McGraw Hill Math Grade 6 Lesson 6.8 Answer Key Estimating Sums and Differences of Fractions and Mixed Numbers

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 6.8 Estimating Sums and Differences of Fractions and Mixed Numbers will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 6.8 Estimating Sums and Differences of Fractions and Mixed Numbers

Exercises Estimate

Question 1.
\(\frac{3}{4}\) + \(\frac{5}{6}\)
Answer:
\(\frac{3}{4}\) + \(\frac{5}{6}\) = 1 \(\frac{7}{12}\)

Explanation:
\(\frac{3}{4}\) + \(\frac{5}{6}\)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.8-Estimating-Sums-and-Differences-of-Fractions-and-Mixed-Numbers- 1

Question 2.
\(\frac{4}{5}\) + \(\frac{1}{7}\)
Answer:
\(\frac{4}{5}\) + \(\frac{1}{7}\) = \(\frac{33}{35}\)

Explanation:
\(\frac{4}{5}\) + \(\frac{1}{7}\)
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Question 3.
\(\frac{1}{3}\) + \(\frac{4}{7}\)
Answer:
\(\frac{1}{3}\) + \(\frac{4}{7}\) = \(\frac{19}{21}\)

Explanation:
\(\frac{1}{3}\) + \(\frac{4}{7}\)
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Question 4.
\(\frac{5}{6}\) + \(\frac{4}{8}\)
Answer:
\(\frac{5}{6}\) + \(\frac{4}{8}\) = 1\(\frac{1}{3}\)

Explanation:
\(\frac{5}{6}\) + \(\frac{4}{8}\)
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Question 5.
1\(\frac{1}{5}\) + 2\(\frac{5}{6}\)
Answer:
1\(\frac{1}{5}\) + 2\(\frac{5}{6}\) = 4\(\frac{1}{30}\)

Explanation:
1\(\frac{1}{5}\) + 2\(\frac{5}{6}\)
= {[(1 × 5) + 1] ÷ 5} + {[(2 × 6) + 5] ÷ 6}
= [(5 + 1) ÷ 5] + [(12 + 5) ÷ 6]
= (6 ÷ 5) + (17 ÷ 6)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.8-Estimating-Sums-and-Differences-of-Fractions-and-Mixed-Numbers- 5

Question 6.
7\(\frac{4}{5}\) + 4\(\frac{1}{3}\)
Answer:
7\(\frac{4}{5}\) + 4\(\frac{1}{3}\) = 12\(\frac{2}{15\)

Explanation:
7\(\frac{4}{5}\) + 4\(\frac{1}{3}\)
= {[(7 × 5) + 4] ÷ 5} + {[(4 × 3) + 1] ÷ 3}
= [(35 + 4) ÷ 5] + [(12 + 1) ÷ 3]
= (39 ÷ 5) + (13 ÷ 3)
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Question 7.
5\(\frac{1}{5}\) – 2\(\frac{3}{5}\)
Answer:
5\(\frac{1}{5}\) – 2\(\frac{3}{5}\) = 2\(\frac{3}{5}\)

Explanation:
5\(\frac{1}{5}\) – 2\(\frac{3}{5}\)
= {[(5 × 5) + 1] ÷ 5} – {[(2 × 5) + 3] ÷ 5}
= [(25 + 1) ÷ 5] – [(10 + 3) ÷ 5]
= (26 ÷ 5) – (13 ÷ 5)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.8-Estimating-Sums-and-Differences-of-Fractions-and-Mixed-Numbers- 7

Question 8.
7\(\frac{1}{2}\) – \(\frac{3}{4}\)
Answer:
7\(\frac{1}{2}\) – \(\frac{3}{4}\) = 6\(\frac{3}{4}\)

Explanation:
7\(\frac{1}{2}\) – \(\frac{3}{4}\)
{[(7 × 2) + 1] ÷ 2} – \(\frac{3}{4}\)
= [(14 + 1) ÷ 2] – \(\frac{3}{4}\)
= (15 ÷ 2) – (3 ÷ 4)
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Question 9.
12\(\frac{1}{8}\) – \(\frac{2}{3}\)
Answer:
12\(\frac{1}{8}\) – \(\frac{2}{3}\) = 11\(\frac{11}{24}\)

