Eureka Math

Engage NY Eureka Math 1st Grade Module 1 Lesson 3 Answer Key

Eureka Math Grade 1 Module 1 Lesson 3 Problem Set Answer Key

Draw one more in the 5-group. In the box, write the numbers to describe the new picture.

Question 1.

1 more than 7 is __.
7 + 1 = ___

Answer:

1 more than 7 is 8
7 + 1 = 8

Question 2.

1 more than 9 is __.
9 + 1 = ___

Answer:

1 more than 9 is 10
9 + 1 = 10

Question 3.

1 more than 6 is __.
6 + 1 = ___

Answer:

1 more than 6 is 7
6 + 1 = 7

Answer: the numbers to describe the new picture above

Question 4.

1 more than 5 is __.
5 + 1 = ___

Answer:

1 more than 5 is 6
5 + 1 = 6

Question 5.

1 more than 8 is __.
8 + 1 = ___
Answer:

1 more than 8 is 9
8 + 1 = 9

Question 6.

__ more than 7.
__ = 7 + 1

Answer:

1 more than 7.
8 = 7 + 1

Question 7.

__ is more than 6.
__ = 6 + 1

Answer:

7 is more than 6.
7 = 6 + 1

Question 8.

__ is more than 5.
__ = 5 + 1

Answer:

1 is more than 5.
6 = 5 + 1

Question 9.
Imagine adding 1 more backpack to the picture. Then, write the numbers to match how many backpacks there will be.

1 more than 7 is __.
__ + 1 = ___

Answer:

1 more than 7 is 8.
7 + 1 = 8

Eureka Math Grade 1 Module 1 Lesson 3 Exit Ticket Answer Key

How many objects do you see? Draw one more. How many objects are there now?

Question 1.

__ is more than 9.
9 + 1 = __

Answer:

1 is more than 9.
9 + 1 = 10

Question 2.

1 more than 6 is __.
__ + 1 = ___

Answer:

1 more than 6 is 7.
6 + 1 = 7

Eureka Math Grade 1 Module 1 Lesson 3 Homework Answer Key

How many objects do you see? Draw one more. How many objects are there now?

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.
Imagine adding 1 more pencil to the picture.
Then, write the numbers to match how many pencils there will be.

Answer:

Question 6.
Imagine adding 1 more flower to the picture.
Then, write the numbers to match how many flowers there will be.

Answer:

Engage NY Eureka Math 1st Grade Module 1 Lesson 1 Answer Key

Eureka Math Grade 1 Module 1 Lesson 1 Sprint Answer Key

* Write the number of dots. Find 1 or 2 groups that make finding the total number of dots easier!

Answer:

B
* Write the number of dots. Find 1 or 2 groups that make finding the total number of dots easier!

Answer:

Eureka Math Grade 1 Module 1 Lesson 1 Problem Set Answer Key

Circle 5, and then make a number bond.

Question 1.

Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum.
5 and 3 are the parts of the number.
The result 8 is the whole.

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Put nail polish on the number of fingernails shown from left to right. Then, fill in the parts. Make the number of fingernails on one hand a part.

Question 5.

Answer:

Question 6.

Answer:

Make a number bond that shows 5 as one part.

Make a number bond for the pictures that shows 5 as one part.

Question 7.

Question 8.

Answer:

Question 9.

Answer:

Question 10.

Answer:

Question 11.

Answer:

Question 12.

Answer:

Eureka Math Grade 1 Module 1 Lesson 1 Exit Ticket Answer Key

Question 1.

Answer:

Question 2.

Answer:

Eureka Math Grade 1 Module 1 Lesson 1 Homework Answer Key

Circle 5, and then make a number bond.

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Make a number bond that shows 5 as one part.

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

Make a number bond for the dominoes.

Question 9.

Answer:

Question 10.

Question 11.

Answer:

Question 12.

Answer:

Circle 5 and count. Then, make a number bond.

Question 13.

Answer:

Question 14.

Answer:

Question 15.

Answer:

Question 16.

Answer:

Engage NY Eureka Math Grade 6 Module 5 Lesson 6 Answer Key

Eureka Math Grade 6 Module 5 Lesson 6 Exploratory Challenge Answer Key

Exploratory Challenge 1: Classroom Wall Paint

Question 1.
The custodians are considering painting our classroom next summer. In order to know how much paint they must buy, the custodians need to know the total surface area of the walls. Why do you think they need to know this, and how can we find the information?
Answer:
All classroom walls are different. Taking overall measurements and then subtracting windows, doors, or other areas will give a good approximation.

Question 2.
Make a prediction of how many square feet of painted surface there are on one wall in the room. If the floor has square tiles, these can be used as a guide.

Estimate the dimensions and the area. Predict the area before you measure. My prediction: _____ ft².

a. Measure and sketch one classroom wall. Include measurements of windows, doors, or anything else that would not be painted.

