McGraw Hill Math Grade 8 Lesson 3.3 Answer Key Adding and Subtracting Fractions with Like Denominators

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 3.3 Adding and Subtracting Fractions with Like Denominators to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 3.3 Adding and Subtracting Fractions with Like Denominators

Exercises Add

Question 1.
\(\frac{3}{4}\) + \(\frac{3}{4}\)
Answer:
\(\frac{6}{4}\) or \(\frac{3}{2}\) or 1\(\frac{1}{2}\),

Explanation:
Given to add \(\frac{3}{4}\) + \(\frac{3}{4}\) as
both have common denominators we add numerators as
\(\frac{3 + 3}{4}\) = \(\frac{6}{4}\) or \(\frac{3}{2}\) as numerator is greater than denominator we write in mixed fraction as \(\frac{1 X 2 + 1}{2}\) = 1\(\frac{1}{2}\).

Question 2.
\(\frac{1}{5}\) + \(\frac{4}{5}\)
Answer:
\(\frac{5}{5}\) or 1,

Explanation:
Given to add \(\frac{1}{5}\) + \(\frac{4}{5}\) as
both have common denominators we add numerators as
\(\frac{1 + 4}{5}\) = \(\frac{5}{5}\) or 1.

Question 3.
\(\frac{5}{8}\) + \(\frac{5}{8}\)
Answer:
\(\frac{10}{8}\) or \(\frac{5}{4}\) or 1\(\frac{1}{4}\),

Explanation:
Given to add \(\frac{5}{8}\) + \(\frac{5}{8}\) as
both have common denominators we add numerators as
\(\frac{5 + 5}{8}\) = \(\frac{10}{8}\) or \(\frac{5}{4}\) as numerator is greater than denominator we write in mixed fraction as \(\frac{1 X 4 + 1}{4}\) = 1\(\frac{1}{4}\).

Question 4.
\(\frac{7}{9}\) + \(\frac{4}{9}\)
Answer:
\(\frac{11}{9}\) or 1\(\frac{2}{9}\),

Explanation:
Given to add \(\frac{7}{9}\) + \(\frac{4}{9}\) as
both have common denominators we add numerators as
\(\frac{7 + 4}{9}\) = \(\frac{11}{9}\) as numerator
is greater than denominator we write in mixed fraction as
\(\frac{1 X 9 + 2}{9}\) = 1\(\frac{2}{9}\).

Question 5.
\(\frac{4}{11}\) + \(\frac{3}{11}\)
Answer:
\(\frac{7}{11}\),

Explanation:
Given to add \(\frac{4}{11}\) + \(\frac{3}{11}\) as
both have common denominators we add numerators as
\(\frac{4 + 3}{11}\) = \(\frac{7}{11}\).

Question 6.
\(\frac{15}{17}\) + \(\frac{5}{17}\)
Answer:
\(\frac{20}{17}\) or 1\(\frac{3}{17}\),

Explanation:
Given to add \(\frac{15}{17}\) + \(\frac{5}{17}\) as
both have common denominators we add numerators as
\(\frac{15 + 5}{17}\) = \(\frac{20}{17}\) as numerator
is greater than denominator we write in mixed fraction as
\(\frac{1 X 17 + 3}{17}\) = 1\(\frac{3}{17}\).

Question 7.
\(\frac{7}{9}\) + \(\frac{14}{9}\)
Answer:
\(\frac{21}{9}\) or \(\frac{7}{3}\) or 2\(\frac{1}{3}\),

Explanation:
Given to add \(\frac{7}{9}\) + \(\frac{14}{9}\) as
both have common denominators we add numerators as
\(\frac{7 + 14}{9}\) = \(\frac{21}{9}\) =
\(\frac{7}{3}\) as numerator is greater than denominator
we write in mixed fraction as \(\frac{2 X 3 + 1}{3}\) = 2\(\frac{1}{3}\).

Question 8.
\(\frac{2}{3}\) + \(\frac{5}{3}\)
Answer:
\(\frac{7}{3}\) or 2\(\frac{1}{3}\),

Explanation:
Given to add \(\frac{2}{3}\) + \(\frac{5}{3}\) as
both have common denominators we add numerators as
\(\frac{2 + 5}{9}\) = \(\frac{7}{9}\)  as numerator is greater than denominator we write in mixed fraction as
\(\frac{2 X 3 + 1}{3}\) = 2\(\frac{1}{3}\).

