lesson 11 homework 5.5 answer key

Common Core Grade 5 Math (Worksheets, Homework, Lesson Plans)

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Grade 5 Homework, Lesson Plans And Worksheets

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Go Math! 5 Student Edition, Grade: 5 Publisher: Houghton Mifflin Harcourt

Go math 5 student edition, title : go math 5 student edition, publisher : houghton mifflin harcourt, isbn : 547352042, isbn-13 : 9780547352046, use the table below to find videos, mobile apps, worksheets and lessons that supplement go math 5 student edition., textbook resources.

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Math Homework Pages and Answers

Topic 1: understand place value.

1-1: Patterns with Exponents and Powers of 10

• Homework Page

1-2: Understand Whole-Number Place Value

1-3: Decimals to Thousandths

1-4: Understand Decimal Place Value

1-5: Compare Decimals

1-6: Round Decimals

Topic 2: Use Models and Strategies to Add and Subtract Decimals

2-2: Estimate Sums and Differences of Decimals

2-3: Use Models to Add and Subtract Decimals

2-4: Use Strategies to Add Decimals

2-5: Use Strategies to Subtract Decimals

2-6: Model with Math

Topic 3: Fluently Multiply Multi-Digit Whole Numbers

3-1: Multiply Greater Numbers by Powers of 10

3-2: Estimate Products

3-3: Multiply by 1-Digit Numbers

3-4: Multiply 2-Digit by 2-Digit Numbers

3-5: Multiply 3-Digit by 2-Digit Numbers

3-6: Multiply Whole Numbers with Zeros

3-7: Practice Multiplying Multi-Digit Numbers

3-8: Solve Word Problems

3-9: Critique Reasoning

Topic 4: Use Models and Strategies to Multiply Decimals

4-1:Multiply Decimals by Powers of 10

4-2: Estimate the Product of a Decimal and a Whole Number 

4-3: Use Models to Multiply a Decimal and a Whole Number

4-4: Multiply a Decimal and a Whole Number

4-5: Use Models to Multiply a Decimal and a Decimal

4-6: Multiply Decimals Using Partial Products

4-7: Use Properties to Multiply Decimals

4-8: Use Number Sense to Multiply Decimals

4-9: Model with Math

Topic 5: Use Models and Strategies to Divide Whole Numbers

Topic 5-1: Use Patterns and Mental Math to Divide

Topic 5-2: Estimate Quotients with 2-Digit Divisors

Topic 5-3: Use Models and Properties to Divide with 2-Digit Divisors

Topic 5-4: Use Partial Quotients to Divide

Topic 5-5: Use Sharing to Divide: Two Digit Divisors

Topic 5-6: Use Sharing to Divide: Greater Dividends

Topic 5-7: Choose a Strategy to Divide 

Lesson 5-8: Make Sense and Persevere

Topic 6: Use Models and Strategies to Divide Decimals

6-1: Patterns for Dividing with Decimals

6-2: Estimate Decimals Quotients

6-3: Use Models to Divide by a 1-Digit Number

6-4: Divide by a 2-digit Whole Number

6-5: Divide by a Decimal

6-6: Reasoning 

Topic 7: Use Equivalent Fractions to Add and Subtract Fractions

7-2: Find Common Denominators

•   Answers

7-3: Add Fractions with Unlike Denominators

7-4: Subtract Fractions with Unlike Denominators

7-5: Add and Subtract Fractions

7-6: Estimate Sums and Differences of Mixed Numbers

7-7: Use Models to Add Mixed Numbers

7-8: Add Mixed Numbers

7-9: Use Models to Subtract Mixed Numbers

7-10: Subtract Mixed Numbers

7-11: Add and Subtract Mixed Numbers

Topic 8: Apply Understanding of Multiplication to Multiply Fractions

8-1: Multiply a Fraction by a Whole Number

8-2: Multiply a Whole Number by a Fraction

8-3: Multiply Fractions and Whole Numbers

8-4: Use Models to Multiply Two Fractions

8-5: Multiply Two Fractions

8-6: Area of a Rectangle

8-7: Multiply Mixed Numbers

Topic 9: Apply Understanding of Division to Divide Fractions

Lesson 9-1: Fractions and Division

Lesson 9-2: Fractions and Mixed Numbers as Quotients

Lesson 9-3: Use Multiplication to Divide

Lesson 9-4: Divide Whole Numbers by Unit Fractions

Lesson 9-5: Divide Unit Fractions by Non-Zero Whole Numbers

Lesson 9-6: Divide Whole Numbers and Unit Fractions

Lesson 9-7: Solve Problems Using Division