Answer:
Student responses will depend on the teacher’s choice of wall.

b. Work with your partners and your sketch of the wall to determine the area that needs paint. Show your sketch and calculations below; clearly mark your measurements and area calculations.
Answer:

c. A gallon of paint covers about 350 ft2. Write an expression that shows the total area of the wall. Evaluate it to find how much paint is needed to paint the wall.
Answer:
Answers will vary based on the size of the wall. Fractional answers are to be expected.

d. How many gallons of paint would need to be purchased to paint the wall?
Answer:
Answers will vary based on the size of the wall. The answer from part (d) should be on exact quantity because gallons of paint are discrete units. Fractional answers from part (c) must be rounded up to the nearest whole gallon.

Exploratory Challenge 2:

Answer:

Eureka Math Grade 6 Module 5 Lesson 6 Problem Set Answer Key

Question 1.
Below is a drawing of a wall that is to be covered with either wallpaper or paint. The wall is 8 ft. high and 16 ft. wide. The window, mirror, and fireplace are not to be painted or papered. The window measures 18 in. wide and 14 ft. high. The fireplace is 5 ft. wide and 3 ft. high, while the mirror above the fireplace is 4 ft. wide and 2 ft. high. (Note: this drawing is not to scale.)

a. How many square feet of wallpaper are needed to cover the wall?
Answer:
Total wall area = 8 ft.× 16 ft. = 128 ft²
Window area = 14 ft.× 1.5 ft. = 21 ft²
Fireplace area= 3 ft.× 51t.= 15 ft²
Mirrorarea= 4 ft.× 2 ft.= 8 ft²
Net wall area to be covered 128 ft² – (21 ft² + 15 ft² + 8 ft²) = 84 ft²

b. The wallpaper is sold in rolls that are 18 in. wide and 33 ft. long. Rolls of solid color wallpaper will be used, so patterns do not have to match up.

i. What is the area of one roll of wallpaper?
Answer:
Area of one roll of wallpaper: 33 ft. × 1.5 ft. = 49.5 ft²

ii. How many rolls would be needed to cover the wall?
Answer:
84 ft² ÷ 49.5 ft² ≈ 1.7; therefore, 2 rolls would need to be purchased.

c. This week, the rolls of wallpaper are on sale for $11. 99/roll. Find the cost of covering the wall with wallpaper.
Answer:
We need two rolls of wallpaper to cover the wall, which will cost $11.99 × 2 = $23.98.

d. A gallon of special textured paint covers 200 ft² and is on sale for $22.99/gallon. The wall needs to be painted twice (the wall needs two coats of paint). Find the cost of using paint to cover the wall.
Answer:
Total wall area = 8 ft. × 16 ft. = 128 ft²
Windowarea = 14 ft. × 1.5 ft.= 21 ft²
Fireplace area = 3 ft.× 5 ft. = 15 ft²
Mirror area = 4 ft. × 2 ft. = 8 ft²
Net wall area to be covered 128 ft² – (21 ft² + 15 ft² + 8 ft²) = 84 ft²
If the wall needs to be painted twice, we need to point a total area of 84 ft² × 2 = 168 ft². One gallon is enough paint for this wall, so the cost will be $22. 99.

Question 2.
A classroom has a length of 30 ft. and a width of 20 ft. The flooring is to be replaced by tiles. If each tile has a length of 36 in. and a width of 24 in., how many tiles are needed to cover the classroom floor?
Answer:
Area of the classroom: 30 ft. × 20 ft. = 600 ft²
Area of each tile 3 ft. × 2 ft. = 6 ft²
Area of the classroom \(\frac{\text { Area of the classroom }}{\text { Area of each tile }}=\frac{600 \mathrm{ft}^{2}}{6 \mathrm{ft}^{2}}\) = 100

100 tiles are needed to cover the classroom floor. Allow for students who say that if the tiles are 3 ft. × 2 ft., and they orient them in a way that corresponds to the 30 ft. × 20 ft. room, then they will have ten rows of ten tiles giving them 100 tiles.

Using this method, the students do not need to calculate the areas and divide. Orienting the
tiles the other way, students could say that they will need 105 tiles as they will need 6\(\frac{2}{3}\) rows of 15 tiles, and since \(\frac{2}{3}\) of a tile cannot be purchased, they will need 7 rows of 15 tiles.

Question 3.
Challenge: Assume that the tiles from Problem 2 are unavailable. Another design is available, but the tiles are square, 18 in. on a side. If these are to be installed, how many must be ordered?
Answer:
Solutions will vary. An even number of tiles fit on the 30 foot length of the room (20 tiles), but the width requires 13\(\frac{1}{3}\) tiles. This accounts for a 20 tile by 13 tile array. 20 × 13 = 260. 260 tiles need to be ordered.

The remaining area is 30 ft. × 0.5 ft. (20 × \(\frac{1}{3}\)) Since 20 of the \(\frac{1}{3}\) tiles are needed, 7 additional tiles must be cut to form \(\frac{21}{3}\). 20 of these will be used with \(\frac{1 {3}\) of 1 tile left over. Using the same logic as above, some students may correctly say they will need 280 tiles.

Question 4.
A rectangular flower bed measures 10 m by 6 m. It has a path 2 m wide around it. Find the area of the path.

