Bridges in Mathematics Grade 4 Home Connections Unit 4 Module 2 Answer Key

Practicing the Bridges in Mathematics Grade 4 Home Connections Answer Key Unit 4 Module 2 will help students analyze their level of preparation.

Bridges in Mathematics Grade 4 Home Connections Answer Key Unit 4 Module 2

Bridges in Mathematics Grade 4 Home Connections Unit 4 Module 2 Session 1 Answer Key

Think Before You Add

Question 1.
Study each problem before you begin to solve it. Think about which strategy would be most efficient (easiest and fastest). Choose your strategy and solve the problem. Use the space below the problems if you need it to do your figuring.
Bridges in Mathematics Grade 4 Home Connections Unit 4 Module 2 Answer Key 1
Answer:
99 + 43 = 142
By adding 99 and 43 we get 142
878 + 121 = 999
By adding 878 and 121 we get 999.
213 + 762 = 975
By adding 213 and 762 we get 975.
232 + 75 = 307
By adding 232 and 75 we get 307.
Bridges-in-Mathematics-Grade-4-Home-Connections-Unit-4-Module-2-Answer-Key-1

Question 2.
Use the standard algorithm for addition to solve the problems below.
Bridges in Mathematics Grade 4 Home Connections Unit 4 Module 2 Answer Key 2
Answer:
189 + 215 = 404
By adding 189 and 215 we get 404.
57 + 84 = 141
By adding 57 and 84 we get 141.
378 + 497 = 875
By adding 378 and 497 we get 875.
764 + 135 = 899
By adding 764 and 135 we get 899.

Question 3.
Look at the problems in item 2. Find a problem that might have been solved faster with another strategy.
a. Which problem did you choose?
Answer: A standard algorithm is a set of steps to complete a process.

b. Which strategy could be faster? Why?
Answer: Standard algorithm could be faster instead of using fingers.

Mixed Review

Question 4.
Use the symbols >, =, or < to compare each pair of fractions. ex: \(\frac{1}{3}\) > \(\frac{1}{4}\)

a. \(\frac{3}{6}\) _______________ \(\frac{2}{3}\)
Answer:
The simplified fraction of \(\frac{3}{6}\) is \(\frac{1}{2}\)
\(\frac{1}{2}\) _______________ \(\frac{2}{3}\)
LCD is 6
\(\frac{3}{6}\) _______________ \(\frac{4}{6}\)
Compare the numerators of the fractions, the number with the greater numerator will be the greatest fraction.
\(\frac{3}{6}\) < \(\frac{4}{6}\)

b. \(\frac{1}{3}\) _______________ \(\frac{1}{4}\)
Answer:
\(\frac{1}{3}\) > \(\frac{1}{4}\)
The denominator with the greater number will be the smallest fraction.

c. \(\frac{3}{4}\) _______________ \(\frac{5}{6}\)
Answer:
\(\frac{3}{4}\) _______________ \(\frac{5}{6}\)
LCD is 12
\(\frac{9}{12}\) _______________ \(\frac{10}{12}\)
Compare the numerators of the fractions, the number with the greater numerator will be the greatest fraction.
\(\frac{9}{12}\) < \(\frac{10}{12}\)

d. \(\frac{2}{3}\) _______________ \(\frac{3}{4}\)
Answer:
\(\frac{2}{3}\) _______________ \(\frac{3}{4}\)
LCD is 12
\(\frac{8}{12}\) > \(\frac{9}{12}\)

e. \(\frac{1}{2}\) _______________ \(\frac{2}{4}\)
Answer:
\(\frac{2}{4}\) = \(\frac{1}{2}\)
\(\frac{1}{2}\) = \(\frac{1}{2}\)
\(\frac{1}{2}\) = \(\frac{2}{4}\)

f. \(\frac{1}{3}\) _______________ \(\frac{2}{4}\)
Answer:
\(\frac{1}{3}\) _______________ \(\frac{2}{4}\)
LCD is 12
\(\frac{4}{12}\) _______________ \(\frac{6}{12}\)
\(\frac{4}{12}\) < \(\frac{6}{12}\)

g. \(\frac{2}{6}\) _______________ \(\frac{1}{3}\)
Answer:
\(\frac{2}{6}\) _______________ \(\frac{1}{3}\)
LCD is 6.
\(\frac{2}{6}\) = \(\frac{2}{6}\)