Lesson 9-8: Repeated Reasoning

Topic 10: Represent and Interpret Data

Lesson 10-1: Analyze Line Plots

Lesson 10-2: Make Line Plots

Lesson 10-3: Solve Word Problems Using Measurement Data

Lesson 10-4: Critique Reasoning 

Topic 11: Understand Volume Concepts

Lesson 11-1: Model Volume

Lesson 11-2: Develop a Formula

Lesson 11-3: Combine Volume of Prisms

Lesson 11-4: Solve Word Problems Using Volume

Lesson 11-5: Use Appropriate Tools

5th Grade Homework Policy

We value your family time. therefore, we will be intentional with any homework we send home. students’ daily homework will be required reading of at least 30 minutes., students will have nightly math homework which supports our learning in class. there are a lot of new math concepts in 5th grade and it is important for students' growth and understanding. additionally, study guides and other assignments may be sent home periodically throughout the year., please note: if a student exhibits off-task behaviors, fails to complete an assignment, or is struggling to understand a concept, an assignment will be sent home for completion..

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Texas Go Math Grade 5 Lesson 5.5 Answer Key Add and Subtract Fractions

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 5.5 Answer Key Add and Subtract Fractions.

Unlock the Problem

Malia bought shell beads and glass beads to weave into designs in her baskets. She bought \(\frac{1}{4}\) pound of shell beads and \(\frac{3}{8}\) pound of glass beads. How many pounds of beads did she buy?

• Underline the question you need to answer.
• Draw a circle around the information you will use.

Add. \(\frac{1}{4}\) + \(\frac{3}{8}\). Write your answer in simplest form.

Another Way

So, Malia bought _________ pound of beads. Answer:

Another Way The least common denominator of 1 4  and  3 8 is 8

So, Malia bought \(\frac{5}{8}\)  pound of beads.

Lesson 5.5 Go Math Grade 5 Answer Key Question 1. Explain how you know whether your answer is reasonable. Answer: Both methods are the same they both give the same answer The least common denominator is the simplest method

When subtracting two fractions with unequal denominators, follow the same steps you follow when adding two fractions. However, instead of adding the fractions, subtract.

Question 2. Explain how you know whether your answer is reasonable. Answer: The fraction solved into simplest form is reasonable which found by least common denominator

Share and Show

Find the sum or difference. Write your answer in simplest form.

Question 1. \(\frac{5}{12}\) + \(\frac{1}{3}\) Answer: \(\frac{5}{12}\) + \(\frac{1}{3}\) = \(\frac{5}{12}\) + \(\frac{4}{12}\) = \(\frac{9}{12}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 2. \(\frac{2}{5}\) + \(\frac{3}{7}\) Answer: \(\frac{2}{5}\) + \(\frac{3}{7}\) = \(\frac{14}{35}\) + \(\frac{15}{35}\) =\(\frac{29}{35}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 3. \(\frac{1}{6}\) + \(\frac{3}{4}\) Answer: \(\frac{1}{6}\) + \(\frac{3}{4}\) = \(\frac{2}{12}\) + \(\frac{9}{12}\) = \(\frac{11}{12}\) Explanation: Step 1: The least common denominator is found Step 2: write equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Lesson 5.5 Answer Key Go Math Grade 5 Question 4. \(\frac{3}{4}\) – \(\frac{1}{8}\) Answer: \(\frac{3}{4}\) – \(\frac{1}{8}\) = \(\frac{6}{8}\) – \(\frac{1}{8}\) = \(\frac{5}{8}\) Explanation: Step 1: The least common denominator is found Step 2: write equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 5. \(\frac{1}{4}\) – \(\frac{1}{7}\) Answer: \(\frac{1}{4}\) – \(\frac{1}{7}\) = \(\frac{7}{28}\) – \(\frac{4}{28}\)= \(\frac{3}{28}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 6. \(\frac{9}{10}\) – \(\frac{1}{4}\) Answer: \(\frac{9}{10}\) – \(\frac{1}{4}\) = \(\frac{18}{20}\) – \(\frac{5}{20}\)= \(\frac{13}{20}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Math Talk Mathematical Processes

