lesson 2 homework 5.1 answer key 5th grade

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5th grade (Eureka Math/EngageNY)

Unit 1: module 1: place value and decimal fractions, unit 2: module 2: multi-digit whole number and decimal fraction operations, unit 3: module 3: addition and subtractions of fractions, unit 4: module 4: multiplication and division of fractions and decimal fractions, unit 5: module 5: addition and multiplication with volume and area, unit 6: module 6: problem solving with the coordinate plane.

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Texas Go Math Grade 5 Lesson 5.1 Answer Key Addition with Unequal Denominators

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 5.1 Answer Key Addition with Unequal Denominators.

Investigate

Hilary is using red fabric to make a tote bag. She uses one piece that is \(\frac{1}{2}\) yard long. She uses another piece that is \(\frac{1}{4}\) yard long. How much red fabric does she use?

Materials; fraction strips; MathBoard

C. Record the sum in simplest form. \(\frac{1}{2}\) + \(\frac{1}{4}\) = ___________ So, Hilary uses ___________ yard of fabric. Answer: So, Hilary uses \(\frac{3}{4}\) yard of fabric. Record the sum in simplest form. \(\frac{1}{2}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\)

Math Talk Mathematical Processes

How can you tell if the sum of the fractions is less than 1? Answer: Fractions greater than 1 have numerators larger than their denominators; those that are less than 1 have numerators smaller than their denominators; the rest are equal to 1.

Draw Conclusions

Go Math Grade 5 Lesson 5.1 Answer Key Question 2. H.O.T. Explain the difference between finding fraction strips with the same denominator for \(\frac{1}{2}\) + \(\frac{1}{3}\) and \(\frac{1}{2}\) + \(\frac{1}{4}\). Answer:

Make Connections

Sometimes, the sum of two fractions is greater than 1. When adding fractions with unequal denominators, you can use the 1-whole strip to help determine If a sum is greater than 1 or less than 1.

Use fraction strips to solve. \(\frac{3}{5}\) + \(\frac{1}{2}\) STEP 1: Work with another student. Place three \(\frac{1}{5}\) fraction strips under the 1-whole strip on your MathBoard. Then place a \(\frac{1}{2}\) fraction strip beside the three \(\frac{1}{5}\) strips.

STEP 2: Find fraction strips, all with the same denominator, that are equivalent to \(\frac{3}{5}\) and \(\frac{1}{2}\). Place the fraction strips under the sum. At the right, draw a picture of the model and write the equivalent fractions. \(\frac{3}{5}\) = __________ \(\frac{1}{2}\) = __________

Share and Show

Use fraction strips to find the sum. Write your answer in simplest form.

Problem Solving

Go Math Practice and Homework Lesson 5.1 Answer Key Question 7. H.O.T. Pose a Problem Write a new problem using different amounts of ingredients Maya used. Each amount should be a fraction with a denominator of 2, 3, or 4. Answer: Maya makes trail mix by combining \(\frac{1}{2}\) cup mixed nuts, \(\frac{1}{3}\) cup of dried fruit, and \(\frac{1}{4}\) cup of chocolate morsels. What is the total amount of ingredients in her trail mix?

Question 9. Explain why you chose the amounts you did for your problem. Answer: in the question asked use different amounts for ingredients and Each amount should be a fraction with a denominator of 2, 3, or 4.

Question 10. Write Math Explain how using fraction strips with equal denominators makes it possible to add fractions with unequal denominators. Answer:

Think of the fruit analogy?

Does it make sense to add two bananas plus one watermelon? The units do not make sense for the sum.

BUT … we could think about changing both of them to common units, servings of fruit. If one banana serves one person and one watermelon serves ten people, then we could convert:

• two bananas + one watermelon =
• = two fruit servings + ten fruit servings =
• = twelve servings of fruit

Daily Assessment Task

Fill in the bubble completely to show your answer.