Question 9.
\(\frac{6}{23}\) + \(\frac{14}{23}\)
Answer:
\(\frac{20}{23}\),

Explanation:
Given to add \(\frac{6}{23}\) + \(\frac{14}{23}\) as
both have common denominators we add numerators as
\(\frac{6 + 14}{23}\) = \(\frac{20}{23}\).

Question 10.
\(\frac{13}{37}\) + \(\frac{24}{37}\)
Answer:
\(\frac{37}{37}\) or 1,

Explanation:
Given to add \(\frac{13}{37}\) + \(\frac{24}{37}\) as
both have common denominators we add numerators as
\(\frac{13 + 24}{37}\) = \(\frac{37}{37}\) = 1.

Question 11.
\(\frac{1}{4}\) + \(\frac{7}{4}\)
Answer:
\(\frac{8}{4}\) or 2,

Explanation:
Given to add \(\frac{1}{4}\) + \(\frac{7}{4}\) as
both have common denominators we add numerators as
\(\frac{1 + 7}{4}\) = \(\frac{8}{4}\) = 2.

Question 12.
\(\frac{3}{11}\) + \(\frac{7}{11}\)
Answer:
\(\frac{10}{11}\),

Explanation:
Given to add \(\frac{3}{11}\) + \(\frac{7}{11}\) as
both have common denominators we add numerators as
\(\frac{3 + 7}{11}\) = \(\frac{10}{11}\).

Question 13.
\(\frac{13}{27}\) + \(\frac{11}{27}\)
Answer:
\(\frac{24}{27}\) or \(\frac{8}{9}\),

Explanation:
Given to add \(\frac{13}{27}\) + \(\frac{11}{27}\) as
both have common denominators we add numerators as
\(\frac{13 + 11}{27}\) = \(\frac{24}{27}\) = \(\frac{8}{9}\).

Question 14.
\(\frac{11}{14}\) + \(\frac{13}{14}\)
Answer:
\(\frac{24}{14}\) or \(\frac{12}{7}\) or 1\(\frac{5}{7}\),

Explanation:
Given to add \(\frac{11}{14}\) + \(\frac{13}{14}\) as
both have common denominators we add numerators as
\(\frac{11 + 13}{14}\) = \(\frac{24}{14}\) =
\(\frac{12}{7}\) as numerator is greater than denominator
we write in mixed fraction as \(\frac{1 X 7 + 5}{7}\) = 1\(\frac{5}{7}\).

Question 15.
\(\frac{13}{24}\) + \(\frac{16}{24}\)
Answer:
\(\frac{29}{24}\) or 1\(\frac{5}{24}\),

Explanation:
Given to add \(\frac{13}{24}\) + \(\frac{16}{24}\) as
both have common denominators we add numerators as
\(\frac{13 + 16}{24}\) = \(\frac{29}{24}\) as numerator is greater than denominator we write in mixed fraction as
\(\frac{1 X 24 + 5}{24}\) = 1\(\frac{5}{24}\).

Question 16.
\(\frac{13}{37}\) + \(\frac{11}{37}\)
Answer:
\(\frac{24}{37}\),

Explanation:
Given to add \(\frac{13}{37}\) + \(\frac{11}{37}\) as
both have common denominators we add numerators as
\(\frac{13 + 11}{37}\) = \(\frac{24}{37}\).

Question 17.
Manny combined \(\frac{1}{7}\) quarts of orange juice, \(\frac{2}{7}\) quarts of lemonade, and \(\frac{5}{7}\) quarts of raspberry tea into one container. How much liquid is now in the container? Express your answer as a mixed number.
Answer:
1\(\frac{1}{7}\) quarts is in the container,

Explanation:
As Manny combined \(\frac{1}{7}\) quarts of orange juice, \(\frac{2}{7}\) quarts of lemonade, and \(\frac{5}{7}\) quarts of raspberry tea into one container.
Liquid is now in the container is \(\frac{1}{7}\) quarts + \(\frac{2}{7}\) quarts + \(\frac{5}{7}\) quarts =
\(\frac{1 + 2 + 5}{7}\) quarts = \(\frac{8}{7}\) quarts ,
as numerator is greater than denominator we write in mixed fraction as
\(\frac{1 X 7 + 1}{7}\) = 1\(\frac{1}{7}\) quarts.

