McGraw Hill Math Grade 6 Lesson 6.7 Answer Key Subtracting Mixed Numbers with Unlike Denominators

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 6.7 Subtracting Mixed Numbers with Unlike Denominators will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 6.7 Subtracting Mixed Numbers with Unlike Denominators

Exercises Subtract

Question 1.
2\(\frac{5}{8}\) – 1\(\frac{1}{4}\)
Answer:
2\(\frac{5}{8}\) – 1\(\frac{1}{4}\) = 1\(\frac{3}{8}\)

Explanation:
2\(\frac{5}{8}\) – 1\(\frac{1}{4}\)
= {[(2 × 8) + 5] ÷ 8} – {[(1 × 4) + 1] ÷ 4}
= [(16 + 5) ÷ 8] – [(4 + 1) ÷ 4]
= (21 ÷ 8) – (5 ÷ 4)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-1

Question 2.
4\(\frac{6}{7}\) – 2\(\frac{3}{4}\)
Answer:
4\(\frac{6}{7}\) – 2\(\frac{3}{4}\) = 2\(\frac{3}{28}\)

Explanation:
4\(\frac{6}{7}\) – 2\(\frac{3}{4}\)
={[(4 × 7) + 6] ÷ 7} – {[(2 × 4) + 3] ÷ 4}
= [(28 + 6) ÷ 7] – [(8 + 3) ÷ 4]
= (34 ÷ 7) – (11 ÷ 4)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-2

Question 3.
5\(\frac{4}{9}\) – 2\(\frac{1}{3}\)
Answer:
5\(\frac{4}{9}\) – 2\(\frac{1}{3}\) = 3\(\frac{1}{9}\)

Explanation:
5\(\frac{4}{9}\) – 2\(\frac{1}{3}\)
= {[(5 × 9) + 4] ÷ 9} – {[(2 × 3) + 1] ÷ 3}
= [(45 + 4) ÷ 9] – [(6 + 1) ÷ 3]
= (49 ÷ 9) – (7 ÷ 3)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-3

Question 4.
7\(\frac{7}{8}\) – 5\(\frac{1}{5}\)
Answer:
7\(\frac{7}{8}\) – 5\(\frac{1}{5}\) = 2\(\frac{27}{40}\)

Explanation:
7\(\frac{7}{8}\) – 5\(\frac{1}{5}\)
= {[(7 × 8) + 7] ÷ 8} – {[(5 × 5) + 1] ÷ 5}
= [(56 + 7) ÷ 8] – [(25 + 1) ÷ 5]
= (63 ÷ 8) – (26 ÷ 5)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-4

Question 5.
2\(\frac{1}{10}\) – 1\(\frac{1}{11}\)
Answer:
2\(\frac{1}{10}\) – 1\(\frac{1}{11}\) = 1\(\frac{1}{110}\)

Explanation:
2\(\frac{1}{10}\) – 1\(\frac{1}{11}\)
= {[(2 × 10) + 1] ÷ 10} – {[(1 × 11) + 1] ÷ 11}
= [(20 + 1) ÷ 10] – [(11 + 1) ÷ 11]
= (21 ÷ 10) – (12 ÷ 11)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-5

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

Explanation:
5\(\frac{3}{4}\) – 4\(\frac{2}{3}\)
= {[(5 × 4) + 3] ÷ 4} – {[(4 × 3) + 2] ÷ 3}
= [(20 + 3) ÷ 4] – [(12 + 2) ÷ 3]
= (23 ÷ 4) – (24 ÷ 3)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-6

Question 7.
7\(\frac{7}{9}\) – 2\(\frac{1}{4}\)
Answer:
7\(\frac{7}{9}\) – 2\(\frac{1}{4}\) = 5\(\frac{19}{36}\)

Explanation:
7\(\frac{7}{9}\) – 2\(\frac{1}{4}\)
= {[(7 × 9) + 7] ÷ 9} – {[(2 × 4) + 1] ÷ 4}
= [(63 + 7) ÷ 9] – [(8 + 1) ÷ 4]
= (70 ÷ 9) – (9 ÷ 4)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-7

Question 8.
5\(\frac{3}{7}\) – 2\(\frac{1}{6}\)
Answer:
5\(\frac{3}{7}\) – 2\(\frac{1}{6}\) = 3\(\frac{11}{42}\)

Explanation:
5\(\frac{3}{7}\) – 2\(\frac{1}{6}\)
= {[(5 × 7) + 3] ÷ 7} – {[(2 × 6) + 1] ÷ 6}
= [(35 + 3) ÷ 7] – [(12 + 1) ÷ 6]
= (38 ÷ 7) – (13 ÷ 6)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-8

Question 9.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.7 Answer Key Subtracting Mixed Numbers with Unlike Denominators 1
Answer:
5\(\frac{1}{4}\) – 2\(\frac{2}{13}\) = 3\(\frac{5}{52}\)