Answer:
Total Area = 14 m × 10 m = 140 m²
Flower bed area = 10 m × 6 m = 60 m²
Area of path: 140 m² – 60 m² = 80 m²

Question 5.
A diagram of Tracy’s deck is shown below, shaded blue. He wants to cover the missing portion of his deck with soil in order to grow a garden.

a. Find the area of the missing portion of the deck. Write the expression and evaluate it.

Answer:
Students should use one of two methods to find the area: finding the dimensions of the garden area (interior rectangle, 6 m????× 2 m or finding the total area minus the sum of the four wooden areas, shown below.

6 m × 2 m = 12 m²
8 × 6 – 7 × 3 – 5 × 1 – 8 × 1 – 2 × 1 = 12 (All linear units are in meters; area is in square metres.)

b. Find the area of the missing portion of the deck using a different method. Write the expression and evaluate it.
Answer:
Students should choose whichever method was not used in part (a).

c. Write two equivalent expressions that could be used to determine the area of the missing portion of the deck.
Answer:
8 × 6 – 7 × 3 – 5 × 1 – 8 × 1 – 2 × 1
6 × 2

d. Explain how each expression demonstrates a different understanding of the diagram.
Answer:
One expression shows the dimensions of the garden area (interior rectangle, 6 m × 2 m), and one shows finding the total area minus the sum of the four wooden areas.

Question 6.
The entire large rectangle below has an area of 3\(\frac{1}{2}\) ft². If the dimensions of the white rectangle are as shown below, write and solve an equation to find the area, A, of the shaded region.

Answer:

Eureka Math Grade 6 Module 5 Lesson 6 Exit Ticket Answer Key

Question 1.
Find the area of the deck around this pool. The deck is the white area in the diagram.

Answer:
Area of Walkway and Pool:
A = bh
A = 90 m × 25 m
A = 2,250 m²

Area of Pool:
A = bh
A = 50 m × 15 m
A = 750 m²

Area of Walkway:
2,250 m² – 750 m² = 1,500 m²

Engage NY Eureka Math Grade 6 Module 5 Lesson 5 Answer Key

Eureka Math Grade 6 Module 5 Lesson 5 Exercise Answer Key

Opening Exercise:

Here is an aerial view of a woodlot.

Question 1.
If AB = 10 units, FE = 8 units, AF = 6 units, and DE = 7 units, find the lengths of the other two sides.
DC =
BC =
Answer:
DC = 2 units
BC = 13 units

Question 2.
If DC = 10 units, FE = 30 units, AF = 28 units, and BC = 54 units, find the lengths of the other two sides.
AB =
DE =
Answer:
AB = 40 units
DE = 26 units

Eureka Math Grade 6 Module 5 Lesson 5 Example Answer Key

Example 1: Decomposing Polygons into Rectangles

The Intermediate School is producing a play that needs a special stage built. A diagram of the stage is shown below (not to scale).

Question a.
On the first diagram, divide the stage into three rectangles using two horizontal lines. Find the dimensions of these rectangles, and calculate the area of each. Then, find the total area of the stage.
Answer:
Dimensions: 2 m by 4m, 2 m by 4 m, and 7 m by 5 m
Area: 2 m × 4m = 8 m², 2 m × 4 m = 8 m², 7 m × 5 m = 35 m²
Total: 8 m² + 8 m² + 35 m² = 51 m²

Question b.
On the second diagram, divide the stage into three rectangles using two vertical lines. Find the dimensions of these rectangles, and calculate the area of each. Then, find the total area of the stage.
Answer:
Dimensions: 2 m by 9 m, 2 m by 9 m, and 3 m by 5 m
Area: 2 m × 9 m = 18 m × 2 m × 9 m = 18 m², 3 m × 5m = 15 m²
Total: 51 m²

Question c.
On the third diagram, divide the stage into three rectangles using one horizontal line and one vertical line. Find the dimensions of these rectangles, and calculate the area of each. Then, find the total area of the stage.

Answer:
Dimensions: 2 m by 9 m, 2 m by 4 m, and 5 m by 5 m
Area: 2 m × 9 m = 18 m², 2 m × 4 m = 8 m², 5 m × 5 m = 25 m²
Total: 51 m²

Question d.
Think of this as a large rectangle with a piece removed.

i. What are the dimensions of the large rectangle and the small rectangle?
Answer:
Dimensions: 9 m by 7 m and 3 m by 4 m

ii. What are the areas of the two rectangles?
Answer:
Area: 9 m × 7 m = 63 m², 3 m × 4 m = 12 m²

iii. What operation Is needed to find the area of the original figure?
Answer:
Subtraction

iv. What is the difference in area between the two rectangles?
Answer:
63 m² – 12 m² = 51 m²

v. What do you notice about your answers to (a), (b), (c), and (d)? 7 m
Answer:
The area is the same.

vi. Why do you think this is true?
Answer:
No matter how we decompose the figure, the total area is the sum of its ports. Even if we take the area around the figure and subtract the part that is not included, the area of the figure remains the same, 51 m².