Question 18.
James, Riley and Nancy surveyed their class about the cafeteria food. James surveyed \(\frac{2}{9}\) of the class, Riley surveyed another \(\frac{5}{9}\) of the class, and Nancy surveyed another \(\frac{1}{9}\) of the class. Were the three of them able to po11 the entire class?
Answer:
No,

Explanation:
As James, Riley and Nancy surveyed their class about the cafeteria food. James surveyed \(\frac{2}{9}\) of the class, Riley surveyed another \(\frac{5}{9}\) of the class, and Nancy surveyed another \(\frac{1}{9}\) of the class.
Now checking the three of them able to po11 the entire class as
\(\frac{2}{9}\) + \(\frac{5}{9}\) + \(\frac{1}{9}\)  = \(\frac{2 + 5 + 1}{9}\) quarts = \(\frac{8}{9}\)
No, the three of them were not able to po11 the entire class as it is
not 1.

Exercises Subtract

Question 1.
\(\frac{3}{4}\) – \(\frac{1}{4}\)
Answer:
\(\frac{2}{4}\) or \(\frac{1}{2}\),

Explanation:
Given to subtract \(\frac{3}{4}\) – \(\frac{1}{4}\) as
both have common denominators we subtract numerators as
\(\frac{3 – 1}{4}\) = \(\frac{2}{4}\) further
can be divided by 2 we get \(\frac{2}{4}\).

Question 2.
\(\frac{3}{3}\) – \(\frac{2}{3}\)
Answer:
\(\frac{1}{3}\),

Explanation:
Given to subtract \(\frac{3}{3}\) – \(\frac{2}{3}\) as
both have common denominators we subtract numerators as
\(\frac{3 – 2}{3}\) = \(\frac{1}{3}\).

Question 3.
\(\frac{8}{9}\) – \(\frac{5}{9}\)
Answer:
\(\frac{3}{9}\) or \(\frac{1}{3}\),

Explanation:
Given to subtract \(\frac{8}{9}\) – \(\frac{5}{9}\) as
both have common denominators we subtract numerators as
\(\frac{8 – 5}{9}\) = \(\frac{3}{9}\) further
can be divided by 3 we get \(\frac{1}{3}\).

Question 4.
\(\frac{7}{8}\) – \(\frac{1}{8}\)
Answer:
\(\frac{6}{8}\) or \(\frac{3}{4}\),

Explanation:
Given to subtract \(\frac{7}{8}\) – \(\frac{1}{8}\) as
both have common denominators we subtract numerators as
\(\frac{7 – 1}{8}\) = \(\frac{6}{8}\) further
can be divided by 2 we get \(\frac{3}{4}\).

Question 5.
\(\frac{5}{7}\) – \(\frac{3}{7}\)
Answer:
\(\frac{2}{7}\),

Explanation:
Given to subtract \(\frac{5}{7}\) – \(\frac{3}{7}\) as
both have common denominators we subtract numerators as
\(\frac{5 – 3}{7}\) = \(\frac{2}{7}\).

Question 6.
\(\frac{1}{2}\) – \(\frac{1}{2}\)
Answer:
0,

Explanation:
Given to subtract \(\frac{1}{2}\) – \(\frac{1}{2}\) as
both have common denominators we subtract numerators as
\(\frac{1 – 1}{2}\) = 0.

Question 7.
\(\frac{2}{3}\) – \(\frac{1}{3}\)
Answer:
\(\frac{1}{3}\),

Explanation:
Given to subtract \(\frac{2}{3}\) – \(\frac{1}{3}\) as
both have common denominators we subtract numerators as
\(\frac{2 – 1}{3}\) = \(\frac{1}{3}\).

Question 8.
\(\frac{5}{3}\) – \(\frac{2}{3}\)
Answer:
1,

Explanation:
Given to subtract \(\frac{5}{3}\) – \(\frac{2}{3}\) as
both have common denominators we subtract numerators as
\(\frac{5 – 2}{3}\) = \(\frac{3}{3}\) further
can be divided by 3 we get 1.

Question 9.
\(\frac{7}{5}\) – \(\frac{4}{5}\)
Answer:
\(\frac{3}{5}\),

Explanation:
Given to subtract \(\frac{7}{5}\) – \(\frac{4}{5}\) as
both have common denominators we subtract numerators as
\(\frac{7 – 4}{5}\) = \(\frac{3}{5}\).