Question 5.
Write the decimal name for each fraction.
a. 5\(\frac{9}{10}\) = ________________
Answer:
5\(\frac{9}{10}\)
\(\frac{9}{10}\) can be written in the decimal form as 0.9
5 + 0.9 = 5.9

b. 6\(\frac{5}{100}\) = ________________
Answer:
6\(\frac{5}{100}\)
\(\frac{5}{100}\) = 0.05
6 + 0.05 = 6.05

c. 2\(\frac{6}{10}\) = ________________
Answer:
2\(\frac{6}{10}\)
\(\frac{6}{10}\) = 0.6
2 + 0.6 = 2.6

d. 8\(\frac{1}{10}\) = ________________
Answer:
8\(\frac{1}{10}\)
\(\frac{1}{10}\) = 0.1
8 + 0.1 = 8.1

e. 1\(\frac{20}{100}\) = _________________
Answer:
1\(\frac{20}{100}\)
\(\frac{20}{100}\) = 0.2
1 + 0.2 = 1.2

f. 3\(\frac{4}{10}\) = _________________
Answer:
3\(\frac{4}{10}\)
\(\frac{4}{10}\) = 0.4
3 + 0.4 = 3.4

g. 9\(\frac{50}{100}\) = _________________
Answer:
9\(\frac{50}{100}\)
\(\frac{50}{100}\) = 0.5
9 + 0.5 = 9.5

Question 6.
CHALLENGE Last year, Monica’s snake was 9.62 inches long. Now her snake is 12.37 inches long. Show your work with numbers, labeled sketches, or words for each question below.
a. How much did Monica’s snake grow in the last year?
Answer:
Given,
Last year, Monica’s snake was 9.62 inches long. Now her snake is 12.37 inches long.
12.37 – 9.62 = 2.75 inches
Monica’s snake grow 2.75 inches last year.

b. How much more does her snake need to grow to be exactly 13 inches?
Answer:
13 – 12.37 = 0.63 inches

Bridges in Mathematics Grade 4 Home Connections Unit 4 Module 2 Session 3 Answer Key

Number Cards

Hayley pulled 6 cards from a regular deck of cards. She arranged the cards into these 3-digit numbers: 348 and 956.

Question 1.
What is the sum of Hayley’s numbers? Use the strategy of your choice and show your work below.
Answer:
Hayley pulled 6 cards from a regular deck of cards.
She arranged the cards into these 3-digit numbers: 348 and 956.
956
+348
1304

Question 2.
What is the difference between Hayley’s numbers? Use the strategy of your choice and show your work below.
Answer:
Hayley pulled 6 cards from a regular deck of cards.
She arranged the cards into these 3-digit numbers: 348 and 956.
Subtract 348 from 956 we get 608.
956
-348
6 0 8

Question 3.
What is the largest 6-digit number Hayley can make with the numbers she chose?
Answer:
The largest 6-digit number Hayley can make with the numbers she chose is 986543.
To get the largest number, place the largest number to the highest place value and follow respectively until the lowest and lowest place value that is “ones”.

Question 4.
What is the smallest 6-digit number Hayley can make with the numbers she chose?
Answer:
The smallest 6-digit number Hayley can make with the numbers she chose is 345689.
To get the smallest number, arrange the number in chronological order from smallest to largest number, that is the highest place value is the smallest number.