Texas Test Prep

Texas Go Math Grade 5 Lesson 5.1 Homework and Practice Answer Key

Use fraction strips to find the sum. Write your answer in the simplest form.

Question 6. \(\frac{1}{4}\) + \(\frac{2}{3}\) = ______________ Answer: \(\frac{11}{12}\) Explanation: To add the given fractions, did the denominator equal first then added \(\frac{1}{4}\) + \(\frac{2}{3}\) = \(\frac{3}{12}\) + \(\frac{8}{12}\) = \(\frac{11}{12}\)

Question 7. \(\frac{1}{3}\) + \(\frac{5}{6}\) = _____________ Answer: \(\frac{7}{6}\) Explanation: To add the given fractions, did the denominator equal first then added \(\frac{1}{3}\) + \(\frac{5}{6}\) = \(\frac{2}{6}\) + \(\frac{5}{6}\) = \(\frac{7}{6}\)

Question 8. \(\frac{3}{5}\) + \(\frac{3}{10}\) = ______________ Answer: \(\frac{9}{10}\) Explanation: To add the given fractions, did the denominator equal first then added \(\frac{3}{5}\) + \(\frac{3}{10}\) = \(\frac{6}{10}\) + \(\frac{3}{10}\) = \(\frac{9}{10}\)

Question 9. \(\frac{1}{8}\) + \(\frac{3}{4}\) = _____________ Answer: \(\frac{7}{8}\) Explanation: To add the given fractions, did the denominator equal first then added \(\frac{1}{8}\) + \(\frac{3}{4}\) = \(\frac{1}{8}\) + \(\frac{6}{8}\) = \(\frac{7}{8}\)

Question 10. \(\frac{7}{10}\) + \(\frac{1}{2}\) = _____________ Answer: \(\frac{7}{8}\) Explanation: To add the given fractions, did the denominator equal first then added \(\frac{7}{10}\) + \(\frac{1}{2}\) = \(\frac{7}{10}\) + \(\frac{5}{10}\) = \(\frac{12}{10}\) = \(\frac{6}{5}\)

Question 11. \(\frac{5}{6}\) + \(\frac{1}{12}\) = _____________ Answer: \(\frac{11}{12}\) Explanation: To add the given fractions, did the denominator equal first then added \(\frac{5}{6}\) + \(\frac{1}{12}\) = \(\frac{10}{12}\) + \(\frac{1}{12}\) = \(\frac{11}{12}\)

Question 12. Cooper is grating cheese for the family taco dinner. He grates \(\frac{1}{2}\) cup of cheddar cheese and \(\frac{3}{4}\) cup of monterey jack cheese. How much cheese does Cooper grate? Answer:  Cooper grate \(\frac{5}{4}\) cup of cheese Explanation: Cooper grates \(\frac{1}{2}\) cup of cheddar cheese and \(\frac{3}{4}\) cup of monterey jack cheese. then total cheese is sum of \(\frac{1}{2}\) cup of cheddar cheese and \(\frac{3}{4}\) cup of monterey jack cheese. So Total cheese is \(\frac{5}{4}\) \(\frac{1}{2}\)+\(\frac{3}{4}\)  =\(\frac{2}{4}\)+\(\frac{3}{4}\)  = \(\frac{5}{4}\)

Lesson 5.1 Answer Key Go Math 5th Grade Question 13. Jasmine has to mix \(\frac{3}{4}\) cup of flour and \(\frac{3}{8}\) cup of cornmeal. She has a container that holds 1 cup. Can Jasmine mix the flour and cornmeal in the container? Explain. Answer: Jasmine Can not mix the flour and cornmeal in the container. Since the total mix of flour and cornmeal is 1/8 cup more than a cup. Explanation: Jasmine has to mix \(\frac{3}{4}\) cup of flour and \(\frac{3}{8}\) cup of cornmeal. So Sum of flour and cornmeal is \(\frac{9}{8}\) if divided into fraction strips of the same denominator the \(\frac{8}{8}\) + \(\frac{1}{8}\). \(\frac{8}{8}\) is equals to 1 cup. So she can not mix in the container.

