Practice questions available in **McGraw Hill Math Grade 6 Answer Key PDF Lesson 6.7 Subtracting Mixed Numbers with Unlike Denominators**Â 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 Ã— 8) + 5] Ã· 8} – {[(1 Ã— 4) + 1] Ã· 4}

= [(16 + 5) Ã· 8] – [(4 + 1) Ã· 4]

= (21 Ã· 8) – (5 Ã· 4)

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 Ã— 7) + 6] Ã· 7} – {[(2 Ã— 4) + 3] Ã· 4}

= [(28 + 6) Ã· 7] – [(8 + 3) Ã· 4]

= (34 Ã· 7) – (11 Ã· 4)

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 Ã— 9) + 4] Ã· 9} – {[(2 Ã— 3) + 1] Ã· 3}

= [(45 + 4) Ã· 9] – [(6 + 1) Ã· 3]

= (49 Ã· 9) – (7 Ã· 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 Ã— 8) + 7] Ã· 8} – {[(5 Ã— 5) + 1] Ã· 5}

= [(56 + 7) Ã· 8] – [(25 + 1) Ã· 5]

= (63 Ã· 8) – (26 Ã· 5)

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 Ã— 10) + 1] Ã· 10} – {[(1 Ã— 11) + 1] Ã· 11}

= [(20 + 1) Ã· 10] – [(11 + 1) Ã· 11]

= (21 Ã· 10) – (12 Ã· 11)

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 Ã— 4) + 3] Ã· 4} – {[(4 Ã— 3) + 2] Ã· 3}

= [(20 + 3) Ã· 4] – [(12 + 2) Ã· 3]

= (23 Ã· 4) – (24 Ã· 3)

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 Ã— 9) + 7] Ã· 9} – {[(2 Ã— 4) + 1] Ã· 4}

= [(63 + 7) Ã· 9] – [(8 + 1) Ã· 4]

= (70 Ã· 9) – (9 Ã· 4)

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 Ã— 7) + 3] Ã· 7} – {[(2 Ã— 6) + 1] Ã· 6}

= [(35 + 3) Ã· 7] – [(12 + 1) Ã· 6]

= (38 Ã· 7) – (13 Ã· 6)

Question 9.

Answer:

5\(\frac{1}{4}\) – 2\(\frac{2}{13}\) = 3\(\frac{5}{52}\)

Explanation:

5\(\frac{1}{4}\) – 2\(\frac{2}{13}\)

= {[(5 Ã— 4) + 1] Ã· 4} – {[(2 Ã— 13) + 2] Ã· 13}

= [(20 + 1) Ã· 4] – [(26 + 2) Ã· 13]

= (21 Ã· 4) – (28 Ã· 13)

Question 10.

Answer:

19\(\frac{6}{7}\) – 3\(\frac{3}{10}\) = 16\(\frac{39}{70}\)

Explanation:

19\(\frac{6}{7}\) – 3\(\frac{3}{10}\)

= {[(19 Ã— 7) + 6] Ã· 7} – {[(3 Ã— 10) + 3] Ã· 10}

= [(133 + 6) Ã· 7] – [(30 + 3) Ã· 10]

= (139 Ã· 7) – (33 Ã· 10)

Question 11.

Answer:

4\(\frac{2}{3}\) – 1\(\frac{1}{8}\) = 3\(\frac{13}{24}\)

Explanation:

4\(\frac{2}{3}\) – 1\(\frac{1}{8}\)

= {[(4 Ã— 3) + 2] Ã· 3} – {[(1 Ã— 8) + 1] Ã· 8}

= [(12 + 2) Ã· 3] – [(8 + 1) Ã· 8]

= (14 Ã· 3) – (9 Ã· 8)

Question 12.

Answer:

10\(\frac{7}{10}\) – 7\(\frac{5}{9}\) = 3\(\frac{13}{90}\)

Explanation:

10\(\frac{7}{10}\) – 7\(\frac{5}{9}\)

= {[(10 Ã— 10) + 7] Ã· 10} – {[(7 Ã— 9) + 5] Ã· 9}

= [(100 + 7) Ã· 10] – [(63 + 5) Ã· 9]

= (107 Ã· 10) – (68 Ã· 9)

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 Ã— 2) + 1] Ã· 2} – {[(3 Ã— 8) + 7] Ã· 8}

= [(16 + 1) Ã· 2] – [(24 + 7) Ã· 8]

= (17 Ã· 2) – (31 Ã· 8)

=

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 Ã— 8) + 7] Ã· 8} – \(\frac{1}{4}\)

= [(8 + 7) Ã· 8] – \(\frac{1}{4}\)

= \(\frac{15}{8}\) – \(\frac{1}{4}\)

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 Ã— 5) + 3] Ã· 5} = {[(1 Ã— 8) + 5] Ã· 8}

=> [(5 + 3) Ã· 5] = [(8 + 5) Ã· 8]

=> (8 Ã· 5) = (13 Ã· 8)

LCD of 5 and 8: 40.

=> [(8 Ã— 8) Ã· 40 = (13 Ã— 5) Ã· 40

=> (64 Ã· 40) = (65 Ã· 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.