Question 5.
Hayley chose 6 more cards. This time she made these numbers: 278 and 421. Hayley says she can add 299 and 400 and get the same sum as 278 and 421. Do you agree or disagree? Why?
Answer:
Yes, because when we are going to add 278 and 421.
578 + 421 = 699
278 + 21 + 421 – 21 = 699
299 + 400 = 699

Question 6.
Hayley says she can find the difference between 278 and 421 by finding the difference between 300 and 443. Do you agree or disagree? Why?
Answer:
Yes, because when we subtract 421 and 278, the difference between them is 143 which is also the difference between 700 and 443.
421 – 278 = 143
(921 + 22) – (278 + 22) = 143
443 – 300 = 143
Therefore, the two are the same.

Review

Question 7.
Add these pairs of fractions. Express the answer for each as a fraction with denominator 100.
\(\frac{5}{10}\) + \(\frac{37}{100}\) = _________________
\(\frac{6}{10}\) + \(\frac{6}{100}\) = _________________
\(\frac{13}{10}\) + \(\frac{87}{100}\) = _________________
\(\frac{4}{10}\) + \(\frac{12}{100}\) = _________________
Answer:
\(\frac{5}{10}\) + \(\frac{37}{100}\)
LCD is 100
\(\frac{5}{10}\) × \(\frac{10}{10}\) + \(\frac{37}{100}\)
= \(\frac{50}{100}\) + \(\frac{37}{100}\)
= \(\frac{87}{100}\)

\(\frac{6}{10}\) + \(\frac{6}{100}\)
\(\frac{6}{10}\) × \(\frac{10}{10}\) + \(\frac{6}{100}\)
\(\frac{60}{100}\) + \(\frac{6}{100}\) = \(\frac{66}{100}\)

\(\frac{13}{10}\) + \(\frac{87}{100}\)
\(\frac{13}{10}\) × \(\frac{10}{10}\) + \(\frac{87}{100}\)
\(\frac{130}{100}\) + \(\frac{87}{100}\) = \(\frac{217}{100}\)

\(\frac{4}{10}\) + \(\frac{12}{100}\)
\(\frac{4}{10}\) × \(\frac{10}{10}\) + \(\frac{12}{100}\)
\(\frac{40}{100}\) + \(\frac{12}{100}\) = \(\frac{52}{100}\)

Question 8.
Place the decimals in their correct places on the number line.
0.4
0.1
0.8
0.25
0.55
0.95
Bridges in Mathematics Grade 4 Home Connections Unit 4 Module 2 Answer Key 3
Answer:
The order of the decimals from lowest to greatest is 0.1, 0.25, 0.4, 0.55, 0.8, 0.95
Bridges-in-Mathematics-Grade-4-Home-Connections-Unit-4-Module-2-Answer-Key-3

Story Problems

Question 9.
There are 137 third graders, 139 fourth graders, and 153 fifth graders at Wood Upper Primary School. How many students are there in all? Show your work using numbers, sketches, or words.
Answer:
Given,
There are 137 third graders, 139 fourth graders, and 153 fifth graders at Wood Upper Primary School.
137 + 139 + 153 = 429
Thus there are 429 students in all.

Question 10.
CHALLENGE Sarah, Rex, and Peter are all friends. One of them lives in a red house, one lives in a blue house, and the other lives in a green house. The person who lives in a green house has more than 3 letters in his name. The person who lives in a red house is not Rex. Which person lives in each house?
Answer:

Name Red Green Blue
Rex No No Yes
Peter No Yes Yes
Sarah Yes No No
  • The person who lives in a green house has more than 3 letters in his name. so this means that Rex is not living in the green house because he has exactly 3 letters in the name.
  • The person who lives in the red house is not rex so this means that Rex is not living in the red house as well.
  • The remaining available house for Rex is the blue house so Rex is living in the blue house.
  • Considering “the person who lives in a green house has more than 3 letters in his name” again. It used the pronoun “his” so this means that a boy is living in the green house. The remaining boy is peter.
  • The remaining available house for Sarah is the red house.

