Practicing the Bridges in Mathematics Grade 4 Home Connections Answer Key Unit 6 Module 3 will help students analyze their level of preparation.
Bridges in Mathematics Grade 4 Home Connections Answer Key Unit 6 Module 3
Bridges in Mathematics Grade 4 Home Connections Unit 6 Module 3 Session 1 Answer Key
Rope Climb Results & Skills Review
Your P.E. teacher has challenged your class to a rope climb! There are 8 blue pieces of tape equally spaced, and wrapped around the rope to mark off the distances. The following results represent the goal levels that were reached by the students in your group.
Question 1.
Display this data on the line plot below. Enter the rest of the goal levels below the heavy line. Make an X above the heavy line to represent each student in your group. Give your finished line plot a good title.
Answer:
Explanation:
In the above image we can observe the data is displayed on the line plot and the rest of the goal levels are entered below the heavy line. To represent each student in group an X is kept above the heavy line. The line plot is represented with the name Rope climbing.
Question 2.
How many students stopped at the goal line 3/8?
Answer:
Two students stopped at the goal line 3/8.
Question 3.
Which goal level did the most students reach?
Answer:
Most of the students reached the goal level 4/8.
Question 4.
How many students touched or even passed 3/8 of the rope?
Answer:
Eight students touched or even passed 3/8 of the rope.
Question 5.
What was the total distance combined for climbing the rope?
Answer:
The total distance combined for climbing the rope is 1.
Question 6.
Solve 216 ÷ 6. Use a ratio table or an array to model and solve the problem.
Answer:
Question 7.
Kevin says that 0.6 is the same as \(\frac{6}{10}\). Do you agree or disagree? Why?
Answer:
I agree with kelvin words because 0.6 is same as the \(\frac{6}{10}\).
The decimal 0.6 can be written in fraction form as \(\frac{6}{10}\).
Question 8.
Write each fraction as a decimal.
\(\frac{4}{10}\) = 0.4
\(\frac{5}{10}\) = _______________
\(\frac{7}{10}\) = _______________
\(\frac{25}{100}\) = _______________
\(\frac{3}{100}\) = ________________
Answer:
\(\frac{4}{10}\) = 0.4
\(\frac{5}{10}\) = 0.5
\(\frac{7}{10}\) = 0.7
\(\frac{25}{100}\) = 0.25
\(\frac{3}{100}\) = 0.03Â
Explanation:
If the denominator value is 10 then we can remove 10 by shifting the decimal point to left side by one place.
If the denominator value is 100 then we can remove 100 by shifting the decimal point to left side by two places.
Question 9.
Write each decimal as a fraction.
0.31 = \(\frac{31}{100}\)
0.9 = ______________
0.1 = ______________
0.36 = ______________
0.75 = _______________
Answer:
0.31 = \(\frac{31}{100}\)
0.9 = \(\frac{9}{10}\)
0.1 = \(\frac{1}{10}\)
0.36 = \(\frac{36}{100}\)
0.75 = \(\frac{75}{100}\)
The above given decimals are converted to fraction forms.
Question 10.
Fill in the blanks with <, >, or =.
\(\frac{2}{3}\) _________________ \(\frac{3}{4}\)
\(\frac{5}{6}\) _________________ \(\frac{10}{12}\)
\(\frac{1}{3}\) _________________ \(\frac{1}{9}\)
\(\frac{4}{10}\) _________________ \(\frac{1}{2}\)
\(\frac{7}{10}\) _________________ \(\frac{75}{100}\)
Answer:
\(\frac{2}{3}\) < \(\frac{3}{4}\)
\(\frac{5}{6}\) = \(\frac{10}{12}\)
\(\frac{1}{3}\) > \(\frac{1}{9}\)
\(\frac{4}{10}\) < \(\frac{1}{2}\)
\(\frac{7}{10}\)< \(\frac{75}{100}\)
Bridges in Mathematics Grade 4 Home Connections Unit 6 Module 3 Session 3 Answer Key
Bakery Bundles
Question 1.
Rachel owns a bakery and sells cookies by the dozen. She sold 16 dozen cookies on Monday. How many cookies did Rachel sell? Show your work.
Answer:
Rachel sold 16 dozen cookies on Monday.
We know that,
1 dozen = 12
16 dozen = X
X = 16 × 12
X = 192
So, 16 dozen is equal to 192.
Rachel sold total 192 cookies on Monday.
Question 2.
For each dozen cookies, Rachel used 1\(\frac{1}{2}\) cups of milk. How many cups of milk did she use for 16 dozen cookies? Show your work.
Answer:
For each dozen cookies, Rachel used 1\(\frac{1}{2}\) cups of milk.
The mixed fraction 1\(\frac{1}{2}\) in fraction form as \(\frac{3}{2}\).
1\(\frac{1}{2}\) = (2 + 1)/2 = \(\frac{3}{2}\)
1 dozen = \(\frac{3}{2}\)
16 dozen = X
X = \(\frac{3}{2}\) × 16
X = 24 cups
Rachel used 24 cups of milk for 16 dozen cookies.
