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

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

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

Solve

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

Question 1.
1\(\frac{1}{4}\) + \(\frac{1}{6}\) ________________
Answer:
1\(\frac{3}{12}\) + \(\frac{2}{12}\) = 1\(\frac{5}{12}\)

Question 2.
1\(\frac{2}{3}\) + \(\frac{1}{2}\) _______________
Answer:
\(\frac{10}{6}\)+\(\frac{4}{6}\)  = 2\(\frac{1}{3}\).

Explanation:
The equivalent fraction of \(\frac{5}{3}\) + \(\frac{1}{2}\) which is \(\frac{10}{6}\) and \(\frac{4}{6}\). So the addition of \(\frac{10}{6}\)+\(\frac{4}{6}\) is \(\frac{10+4}{6}\)
= \(\frac{14}{6}\)
= 2\(\frac{2}{6}\)
= 2\(\frac{1}{3}\).

Question 3.
2\(\frac{1}{4}\) + 3\(\frac{5}{8}\) _______________
Answer:
\(\frac{18}{8}\)+\(\frac{29}{8}\) = 5\(\frac{7}{8}\).

Explanation:
The equivalent fraction of \(\frac{9}{4}\) + \(\frac{29}{8}\) which is \(\frac{18}{8}\) and \(\frac{29}{8}\). So the addition of \(\frac{18}{8}\)+\(\frac{29}{8}\) is \(\frac{18+29}{8}\)
= \(\frac{47}{8}\)
= 5\(\frac{7}{8}\).

Question 4.
2\(\frac{1}{9}\) + \(\frac{5}{6}\) ________________
Answer:
\(\frac{114}{54}\)+\(\frac{45}{54}\) = 2\(\frac{1}{9}\).

Explanation:
The equivalent fraction of \(\frac{19}{9}\) + \(\frac{5}{6}\) which is \(\frac{114}{54}\) and \(\frac{45}{54}\). So the addition of \(\frac{114}{54}\)+\(\frac{45}{54}\) is \(\frac{114+45}{54}\)
= \(\frac{159}{54}\)
= 2\(\frac{6}{54}\) = 2\(\frac{1}{9}\).

Question 5.
4\(\frac{3}{8}\) + 2\(\frac{1}{6}\) ________________
Answer:
\(\frac{210}{48}\)+\(\frac{104}{48}\) = 6\(\frac{13}{16}\).

Explanation:
The equivalent fraction of \(\frac{35}{8}\) + \(\frac{13}{6}\) which is \(\frac{210}{48}\) and \(\frac{104}{48}\). So the addition of \(\frac{210}{48}\)+\(\frac{104}{48}\) is \(\frac{210+104}{8}\)
= \(\frac{314}{48}\) =6\(\frac{26}{48}\)
= 6\(\frac{13}{16}\).

Question 6.
1\(\frac{1}{5}\) + 1\(\frac{1}{2}\) ________________
Answer:

\(\frac{18}{8}\)+\(\frac{29}{8}\) = 5\(\frac{7}{8}\).

Explanation:
The equivalent fraction of \(\frac{9}{4}\) + \(\frac{29}{8}\) which is \(\frac{18}{8}\) and \(\frac{29}{8}\). So the addition of \(\frac{18}{8}\)+\(\frac{29}{8}\) is \(\frac{18+29}{8}\)
= \(\frac{47}{8}\)

Question 7.
McGraw Hill Math Grade 5 Chapter 6 Lesson 6 Answer Key Adding Mixed Numbers with Unlike Denominators 25
Answer:
\(\frac{49}{14}\)+\(\frac{20}{14}\)= 1\(\frac{7}{16}\).

Explanation:
The equivalent fraction of \(\frac{7}{2}\) + \(\frac{10}{7}\) which is \(\frac{49}{14}\) and \(\frac{20}{14}\). So the addition of \(\frac{49}{14}\)+\(\frac{20}{14}\) is \(\frac{49+20}{48}\)
= \(\frac{69}{48}\) = 1\(\frac{21}{48}\)
= 1\(\frac{7}{16}\).

Question 8.
McGraw Hill Math Grade 5 Chapter 6 Lesson 6 Answer Key Adding Mixed Numbers with Unlike Denominators 26
Answer:
\(\frac{35}{15}\)+\(\frac{24}{15}\) = 3\(\frac{14}{15}\).

Explanation:
The equivalent fraction of \(\frac{7}{3}\) + \(\frac{8}{5}\) which is \(\frac{35}{15}\) and \(\frac{24}{15}\). So the addition of \(\frac{35}{15}\)+\(\frac{24}{15}\) is \(\frac{35+24}{15}\)
= \(\frac{59}{15}\)
= 3\(\frac{14}{15}\).

Question 9.
McGraw Hill Math Grade 5 Chapter 6 Lesson 6 Answer Key Adding Mixed Numbers with Unlike Denominators 27
Answer:
\(\frac{376}{56}\)+\(\frac{344}{56}\) = 12\(\frac{6}{7}\).

Explanation:
The equivalent fraction of \(\frac{47}{7}\) + \(\frac{43}{8}\) which is \(\frac{376}{56}\) and \(\frac{344}{56}\). So the addition of \(\frac{376}{56}\)+\(\frac{344}{56}\) is \(\frac{376+344}{56}\)
= \(\frac{720}{56}\) = \(\frac{90}{7}\)
= 12\(\frac{6}{7}\).

Question 10.
McGraw Hill Math Grade 5 Chapter 6 Lesson 6 Answer Key Adding Mixed Numbers with Unlike Denominators 28
Answer:
\(\frac{150}{42}\)+\(\frac{175}{42}\) = 7\(\frac{31}{42}\).

Explanation:
The equivalent fraction of \(\frac{25}{7}\) + \(\frac{25}{6}\) which is \(\frac{150}{42}\) and \(\frac{175}{42}\). So the addition of \(\frac{150}{42}\)+\(\frac{175}{42}\) is \(\frac{150+175}{42}\)
= \(\frac{325}{42}\)
= 7\(\frac{31}{42}\).

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