Bridges in Mathematics Grade 4 Home Connections Unit 4 Module 2 Session 5 Answer Key

Thinking About Subtraction

Question 1.
Look at each subtraction problem below. Think about which strategy makes the most sense for each problem. Solve each problem.
a.
Bridges in Mathematics Grade 4 Home Connections Unit 4 Module 2 Answer Key 4
What strategy did you use? Why did you use this strategy?
Answer:
Bridges-in-Mathematics-Grade-4-Home-Connections-Unit-4-Module-2-Answer-Key-4
By subtracting 4859 from 4875 we get 16.
You use a standard algorithm strategy to find the answer.

b.
Bridges in Mathematics Grade 4 Home Connections Unit 4 Module 2 Answer Key 5
What strategy did you use? Why did you use this strategy?
Answer:
Bridges-in-Mathematics-Grade-4-Home-Connections-Unit-4-Module-2-Answer-Key-5
By subtracting the same number from the given number we get the result 0.

c.
Bridges in Mathematics Grade 4 Home Connections Unit 4 Module 2 Answer Key 6
What strategy did you use? Why did you use this strategy?
Answer:
Bridges-in-Mathematics-Grade-4-Home-Connections-Unit-4-Module-2-Answer-Key-6
By subtracting 424 from 699 is 275.
You use a standard algorithm strategy to find the answer.

Question 2.
Fill in the blanks in the equations below.
498 – 323 = 500 – ______________
68 – ____________ = 70 – 55
1003 – 498 = _____________ – 495
Answer:
498 – 323 = 500 – 325
68 – 53 = 70 – 55
1003 – 498 = 1000 – 495

Review

Question 3.
Jenny got a box of 15 stickers for her birthday. Use this information as you solve each problem below. Use numbers, labeled sketches, or words to show your thinking.
a. Jenny used 5 stickers on a thank-you card. What fraction of the box did she use?
Answer:
Jenny got a box of 15 stickers for her birthday.
Jenny used 5 stickers on a thank-you card.
5/15 = 1/3
She used 1/3rd fraction of the box.

b. Jenny gave her brother 4 stickers. What fraction does she have left out of her box of 15?
Answer:
Jenny got a box of 15 stickers for her birthday.
Jenny gave her brother 4 stickers.
4/15
4/15 fraction of stickers left out of her box.

Question 4.
After she gave some stickers to her brother, Jenny’s dog ate 3 of her stickers.
a. Now what fraction does Jenny have left of her original box of 15 stickers?
Answer:
4/15 – 3/15 = 1/15
1/15 fraction of stickers is left of her original box of 15 stickers.

b. What fraction of the stickers went to Jenny’s brother and her dog?
Answer:
11/15 + 3/15 = 14/15
So, a 14/15 fraction of the stickers went to Jenny’s brother and her dog.

Question 5.
The third-grade gymnastics team has 279 points. In order to place in the top three teams, they’ll need a score of 425 or more. How many more points do they need to earn in order to rank in the top three?
Answer:
Given,
The third-grade gymnastics team has 279 points.
In order to place in the top three teams, they’ll need a score of 425 or more.
425 – 279 = 146
They need 146 more points to earn in order to rank in the top three.

Question 6.
CHALLENGE Brendan needs to mail a 12-page letter to his friend in Texas. It costs $1.38 to mail all 12 sheets together. A 6-page letter costs 68¢ to mail. A 4-page letter costs 45¢ to mail. Envelopes costs 3¢ each. What is the least expensive way to mail his 12 pages?
Answer:
Given,
Brendan needs to mail a 12-page letter to his friend in Texas. It costs $1.38 to mail all 12 sheets together. A 6-page letter costs 68¢ to mail
6-page × 2 = 12-page
68 × 2 = 136¢
136¢ = 1.36$
136 + 3 = 139¢ = 1.39$
A 4-page letter costs 45¢ to mail.
45 × 3 = 135¢
135¢ = 1.35$
135 + 3 = 138¢ = 1.38$
A 4-page letter is the least expensive way to mail his 12 pages.

Leave a Comment

Scroll to Top
Scroll to Top