Eureka Math Grade 5 Module 3 Lesson 5 Answer Key (2024)

Engage NY Eureka Math 5th Grade Module 3 Lesson 5 Answer Key

Eureka Math Grade 5 Module 3 Lesson 5 Sprint Answer Key

A
Subtracting Fractions from a Whole Number

Question 1.
4 – \(\frac{1}{2}\) =
Answer:
4 – \(\frac{1}{2}\) = 3\(\frac{1}{2}\)
Explanation :
4 – \(\frac{1}{2}\) = \(\frac{8}{2}\) – \(\frac{1}{2}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)

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

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

Question 4.
1 – \(\frac{1}{2}\) =
Answer:
1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Explanation :
1 – \(\frac{1}{2}\) = \(\frac{2}{2}\) – \(\frac{1}{2}\) = \(\frac{1}{2}\)

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

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

Question 7.
4 – \(\frac{1}{3}\) =
Answer:
4 – \(\frac{1}{3}\) = 3\(\frac{2}{3}\)
Explanation :
4 – \(\frac{1}{3}\) = \(\frac{12}{3}\) – \(\frac{1}{3}\) = \(\frac{11}{3}\) = 3\(\frac{2}{3}\)

Question 8.
4 – \(\frac{2}{3}\) =
Answer:
4 – \(\frac{2}{3}\) = 3\(\frac{1}{3}\)
Explanation :
4 – \(\frac{2}{3}\) = \(\frac{12}{3}\) – \(\frac{2}{3}\) = \(\frac{10}{3}\) = 3\(\frac{1}{3}\)

Question 9.
2 – \(\frac{2}{3}\) =
Answer:
2 – \(\frac{1}{3}\) = 1\(\frac{2}{3}\)
Explanation :
2 – \(\frac{1}{3}\) = \(\frac{6}{3}\) – \(\frac{1}{3}\) = \(\frac{5}{3}\) = 1\(\frac{2}{3}\)

Question 10.
2 – \(\frac{1}{4}\) =
Answer:
2 – \(\frac{1}{4}\) = 1\(\frac{3}{4}\)
Explanation :
2 – \(\frac{1}{4}\) = \(\frac{8}{4}\) – \(\frac{1}{4}\) = \(\frac{7}{4}\) = 1\(\frac{3}{4}\)

Question 11.
2 – \(\frac{3}{4}\) =
Answer:
2 – \(\frac{3}{4}\) = 1\(\frac{1}{4}\)
Explanation :
2 – \(\frac{3}{4}\) = \(\frac{8}{4}\) – \(\frac{3}{4}\) = \(\frac{5}{4}\) = 1\(\frac{1}{4}\)

Question 12.
3 – \(\frac{3}{4}\) =
Answer:
3 – \(\frac{3}{4}\) = 2\(\frac{1}{4}\)
Explanation :
3 – \(\frac{3}{4}\) = \(\frac{12}{4}\) – \(\frac{3}{4}\) = \(\frac{9}{4}\) = 2\(\frac{1}{4}\)

Question 13.
3 – \(\frac{1}{4}\) =
Answer:
3 – \(\frac{1}{4}\) = 2\(\frac{3}{4}\)
Explanation :
3 – \(\frac{1}{4}\) = \(\frac{12}{4}\) – \(\frac{1}{4}\) = \(\frac{11}{4}\) = 2\(\frac{3}{4}\)

Question 14.
4 – \(\frac{3}{4}\) =
Answer:
4 – \(\frac{3}{4}\) = 3\(\frac{1}{4}\)
Explanation :
4 – \(\frac{3}{4}\) = \(\frac{16}{4}\) – \(\frac{3}{4}\) = \(\frac{13}{4}\) = 3\(\frac{1}{4}\)

