McGraw Hill Math Grade 5 Chapter 6 Lesson 5 Answer Key Adding Fractions with Unlike Denominators

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

McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 5 Adding Fractions with Unlike Denominators

Solve

Find equivalent fractions and then add. Change improper fractions to mixed numbers.

Question 1.
\(\frac{1}{4}\) + \(\frac{1}{5}\) ______________
Answer:
\(\frac{5}{20}\) + \(\frac{4}{20}\) = \(\frac{9}{20}\)

Question 2.
\(\frac{2}{9}\) + \(\frac{1}{6}\) _______________
Answer:
\(\frac{2}{9}\) + \(\frac{1}{6}\) = \(\frac{11}{18}\).

Explanation:
To get the equivalent fraction we will multiply both numerator and denominator by 2 which is \(\frac{4}{18}\) and we will multiply other addend by 3 which is \(\frac{3}{18}\). So the addition of \(\frac{4}{18}\) + \(\frac{3}{18}\)
= \(\frac{4+7}{18}\)
= \(\frac{11}{18}\).

Question 3.
\(\frac{3}{8}\) + \(\frac{1}{3}\) ________________
Answer:
\(\frac{3}{8}\) + \(\frac{1}{3}\) = \(\frac{11}{18}\).

Explanation:
To get the equivalent fraction we will multiply \(\frac{3}{8}\) both numerator and denominator by 3 which is \(\frac{9}{24}\) and we will multiply other addend \(\frac{1}{3}\) by 8 which is \(\frac{8}{24}\). So the addition of \(\frac{9}{24}\) + \(\frac{8}{24}\)
= \(\frac{9+8}{24}\)
= \(\frac{17}{24}\).

Add. Change any improper fractions to mixed numbers.

Question 4.
McGraw Hill Math Grade 5 Chapter 6 Lesson 5 Answer Key Adding Fractions with Unlike Denominators 21
Answer:
\(\frac{1}{2}\) + \(\frac{3}{7}\) = \(\frac{13}{14}\).

Explanation:
By adding \(\frac{1}{2}\) + \(\frac{3}{7}\) the sum will be \(\frac{7+6}{14}\)
= \(\frac{13}{14}\).

Question 5.
McGraw Hill Math Grade 5 Chapter 6 Lesson 5 Answer Key Adding Fractions with Unlike Denominators 22
Answer:
\(\frac{2}{3}\) + \(\frac{3}{5}\) = \(\frac{16}{15}\).

Explanation:
By adding \(\frac{2}{3}\) + \(\frac{3}{5}\) the sum will be \(\frac{10+6}{15}\)
= \(\frac{16}{15}\).

Question 6.
McGraw Hill Math Grade 5 Chapter 6 Lesson 5 Answer Key Adding Fractions with Unlike Denominators 23
Answer:
\(\frac{1}{7}\) + \(\frac{1}{8}\) = \(\frac{15}{56}\).

Explanation:
By adding \(\frac{1}{7}\) + \(\frac{1}{8}\) the sum will be \(\frac{8+7}{56}\)
= \(\frac{15}{56}\).

Question 7.
McGraw Hill Math Grade 5 Chapter 6 Lesson 5 Answer Key Adding Fractions with Unlike Denominators 24
Answer:
\(\frac{1}{7}\) + \(\frac{1}{6}\) = \(\frac{13}{42}\).

Explanation:
By adding \(\frac{1}{7}\) + \(\frac{1}{6}\) the sum will be \(\frac{7+6}{42}\)
= \(\frac{13}{42}\).

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