Question 3.
A customer ordered 28 cupcakes. What are all the different rectangular arrangements Rachel could use to package the cupcakes? Use labeled sketches to show the possible arrangements below.
Answer:
Question 4.
Rachel’s assistant says that \(\frac{3}{5}\) cup or oil is more than \(\frac{2}{3}\) cup or oil. Is he correct? Explain your reasoning.
Answer:
No, Rachel’s assistant is not correct.
\(\frac{3}{5}\) cup or oil is equal to 0.6.
\(\frac{2}{3}\) cup or oil is equal to 0.67.
So, \(\frac{3}{5}\) cup or oil is not more than \(\frac{2}{3}\) cup or oil.
Question 5.
Rachel uses \(\frac{4}{5}\) cup of cocoa for her brownies. Write two fractions that are equivalent to \(\frac{4}{5}\).
Answer:
Rachel uses \(\frac{4}{5}\) cup of cocoa for her brownies.
The two fractions that are equivalent to \(\frac{4}{5}\) are as follows.
\(\frac{4}{5}\) = \(\frac{8}{10}\), \(\frac{12}{15}\)Â
Question 6.
A large order of 240 cookies was placed. How many cookies would go in each box if Rachel put them in the different numbers of boxes listed below? Show your work for each.
a. 24 boxes?
Answer:
A large order of 240 cookies was placed.
24 boxes = 240 cookies
1 box = X
24 × X = 240 × 1
24X = 240
X = 240/24
X = 10 cookies
Rachel can put 10 cookies in each box in the given 24 boxes.
b. 12 boxes?
Answer:
A large order of 240 cookies was placed.
12 boxes = 240 cookies
1 box = X
12 × X = 240 × 1
12X = 240
X = 240/12
X = 20 cookies
Rachel can put 20 cookies in each box in the given 12 boxes.
c. 6 boxes?
Answer:
A large order of 240 cookies was placed.
6 boxes = 240 cookies
1 box = X
6 × X = 240 × 1
6X = 240
X = 240/6
X = 40 cookies
Rachel can put 40 cookies in each box in the given 6 boxes.
Question 7.
CHALLENGE Rachel had \(\frac{1}{4}\) gallon of milk left in her bakery. She needed to make 4 desserts for an order that afternoon. Use the table to help Rachel decide which dessert she can make 4 of with the milk she has left. Use equations, labeled sketches, or words to prove that your choice will work.
Answer:
Bridges in Mathematics Grade 4 Home Connections Unit 6 Module 3 Session 5 Answer Key
Danny’s Data
Danny collected data about the length of 12 worms he found while he was digging in his yard. Use the data shown on Danny’s line plot to answer the questions.
Question 1.
What is the range of the data? (The range is difference between the length of the longest and the shortest worm.) Show your work.
Answer:
The length of the longest worm is 5\(\frac{7}{8}\).
Convert the mixed fraction into fraction.
5\(\frac{7}{8}\) = ((5 x 8) + 7)/8 = (40 + 7)/8 = 47/8
The length of the shortest worm is 4\(\frac{2}{8}\).
Convert the mixed fraction into fraction.
4\(\frac{2}{8}\) = ((4 x 8) + 2)/8 = (32 + 2)/8 = 34/8
Range = 47/8 – 34/8 = 13/8 or 1\(\frac{5}{8}\)
The range of the data is 13/8.
Question 2.
What is the median (the middle value in the set of X’s) of the data?
Answer:
Median is the middle value in the set of X’s of the given data.
The middle value is 5.
Median = 5
Question 3.
What is the mode (the most common worm length) of the data?
Answer:
Mode is the most common worm length of the given data.
The most common worm length is 5\(\frac{3}{8}\).
Mode = 5\(\frac{3}{8}\)
Question 4.
What fraction of the 12 worms were less than 5 inches long?
Answer:
The number of worms less than 5 inches are 4.
Total number of worms are 12.
So, 4/12 worms are less than 5 inches long.
Question 5.
What fraction of the worms were longer than 4\(\frac{7}{8}\) inches but shorter than 5\(\frac{5}{8}\) inches?
Answer:
The worms that are longer than 4\(\frac{7}{8}\) inches but shorter than 5\(\frac{5}{8}\) inches are 7.
Total number of worms are 12.
So, 7/12 worms are were longer than 4\(\frac{7}{8}\) inches but shorter than 5\(\frac{5}{8}\) inches.
Question 6.
What fraction of the worms were more than 5 inches long?
Answer:
The number of worms more than 5 inches are 7.
Total number of worms are 12.
So, 7/12 worms are less than 5 inches long.
Question 7.
If Danny laid the two shortest worms end to end, how long would they be together? Show your work.
Answer:
The first shortest worm is 4\(\frac{2}{8}\).
The first shortest worm is 4\(\frac{5}{8}\).
The two shortest worms should be together for the length of the shortest worm 4\(\frac{2}{8}\).
Question 8.
If Danny put the longest and shortest worm end to end, how long would they be together? Show your work.
