## Engage NY Eureka Math 5th Grade Module 3 Lesson 8 Answer Key

### Eureka Math Grade 5 Module 3 Lesson 8 Problem Set Answer Key

Question 1.
a. 2 + 1$$\frac{1}{5}$$ =
b. 2 – 1$$\frac{3}{8}$$ =
c. 5$$\frac{2}{5}$$ + 2$$\frac{3}{5}$$=
d. 4 – 2$$\frac{2}{7}$$ =
e. 9$$\frac{3}{4}$$ + 8 =
f. 17 – 15$$\frac{2}{3}$$ =
g. 15 + 17 $$\frac{2}{3}$$ =
h. 100 – 20$$\frac{7}{8}$$ =
a.
2 + 1$$\frac{1}{5}$$ = 2 + $$\frac{6}{5}$$
lcm of 5 and 1 is 5
$$\frac{10}{5}$$  + $$\frac{6}{5}$$
$$\frac{16}{5}$$
3$$\frac{1}{5}$$

b.
2 – 1$$\frac{3}{8}$$
2 – $$\frac{11}{8}$$
lcm is 8
2 – $$\frac{11}{8}$$
$$\frac{16}{8}$$ – $$\frac{11}{8}$$
$$\frac{5}{8}$$

c.
5$$\frac{2}{5}$$ + 2$$\frac{3}{5}$$
$$\frac{27}{5}$$ + $$\frac{13}{5}$$
$$\frac{40}{5}$$ = 8

d.
4 – 2$$\frac{2}{7}$$ = 4 – $$\frac{16}{7}$$
lcm of 1 and 7 is 7
$$\frac{28}{7}$$ – $$\frac{16}{7}$$ =$$\frac{12}{7}$$
e.
9$$\frac{3}{4}$$ + 8 = $$\frac{39}{4}$$ + 8
lcm of 1 and 4 is 4 .
$$\frac{39}{4}$$ + $$\frac{32}{4}$$  = $$\frac{71}{4}$$ =17 $$\frac{2}{4}$$
f.
17 – 15$$\frac{2}{3}$$ = 17 – $$\frac{47}{3}$$
lcm is 3
$$\frac{51}{3}$$ – $$\frac{47}{3}$$ = $$\frac{4}{3}$$
g.
15 + 17 $$\frac{2}{3}$$ = 15 + $$\frac{53}{3}$$
lcm is 3
$$\frac{45}{3}$$ + $$\frac{53}{3}$$ = $$\frac{98}{3}$$ = 32$$\frac{1}{3}$$
h.
100 – 20$$\frac{7}{8}$$ = 100 – $$\frac{167}{8}$$
lcm is 8
$$\frac{800}{8}$$ – $$\frac{167}{8}$$ = $$\frac{733}{8}$$ = 91$$\frac{5}{8}$$

Question 2.
Calvin had 30 minutes in time-out. For the first 23$$\frac{1}{3}$$ minutes, Calvin counted spots on the ceiling. For the rest of the time, he made faces at his stuffed tiger. How long did Calvin spend making faces at his tiger?
Number of Minutes of Time-out =30 minutes .
Fraction of Minutes did calvin counted spots on ceiling = 23$$\frac{1}{3}$$ minutes
Fraction of Minutes did calvin did faces at his stuffed tiger = x
30 minutes = 23$$\frac{1}{3}$$ + x
lcm is 3
30 minutes = $$\frac{70}{3}$$ + x
$$\frac{90}{3}$$ = $$\frac{70}{3}$$ + x
x = $$\frac{90}{3}$$ – $$\frac{70}{3}$$
x = $$\frac{20}{3}$$
Therefore, Fraction of Minutes did calvin did faces at his stuffed tiger = x = $$\frac{20}{3}$$ .