Area of Rectangle 1: b . h
2m . 4m = 8m²
Area of Rectangle 2: b . h
2 m 4m = 8 m²
Area of Rectangle 3: b . h
7m 5m 35 m²
Area of Polygon: 8m² + 8m² + 35 m² = 51 m²

Area of Rectangle 1: b . h
9m . 2m = 18 m²
Area of Rectangle 2: b . h
9m . 2m =18 m²
Area of Rectangle 3: b . h
3m . 5m = 15 m²
Area of Polygon: 18 m² + 18 m² + 15 m² = 51 m²

Area of Rectangle: b . h
9m . 7 m = 63 m²
Area of Missing Rectangle 2: b . h
3m . 4m = 12 m²
Area of Polygon: 63 m² – 12 m² = 51 m²

Example 2: Decomposing Polygons into Rectangles and Triangles

Parallelogram ABCD is part of a large solar power collector. The base measures 6 m and the height is 4 m.

Question a.
Draw a diagonal from A to C. Find the area of both triangles ABC and ACD.
Answer:

Triangle ABC
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (6 m) (4 m)
A = 12 m².

Triangle ACD
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (6m)(4m)
A = 12 m²

Question b.
Find the area of both triangles ABD and BCD. Then find the area of the trapezoid.
Answer:

Triangle ABD
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (8m)(5m)
A = 20 m²

Triangle BCD
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (4 m) (5 m)
A = 10 m²

A = 20 m² + 10 m² = 30 m²

Question c.
How else could we find this area?
Answer:
We could consider the rectangle that surrounds the trapezoid. Find the area of that rectangle, and then subtract the area of both right triangles.

Area of Rectangle;
A = bh
A = (8 m) (5 m)
A = 40 m²

Triangle 1
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (3 m) (5 m)
A = 7.5 m²

Triangle 2
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (1 m) (5 m)
A = 2.5 m²

A = 40 m² – 7.5 m² – 2.5 m² = 30 m² (or)

A = 40 m² – (7.5 m² + 2.5 m²) = 30 m²

Eureka Math Grade 6 Module 5 Lesson 5 Problem Set Answer Key

Question 1.
If AB = 20 units, FE = 12 units, AF = 9 units, and DE = 12 units, find the length of the other two sides. Then, find the area of the irregular polygon.

Answer:
CD = 8 units, BC = 21 units, Area = 276 square units.

Question 2.
If DC = 1.9 cm, FE = 5.6 cm, AF = 4.8 cm, and BC = 10.9 cm, find the length of the other two sides. Then, find the area of the irregular polygon.

Answer:
AB = 7.5 cm, DE = 6.1 cm, Area = 47.59 cm²

Question 3.
Determine the area of the trapezoid below. The trapezoid is not drawn to scale.

Answer:
Area of Triangle 1
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) × 22 m × 13 m
A = 198 m²

Area of Triangle 2
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) × 3 m × 18 m
A = 27 m²

Area of Trapezoid = Area of Triangle 1 + Area of Triangle 2
Area = 198 m² + 27 m² = 225 m²

Question 4.
Determine the area of the shaded isosceles trapezoid below. The image is not drawn to scale.

Answer:
Area of Rectangle
A = bh
A = 18 m × 12 m
A = 216 m²

Area of Triangles 1 and 2
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) × 7.5 m × 12 m
A = 45 m²

Area of Trapezoid = Area of Rectangle – Area of Triangle 1 – Area of Triangle 2
A = 216 m² – 45 m² – 45 m² = 126 m²

Question 5.
Here is a sketch of a wall that needs to be painted:

a. The windows and door will not be painted. Calculate the area of the wall that will be painted.
Answer:
Whole wall: 12 ft. × 8 ft. = 96 ft²
Window: 2 ft. × 2 ft. = 4 ft²
There are two identical windows, 4 ft² × 2 = 8 ft²
Door: 6 ft. × 3 ft. = 18 ft²
96 ft² – 8 ft² – 18 ft² = 70 ft²

b. If a quart of Extra-Thick Gooey Sparkle paint covers 30 ft². how many quarts must be purchased for the painting job?
Answer:
70 ÷ 30 = 2\(\frac{1}{3}\)
Therefore, 3 quarts must be purchased.

Question 6.
The figure below shows a floor plan of a new apartment. New carpeting has been ordered, which will cover the living room and bedroom but not the kitchen or bathroom. Determine the carpeted area by composing or decomposing in two different ways, and then explain why they are equivalent.

Answer:
Answers will vary. Sample student responses are shown.
Bedroom: 15 ft. × 25 ft. = 375 ft²
Living room: 35 ft. × 20 ft. = 700 ft²
Sum of bedroom and living room: 375 ft² + 700 ft² = 1,075 ft²

Alternatively, the whole apartment is 45 ft. × 35 ft. = 1,575 ft²
Subtracting the kitchen and bath (300 ft² and 200 ft²) still gives 1,075 ft².
The two areas are equivalent because they both represent the area of the living room and bedroom.