Explanation:
5\(\frac{1}{4}\) – 2\(\frac{2}{13}\)
= {[(5 × 4) + 1] ÷ 4} – {[(2 × 13) + 2] ÷ 13}
= [(20 + 1) ÷ 4] – [(26 + 2) ÷ 13]
= (21 ÷ 4) – (28 ÷ 13)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-9

Question 10.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.7 Answer Key Subtracting Mixed Numbers with Unlike Denominators 2
Answer:
19\(\frac{6}{7}\) – 3\(\frac{3}{10}\) = 16\(\frac{39}{70}\)

Explanation:
19\(\frac{6}{7}\) – 3\(\frac{3}{10}\)
= {[(19 × 7) + 6] ÷ 7} – {[(3 × 10) + 3] ÷ 10}
= [(133 + 6) ÷ 7] – [(30 + 3) ÷ 10]
= (139 ÷ 7) – (33 ÷ 10)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-10

Question 11.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.7 Answer Key Subtracting Mixed Numbers with Unlike Denominators 3
Answer:
4\(\frac{2}{3}\) – 1\(\frac{1}{8}\) = 3\(\frac{13}{24}\)

Explanation:
4\(\frac{2}{3}\) – 1\(\frac{1}{8}\)
= {[(4 × 3) + 2] ÷ 3} – {[(1 × 8) + 1] ÷ 8}
= [(12 + 2) ÷ 3] – [(8 + 1) ÷ 8]
= (14 ÷ 3) – (9 ÷ 8)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-11

Question 12.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.7 Answer Key Subtracting Mixed Numbers with Unlike Denominators 4
Answer:
10\(\frac{7}{10}\) – 7\(\frac{5}{9}\) = 3\(\frac{13}{90}\)

Explanation:
10\(\frac{7}{10}\) – 7\(\frac{5}{9}\)
= {[(10 × 10) + 7] ÷ 10} – {[(7 × 9) + 5] ÷ 9}
= [(100 + 7) ÷ 10] – [(63 + 5) ÷ 9]
= (107 ÷ 10) – (68 ÷ 9)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-12

Question 13.
Mario is selling lemonade to raise money for his school. He started with 8\(\frac{1}{2}\) gallons of lemonade and has, so far, sold 3\(\frac{7}{8}\) gallons. How many gallons does he have left to sell?
Answer:
Number of gallons of lemonade he left to sell = 4\(\frac{5}{8}\)

Explanation:
Number of gallons of lemonade he started with = 8\(\frac{1}{2}\)
Number of gallons of lemonade he sold = 3\(\frac{7}{8}\)
Number of gallons of lemonade he left to sell = Number of gallons of lemonade he started with – Number of gallons of lemonade he sold
= 8\(\frac{1}{2}\) – 3\(\frac{7}{8}\)
= {[(8 × 2) + 1] ÷ 2} – {[(3 × 8) + 7] ÷ 8}
= [(16 + 1) ÷ 2] – [(24 + 7) ÷ 8]
= (17 ÷ 2) – (31 ÷ 8)
= McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-13

Question 14.
Bethany is making strawberry shortcake for her entire family. The recipe calls for 1\(\frac{3}{5}\) pounds of strawberries. If Bethany purchases 1\(\frac{7}{8}\) pounds, but drops \(\frac{1}{4}\) pound of strawberries on her way home, will she have enough to complete the recipe?
Answer:
Yes, she has enough to complete the recipe.

Explanation:
Number of pounds of recipe calls of strawberries = 1\(\frac{3}{5}\)
Number of pounds Bethany purchases = 1\(\frac{7}{8}\)
Number of pounds of strawberries Bethany drops on her way home = \(\frac{1}{4}\)
Number of pounds of strawberries Bethany is left = Number of pounds Bethany purchases – Number of pounds of strawberries Bethany drops on her way home
= 1\(\frac{7}{8}\) – \(\frac{1}{4}\)
= {[(1 × 8) + 7] ÷ 8} – \(\frac{1}{4}\)
= [(8 + 7) ÷ 8] – \(\frac{1}{4}\)
= \(\frac{15}{8}\) – \(\frac{1}{4}\)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.7-Subtracting-Mixed-Numbers-with-Unlike-Denominators-Exercises-Subtract-14
Number of pounds of recipe calls of strawberries = 1\(\frac{3}{5}\)
Number of pounds of strawberries Bethany is left = 1\(\frac{5}{8}\)
Equating:
=> 1\(\frac{3}{5}\) = 1\(\frac{5}{8}\)
=> {[(1 × 5) + 3] ÷ 5} = {[(1 × 8) + 5] ÷ 8}
=> [(5 + 3) ÷ 5] = [(8 + 5) ÷ 8]
=> (8 ÷ 5) = (13 ÷ 8)
LCD of 5 and 8: 40.
=> [(8 × 8) ÷ 40 = (13 × 5) ÷ 40
=> (64 ÷ 40) = (65 ÷ 40)
=> 1.6 = 1.625.
=> Number of pounds of strawberries Bethany is left with him are more than the Number of pounds of recipe calls of strawberries.

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