Question 3.
Linda planned to spend 9 hours practicing piano this week. By Tuesday, she had spent 2$$\frac{1}{2}$$ hours practicing. How much longer does she need to practice to reach her goal?
Number of Hours required to practice = 9 hours .
Fraction of hours practiced till Tuesday = 2$$\frac{1}{2}$$ hours.
Fraction of hours needed to practice to reach goal = x
9 hours = 2$$\frac{1}{2}$$  + x
lcm is 2
9 = $$\frac{5}{2}$$ + x
$$\frac{18}{2}$$ = $$\frac{5}{2}$$ + x
x = $$\frac{18}{2}$$ – $$\frac{5}{2}$$
x = $$\frac{13}{2}$$ .
Therefore Fraction of hours needed to practice to reach goal =x = $$\frac{13}{2}$$ .

Question 4.
Gary says that 3-1$$\frac{1}{3}$$ will be more than 2, since 3 – 1 is 2. Draw a picture to prove that Gary is wrong.
Gray is wrong
Explanation :
3-1$$\frac{1}{3}$$ = 3 – $$\frac{4}{3}$$
lcm is 3
$$\frac{9}{3}$$ – $$\frac{4}{3}$$  = $$\frac{5}{3}$$

### Eureka Math Grade 5 Module 3 Lesson 8 Exit Ticket Answer Key

a. 5 + 1$$\frac{7}{8}$$ =
b. 3 – 1$$\frac{3}{4}$$ =
c. 7$$\frac{3}{8}$$ + 4=
d. 4 – 2$$\frac{3}{7}$$ =
a.
5 + 1$$\frac{7}{8}$$ = 5 + $$\frac{15}{8}$$
lcm of 1 and 8 is 8
$$\frac{40}{8}$$ + $$\frac{15}{8}$$  = $$\frac{55}{8}$$ = 6$$\frac{7}{8}$$ .
b.
3 – 1$$\frac{3}{4}$$ = 3 – $$\frac{7}{4}$$
lcm is 4
$$\frac{12}{4}$$ – $$\frac{7}{4}$$ = $$\frac{5}{4}$$ =1$$\frac{1}{4}$$
c.
7$$\frac{3}{8}$$ + 4= $$\frac{59}{8}$$ + 4
lcm of 1 and 8 is 8
$$\frac{59}{8}$$ +$$\frac{32}{8}$$ = $$\frac{91}{8}$$ =11$$\frac{3}{8}$$ .
d.
4 – 2$$\frac{3}{7}$$ = 4 – $$\frac{17}{7}$$
lcm of 1 and 7 is 7
$$\frac{28}{7}$$ – $$\frac{17}{7}$$ = $$\frac{11}{7}$$ =1$$\frac{4}{7}$$ .