Eureka Math Grade 6 Module 5 Lesson 5 Exit Ticket Answer Key

Question 1.
Find the missing dimensions of the figure below, and then find the area. The figure is not drawn to scale.

Answer:
The solution can be any of these below.

Question 2.
Find the area of the parallelogram below by decomposing it into two triangles. The figure is not drawn to scale.

Answer:
Area of Triangle 1
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) × 8 mi.× 10 mi.
A = 40 mi²

Area of Triangle 2
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) × 8 mi. × 10 mi.
A = 40 mi²

Area of Parallelogram = Area of Triangle 1 + Area of Triangle 2
A = 40 mi² + 40 mi² = 80 mi²
The area of the parallelogram is 80 mi².

Engage NY Eureka Math Grade 6 Module 5 Lesson 4 Answer Key

Eureka Math Grade 6 Module 5 Lesson 4 Exercise Answer Key

Opening Exercise:

Draw and label the altitude of each triangle below.

Question a.

Answer:

Question b.

Answer:

Question c.

Answer:

Exploratory Challenge/Exercises 1 – 5:

Question 1.
Use rectangle X and the triangle with the altitude inside (triangle X) to show that the area formula for the triangle is A = \(\frac{1}{2}\) × base × height.

a. Step One: Find the area of rectangle X.
Answer:
A = 3 in. × 2.5 in. = 7.5 in²

b. Step Two: What is half the area of rectangle X?
Answer:
Half of the area of the rectangle is 7.5 in² ÷ 2= 3.75 in²

c. Step Three: Prove, by decomposing triangle X, that it is the same as half of rectangle X. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle X. What conclusions can you make about the triangle’s area compared to the rectangle’s area?
Answer:
Because the triangle fits inside half of the rectangle, we know the triangle’s area is half of the rectangle’s area.

Question 2.
Use rectangle Y and the triangle with a side that ¡s the altitude (triangle Y) to show the area formula for the triangle is A = \(\frac{1}{2}\) × base × height.

a. Step One: Find the area of rectangle Y.
Answer:
A = 3 in. × 3 in. = 9 in²

b. Step Two: What is half the area of rectangle Y?
Answer:
Half the area of the rectangle is 9 in² ÷ 2 = 4.5 in².

c. Step Three: Prove, by decomposing triangle Y, that it is the same as half of rectangle Y. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle Y. What conclusions can you make about the triangle’s area compared to the rectangle’s area?
Answer:
The right triangle also fits in exactly half of the rectangle, so the triangle’s area is • Students may struggle once again half the size of the rectangle’s area. with this step since they have yet to see an obtuse

Question 3.
Use rectangle Z and the triangle with the altitude outside (triangle Z) to show that the area angle. formula for the triangle is A = \(\frac{1}{2}\) × base × height.

a. Step One: Find the area of rectangle Z.
Answer:
A= 3 in. × 2.5 in.=7.5 in²

b. Step Two: What is half the area of rectangle Z?
Answer:
Half of the area of the rectangle is 7.5 in² ÷2 = 3.75 in².

c. Step Three: Prove, by decomposing triangle Z, that it is the same as half of rectangle Z. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle Z. What conclusions can you make about the triangle’s area compared to the rectangle’s area?
Answer:
Similar to the other two triangles, when the altitude is outside the triangle, the area of the triangle is exactly half of the area of the rectangle.

Question 4.
When finding the area of a triangle, does it matter where the altitude is located?
Answer:
It does not matter where the altitude is located. To find the area of a triangle, the formula is always
A = \(\frac{1}{2}\) × base × height.

Question 5.
How can you determine which part of the triangle is the base and which is the height?
Answer:
The base and the height of any triangle form a right angle because the altitude is always perpendicular to the base.

Exercises 6 – 8:

Calculate the area of each triangle. Figures are not drawn to scale.

Question 6.

Answer:
A = \(\frac{1}{2}\) × (24 in.) (8 in.) = 96 in²

Question 7.

Answer:

Question 8.
Draw three triangles (acute, right, and obtuse) that have the same area. Explain how you know they have the same area.
Answer:
Answers will vary.

Eureka Math Grade 6 Module 5 Lesson 4 Problem Set Answer Key

Calculate the area of each figure below. Figures are not drawn to scale.

Question 1.

Answer:
A = \(\frac{1}{2}\) (21 in.) (8 in.) = 84 in²

Question 2.

Answer:
A = \(\frac{1}{2}\) (72 m) (21 m) = 756 m²

Question 3.

Answer:
A = \(\frac{1}{2}\) (75.8 km) (29.2 km) = 1,106.68 km²

Question 4.

Answer:
A = \(\frac{1}{2}\) (5 m) (12 m) = 30 m²
A = \(\frac{1}{2}\) (7 m) (29 m) = 101.5 m²
A = (12 m) (19 m) = 228 m²
A = 30 m² + 30 m² + 101.5 m² + 228 m²
A = 389.5 m²

Question 5.
The Andersons are going on a long sailing trip during the summer. However, one of the sails on their sailboat ripped, and they have to replace it. The sail is pictured below. If the sailboat sails are on sale for $2 per square foot, how much will the new sail cost?