Lesson Check

Question 14. Julio spent \(\frac{1}{10}\) of his weekly allowance on a set of markers and \(\frac{2}{5}\) of it on a book. What fraction of Julio’s allowance is this altogether? (A) \(\frac{1}{2}\) (B) \(\frac{3}{10}\) (C) \(\frac{3}{5}\) (D) \(\frac{1}{5}\) Answer: \(\frac{1}{2}\) Explanation: \(\frac{1}{10}\) + \(\frac{2}{5}\)  = \(\frac{1}{10}\) + \(\frac{4}{10}\) =\(\frac{5}{10}\) = \(\frac{1}{2}\)

Question 17. An apple was cut into 8 equal-size pieces. Stacy ate \(\frac{1}{4}\) of the apple. Tony ate \(\frac{3}{8}\) of the apple. What part of the apple did Stacy and Tony eat in all? (A) \(\frac{1}{2}\) (B) \(\frac{5}{8}\) (C) \(\frac{3}{4}\) (D) \(\frac{1}{4}\) Answer: B Explanation: Equal the all denominators: Stacy ate \(\frac{1}{4}\) of the apple = \(\frac{1}{4}\) = \(\frac{2}{8}\) Tony ate \(\frac{3}{8}\) of the apple = \(\frac{3}{8}\) Stacy and Tony ate = \(\frac{3}{8}\)  + \(\frac{2}{8}\)= \(\frac{5}{8}\)

Question 18. Multi-Step Last weekend, Beatrice walked her poodle \(\frac{2}{3}\) mile on Saturday and \(\frac{5}{6}\) mile on Sunday. Fiona walked her beagle \(\frac{1}{3}\) mile on Saturday and \(\frac{1}{2}\) mile on Sunday. How much farther did the poodle walk last weekend than the beagle? (A) \(\frac{1}{2}\) mile (B) 1\(\frac{1}{3}\) miles (C) \(\frac{2}{3}\) mile (D) 1\(\frac{1}{2}\) miles Answer: C Explanation: Last weekend, Beatrice walked her poodle \(\frac{2}{3}\) mile on Saturday and \(\frac{5}{6}\) mile on Sunday. Fiona walked her beagle \(\frac{1}{3}\) mile on Saturday and \(\frac{1}{2}\) mile on Sunday. \(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{4}{6}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\) = \(\frac{3}{2}\) \(\frac{1}{3}\) + \(\frac{1}{2}\) = \(\frac{2}{6}\) + \(\frac{3}{6}\) = \(\frac{5}{6}\) \(\frac{3}{2}\) – \(\frac{5}{6}\) = \(\frac{9}{6}\) – \(\frac{5}{6}\) = \(\frac{2}{3}\)

Question 19. Multi-Step Rick worked in his garden on Friday. He pulled weeds for \(\frac{5}{6}\) hour, planted seeds for \(\frac{1}{2}\) hour, and watered for \(\frac{1}{6}\) hour. How much time did Rick spend working in his garden on Friday? (A) \(\frac{1}{2}\) hour (B) 1 hour (C) 1\(\frac{1}{3}\) hours (D) 1\(\frac{1}{2}\) hours Answer: D Explanation: \(\frac{5}{6}\) +\(\frac{1}{2}\) + \(\frac{1}{6}\) = \(\frac{5}{6}\)  +\(\frac{3}{6}\) +\(\frac{1}{6}\) = \(\frac{9}{6}\) = \(\frac{6}{6}\) +\(\frac{3}{6}\) = 1 + \(\frac{1}{2}\)

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Math Expressions Grade 5 Unit 7 Lesson 2 Answer Key Simplify Expressions

Solve the questions in Math Expressions Grade 5 Homework and Remembering Answer Key Unit 7 Lesson 2 Answer Key Simplify Expressions to attempt the exam with higher confidence. https://mathexpressionsanswerkey.com/math-expressions-grade-5-unit-7-lesson-2-answer-key/

