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)

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)

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)

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)

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)

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)

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)

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)

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 × 4) + 1] ÷ 4} – {[(2 × 13) + 2] ÷ 13}

= [(20 + 1) ÷ 4] – [(26 + 2) ÷ 13]

= (21 ÷ 4) – (28 ÷ 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 × 7) + 6] ÷ 7} – {[(3 × 10) + 3] ÷ 10}

= [(133 + 6) ÷ 7] – [(30 + 3) ÷ 10]

= (139 ÷ 7) – (33 ÷ 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 × 3) + 2] ÷ 3} – {[(1 × 8) + 1] ÷ 8}

= [(12 + 2) ÷ 3] – [(8 + 1) ÷ 8]

= (14 ÷ 3) – (9 ÷ 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 × 10) + 7] ÷ 10} – {[(7 × 9) + 5] ÷ 9}

= [(100 + 7) ÷ 10] – [(63 + 5) ÷ 9]

= (107 ÷ 10) – (68 ÷ 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 × 2) + 1] ÷ 2} – {[(3 × 8) + 7] ÷ 8}

= [(16 + 1) ÷ 2] – [(24 + 7) ÷ 8]

= (17 ÷ 2) – (31 ÷ 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 × 8) + 7] ÷ 8} – \(\frac{1}{4}\)

= [(8 + 7) ÷ 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 × 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.