### Eureka Math Grade 5 Module 3 Lesson 8 Homework Answer Key

Question 1.
a. 3 + 1$$\frac{1}{4}$$ =
b. 2 – 1$$\frac{5}{8}$$ =
c. 5$$\frac{2}{5}$$ + 2 $$\frac{3}{5}$$ =
d. 4 – 2$$\frac{5}{7}$$ =
e. 8$$\frac{4}{5}$$ + 7 =
f. 18 – 15$$\frac{3}{4}$$ =
g. 16 + 18$$\frac{5}{6}$$ =
h. 100 -50$$\frac{3}{8}$$ =
a.
3 + 1$$\frac{1}{4}$$ = 3 + $$\frac{5}{4}$$
lcm of 1 and 4 is 4
$$\frac{12}{4}$$ + $$\frac{5}{4}$$ = $$\frac{17}{4}$$ = 4$$\frac{1}{4}$$ .
b.
2 – 1$$\frac{5}{8}$$ = 2 – $$\frac{13}{8}$$
lcm is 8
$$\frac{16}{8}$$  – $$\frac{13}{8}$$ =$$\frac{3}{8}$$ .
c.
5$$\frac{2}{5}$$ + 2 $$\frac{3}{5}$$ = $$\frac{27}{5}$$ + $$\frac{13}{5}$$ = $$\frac{40}{5}$$ = 8.
d.
4 – 2$$\frac{5}{7}$$ = 4 – $$\frac{19}{7}$$
lcm of 1 and 7 is 7 .
$$\frac{28}{7}$$  – $$\frac{19}{7}$$ = $$\frac{9}{7}$$ =1$$\frac{2}{7}$$
e.
8$$\frac{4}{5}$$ + 7 = $$\frac{44}{5}$$ + 7
lcm of  5 and 7 is  35.
$$\frac{308}{35}$$ + $$\frac{245}{35}$$ = $$\frac{553}{35}$$ =15 $$\frac{28}{35}$$
f.
18 – 15$$\frac{3}{4}$$ = 18 – $$\frac{63}{4}$$
lcm of 1 and 4 is 4.
18 – $$\frac{63}{4}$$  = $$\frac{72}{4}$$ – $$\frac{63}{4}$$  = $$\frac{9}{4}$$ =2$$\frac{1}{4}$$
g.
16 + 18$$\frac{5}{6}$$ = 16 + $$\frac{113}{6}$$
lcm is 6
$$\frac{96}{6}$$ + $$\frac{113}{6}$$  = $$\frac{209}{6}$$ = 34$$\frac{5}{6}$$
h.
100 -50$$\frac{3}{8}$$ = 100 –$$\frac{403}{8}$$
lcm of 1 and 8 is 8 .
$$\frac{800}{8}$$ – $$\frac{403}{8}$$ = $$\frac{397}{8}$$ = 49$$\frac{5}{8}$$ .

Question 2.
The total length of two ribbons is 13 meters. If one ribbon is 7$$\frac{5}{8}$$ meters long, what is the length of the other ribbon?
The total length of two ribbons = 13 meters.
Fraction of length of one ribbon = 7$$\frac{5}{8}$$
Fraction of length of other ribbon = x
13 = 7$$\frac{5}{8}$$ + x
13 = $$\frac{61}{8}$$ + x
lcm of 1 and 8 is 8
$$\frac{104}{8}$$ = $$\frac{61}{8}$$ + x
x = $$\frac{104}{8}$$ – $$\frac{61}{8}$$
x = $$\frac{43}{8}$$ = 5 $$\frac{3}{8}$$
Therefore, Fraction of length of other ribbon = x  = $$\frac{43}{8}$$ = 5 $$\frac{3}{8}$$ .

Question 3.
It took Sandy two hours to jog 13 miles. She ran 7$$\frac{1}{2}$$ miles in the first hour. How far did she run during the second hour?
Distance traveled by sandy in 2 hours = 13 miles.
Distance traveled in one hour = 7$$\frac{1}{2}$$ miles
Distance traveled in second hour = x
13 miles = 7$$\frac{1}{2}$$ + x miles.
13 = $$\frac{15}{2}$$ + x
lcm of 1 and 2 is 2 .
$$\frac{26}{2}$$ = $$\frac{15}{2}$$ + x
x = $$\frac{26}{2}$$ – $$\frac{15}{2}$$
x = $$\frac{11}{2}$$ = 5$$\frac{1}{2}$$ .
Therefore, Distance traveled in second hour = x = $$\frac{11}{2}$$ = 5$$\frac{1}{2}$$

Question 4.
Andre says that 5$$\frac{3}{4}$$ + 2$$\frac{1}{4}$$ = 7$$\frac{1}{2}$$ because 7$$\frac{4}{8}$$ = 7 $$\frac{1}{2}$$. Identify his mistake. Draw a picture to prove that he is wrong.
5$$\frac{3}{4}$$ + 2$$\frac{1}{4}$$ = 7$$\frac{1}{2}$$
5$$\frac{3}{4}$$ + 2$$\frac{1}{4}$$ = $$\frac{23}{4}$$ + $$\frac{9}{4}$$ = $$\frac{32}{4}$$ = 8
that means andre subtracted $$\frac{3}{4}$$ and $$\frac{1}{4}$$ instead of adding