Math Expressions Common Core Grade 5 Unit 7 Lesson 2 Answer Key Simplify Expressions

Math Expressions Grade 5 Unit 7 Lesson 2 Homework

Unit 7 Lesson 2 Simplify Expressions  Question 1. Follow the Order of Operations to simplify 27 ÷ (3 • 3) + 17 Step I Perform operations inside parentheses. ___________ Step 2 Multiply and divide from left to right. ___________ Step 3 Add and subtract from left to right. __________________ Answer:

Step 1: 27 ÷ (9) + 17

Step 2: 3 + 17

Simplify. Follow the Order of Operations.

Math Expressions Grade 5 Simplify  Question 2. 54 – 200 ÷ 4 Answer:

54 – 200 ÷ 4

54 – 50

Unit 7 Answer Key Math Expressions Lesson 2 Simplify  Question 3. 0.8 ÷ (0.07 – 0.06) Answer:

0.8 ÷ ( 0.01 )

Unit 7 Lesson 2 Math Expressions Question 4. 3 • 8 – 6 ÷ 2 Answer:

3 • 8 – 6 ÷ 2

3 . 8 – 3

24 – 3

Unit 7 Lesson 2 Answer Key Math Expressions Question 5. (\(\frac{3}{8}\) + \(\frac{1}{4}\)) • 16 Answer:

3/8 + 1/4 . 16

Math Expressions Common Core Grade 5 Simplify  Question 6. 64 + 46 – 21 + 29 Answer:

64 + 46 – 21 + 29

139 – 21

Simplify Answers Math Expressions Unit 7 Lesson 2 Question 7. 72 ÷ (7 – 1) • 3 Answer:

72 ÷ (7 – 1) • 3

Question 8. 0.8 – 0.5 ÷ 5 + 0.2 Answer:

0.8 – 0.5 ÷ 5 + 0.2

0.8 + 0.2 – 0.1

1 – 0.1

Question 9. \(\frac{5}{6}\) – 4 • \(\frac{1}{12}\)

5/6 – 4 . 1/12

5/6 – 1/3

Question 10. 5 • 15 ÷ 3 Answer:

Question 11. 32 ÷ (2.3 + 1.7) • 3 Answer:

32 ÷ (2.3 + 1.7) • 3

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

1 1/2 – 3/4 x 1/4

Question 13. (6.3 – 5.1) • (0.7 + 0.3) Answer:

(6.3 – 5.1) • (0.7 + 0.3)

Question 14. 12 ÷ 0.1 + 12 ÷ 0.01 Answer:

Question 15. \(\frac{1}{2}\) • \(\frac{1}{2}\) ÷ \(\frac{1}{2}\) Answer:

1/2 x 1/2 ÷  1/2

Question 16. 10 – 4 + 2 – 1 Answer:

10 – 4 + 2 – 1

10 – 6 – 1

Math Expressions Grade 5 Unit 7 Lesson 2 Remembering

Write an equation to solve the problem. Draw a model if you need to.

Question 4. The students of Turner Middle School are going on a field trip. There are 540 students attending. A bus can hold 45 students. How many buses are needed for the field trip? Answer:

Given, Total number of students = 540

Number of students a bus can hold = 45

Now, 540/45

Therefore, 12 busses are needed for the field trip.

Question 5. The area of a rectangular court is 433.37 square meters, and the length of the court is 28.7 meters. What is the width of the court? Answer:

Area of the rectangular court = 433.37 sq.m

Length = 28.7 sq. m

Area = length x width

433.37 = 28.7 x n

n = 433.37/28.7

n = 15.1 meters.

Write the computation in words.