Explain why it is important to check your answer for reasonableness. Answer:

Problem Solving

Practice: Copy and Solve Find the sum or difference. Write your answer in simplest form.

Question 7. \(\frac{1}{3}\) + \(\frac{4}{18}\) Answer: \(\frac{1}{3}\) + \(\frac{4}{18}\) = \(\frac{6}{18}\) + \(\frac{4}{18}\) =\(\frac{10}{18}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 8. \(\frac{3}{5}\) + \(\frac{1}{3}\) Answer: \(\frac{3}{5}\) + \(\frac{1}{3}\) = \(\frac{9}{15}\) + \(\frac{5}{15}\) = \(\frac{14}{15}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 9. \(\frac{3}{10}\) + \(\frac{1}{6}\) Answer: \(\frac{3}{10}\) + \(\frac{1}{6}\) = \(\frac{9}{30}\) + \(\frac{5}{30}\) = \(\frac{14}{30}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 10. \(\frac{1}{2}\) + \(\frac{4}{9}\) Answer: \(\frac{1}{2}\) + \(\frac{4}{9}\) = \(\frac{9}{18}\) + \(\frac{8}{18}\) = \(\frac{17}{18}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Go Math Grade 5 Lesson 5.5 Answer Key Question 11. \(\frac{1}{2}\) – \(\frac{3}{8}\) Answer: \(\frac{1}{2}\) – \(\frac{3}{8}\) = \(\frac{4}{8}\) – \(\frac{3}{8}\) = \(\frac{1}{8}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 12. \(\frac{5}{7}\) – \(\frac{2}{3}\) Answer: \(\frac{5}{7}\) – \(\frac{2}{3}\) = \(\frac{15}{21}\) – \(\frac{14}{21}\) = \(\frac{1}{21}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 13. \(\frac{4}{9}\) – \(\frac{1}{6}\) Answer: \(\frac{4}{9}\) – \(\frac{1}{6}\) = \(\frac{8}{18}\) – \(\frac{3}{18}\) = \(\frac{5}{18}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 14. \(\frac{11}{12}\) – \(\frac{7}{15}\) Answer: \(\frac{11}{12}\) – \(\frac{7}{15}\) = \(\frac{55}{60}\) – \(\frac{28}{60}\) = \(\frac{27}{60}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

H.O.T. Algebra Find the unknown number.

Question 15. \(\frac{9}{10}\) – ☐ = \(\frac{1}{5\) ☐ = ___________ Answer: \(\frac{9}{10}\) – \(\frac{7}{10}\)= \(\frac{1}{5\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 16. \(\frac{5}{12}\) + ☐ = \(\frac{1}{2}\) ☐ = ____________ Answer: \(\frac{5}{12}\) + \(\frac{1}{12}\) =\(\frac{6}{12}\) =\(\frac{1}{2}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Go Math Grade 5 Lesson 5.5 Practice Answer Key Question 17. Sara is making a key chain using the bead design shown. What fraction of the beads in her design are either blue or red? Answer: Explanation: Let us consider dark black as red and light black-as-blue The number of beads is 15 Number of red beads are \(\frac{5}{15}\) Number of black beads are \(\frac{6}{15}\)

Question 19. Write Math Jamie had \(\frac{4}{5}\) of a spool of twine. He then used \(\frac{1}{2}\) of a spool of twine to make friendship knots. He claims to have \(\frac{3}{10}\) of the original spool of twine leftover. How you know whether Jamie’s claim is reasonable. Answer: Yes. Jamie’s claim is reasonable. Explanation: Jamie had \(\frac{4}{5}\) of a spool of twine. He then used \(\frac{1}{2}\) of a spool of twine to make friendship knots.  So \(\frac{3}{10}\) of the original spool of twine leftover. Since \(\frac{4}{5}\) –\(\frac{1}{2}\) = \(\frac{8}{10}\) – \(\frac{5}{10}\) = \(\frac{3}{10}\) He claims to have \(\frac{3}{10}\) of the original spool of twine leftover. So it is equla to what he leftover. So his claim is reasonabale.