Question 10.
\(\frac{4}{7}\) – \(\frac{1}{7}\)
Answer:
\(\frac{3}{7}\),

Explanation:
Given to subtract \(\frac{4}{7}\) – \(\frac{1}{7}\) as
both have common denominators we subtract numerators as
\(\frac{4 – 1}{7}\) = \(\frac{3}{7}\).

Question 11.
\(\frac{9}{7}\) – \(\frac{6}{7}\)
Answer:
\(\frac{3}{7}\),

Explanation:
Given to subtract \(\frac{9}{7}\) – \(\frac{6}{7}\) as
both have common denominators we subtract numerators as
\(\frac{9 – 6}{7}\) = \(\frac{3}{7}\).

Question 12.
\(\frac{11}{9}\) – \(\frac{2}{9}\)
Answer:
1,

Explanation:
Given to subtract \(\frac{11}{9}\) – \(\frac{2}{9}\) as
both have common denominators we subtract numerators as
\(\frac{11 – 2}{9}\) = \(\frac{9}{9}\) further
can be divided by 9 we get 1.

Question 13.
\(\frac{6}{5}\) – \(\frac{3}{5}\)
Answer:
\(\frac{3}{5}\),

Explanation:
Given to subtract \(\frac{6}{5}\) – \(\frac{3}{5}\) as
both have common denominators we subtract numerators as
\(\frac{6 – 3}{5}\) = \(\frac{3}{5}\).

Question 14.
\(\frac{12}{11}\) – \(\frac{9}{11}\)
Answer:
\(\frac{3}{11}\),

Explanation:
Given to subtract \(\frac{12}{11}\) – \(\frac{9}{11}\) as
both have common denominators we subtract numerators as
\(\frac{12 – 9}{11}\) = \(\frac{3}{11}\).

Question 15.
\(\frac{31}{35}\) – \(\frac{1}{35}\)
Answer:
\(\frac{30}{35}\) or \(\frac{6}{7}\),

Explanation:
Given to subtract \(\frac{31}{35}\) – \(\frac{1}{35}\) as
both have common denominators we subtract numerators as
\(\frac{31 – 1}{35}\) = \(\frac{30}{35}\) further
can be divided by 5 we get \(\frac{6}{7}\).

Question 16.
\(\frac{21}{22}\) – \(\frac{17}{22}\)
Answer:
\(\frac{4}{22}\) or \(\frac{2}{11}\),

Explanation:
Given to subtract \(\frac{21}{22}\) – \(\frac{17}{22}\) as
both have common denominators we subtract numerators as
\(\frac{21 – 17}{22}\) = \(\frac{4}{22}\) further
can be divided by 2 we get \(\frac{2}{11}\).

Question 17.
Kira made \(\frac{19}{16}\) quarts of grape juice and served \(\frac{3}{8}\) quarts for dinner. How much juice does she have left?
Answer:
Kira is left with \(\frac{13}{16}\) quarts of juice,

Explanation:
Given Kira made \(\frac{19}{16}\) quarts of grape juice and
served \(\frac{3}{8}\) quarts for dinner.
So juice does she have left is \(\frac{19}{16}\) quarts – \(\frac{3}{8}\) quarts as we have to make both common denominators we multiply \(\frac{3}{8}\) quarts by 2 we get \(\frac{3 X 2}{8 X 2}\) quarts = \(\frac{6}{16}\) quarts,
now we subtract numerators as \(\frac{19 – 6}{16}\) quarts = \(\frac{13}{16}\) quarts.

Question 18.
Ellen bought \(\frac{19}{16}\) pounds of flour from the store. On her way home, she spilled \(\frac{11}{16}\) pounds of flour, If she needs \(\frac{7}{16}\) pounds of flour to make bread, will she have enough flour?
Answer:
No, Ellen will not have enough flour,

Explanation:
Given Ellen bought \(\frac{19}{16}\) pounds of flour from the store. On her way home, she spilled \(\frac{11}{16}\) pounds of flour now she is left with \(\frac{19}{16}\) – \(\frac{11}{16}\) as both have common denominators we subtract numerators as
\(\frac{19 – 11}{16}\) = \(\frac{8}{16}\), As
she needs \(\frac{7}{16}\) pounds of flour to make bread,
\(\frac{8}{16}\) ≠ \(\frac{7}{16}\) Ellen will not have enough flour.

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