Question 6. 5 ÷ \(\frac{1}{8}\) _____________________ Answer:

5 divided by 1/8

Question 7. 2.4 ÷ 0.6 + 0.2 __________________________ Answer:

Divide the sum of  0.6 and 0.2 by 2.4

Question 8. Stretch Your Thinking Write step-by-step instructions for simplifying the following expression. 10 • [60 ÷ (11 + 4)] – 3 Answer:

Step 1 : Add 11 + 4

10. [60 ÷ 15]-3

Step 2: Divide 60 by 15

10. [ 4 ] – 3

Step 3: Multiply 10 by 4

40 – 3

Step 4: Subtract 3 from 40

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Eureka Math Grade 5 Module 5 Lesson 15 Answer Key

Engage ny eureka math 5th grade module 5 lesson 15 answer key, eureka math grade 5 module 5 lesson 15 problem set answer key.

Question 1. The length of a flowerbed is 4 times as long as its width. If the width is “\(\frac{3}{8}\) meter, what is the area? Answer:

The width of the flower bed = 3/8 meters

The length of the flower bed is 4 times as long as its width

Which means, 3/8 x 4 = 12/ 8 = 3/2

Now, Area = length x width

A= 3/2 x 3/8

Therefore, the area of the flowerbed is 9/16 m 2 .

The measurement of basil plot on each side = 5/9 yard

A= 5/8 x 5/8 = 25/64

Therefore, the area of the basil plot = 25/64 yd 2

The measurement of basil plot in each side = 5/9 yard

=  15/8 or 1 7/8 feet

Now, the length of one side

2 feet + 2 feet + 1 7/8 feet

Perimetre = 4 x sides

= 4 x 5 7/8

= 20 + 3 4/8

Therefore, the perimetre of the fence = 23 1/2 feet.

The total area that the fence encloses

= length x width

= 5 7/8 x 5 7/8

=  ( 5 x 5 ) + ( 5 x 7/8 ) +( 5 x 7/8 ) + ( 7/8 x 7/8 )

= 25 + 4 3/8 + 4 3/8 + 49/64

= 33 + 48/64 + 49/64

= 33 + 97/64

Therefore, the area that fence encloses = 34 33/64 square feet.

Question 3. Janet bought 5 yards of fabric 2\(\frac{1}{4}\)-feet wide to make curtains. She used \(\frac{1}{3}\) of the fabric to make a long set of curtains and the rest to make 4 short sets. a. Find the area of the fabric she used for the long set of curtains. b. Find the area of the fabric she used for each of the short sets. Answer:

Area = length x width

= 5 x 2 1/4

= 11 1/4 square feet

Therefore, the area of long set is 11 1/4 square feet

15 feet – 5 feet = 10 feet

1/4 of 10 is 2 1/2 feet

Now, the area of each set of shorter curtain

A = 2 1/2 x 2 1/2

( 2 x 2 ) + ( 2 x 1/2 ) + ( 2 x 1/2 ) + ( 1/2 x 1/2 )

= 4 + 1/2 + 1 + 1/8

= 5 + 1/2 +1/8

= 5 5/8 square feet

Therefore, the area of each set of shorter curtain is 5 5/8 square feet.

Question 4. Some wire is used to make 3 rectangles: A, B, and C. Rectangle bs dimensions are \(\frac{3}{5}\) cm larger than Rectangle A’s dimensions, and Rectangle C’s dimensions are \(\frac{3}{5}\) cm larger than Rectangle B’s dimensions. Rectangle A is 2 cm by 3\(\frac{1}{5}\) cm. a. What is the total area of all three rectangles? b. If a 40-cm coil of wire was used to form the rectangles, how much wire is left? Answer:

The area of rectangle A =

A = 2 X 3 1/5

= 6 2/5 square centimetres

Area of of rectangle B =

A = 2 3/5 cm x 3 4/5 cm

= 6 + 8/5 + 9/5 + 12/25

= 9 + 21/5 + 12/25

= 9 22/25 square centimetres

Area of rectangle C =

A =  3 1/5 cm x 4 2/5 cm

= 12 + 6/5 + 4/5 + 2/25

= 14 2/25 square centimetres

Total area of three rectangles =

= 6 2/5 + 9 22/25 + 14 2/25

= 29 + 2/5 + 24/25

= 29 +10/25 +24/25

= 30 9/25 square centimetres

Therefore, the area of three rectangles = 30 9/25 square centimetres.