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 20. Apply Students are voting for a new school mascot. So far, the results show that \(\frac{3}{10}\) of the students voted for “Fightin’ Titan,” \(\frac{1}{2}\) of the students voted for “Nifty Knight,” and the rest of the students have not voted yet. What fraction of the student population has not voted yet? (A) \(\frac{3}{10}\) (B) \(\frac{2}{5}\) (C) \(\frac{1}{5}\) (D) \(\frac{4}{5}\) Answer: (C) \(\frac{1}{5}\) Explanation: So far, the results show that \(\frac{3}{10}\) of the students voted for “Fightin’ Titan,” \(\frac{1}{2}\) of the students voted for “Nifty Knight,” Then \(\frac{8}{10}\) voted. Since \(\frac{3}{10}\) +\(\frac{1}{2}\) = \(\frac{8}{10}\) So \(\frac{1}{5}\) of the students have not voted yet. Since 1- \(\frac{8}{10}\) = \(\frac{1}{5}\)

Question 21. Tina spent \(\frac{3}{5}\) of her paycheck on a trip to the beach. She spent \(\frac{3}{8}\) of her paycheck on new clothes for the trip. What fraction of her paycheck did Tina spend on the trip and clothes together? (A) \(\frac{9}{40}\) (B) \(\frac{3}{4}\) (C) \(\frac{7}{8}\) (D) \(\frac{39}{40}\) Answer: (D) \(\frac{39}{40}\) Explanation: Tina spent \(\frac{3}{5}\) of her paycheck on a trip to the beach. She spent \(\frac{3}{8}\) of her paycheck on new clothes for the trip. So Tortal Spent is \(\frac{39}{40}\) \(\frac{3}{5}\) + \(\frac{3}{8}\)  = \(\frac{39}{40}\)

Question 22. Multi-Step On Friday, \(\frac{1}{6}\) of band practice was spent trying on uniforms. The band spent \(\frac{1}{4}\) of practice on marching. What fraction of practice time was left for playing music? (A) \(\frac{5}{12}\) (B) \(\frac{1}{2}\) (C) \(\frac{7}{12}\) (D) \(\frac{1}{4}\) Answer: (C) \(\frac{7}{12}\) Explanation: \(\frac{1}{6}\) of band practice was spent trying on uniforms. The band spent \(\frac{1}{4}\) of practice on marching. So Total time spent is \(\frac{5}{12}\). So Time left is \(\frac{7}{12}\) \(\frac{1}{6}\) + \(\frac{1}{4}\) = \(\frac{5}{12}\) 1-\(\frac{5}{12}\) =\(\frac{7}{12}\)

