# 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 DenominatorsÂ 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.

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