Answer:
A = \(\frac{1}{2}\) bh
= \(\frac{1}{2}\) (8 ft.) (12 ft.)
= 48 ft²
\(\frac{2 \text { dollars }}{\mathrm{ft}^{2}}\) × 48 ft² = 96 dollars(or $96)
The cost of the new sail is $96.

Question 6.
Darnell and Donovan are both trying to calculate the area of an obtuse triangle. Examine their calculations below.

Which student calculated the area correctly? Explain why the other student is not correct.
Answer:
Donovan calculated the area correctly. Although Darnell did use the altitude of the triangle, he used the length between the altitude and the base rather than the length of the actual base.

Question 7.
Russell calculated the area of the triangle below. His work is shown.

A = \(\frac{1}{2}\) bh
A = 150.5 cm²
Although Russell was told his work is correct, he had a hard time explaining why it is correct. Help Russell explain why his calculations are correct.
Answer:
The formula for the area of triangle is A = bit. Russell followed this formula because 7 cm is the height of the triangle, and 43 cm is the base of the triangle.

Question 8.
The larger triangle below has a base of 10.14 m; the gray triangle has an area of 40.325 m².

a. Determine the area of the larger triangle if it has a height of 12.2 m.
Answer:
A = \(\frac{1}{2}\) (10.14 m) (12.2 m)
= 61.854 m²

b. Let A be the area of the unshaded (white) triangle in square meters. Write and solve an equation to determine the value of A, using the areas of the larger triangle and the gray triangle.
Answer:
40.325m² + A = 61.854m²
40.325m² + A – 40.325 m² = 61.854 m² – 40.325 m²
A = 21.529 m²

Eureka Math Grade 6 Module 5 Lesson 4 Exit Ticket Answer Key

Find the area of each triangle. Figures are not drawn to scale.

Question 1.

Answer:
A = \(\frac{1}{2}\) (12.6 cm) (16.8 cm) = 105.84 cm²

Question 2.

Answer:
A = \(\frac{1}{2}\) (28 in.) (15 in.) = 210 in²

Question 3.

Answer:
A = \(\frac{1}{2}\) (12 ft.) (21 ft.) = 126 ft²

Engage NY Eureka Math Grade 6 Module 5 Lesson 3 Answer Key

Eureka Math Grade 6 Module 5 Lesson 3 Exercise Answer Key

Exercises:

Question 1.
Work with a partner on the exercises below. Determine if the area formula A = \(\frac{1}{2}\) bh is always correct. You may use a calculator, but be sure to record your work on your paper as well. Figures are not drawn to scale.

Answer:

Question 2.
Can we use the formula A = \(\frac{1}{2}\) × base × height to calculate the area of triangles that are not right triangles? Explain your thinking.
Answer:
Yes, the formula A = \(\frac{1}{2}\) × base × height can be used for more than just right triangles. We just need to be able to determine the height when it is not necessarily the length of one of the sides.

Question 3.
Examine the given triangle and expression.
\(\frac{1}{2}\) (11 ft.) (4ft.)
Explain what each part of the expression represents according to the triangle.
Answer:
11 ft. represents the base of the triangle because 8 ft. + 3 ft. = 11 ft.
4 ft. represents the altitude of the triangle because this length is perpendicular to the base.

Question 4.
Joe found the area of a triangle by writing A = \(\frac{1}{2}\) (11 in.)(4 in.), while Kaitlyn found the area by writing A = \(\frac{1}{2}\) (3 in.)(4 in.) + \(\frac{1}{2}\) (8 in.)(4 in.). Explain how each student approached the problem.
Answer:
Joe combined the two bases of the triangle first and then calculated the area of the entire triangle, whereas Kaitlyn calculated the area of two smaller right triangles and then added these areas together.

Question 5.
The triangle below has an area of 4.76 sq. in. If the base is 3.4 in., let h be the height in inches.

a. Explain how the equation 4.76 in² = (3.4 in. )h represents the situation.
Answer:
The equation shows the area, 4. 76 in², is one-half the base, 3.4 in times the height, in inches, h.

b. Solve the equation.
Answer:
4.76 in² = \(\frac{1}{2}\) (3.4 in.)h
4.76 in² = (1.7 in.)h
4.76 in² + 1.7 in. = (1.7 in.)h + 1.7 in.
2.8 in. = h

Eureka Math Grade 6 Module 5 Lesson 3 Problem Set Answer Key

Calculate the area of each shape below. Figures are not drawn to scale.

Question 1.

Answer:
A = \(\frac{1}{2}\) (3.3 in.)(4.4 in.) = 7.26 in²
A = \(\frac{1}{2}\) (6. 1 in.)(4.4 in.) 13.42 in²
A = 7.26 in² + 13.42 in² = 20.68 in²
or
A = \(\frac{1}{2}\) (9.4 in. )(4. 4 in.) = 20.68 in²

Question 2.