Perimeter of rectangle A =

2 + 2 + 3 1/5 + 3 1/5

= 6 2/5  centimetres

Perimeter of rectangle B =

2 3/5 + 2 3/5 + 3 4/5 + 3 4/5

= 12 4/5 centimetres

Perimeter of rectangle C =

3 1/5 + 3 1/5 + 4 2/5 + 4 2/5

= 15 1/5 centimetres

Total perimeter =

1 2/5 cm + 12 4/5 cm + 15 1/5 cm

= 38 2/5 cm

Now, according to the given condition :

40 cm – 38 2/5 cm

Therefore, the leftover wire length = 1 3/5 cm.

Eureka Math Grade 5 Module 5 Lesson 15 Exit Ticket Answer Key

Wheat grass is grown in planters that are 3\(\frac{1}{2}\) inch by 1\(\frac{3}{4}\) inch. If there is a 6 × 6 array of these planters with no space between them, what is the area covered by the planters? Answer:

The area of planter =

3 1/2 x 1 3/4

( 3 x  1 ) + ( 3 x 3/4 ) + ( 1 x 1/2 ) + ( 1/2 x 3/4 )

= 3 + 9/4 + 1/2 + 3/8

= 24 + 18/8 + 4/8 + 3/8

=49/8 = 6 1/8

According to given condition

= 36 x  6 1/8

= 200 + 1/2

Therefore, the area covered by the planters = 200 1/2 square inches.

Eureka Math Grade 5 Module 5 Lesson 15 Homework Answer Key

Question 1. The width of a picnic table is 3 times its length. If the length is \(\frac{5}{6}\)-yd long, what is the area of the picnic table in square feet? Answer:

1 yard = 3 feet

The area of the picnic table =

5/6 x 3= 15/6 = 5/2 = 2 1/2 feet

Area = 2 1/2 x 7 1/2

A = 5/2 x 15 1/2

= 18 3/4  square feet

Therefore, the area of the table = 2 1/2 square yards.

The area of window A =

= 6 1/4 x  5 3/4

= 30 x  1 1/4 + 4 2/4 + 3/16

= 35 + 3/4 + 3/16

= 35 + 15/16

= 35 15/16 square feet

The area of window B =

= 12 1/2 square feet

The area of D  =

4 x 8 = 32 square feet

The total area of the wall =

52 1/2 x 33

= 1716 + 33/2

= 1732 1/2 square feet

Now, the total area of A,B,C A and D

= 35 15/16 +12 1/2 + 32 + 9 1/2

= 22 + 35 15/16 +32

Now, the area of the painted wall =

1732 1/2 sq. ft – 89 15/16

= 1642 9/16 square feet

Therefor, the area of the wall painted = 1642 9/16 square feet.

The area of the smallest rectangle =

4 1/2 x 7 3/4

= 24 + 3 + 3 1/2 + 3/8

= 34 7/8 square inches

4 1/2 + 2 1/4

= 4 2/4 + 2 1/4

7 3/4 + 2 1/4 = 10

36 3/4 x 10

= 60 + 7 2/4

= 67 1/2 square inches

6 3/4 + 2 1/4 =9

9 x 12 1/4 =

= 108 + 2 1/4 = 110 1/4 square inches

d. 9 + 2 1/4 = 11 1/4

12 1/4 + 2 1/4 = 14 2/4

11 1/4 x 14 2/4

= 154 + 3 2/4 + 5 1/2 + 1/8

= 163 1/8 square inches.

So, the total area of the figure = 110 + 34 + 67 + 163

= 374 + 1 3/4

= 375 3/4 square inches

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