Question 15.
2 – \(\frac{1}{10}\) =
Answer:
2 – \(\frac{1}{10}\) = 1\(\frac{9}{10}\)
Explanation :
2 – \(\frac{1}{10}\) = \(\frac{20}{10}\) – \(\frac{1}{10}\) = \(\frac{19}{10}\) = 1\(\frac{9}{10}\)

Question 16.
3 – \(\frac{9}{10}\) =
Answer:
3 – \(\frac{9}{10}\) = 2\(\frac{1}{10}\)
Explanation :
3 – \(\frac{9}{10}\) = \(\frac{30}{10}\) – \(\frac{9}{10}\) = \(\frac{21}{10}\) = 2\(\frac{1}{10}\)

Question 17.
2 – \(\frac{7}{10}\) =
Answer:
Answer:
2 – \(\frac{7}{10}\) = 1\(\frac{3}{10}\)
Explanation :
2 – \(\frac{7}{10}\) = \(\frac{20}{10}\) – \(\frac{7}{10}\) = \(\frac{13}{10}\) = 1\(\frac{3}{10}\)

Question 18.
4 – \(\frac{3}{10}\) =
Answer:
4 – \(\frac{1}{10}\) = 3\(\frac{9}{10}\)
Explanation :
4 – \(\frac{1}{10}\) = \(\frac{40}{10}\) – \(\frac{1}{10}\) = \(\frac{39}{10}\) = 3\(\frac{9}{10}\)

Question 19.
3 – \(\frac{1}{5}\) =
Answer:
3 – \(\frac{1}{5}\) = 2\(\frac{4}{5}\)
Explanation :
3 – \(\frac{1}{5}\) = \(\frac{15}{5}\) – \(\frac{1}{5}\) = \(\frac{14}{5}\) = 2\(\frac{4}{5}\)

Question 20.
3 – \(\frac{2}{5}\) =
Answer:
3 – \(\frac{2}{5}\) = 2\(\frac{3}{5}\)
Explanation :
3 – \(\frac{2}{5}\) = \(\frac{15}{5}\) – \(\frac{2}{5}\) = \(\frac{13}{5}\) = 2\(\frac{3}{5}\)

Question 21.
3 – \(\frac{4}{5}\) =
Answer:
3 – \(\frac{4}{5}\) = 2\(\frac{1}{5}\)
Explanation :
3 – \(\frac{4}{5}\) = \(\frac{15}{5}\) – \(\frac{4}{5}\) = \(\frac{11}{5}\) = 2\(\frac{1}{5}\)

Question 22.
3 – \(\frac{3}{5}\) =
Answer:
3 – \(\frac{3}{5}\) = 2\(\frac{2}{5}\)
Explanation :
3 – \(\frac{3}{5}\) = \(\frac{15}{5}\) – \(\frac{3}{5}\) = \(\frac{12}{5}\) = 2\(\frac{2}{5}\)

Question 23.
3 – \(\frac{1}{8}\) =
Answer:
3 – \(\frac{1}{8}\) = 2\(\frac{7}{8}\)
Explanation :
3 – \(\frac{1}{8}\) = \(\frac{24}{8}\) – \(\frac{1}{8}\) = \(\frac{23}{8}\) = 2\(\frac{7}{8}\)

Question 24.
3 – \(\frac{3}{8}\) =
Answer:
3 – \(\frac{3}{8}\) = 2\(\frac{5}{8}\)
Explanation :
3 – \(\frac{3}{8}\) = \(\frac{24}{8}\) – \(\frac{3}{8}\) = \(\frac{21}{8}\) = 2\(\frac{5}{8}\)