Texas Test Prep

Texas Go Math Grade 5 Lesson 5.5 Homework and Practice Answer Key

Question 1. \(\frac{1}{5}\) + \(\frac{1}{2}\) ____________ Answer: \(\frac{1}{5}\) + \(\frac{1}{2}\) = \(\frac{2}{10}\) + \(\frac{5}{10}\) = \(\frac{7}{10}\) Explanation: Step 1: The least common denominator is found Step 2: write equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Go Math Grade 5 Lesson 5.5 Answer Key Question 2. \(\frac{2}{3}\) + \(\frac{1}{6}\) ____________ Answer: \(\frac{2}{3}\) + \(\frac{1}{6}\) = \(\frac{4}{6}\) + \(\frac{1}{6}\) = \(\frac{5}{6}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 3. \(\frac{1}{4}\) + \(\frac{2}{3}\) ____________ Answer: \(\frac{1}{4}\) + \(\frac{2}{3}\) = \(\frac{3}{12}\) + \(\frac{8}{12}\) = \(\frac{11}{12}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 4. \(\frac{3}{4}\) + \(\frac{1}{8}\) ____________ Answer: \(\frac{3}{4}\) + \(\frac{1}{8}\) = \(\frac{6}{8}\) + \(\frac{1}{8}\) = \(\frac{7}{8}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 5. \(\frac{2}{9}\) + \(\frac{1}{3}\) ____________ Answer: \(\frac{2}{9}\) + \(\frac{1}{3}\) = \(\frac{2}{9}\) + \(\frac{3}{9}\) = \(\frac{5}{9}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 6. \(\frac{1}{2}\) + \(\frac{2}{6}\) ____________ Answer: \(\frac{1}{2}\) + \(\frac{2}{6}\) = \(\frac{3}{6}\) + \(\frac{2}{6}\) = \(\frac{5}{6}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 7. \(\frac{3}{10}\) + \(\frac{1}{3}\) ____________ Answer: \(\frac{3}{10}\) + \(\frac{1}{3}\) = \(\frac{9}{30}\) + \(\frac{10}{30}\) = \(\frac{19}{30}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 8. \(\frac{4}{18}\) + \(\frac{2}{6}\) ____________ Answer: \(\frac{4}{18}\) + \(\frac{2}{6}\) = \(\frac{4}{18}\) + \(\frac{6}{18}\) = \(\frac{10}{18}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 9. \(\frac{6}{12}\) – \(\frac{1}{3}\) ____________ Answer: \(\frac{6}{12}\) – \(\frac{1}{3}\) = \(\frac{6}{12}\) – \(\frac{4}{12}\) = \(\frac{2}{12}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 10. \(\frac{3}{4}\) – \(\frac{1}{6}\) ____________ Answer: \(\frac{3}{4}\) – \(\frac{1}{6}\) = \(\frac{9}{12}\) – \(\frac{2}{12}\) = \(\frac{7}{12}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 11. \(\frac{5}{7}\) – \(\frac{1}{2}\) ____________ Answer: \(\frac{5}{7}\) – \(\frac{1}{2}\) = \(\frac{10}{14}\) – \(\frac{7}{14}\) = \(\frac{3}{14}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 12. \(\frac{8}{9}\) – \(\frac{2}{3}\) ____________ Answer: \(\frac{8}{9}\) – \(\frac{2}{3}\) = \(\frac{8}{9}\) – \(\frac{6}{9}\) = \(\frac{2}{9}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 13. \(\frac{5}{9}\) – \(\frac{1}{6}\) ____________ Answer: \(\frac{5}{9}\) – \(\frac{1}{6}\) = \(\frac{10}{18}\) – \(\frac{3}{18}\) = \(\frac{7}{18}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 14. \(\frac{2}{3}\) – \(\frac{1}{4}\) ____________ Answer: \(\frac{2}{3}\) – \(\frac{1}{4}\) = \(\frac{8}{12}\) – \(\frac{3}{12}\) = \(\frac{5}{12}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 15. \(\frac{7}{14}\) – \(\frac{2}{7}\) ____________ Answer: \(\frac{7}{14}\) – \(\frac{2}{7}\) = \(\frac{7}{14}\) – \(\frac{4}{14}\) = \(\frac{3}{14}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 16. \(\frac{5}{6}\) – \(\frac{3}{4}\) ____________ Answer: \(\frac{5}{6}\) – \(\frac{3}{4}\) = \(\frac{10}{12}\) – \(\frac{9}{12}\) = \(\frac{1}{12}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Find the unknown number.