Answer:
The shortest worm is 4\(\frac{2}{8}\).
The longest worm is 5\(\frac{7}{8}\).
The longest and shortest worm should be together for the length of the shortest worm 4\(\frac{2}{8}\).
Question 9.
Add or subtract these mixed numbers. Show your work.
a. 2\(\frac{1}{3}\) + 4\(\frac{2}{3}\) = _________________
Answer:
2\(\frac{1}{3}\) = ((2 x 3) + 1)/3 = (6 + 1)/3 = 7/3
The mixed fraction 2\(\frac{1}{3}\) in fraction form as \(\frac{7}{3}\).
4\(\frac{2}{3}\) = ((4 x 3) + 2)/3 = (12 + 2)/3 = 14/3
The mixed fraction 4\(\frac{2}{3}\) in fraction form as \(\frac{14}{3}\).
7/3 + 14/3 = 21/3 = 7
2\(\frac{1}{3}\) + 4\(\frac{2}{3}\) = 7
b. 16\(\frac{5}{8}\) – 4\(\frac{3}{8}\) = ________________
Answer:
16\(\frac{5}{8}\) = ((16 x 8) + 5)/8 = (128 + 5)/8 = 133/8
The mixed fraction 16\(\frac{5}{8}\) in fraction form as \(\frac{133}{8}\).
4\(\frac{3}{8}\) = ((4 x 8) + 3)/8 = (32 + 3)/8 = 35/8
The mixed fraction 4\(\frac{3}{8}\) in fraction form as \(\frac{35}{8}\).
133/8 – 35/8 = 98/8
16\(\frac{5}{8}\) – 4\(\frac{3}{8}\) = 98/8
c. 8\(\frac{4}{7}\) – 3\(\frac{5}{7}\) = ________________
Answer:
8\(\frac{4}{7}\) = ((8 x 7) + 4)/7 = (56 + 4)/7 = 60/7
The mixed fraction 8\(\frac{4}{7}\) in fraction form as \(\frac{60}{7}\).
3\(\frac{5}{7}\) = ((3 x 7) + 5)/7 = (21 + 5)/7 = 26/7
The mixed fraction 3\(\frac{5}{7}\) in fraction form as \(\frac{26}{7}\).
60/7 – 26/7 = 34/7
8\(\frac{4}{7}\) – 3\(\frac{5}{7}\) =34/7
d. 14\(\frac{5}{9}\) + 6\(\frac{7}{9}\) = ________________
Answer:
14\(\frac{5}{9}\) = ((14 x 9) + 5)/9 = (126 + 5)/9= 131/9
The mixed fraction 14\(\frac{5}{9}\) in fraction form as \(\frac{131}{9}\).
6\(\frac{7}{9}\) = ((6 x 9) + 7)/9 = (54 + 7)/9 = 61/9
The mixed fraction 6\(\frac{7}{9}\) in fraction form as \(\frac{61}{9}\).
131/9 + 61/9 = 192/9
14\(\frac{5}{9}\) + 6\(\frac{7}{9}\) = 192/9
e. 20\(\frac{1}{8}\) – 19\(\frac{7}{8}\) = _________________
Answer:
20\(\frac{1}{8}\) = ((20 x 8) + 1)/8 = (160 + 1)/8 = 161/8
The mixed fraction 20\(\frac{1}{8}\) in fraction form as \(\frac{161}{8}\).
19\(\frac{7}{8}\) = ((19 x 8) + 7)/8 = (152 + 7)/8 = 159/8
The mixed fraction 19\(\frac{7}{8}\) in fraction form as \(\frac{159}{8}\).
161/8 – 159/8 = 2/8
20\(\frac{1}{8}\) – 19\(\frac{7}{8}\) = 2/8
Question 10.
CHALLENGE Danny found one more worm and wanted to add the data to his line plot. He wondered how it would affect the original mode, median and range.
a. What is a length the worm could be that would not change the mode?
Answer:
Mode is the most common worm length. As the current mode is 5\(\frac{3}{8}\), so, the length of new worm should also be 5\(\frac{3}{8}\) in order to not to change the current mode.
b. What is a length that would change the mode?
Answer:
A length of 4\(\frac{7}{8}\) or 5\(\frac{2}{8}\) can change the mode.
c. What is a length the worm could be that would not change the range?
Answer:
The length of the worm that would not change the range is greater than or equal to 4\(\frac{2}{8}\) and less than or equal to 5\(\frac{7}{8}\) in order to not to change the range.
d. What is a length that would change the range?
Answer:
The length of the worm that would change the range is less than 4\(\frac{2}{8}\) or greater than 5\(\frac{7}{8}\).
e. What is a length the worm could be that would not change the median?
Answer:
The length of the worm would be greater than or equal to 4\(\frac{2}{8}\) and less than or equal to 5\(\frac{7}{8}\) in order to not to change the median.
f. What is a length that would change the median?
Answer:
The length of the worm would be less than 4\(\frac{2}{8}\) and greater than 5\(\frac{7}{8}\) in order to change the median.