Practice the questions of **McGraw Hill Math Grade 8 Answer Key**** PDF** **Lesson 3.4 Adding or Subtracting Fractions with Unlike Denominators** to secure good marks & knowledge in the exams.

## McGraw-Hill Math Grade 8 Answer Key Lesson 3.4 Adding or Subtracting Fractions with Unlike Denominators

**Exercises Add or Subtract**

Question 1.

\(\frac{3}{4}\) + \(\frac{2}{5}\)

Answer:

\(\frac{23}{20}\) or 1\(\frac{3}{20}\),

Explanation:

Given to add \(\frac{3}{4}\) + \(\frac{2}{5}\) as

both don’t have common denominators first we multiply \(\frac{3}{4}\) by 5 and \(\frac{2}{5}\) by 4 we get

\(\frac{3 X 5}{4 X 5}\) = \(\frac{15}{20}\) and

\(\frac{2 X 4}{5 X 4}\) = \(\frac{8}{20}\)

now we add numerators as \(\frac{15 + 8}{20}\) = \(\frac{23}{20}\) as numerator is greater than denominator we write in mixed fraction as \(\frac{1 X 20 + 3}{20}\) = 1\(\frac{3}{20}\).

Question 2.

\(\frac{5}{7}\) + \(\frac{1}{3}\)

Answer:

\(\frac{22}{21}\) or 1\(\frac{1}{21}\),

Explanation:

Given to add \(\frac{5}{7}\) + \(\frac{1}{3}\) as

both don’t have common denominators first we multiply \(\frac{5}{7}\) by 3 and \(\frac{1}{3}\) by 7 we get

\(\frac{5 X 3}{7 X 3}\) = \(\frac{15}{21}\) and

\(\frac{1 X 7}{3 X 7}\) = \(\frac{7}{21}\)

now we add numerators as \(\frac{15 + 7}{21}\) = \(\frac{22}{21}\) as numerator is greater than denominator we write in mixed fraction as \(\frac{1 X 21 + 1}{21}\) = 1\(\frac{1}{21}\).

Question 3.

\(\frac{11}{20}\) + \(\frac{2}{3}\)

Answer:

\(\frac{73}{60}\) or 1\(\frac{13}{60}\),

Explanation:

Given to add \(\frac{11}{20}\) + \(\frac{2}{3}\) as

both don’t have common denominators first we multiply \(\frac{11}{20}\) by 3 and \(\frac{2}{3}\) by 20 we get

\(\frac{11 X 3}{20 X 3}\) = \(\frac{33}{60}\) and

\(\frac{2 X 20}{3 X 20}\) = \(\frac{40}{60}\)

now we add numerators as \(\frac{33 + 40}{60}\) = \(\frac{73}{60}\) as numerator is greater than denominator we write in mixed fraction as \(\frac{1 X 60 + 13}{60}\) = 1\(\frac{13}{60}\).

Question 4.

\(\frac{1}{4}\) – \(\frac{1}{13}\)

Answer:

\(\frac{9}{52}\),

Explanation:

Given to subtract \(\frac{1}{4}\) – \(\frac{1}{13}\) as

both don’t have common denominators first we multiply \(\frac{1}{4}\) by 13 and \(\frac{1}{13}\) by 4 we get

\(\frac{1 X 13}{4 X 13}\) = \(\frac{13}{52}\) and

\(\frac{1 X 4}{13 X 4}\) = \(\frac{4}{52}\)

now we subtract numerators as \(\frac{13 – 4}{52}\) = \(\frac{9}{52}\).

Question 5.

\(\frac{1}{4}\) – \(\frac{1}{9}\)

Answer:

\(\frac{5}{36}\),

Explanation:

Given to subtract \(\frac{1}{4}\) – \(\frac{1}{9}\) as

both don’t have common denominators first we multiply \(\frac{1}{4}\) by 9 and \(\frac{1}{9}\) by 4 we get

\(\frac{1 X 9}{4 X 9}\) = \(\frac{9}{36}\) and

\(\frac{1 X 4}{9 X 4}\) = \(\frac{4}{36}\)

now we subtract numerators as \(\frac{9 – 4}{36}\) = \(\frac{5}{36}\).

Question 6.

\(\frac{2}{21}\) + \(\frac{1}{3}\)

Answer:

\(\frac{27}{63}\) or \(\frac{9}{21}\) or

\(\frac{3}{7}\),

Explanation:

Given to add \(\frac{2}{21}\) + \(\frac{1}{3}\) as

both don’t have common denominators first we multiply \(\frac{2}{21}\) by 3 and \(\frac{1}{3}\) by 21 we get

\(\frac{2 X 3}{21 X 3}\) = \(\frac{6}{63}\) and

\(\frac{1 X 21}{3 X 21}\) = \(\frac{21}{63}\)

now we add numerators as \(\frac{6 + 21}{63}\) = \(\frac{27}{63}\) further can be divided by 3 we get \(\frac{9}{21}\) again further can be divided by 3 we get \(\frac{3}{7}\).

