simplify the result by dividing the numerator and denominator by 3: $\frac{15}{4}$ . Convert back to a mixed number: $3\frac{3}{4}$ .

Start this quiz