All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 7 Subtracting Fractions with Unlike Denominators are as per the latest syllabus guidelines.
McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 7 Subtracting Fractions with Unlike Denominators
Solve
Find equivalent fractions and subtract. Write the difference in simplest form.
Question 1.
\(\frac{2}{9}\) – \(\frac{1}{6}\) _____________
Answer:
\(\frac{4}{18}\) – \(\frac{3}{18}\) = \(\frac{1}{18}\)
Question 2.
\(\frac{1}{2}\) – \(\frac{1}{3}\) _______________
Answer:
\(\frac{3}{6}\)–\(\frac{2}{6}\) = \(\frac{1}{6}\).
Explanation:
The equivalent fraction of \(\frac{1}{2}\) – \(\frac{1}{3}\) which is \(\frac{3}{6}\) and \(\frac{2}{6}\). So the addition of \(\frac{3}{6}\)–\(\frac{2}{6}\) is \(\frac{3-2}{6}\)
= \(\frac{1}{6}\).
Question 3.
\(\frac{3}{8}\) – \(\frac{1}{3}\) _______________
Answer:
\(\frac{3}{6}\)–\(\frac{2}{6}\) = \(\frac{1}{6}\).
Explanation:
The equivalent fraction of \(\frac{1}{2}\) – \(\frac{1}{3}\) which is \(\frac{3}{6}\) and \(\frac{2}{6}\). So the addition of \(\frac{3}{6}\)–\(\frac{2}{6}\) is \(\frac{3-2}{6}\)
= \(\frac{1}{6}\).
Question 4.
\(\frac{3}{4}\) – \(\frac{1}{5}\) _______________
Answer:
\(\frac{15}{20}\)–\(\frac{4}{20}\) = \(\frac{11}{20}\).
Explanation:
The equivalent fraction of \(\frac{3}{4}\) – \(\frac{1}{5}\) which is \(\frac{15}{20}\) and \(\frac{4}{20}\). So the addition of \(\frac{15}{20}\)–\(\frac{4}{20}\) is \(\frac{15-4}{20}\)
= \(\frac{11}{20}\).
Question 5.
Answer:
\(\frac{7}{14}\)–\(\frac{6}{14}\) = \(\frac{1}{14}\).
Explanation:
The equivalent fraction of \(\frac{1}{2}\) – \(\frac{3}{7}\) which is \(\frac{7}{14}\) and \(\frac{6}{14}\). So the addition of \(\frac{7}{14}\)–\(\frac{6}{14}\) is \(\frac{7-6}{14}\)
= \(\frac{1}{14}\).
Question 6.
Answer:
\(\frac{8}{12}\)–\(\frac{3}{12}\) = \(\frac{5}{12}\).
Explanation:
The equivalent fraction of \(\frac{2}{3}\) – \(\frac{1}{4}\) which is \(\frac{8}{12}\) and \(\frac{3}{12}\). So the addition of \(\frac{8}{12}\)–\(\frac{3}{12}\) is \(\frac{8-3}{12}\)
= \(\frac{5}{12}\).
Question 7.
Answer:
\(\frac{11}{99}\)–\(\frac{9}{99}\) = \(\frac{2}{99}\).
Explanation:
The equivalent fraction of \(\frac{1}{9}\) – \(\frac{1}{11}\) which is \(\frac{11}{99}\) and \(\frac{9}{99}\). So the addition of \(\frac{11}{99}\)–\(\frac{9}{99}\) is \(\frac{11-9}{99}\)
= \(\frac{2}{99}\).
Question 8.
Answer:
\(\frac{7}{42}\)–\(\frac{6}{42}\) = \(\frac{1}{42}\).
Explanation:
The equivalent fraction of \(\frac{1}{6}\) – \(\frac{1}{7}\) which is \(\frac{7}{42}\) and \(\frac{6}{42}\). So the addition of \(\frac{7}{42}\)–\(\frac{6}{42}\) is \(\frac{7-6}{42}\)
= \(\frac{1}{42}\).