Question 7.

\(\frac{34}{11}\) – \(\frac{1}{3}\)

Answer:

\(\frac{91}{33}\) or 2\(\frac{25}{33}\),

Explanation:

Given to subtract \(\frac{34}{11}\) – \(\frac{1}{3}\) as

both don’t have common denominators first we multiply \(\frac{34}{11}\) by 3 and \(\frac{1}{3}\) by 11 we get

\(\frac{34 X 3}{11 X 3}\) = \(\frac{102}{33}\) and

\(\frac{1 X 11}{3 X 11}\) = \(\frac{11}{33}\)

now we subtract numerators as \(\frac{102 – 11}{33}\) = \(\frac{91}{33}\) as numerator is greater than denominator we write in mixed fraction as \(\frac{2 X 33 + 25}{33}\) = 2\(\frac{25}{33}\).

Question 8.

\(\frac{11}{7}\) + \(\frac{1}{6}\)

Answer:

\(\frac{73}{42}\) or 1\(\frac{31}{42}\),

Explanation:

Given to add \(\frac{11}{7}\) + \(\frac{1}{6}\) as

both don’t have common denominators first we multiply \(\frac{11}{7}\) by 6 and \(\frac{1}{6}\) by 7 we get

\(\frac{11 X 6}{7 X 6}\) = \(\frac{66}{42}\) and

\(\frac{1 X 7}{6 X 7}\) = \(\frac{7}{42}\)

now we add numerators as \(\frac{66 + 7}{42}\) = \(\frac{73}{42}\) as numerator is greater than denominator we write in mixed fraction as \(\frac{1 X 42 + 31}{42}\) = 1\(\frac{31}{42}\).

Question 9.

\(\frac{11}{15}\) – \(\frac{2}{3}\)

Answer:

\(\frac{3}{45}\) or \(\frac{1}{15}\),

Explanation:

Given to subtract \(\frac{11}{15}\) – \(\frac{2}{3}\) as

both don’t have common denominators first we multiply \(\frac{11}{15}\) by 3 and \(\frac{2}{3}\) by 15 we get

\(\frac{11 X 3}{15 X 3}\) = \(\frac{33}{45}\) and

\(\frac{2 X 15}{3 X 15}\) = \(\frac{30}{45}\)

now we subtract numerators as \(\frac{33 – 30}{45}\) = \(\frac{3}{45}\) further can be divided by 3 we get \(\frac{1}{15}\).

Question 10.

\(\frac{5}{8}\) – \(\frac{2}{7}\)

Answer:

\(\frac{19}{56}\),

Explanation:

Given to subtract \(\frac{5}{8}\) – \(\frac{2}{7}\) as

both don’t have common denominators first we multiply \(\frac{5}{8}\) by 7 and \(\frac{2}{7}\) by 8 we get

\(\frac{5 X 7}{8 X 7}\) = \(\frac{35}{56}\) and

\(\frac{2 X 8}{7 X 8}\) = \(\frac{16}{56}\)

now we subtract numerators as \(\frac{35 – 16}{56}\) = \(\frac{19}{56}\).

Question 11.

\(\frac{14}{11}\) + \(\frac{3}{7}\)

Answer:

\(\frac{131}{77}\) or 1\(\frac{54}{77}\),

Explanation:

Given to add \(\frac{14}{11}\) + \(\frac{3}{7}\) as

both don’t have common denominators first we multiply \(\frac{14}{11}\) by 7 and \(\frac{3}{7}\) by 11 we get

\(\frac{14 X 7}{11 X 7}\) = \(\frac{98}{77}\) and

\(\frac{3 X 11}{7 X 11}\) = \(\frac{33}{77}\)

now we add numerators as \(\frac{98 + 33}{77}\) = \(\frac{131}{77}\) as numerator is greater than denominator we write in mixed fraction as \(\frac{1 X 77 + 54}{77}\) = 1\(\frac{54}{77}\).

Question 12.

\(\frac{11}{12}\) – \(\frac{3}{5}\)

Answer:

\(\frac{47}{60}\),

Explanation:

Given to subtract \(\frac{11}{12}\) – \(\frac{3}{15}\) as

both don’t have common denominators first we multiply \(\frac{11}{12}\) by 15 and \(\frac{3}{15}\) by 12 we get

\(\frac{11 X 15}{12 X 15}\) = \(\frac{165}{180}\) and

\(\frac{2 X 12}{15 X 12}\) = \(\frac{24}{180}\)

now we subtract numerators as \(\frac{165 – 24}{180}\) = \(\frac{141}{180}\) further can be divided by 3 we get \(\frac{47}{60}\).