Explanation:
12\(\frac{1}{8}\) – \(\frac{2}{3}\)
= {[(12 × 8) + 1] ÷ 8} – \(\frac{2}{3}\)
= [(96 + 1) ÷ 8] – \(\frac{2}{3}\)
= (97 ÷ 8) – (2 ÷ 3)
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Question 10.
4\(\frac{4}{7}\) + 4\(\frac{4}{7}\)
Answer:
4\(\frac{4}{7}\) + 4\(\frac{4}{7}\) = 9\(\frac{1}{7}\)

Explanation:
4\(\frac{4}{7}\) + 4\(\frac{4}{7}\)
{[(4 × 7) + 4] ÷ 7} + {[(4 × 7) + 4] ÷ 7}
= [(28 + 4) ÷ 7] + [(28 + 4) ÷ 7]
= (32 ÷ 7) + (32 ÷ 7)
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Question 11.
13\(\frac{5}{8}\) – 12\(\frac{1}{4}\)
Answer:
13\(\frac{5}{8}\) – 12\(\frac{1}{4}\) = 1\(\frac{3}{8}\)

Explanation:
13\(\frac{5}{8}\) – 12\(\frac{1}{4}\)
{[(13 × 8) + 5] ÷ 8} – {[(12 × 4) + 1] ÷ 4}
= [(104 + 5) ÷ 8] – [(48 + 1) ÷ 4]
= (109 ÷ 8) – (49 ÷ 4)
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Question 12.
17\(\frac{1}{7}\) – 13\(\frac{3}{4}\)
Answer:
17\(\frac{1}{7}\) – 13\(\frac{3}{4}\) = 3\(\frac{11}{28}\)

Explanation:
17\(\frac{1}{7}\) – 13\(\frac{3}{4}\)
{[(17 × 7) + 1] ÷ 7} – {[(13 × 4) + 3] ÷ 4}
= [(119 + 1) ÷ 7] – [(52 + 3) ÷ 4]
= (120 ÷ 7) – (55 ÷ 4)
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Question 13.
Leslie had 25\(\frac{4}{9}\) ounces of cat food left in a bag. If she feeds each of her two cats 1\(\frac{7}{8}\) ounces of food, about how much cat food will she have left?
Answer:
Number of ounces of cat food left with her = 21\(\frac{25}{36}\)

Explanation:
Number of ounces of cat food Leslie had left in a bag = 25\(\frac{4}{9}\)
Number of ounces of cat food she feeds each of her two cats = 1\(\frac{7}{8}\)
Number of ounces of cat food left with her = Number of ounces of cat food Leslie had left in a bag – Number of ounces of cat food she feeds each of her two cats
= 25\(\frac{4}{9}\)  – 2(1\(\frac{7}{8}\))
= {[(25 × 9 ) + 4] ÷ 9} – 2{[(1 × 8) + 7] ÷ 8}
= [(225 + 4) ÷ 9] – 2[(8 + 7) ÷ 8]
= (229 ÷ 9) – 2(15 ÷ 8)
= (229 ÷ 9) – (15 ÷ 4)
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Question 14.
James was gathering wood for the fireplace. He already had 1\(\frac{4}{5}\) cords of wood and he gathered another 2\(\frac{1}{3}\) cords today. About how many cords of wood does James have now?
Answer:
Number of cords of wood he has now = 4\(\frac{2}{15}\)

Explanation:
Number of cords of wood he already had = 1\(\frac{4}{5}\)
Number of cords of wood he again gathered today = 2\(\frac{1}{3}\)
Number of cords of wood he has now = Number of cords of wood he already had + Number of cords of wood he again gathered today
= 1\(\frac{4}{5}\) + 2\(\frac{1}{3}\)
= {[(1 × 5) + 4] ÷ 5} + {[(2 × 3) + 1] ÷ 3}
= [(5 + 4) ÷ 5] + [(6 + 1) ÷ 3]
= (9 ÷ 5) + (7 ÷ 3)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.8-Estimating-Sums-and-Differences-of-Fractions-and-Mixed-Numbers- 14

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