Question 17. \(\frac{7}{12}\) – ☐ = \(\frac{1}{6}\) ☐ = _____________ Answer: \(\frac{7}{12}\) – \(\frac{5}{12}\) = \(\frac{1}{6}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 18. \(\frac{5}{18}\) + ☐ = \(\frac{1}{2}\) ☐ = _____________ Answer: \(\frac{5}{18}\) + \(\frac{4}{18}\) = \(\frac{1}{2}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 19. \(\frac{7}{10}\) – ☐ = \(\frac{2}{5}\) ☐ = ______________ Answer: \(\frac{7}{10}\) – \(\frac{3}{10}\) = \(\frac{2}{5}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 20. ☐ + \(\frac{1}{9}\) = \(\frac{1}{3}\) ☐ = _______________ Answer: \(\frac{2}{9}\) + \(\frac{1}{9}\) = \(\frac{1}{3}\) Explanation: Step 1: The least common denominator is found Step 2: written equivalent fractions with equal denominators Step 3: write the answer in simplest form.

Question 21. There are 12 students in the pep squad. Three students are wearing white shirts. Six students are wearing blue shirts. What fraction of the students in the pep squad are wearing either white or blue shirts? Answer: \(\frac{1}{4}\)  wearing the white shirts and \(\frac{1}{2}\) wearing the blue shirts. Explanation: There are 12 students in the pep squad. Three students are wearing white shirts. \(\frac{3}{12}\)  = \(\frac{1}{4}\) Six students are wearing blue shirts. \(\frac{6}{12}\)  = \(\frac{1}{2}\)

Question 22. Tiffany ran \(\frac{5}{6}\) mile. Shayne ran \(\frac{3}{4}\) mile. Who ran farther? How much farther? Answer:

Lesson Check

Question 23. Mr. Benson spent \(\frac{2}{5}\) of the monthly budget on rent and \(\frac{3}{10}\) of the budget on food. What fraction of Mr. Benson’s budget was spent on rent and food? (A) \(\frac{1}{3}\) (B) \(\frac{3}{10}\) (C) \(\frac{7}{10}\) (D) \(\frac{1}{2}\) Answer: (C) \(\frac{7}{10}\) Explanation: Mr. Benson spent \(\frac{2}{5}\) of the monthly budget on rent and \(\frac{3}{10}\) of the budget on food. Sum of \(\frac{2}{5}\) and \(\frac{3}{10}\)  is \(\frac{7}{10}\) . Since \(\frac{3}{10}\)+ \(\frac{2}{5}\)= \(\frac{7}{10}\) .

Question 24. The Ortega family made \(\frac{15}{16}\) pound of confetti for the annual Fiesta celebration in San Antonio. They used \(\frac{1}{4}\) pound to make confetti filled eggs. How much confetti is left to use next year? (A) \(\frac{11}{16}\) pound (B) \(\frac{9}{16}\) pound (C) \(\frac{4}{5}\) pound (D) \(\frac{3}{4}\) pound Answer: (A) \(\frac{11}{16}\) pound Explanation: The Ortega family made \(\frac{15}{16}\) pound of confetti for the annual Fiesta celebration in San Antonio. They used \(\frac{1}{4}\) pound to make confetti filled eggs. confetti is left to use next year is \(\frac{11}{16}\) pound. Since

\(\frac{15}{16}\) – \(\frac{1}{4}\) = \(\frac{11}{16}\) pound

Question 25. If Rory measures the lemon juice and the vanilla extract into one spoon before adding them to the blender, how much liquid will be in the spoon? (A) \(\frac{5}{8}\) teaspoon (B) \(\frac{1}{5}\) teaspoon (C) \(\frac{1}{4}\) teaspoon (D) \(\frac{3}{8}\) teaspoon Answer: (A) \(\frac{5}{8}\) teaspoon Explanation: Sum of lemon juice and the vanilla extract is \(\frac{1}{2}\) teaspoon + \(\frac{1}{8}\) teaspoon = \(\frac{5}{8}\) teaspoon

Question 26. Multi-Step Rory has \(\frac{5}{8}\) cup of milk. How much milk does she have left after she doubles the recipe for the smoothie? (A) \(\frac{3}{8}\) cup (B) \(\frac{1}{8}\) cup (C) \(\frac{3}{4}\) cup (D) \(\frac{1}{2}\) cup Answer: (B) \(\frac{1}{8}\) cup Explanation: she doubles the recipe for the smoothie. So it is \(\frac{1}{2}\) cup. Since \(\frac{1}{4}\) cup + \(\frac{1}{4}\) cup  = \(\frac{1}{2}\) cup. Rory has \(\frac{5}{8}\) cup of milk. She left \(\frac{1}{8}\) cup of milk. Since \(\frac{5}{8}\)  –\(\frac{1}{2}\) cup. = \(\frac{1}{8}\)

