McGraw Hill Math Grade 5 Chapter 6 Lesson 8 Answer Key Subtracting Mixed Numbers with Unlike Denominators

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All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 8 Subtracting Mixed Numbers with Unlike Denominators are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 8 Subtracting Mixed Numbers with Unlike Denominators

Solve

Find equivalent fractions and subtract. Simplify improper fractions.

Question 1.
1\(\frac{1}{4}\) – \(\frac{1}{6}\) ______________
Answer:
1\(\frac{3}{12}\) – \(\frac{2}{12}\) = 1\(\frac{1}{12}\)

Question 2.
1\(\frac{2}{3}\) – \(\frac{1}{2}\) ______________
Answer:
1\(\frac{2}{3}\) – \(\frac{1}{2}\) = \(\frac{7}{6}\).

Explanation:
The difference between 1\(\frac{2}{3}\) – \(\frac{1}{2}\) is \(\frac{5}{3}\) – \(\frac{1}{2}\)
= \(\frac{10-3}{6}\)
= \(\frac{7}{6}\).

Question 3.
1\(\frac{7}{8}\) – 1\(\frac{3}{6}\) ______________
Answer:
1\(\frac{7}{8}\) – 1\(\frac{3}{6}\) = \(\frac{3}{8}\).

Explanation:
The difference between 1\(\frac{7}{8}\) – 1\(\frac{3}{6}\) is \(\frac{15}{8}\) – \(\frac{9}{6}\)
= \(\frac{90-72}{48}\)
= \(\frac{18}{48}\)
= \(\frac{3}{8}\).

Question 4.
Answer:

Question 5.
2\(\frac{1}{9}\) – \(\frac{5}{6}\) _______________
Answer:
2\(\frac{1}{9}\) – \(\frac{5}{6}\) = \(\frac{23}{18}\).

Explanation:
The difference between 2\(\frac{1}{9}\) – \(\frac{5}{6}\) is \(\frac{19}{9}\) – \(\frac{5}{6}\)
= \(\frac{114-45}{54}\)
= \(\frac{69}{54}\)
= \(\frac{23}{18}\).

Question 6.
3\(\frac{1}{2}\) – 2\(\frac{1}{5}\) _______________
Answer:
3\(\frac{1}{2}\) – 2\(\frac{1}{5}\) = \(\frac{24}{5}\).

Explanation:
The difference between 3\(\frac{1}{2}\) – 2\(\frac{1}{5}\) is \(\frac{7}{2}\) – \(\frac{11}{5}\)
= \(\frac{70-22}{10}\)
= \(\frac{48}{10}\)
= \(\frac{24}{5}\).

Question 7.
McGraw Hill Math Grade 5 Chapter 6 Lesson 8 Answer Key Subtracting Mixed Numbers with Unlike Denominators 33
Answer:
3\(\frac{1}{2}\) – 1\(\frac{3}{7}\) = \(\frac{29}{14}\).

Explanation:
The difference between 3\(\frac{1}{2}\) – 1\(\frac{3}{7}\) is \(\frac{7}{2}\) – \(\frac{10}{7}\)
= \(\frac{49-20}{14}\)
= \(\frac{29}{14}\).

Question 8.
McGraw Hill Math Grade 5 Chapter 6 Lesson 8 Answer Key Subtracting Mixed Numbers with Unlike Denominators 34
Answer:
2\(\frac{1}{3}\) – 1\(\frac{2}{5}\) = \(\frac{14}{15}\).

Explanation:
The difference between 2\(\frac{1}{3}\) – 1\(\frac{2}{5}\) is \(\frac{7}{3}\) – \(\frac{7}{5}\)
= \(\frac{35-21}{15}\)
= \(\frac{14}{15}\).

Question 9.
McGraw Hill Math Grade 5 Chapter 6 Lesson 8 Answer Key Subtracting Mixed Numbers with Unlike Denominators 35
Answer:
5\(\frac{6}{7}\) – 4\(\frac{5}{6}\) = \(\frac{43}{42}\).

Explanation:
The difference between 5\(\frac{6}{7}\) – 4\(\frac{5}{6}\) is \(\frac{41}{7}\) – \(\frac{29}{6}\)
= \(\frac{246-203}{42}\)
= \(\frac{43}{42}\).

Question 10.
McGraw Hill Math Grade 5 Chapter 6 Lesson 8 Answer Key Subtracting Mixed Numbers with Unlike Denominators 36
Answer:
6\(\frac{7}{8}\) – 1\(\frac{2}{3}\) = \(\frac{121}{24}\).

Explanation:
The difference between 6\(\frac{7}{8}\) – 1\(\frac{2}{3}\) is \(\frac{55}{8}\) – \(\frac{5}{3}\)
= \(\frac{165-40}{24}\)
= \(\frac{121}{24}\).

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