Answer:
A = \(\frac{1}{2}\) (8m)(14m)= 56m²
A = \(\frac{1}{2}\) (16m)(14m)= 112m²
A = 56m² ÷ 112 m² = 168m²
or
A = \(\frac{1}{2}\) (24 m)(14 m) = 168m²

Question 3.

Answer:
A = \(\frac{1}{2}\) (5 ft) (12 ft) = 30 ft²
A = (12 ft.)(12 ft.) = 144 ft²
A = \(\frac{1}{2}\) (5 ft.)(12 ft.) = 30 ft²
A = 30ft² + 144ft² + 30ft² = 204 ft²

Question 4.

Answer:
A = \(\frac{1}{2}\) (48 km) (7 km) = 168km²
A = (35 km)(48 km) = 1,680 km²
A = \(\frac{1}{2}\) (48 km) (7 km) = 168km²
A = 168 km² + 1,680 km² + 168 km² = 2,016km²

Question 5.
Immanuel is building a fence to make an enclosed play area for his dog. The enclosed area will be in the shape of a triangle with a base of 48 m and an altitude of 32 m. How much space does the dog have to play?
Answer:
A = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) (48 m) (32 m) = 768 m²
The dog has 768 m² in which to play.

Question 6.
Chauncey is building a storage bench for his son’s playroom. The storage bench will fit into the comer and against two walls to form a triangle. Chauncey wants to buy a triangular shaped cover for the bench.
If the storage bench is 2\(\frac{1}{2}\) ft. along one wall and 4\(\frac{1}{4}\) ft. along the other wall, how big will the cover have to be to cover the entire bench?

Answer:
A = \(\frac{1}{2}\) (2\(\frac{1}{2}\) ft.) (4\(\frac{1}{4}\) ft.)

= \(\frac{1}{2}\) (\(\frac{5}{2}\) ft) (\(\frac{17}{4}\) ft.)

= \(\frac{85}{16}\)ft² = 5\(\frac{5}{16}\) ft²

Chauncey would have to buy a cover that has an area of 5\(\frac{5}{16}\) ft² to cover the entire scale. bench.

Question 7.
Examine the triangle to the right.

a. Write an expression to show how you would calculate the area.
Answer:
\(\frac{1}{2}\) (7 in.) (4 in.) + \(\frac{1}{2}\) (3 in.) (4in.) or \(\frac{1}{2}\) (10 in.) (4in.)

b. Identify each part of your expression as it relates to the triangle.
Answer:
If students wrote the first expression, then 7 in. and 3 in. represent the two parts of the base, and 4 in. is the height, or the altitude, of the triangle.

If students wrote the second expression, then 10 in. represents the base because 7 in. + 3 in. = 10 in., and 4 in. represents the height, or the altitude, of the triangle.

Question 8.
The floor of a triangular room has an area of 32\(\frac{1}{2}\) sq. m. If the triangle’s altitude is 7\(\frac{1}{2}\) m, write an equation to determine the length of the base, b, in meters. Then solve the equation.
Answer:

Therefore, the base is 8\(\frac{2}{3}\) m.

Eureka Math Grade 6 Module 5 Lesson 3 Exit Ticket Answer Key

Calculate the area of each triangle using two different methods. Figures are not drawn to scale.

Question 1.

Answer:
1
A = \(\frac{1}{2}\) (3ft.) (7 ft.)= 10.5 ft²
A = \(\frac{1}{2}\) (12 ft.) (7 ft.) = 42 ft²
A= 10.5 ft² + 42 ft² = 525 ft²
OR
A = \(\frac{1}{2}\) (15 ft.) (7 ft.) = 52.5 ft²

Question 2.

Answer:
A = \(\frac{1}{2}\) (9 in.)(18 in.) = 81 in²
A = \(\frac{1}{2}\) (32 in. )(18 in) = 288 in²
A = 81 in² + 288 in² = 369 in²
OR
A = \(\frac{1}{2}\) (41 in.)(18 in.)= 369 in²

Eureka Math Grade 6 Module 5 Lesson 3 Multiplication of Decimals Answer Key

Multiplication of Decimals – Round 1

Directions: Evaluate each expression.