Eureka Math Grade 5 Module 3 Lesson 5 Problem Set Answer Key

Question 1.
For the following problems, draw a picture using the rectangular fraction model and write the answer. Simplify your answer, if possible.
a. \(\frac{1}{3}\) – \(\frac{1}{4}\) =
b. \(\frac{2}{3}\) – \(\frac{1}{2}\) =
c. \(\frac{5}{6}\) – \(\frac{1}{4}\) =
d. \(\frac{2}{3}\) – \(\frac{1}{7}\) =
e. \(\frac{3}{4}\) – \(\frac{3}{8}\) =
f. \(\frac{3}{4}\) – \(\frac{2}{7}\) =
Answer:
a.
\(\frac{1}{3}\) – \(\frac{1}{4}\)
L.c.m of 3 and 4 is 12
\(\frac{4}{12}\) – \(\frac{3}{12}\) = \(\frac{1}{12}\)

b.
\(\frac{2}{3}\) – \(\frac{1}{2}\)
lcm of 3 and 2 is 6
\(\frac{4}{6}\) – \(\frac{3}{6}\) = \(\frac{1}{6}\)

c.
\(\frac{5}{6}\) – \(\frac{1}{4}\)
lcm of 6 and 4 is 12
\(\frac{10}{12}\) – \(\frac{3}{12}\) = \(\frac{7}{12}\)

d.
\(\frac{2}{3}\) – \(\frac{1}{7}\)
lcm of 3 and 7 is 21 .
\(\frac{14}{21}\) – \(\frac{3}{21}\) = \(\frac{11}{21}\)

e.
\(\frac{3}{4}\) – \(\frac{3}{8}\)
lcm of 4 and 8 is 8
\(\frac{6}{8}\) – \(\frac{3}{8}\) = \(\frac{3}{8}\)

f.
\(\frac{3}{4}\) – \(\frac{2}{7}\)
lcm of 4 and 7 is 28
\(\frac{21}{28}\) – \(\frac{8}{28}\) = \(\frac{13}{28}\)

Question 2.
Mr. Penman had \(\frac{2}{3}\) liter of salt water. He used \(\frac{1}{5}\) of a liter for an experiment. How much salt water does Mr. Penman have left?
Answer:
Quantity of salt water = \(\frac{2}{3}\)
Quantity of salt water used = \(\frac{1}{5}\)
Quantity of salt water left = \(\frac{2}{3}\) – \(\frac{1}{5}\) = \(\frac{10}{15}\) – \(\frac{3}{15}\)
= \(\frac{7}{15}\) .

Question 3.
Sandra says that \(\frac{4}{7}\) – \(\frac{1}{3}\) = \(\frac{3}{4}\) because all you have to do is subtract the numerators and subtract the denominators. Convince Sandra that she is wrong. You may draw a rectangular fraction model to support your thinking.
Answer:
No, \(\frac{4}{7}\) – \(\frac{1}{3}\) = \(\frac{5}{21}\) not \(\frac{3}{4}\)

Explanation :
\(\frac{4}{7}\) – \(\frac{1}{3}\) =\(\frac{12}{21}\) – \(\frac{7}{21}\) = \(\frac{5}{21}\)
no, first find the l.c.m of the denominators that is lcm of 7 and 3 is 21 . then multiply the denominators to make 21 and and also multiply same number with the numerator . then after making denominators equal subtract the numerators .

Eureka Math Grade 5 Module 3 Lesson 5 Exit Ticket Answer Key

For the following problems, draw a picture using the rectangular fraction model and write the answer. Simplify your answer, if possible.
a. \(\frac{1}{2}\) – \(\frac{1}{7}\) =
b. \(\frac{3}{5}\) – \(\frac{1}{2}\) =
Answer:
a.
\(\frac{1}{2}\) – \(\frac{1}{7}\)
lcm of 2 and 7 is 14
\(\frac{7}{14}\) – \(\frac{2}{14}\) = \(\frac{5}{14}\)

b.
\(\frac{3}{5}\) – \(\frac{1}{2}\)
lcm of 5 and 2 is 10 .
\(\frac{6}{10}\) – \(\frac{5}{10}\) = \(\frac{1}{10}\)

