# 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$$\frac{1}{4}$$
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$$\frac{3}{4}$$
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$$\frac{1}{3}$$
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$$\frac{1}{5}$$
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$$\frac{1}{11}$$
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$$\frac{2}{3}$$
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$$\frac{1}{4}$$
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$$\frac{1}{6}$$
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.

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.

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.

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.

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?
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?
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.