Question 27. Multi-Step Torn has \(\frac{7}{8}\) cup of olive oil. He uses \(\frac{1}{2}\) cup to make salad dressing and \(\frac{1}{4}\) cup to make tomato sauce. How much olive oil does Torn have left? (A) \(\frac{5}{4}\) cups (B) \(\frac{5}{8}\) cup (C) \(\frac{3}{8}\) cup (D) \(\frac{1}{8}\) cup Answer: (D) \(\frac{1}{8}\) cup Explanation: \(\frac{1}{2}\) + \(\frac{1}{4}\)  = \(\frac{3}{4}\) and \(\frac{7}{8}\) cup – \(\frac{3}{4}\) cup = \(\frac{1}{8}\) cup

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Eureka Math Grade 5 Module 5 Lesson 17 Answer Key

Engage ny eureka math 5th grade module 5 lesson 17 answer key, eureka math grade 5 module 5 lesson 17 problem set answer key.

Question 1. Draw a parallelogram in each box with the attributes listed. a. No right angles. b. At least 2 right angles. c. Equal sides with no right angles. d. All sides equal with at least 2 right angles. Answer:

Question 2. Use the parallelograms you drew to complete the tasks below. a. Measure the angles of the parallelogram with your protractor, and record the measurements on the figures. b. Use a marker or crayon to circle pairs of angles inside each parallelogram with a sum equal to 180°. Use a different color for each pair. Answer:

Question 3. Draw another parallelogram below. a. Draw the diagonals, and measure their lengths. Record the measurements to the side of your figure. b. Measure the length of each of the four segments of the diagonals from the vertices to the point of intersection of the diagonals. Color the segments that have the same length the same color. What do you notice? Answer:

The diagonals bisect with each other.

Each diagonal  cuts the parallelogram into two equal parts.

Question 4. List the properties that are shared by all of the parallelograms that you worked with today. a. When can a quadrilateral also be called a parallelogram? b. When can a trapezoid also be called a parallelogram? Answer:

Properties of parallelogram :

It has 4 sides, and their diagonals bisect with each other.

They have 2 sets of opposite sides in parallel and opposite angles and sides are equal in measure.

A quadrilateral is also called a parallelogram when a quadrilateral has 2 sets of opposite sides  parallel

A trapezoid is also called a parallelogram when, It have 2 parallel sets of opposite sides pararllel.

Eureka Math Grade 5 Module 5 Lesson 17 Exit Ticket Answer Key

Question 1. Draw a parallelogram. Answer:

Question 2 When is a trapezoid also  called a parallelogram?

A trapezoid is also called a parallelogram when, It has 2 parallel sets of opposite sides.

Eureka Math Grade 5 Module 5 Lesson 17 Homework Answer Key

∠B = 120 Degrees, ∠C =  60 degrees  and ∠D = 120 degress

XY = 6 CM  and  YZ = 3 cm

∠XYZ =67  ° ∠YZW = 113° ∠ZWX = 67°

Question 3. Jack measured some segments in Problem 2. He found that \(\overline{W Y}\) = 8 cm and \(\overline{M Z}\) = 3 cm. Give the lengths of the following segments: WM = __________ cm MY = __________ cm XM = __________ cm XZ = __________ cm Answer:

WM = 4 cm MY = 4 cm XM = 3 cm XZ = 6 cm

Question 4. Using the properties of shapes, explain why all parallelograms are trapezoids. Answer:

Parallelogram have 2 sets of parallel lines

Trapezoid have at least one set of parallel lines.

Both shapes have diagnols which bisect with each other.

Question 5. Teresa says that because the diagonals of a parallelogram bisect each other, if one diagonal is 4.2 cm, the other diagonal must be half that length. Use words and pictures to explain Teresa’s error. Answer:

The two diagonals do not have that relatonship. They can be equal or different, + varies.

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