📝 Add and Subtract Fractions in Algebra (7 MCQs)

📖 From Digital SAT Algebra • 1. Basics of Algebra • 7 questions available

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What is Add and Subtract Fractions in Algebra?

Definition:
Adding and subtracting fractions in algebra involves combining fractions by ensuring they have a common denominator, then adding or subtracting the numerators while keeping the denominator unchanged, and finally simplifying the result, which applies to both numeric fractions and algebraic fractions with variables.

Working:
Find a common denominator (usually the LCM of denominators), rewrite each fraction as an equivalent fraction with that denominator, then add/subtract the numerators; for algebraic fractions, factor denominators to find LCD, and combine like terms in the numerator.

Example:
Add $\frac{2}{5} + \frac{1}{3}$ .
Solution: LCM of 5 and 3 is 15; rewrite $\frac{2}{5} = \frac{6}{15}$ , $\frac{1}{3} = \frac{5}{15}$ ; sum $\frac{6+5}{15} = \frac{11}{15}$ .

Reason:
This operation is foundational for solving equations with fractions, combining rational expressions, and simplifying complex algebraic fractions, ensuring accurate results in many mathematical and practical applications.

Easy

Medium

Hard

📝 All Add and Subtract Fractions in Algebra MCQs

Q1. A student simplifies $\frac{5}{6}-\frac{1}{4}$ by subtracting numerators and denominators separately, obtaining $\frac{4}{2}$ . Which expression correctly represents the result and why?

\frac{2}{12}

, because denominators are multiplied

\frac{7}{12}

, because the fractions must first be rewritten with a common denominator ✅

\frac{4}{2}

, because both numerators and denominators can be subtracted

\frac{1}{2}

, because the denominators cancel

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The denominators are different, so the fractions must be rewritten using a common denominator. Since $\frac{5}{6}=\frac{10}{12}$ and $\frac{1}{4}=\frac{3}{12}$ , subtraction gives $\frac{7}{12}$ . The incorrect choices reflect common errors involving separate numerator-denominator operations or cancellation.

Q2. Two students calculate $\frac{3}{5}+\frac{7}{10}$ . Student A obtains $\frac{10}{15}$ , while Student B obtains $\frac{13}{10}$ . Which evaluation is correct?

A.Student A, because numerators and denominators should both be added

B.Student B, because both fractions can be expressed with denominator

10

before adding ✅

C.Both students, because different equivalent forms are possible

D.Neither, because fractions cannot be added when their denominators differ

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Student B is correct because $\frac{3}{5}=\frac{6}{10}$ , so $\frac{6}{10}+\frac{7}{10}=\frac{13}{10}$ . Student A demonstrates the misconception of adding numerators and denominators directly rather than creating equivalent fractions with a common denominator.

Q3. A recipe requires $\frac{3}{4}$ cup of flour and $\frac{2}{3}$ cup of another ingredient. If a measuring container holds exactly $\frac{1}{6}$ cup, how many full container portions are needed to measure the combined amount?

A.7 portions

B.8 portions

C.9 portions ✅

D.10 portions

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: First add the amounts: $\frac{3}{4}+\frac{2}{3}=\frac{9}{12}+\frac{8}{12}=\frac{17}{12}$ cups. Dividing by $\frac{1}{6}$ gives $\frac{17}{12}\times6=\frac{17}{2}=8.5$ . Therefore, $9$ full portions are required to measure at least the total amount.

Q4. A student claims that $\frac{7}{8}-\frac{2}{3}=\frac{5}{5}=1$ . Which single change best explains the mistake?

A.The student should have multiplied the numerators instead of subtracting them

B.The student treated the denominators as if they could be subtracted directly ✅

C.The student should have changed both fractions to improper fractions

D.The student forgot that subtraction always produces a negative result

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The student incorrectly subtracts denominators as well as numerators. A common denominator is required: $\frac{7}{8}=\frac{21}{24}$ and $\frac{2}{3}=\frac{16}{24}$ , giving $\frac{5}{24}$ . The reasoning fails because denominators describe equal-sized parts and cannot be directly subtracted.

Q5. A number line has points $A=-\frac{3}{4}$ , $B=-\frac{1}{4}$ , $C=\frac{1}{4}$ , and $D=\frac{3}{4}$ . Which expression represents moving from $A$ to $D$ by first moving from $A$ to $B$ and then from $B$ to $D$ ?

-\frac{3}{4}+\frac{1}{2}+\frac{1}{2}=\frac{1}{4}

-\frac{3}{4}+\frac{1}{2}+1=\frac{3}{4}

✅

\frac{3}{4}-\frac{1}{4}+\frac{1}{2}=1

-\frac{3}{4}-\frac{1}{2}+1=\frac{3}{4}

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The movement from $A$ to $B$ is $+\frac{1}{2}$ , while the movement from $B$ to $D$ is $+1$ . Thus $-\frac{3}{4}+\frac{1}{2}+1=\frac{3}{4}$ . This requires interpreting directed movement on a number line rather than simply combining fraction symbols.

Q6. A store records a morning balance change of $-\frac{5}{6}$ thousand dollars and an afternoon change of $\frac{3}{4}$ thousand dollars. A manager says the total change is $-\frac{2}{10}$ thousand dollars because $5-3=2$ and $6-4=2$ . What is the correct total change?

-\frac{1}{12}

thousand dollars ✅

\frac{1}{12}

thousand dollars

-\frac{2}{10}

thousand dollars

\frac{19}{12}

thousand dollars

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The signed changes must be combined using a common denominator. Since $-\frac{5}{6}=-\frac{10}{12}$ and $\frac{3}{4}=\frac{9}{12}$ , their sum is $-\frac{1}{12}$ . The manager's method incorrectly subtracts denominators and ignores the need to preserve the signs.

Q7. Suppose $x=\frac{2}{3}$ and $y=\frac{5}{8}$ . A student evaluates $x-y+\frac{1}{4}$ as $\frac{2}{3}-\frac{5}{8}+\frac{1}{4}$ and gets $\frac{7}{24}$ . Which conclusion is correct?

A.The result is correct because all denominators were included ✅

B.The result is incorrect; the correct value is

\frac{5}{24}

C.The result is incorrect; the correct value is

\frac{11}{24}

D.The result is correct because

\frac{1}{4}

cancels part of

\frac{5}{8}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Using denominator $24$ , $\frac{2}{3}=\frac{16}{24}$ , $\frac{5}{8}=\frac{15}{24}$ , and $\frac{1}{4}=\frac{6}{24}$ . Therefore $\frac{16}{24}-\frac{15}{24}+\frac{6}{24}=\frac{7}{24}$ , so the student's stated result is actually correct. However, among the listed conclusions, none identifies that correctly; therefore the question exposes an inconsistency and should be revised before assessment use.

🔗 Related Topics (MCQs)

📝 Place value of whole numbers MCQs 📝 How to round whole numbers MCQs 📝 Identify Multiples and divisibility Tests and rules MCQs 📝 Prime factorization and (LCM) Least Common Multiples MCQs 📝 Variables and algebraic symbols MCQs

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