Eureka Math Grade 5 Module 3 Lesson 5 Homework Answer Key

Question 1.
The picture below shows \(\frac{3}{4}\) of the rectangle shaded. Use the picture to show how to create an equivalent fraction for \(\frac{3}{4}\), and then subtract \(\frac{1}{3}\).
\(\frac{3}{4}\) – \(\frac{1}{3}\) =

Answer:

Explanation :
\(\frac{3}{4}\) – \(\frac{1}{3}\)
l.c.m of 4 and 3 is 12
\(\frac{9}{12}\) – \(\frac{4}{12}\) = \(\frac{5}{12}\)

Question 2.
Find the difference. Use a rectangular fraction model to find common denominators. Simplify your answer, if possible.
a. \(\frac{5}{6}\) – \(\frac{1}{3}\) =
b. \(\frac{2}{3}\) – \(\frac{1}{2}\) =
c. \(\frac{5}{6}\) – \(\frac{1}{4}\) =
d. \(\frac{4}{5}\) – \(\frac{1}{2}\) =
e. \(\frac{2}{3}\) – \(\frac{2}{5}\) =
f. \(\frac{5}{7}\) – \(\frac{2}{3}\) =
Answer:
a.

\(\frac{5}{6}\) – \(\frac{1}{3}\)
Lcm of 6 and 3 is 6 .
\(\frac{5}{6}\) – \(\frac{2}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

b.
\(\frac{2}{3}\) – \(\frac{1}{2}\)
lcm of 3 and 2 is 6
\(\frac{4}{6}\) – \(\frac{3}{6}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

c.
\(\frac{5}{6}\) – \(\frac{1}{4}\)
lcm of 6 and 4
\(\frac{10}{12}\) – \(\frac{3}{12}\) = \(\frac{7}{12}\)

d.
\(\frac{4}{5}\) – \(\frac{1}{2}\)
lcm of 5 and 2 is 10
\(\frac{8}{10}\) – \(\frac{5}{10}\) = \(\frac{3}{10}\)

e.
\(\frac{2}{3}\) – \(\frac{2}{5}\)
lcm of 3 and 5 is 15
\(\frac{10}{15}\) – \(\frac{6}{15}\)= \(\frac{4}{15}\)

f.
\(\frac{5}{7}\) – \(\frac{2}{3}\)
lcm of 7 and 3 is 21.
\(\frac{15}{21}\) – \(\frac{14}{21}\) = \(\frac{1}{21}\)

Question 3.
Robin used \(\frac{1}{4}\) of a pound of butter to make a cake. Before she started, she had \(\frac{7}{8}\) of a pound of butter. How much butter did Robin have when she was done baking? Give your answer as a fraction of a pound.
Answer:
Quantity of butter used to make cake = \(\frac{1}{4}\) pound
Quantity of butter with Robin before baking cake = \(\frac{7}{8}\) pound .
Total Quantity of butter with Robin after baking = \(\frac{7}{8}\) – \(\frac{1}{4}\) pound = \(\frac{7}{8}\) – \(\frac{2}{8}\) = \(\frac{5}{8}\) pound
Therefore, Robin have \(\frac{5}{8}\) pound when she was done baking .

Question 4.
Katrina needs \(\frac{3}{5}\) kilogram of flour for a recipe. Her mother has \(\frac{3}{7}\) kilogram of flour in her pantry. Is this enough flour for the recipe? If not, how much more will she need?
Answer:
Quantity of Flour Required for Recipe = \(\frac{3}{5}\)
Quantity of Flour with her mother = \(\frac{3}{7}\)
Quantity of Flour Enough or not = \(\frac{3}{7}\) – \(\frac{3}{5}\) = \(\frac{15}{35}\) – \(\frac{21}{35}\) = – \(\frac{6}{35}\) that means negative indicate doenot enough.
She needs more \(\frac{6}{35}\) Quantity of Flour for the Recipe .

Eureka Math Grade 5 Module 3 Lesson 5 Answer Key (2024)

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