Question 1.
5 × 1 =
Answer:
5

Question 2.
5 × 0.1 =
Answer:
0.5

Question 3.
5 × 0.01 =
Answer:
0.05

Question 4.
5 × 0.001 =
Answer:
0.005

Question 5.
2 × 4 =
Answer:
8

Question 6.
0.2 × 4 =
Answer:
0.8

Question 7.
0.02 × 4 =
Answer:
0.08

Question 8.
0.002 × 4 =
Answer:
0.008

Question 9.
3 × 3 =
Answer:
9

Question 10.
3 × 0.3 =
Answer:
0.9

Question 11.
3 × 0.03 =
Answer:
0.09

Question 12.
0.1 × 0.8 =
Answer:
0.08

Question 13.
0.1 × 0.08 =
Answer:
0.008

Question 14.
0.01 × 0.8 =
Answer:
0.008

Question 15.
0.01 × 0.08 =
Answer:
0.0008

Question 16.
0.3 × 0.2 =
Answer:
0.6

Question 17.
0.03 × 0.2 =
Answer:
0.006

Question 18.
0.02 × 0.3 =
Answer:
0.006

Question 19.
0.02 × 0.03 =
Answer:
0.0006

Question 20.
0.2 × 0.2 =
Answer:
0.04

Question 21.
0.02 × 0.2 =
Answer:
0.004

Question 22.
0.2 × 0.02 =
Answer:
0.004

Question 23.
5 × 3 =
Answer:
15

Question 24.
5 × 0.3 =
Answer:
1.5

Question 25.
0.5 × 3 =
Answer:
1.5

Question 26.
0.3 × 0.5 =
Answer:
0.15

Question 27.
9 × 2 =
Answer:
18

Question 28.
0.2 × 9 =
Answer:
1.8

Question 29.
0.9 × 2 =
Answer:
1.8

Question 30.
0.2 × 0.9 =
Answer:
0.18

Question 31.
4 × 0.4 =
Answer:
1.6

Question 32.
0.4 × 0.4 =
Answer:
0.16

Question 33.
0.04 × 0.4 =
Answer:
0.016

Question 34.
0.8 × 0.6 =
Answer:
0.48

Question 35.
0.8 × 0.06 =
Answer:
0.048

Question 36.
0.006 × 0.8 =
Answer:
0.0048

Question 37.
0.006 × 0.08 =
Answer:
0.00048

Question 38.
0.7 × 0.9 =
Answer:
0.63

Question 39.
0.07 × 0.9 =
Answer:
0.063

Question 40.
0.9 × 0.007 =
Answer:
0.0063

Question 41.
0.09 × 0.007 =
Answer:
0.00063

Question 42.
1.2 × 0.7 =
Answer:
0.84

Question 43.
1.2 × 0.07 =
Answer:
0.084

Question 44.
0.007 × 0.12 =
Answer:
0.00084

Multiplication of Decimals – Round 1

Directions: Evaluate each expression

Question 1.
9 × 1 =
Answer:
9

Question 2.
0.9 × 1 =
Answer:
0.9

Question 3.
0.09 × 1 =
Answer:
0.09

Question 4.
0.009 × 1 =
Answer:
0.009

Question 5.
2 × 2 =
Answer:
4

Question 6.
2 × 0.2 =
Answer:
0.4

Question 7.
2 × 0.02 =
Answer:
0.04

Question 8.
2 × 0.002 =
Answer:
0.004

Question 9.
3 × 2 =
Answer:
6

Question 10.
0.3 × 2=
Answer:
0.6

Question 11.
2 × 0.03 =
Answer:
0.06

Question 12.
0.7 × 0.1 =
Answer:
0.07

Question 13.
0.07 × 0.1 =
Answer:
0.007

Question 14.
0.01 × 0.7 =
Answer:
0.007

Question 15.
0.01 × 0.07 =
Answer:
0.0007

Question 16.
0.2 × 0.4 =
Answer:
0.08

Question 17.
0.02 × 0.4 =
Answer:
0.008

Question 18.
0.4 × 0.02 =
Answer:
0.008

Question 19.
0.04 × 0.02 =
Answer:
0.0008

Question 20.
0.1 × 0.1 =
Answer:
0.01

Question 21.
0.01 × 0.1 =
Answer:
0.0001

Question 22.
0.1 × 0.01 =
Answer:
0.001

Question 23.
3 × 4 =
Answer:
12

Question 24.
3 × 0.4 =
Answer:
1.2

Question 25.
0.3 × 4 =
Answer:
1.2

Question 26.
0.4 × 0.3 =
Answer:
0.12

Question 27.
7 × 7 =
Answer:
49

Question 28.
7 × 0.7 =
Answer:
4.9

Question 29.
0.7 × 7 =
Answer:
4.9

Question 30.
0.7 × 0.7 =
Answer:
0.49

Question 31.
2 × 0.8 =
Answer:
1.6

Question 32.
0.2 × 0.8 =
Answer:
0.16

Question 33.
0.02 × 0.8 =
Answer:
0.016

Question 34.
0.6 × 0.5 =
Answer:
0.3

Question 35.
0.6 × 0.05 =
Answer:
0.03

Question 36.
0.005 × 0.6 =
Answer:
0.003

Question 37.
0.005 × 0.06 =
Answer:
0.0003

Question 38.
0.9 × 0.9 =
Answer:
0.81

Question 39.
0.09 × 0.9 =
Answer:
0.081

Question 40.
0.009 × 0.9 =
Answer:
0.0081

Question 41.
0.009 × 0.09 =
Answer:
0.00081

Question 42.
1.3 × 0.6 =
Answer:
0.78

Question 43.
1.3 × 0.06 =
Answer:
0.078

Question 44.
0.006 × 1.3 =
Answer:
0.0078

Engage NY Eureka Math 6th Grade Module 3 Lesson 2 Answer Key

Eureka Math Grade 6 Module 3 Lesson 2 Example Answer Key

Example 1:
Take It to the Bank
Read Example 1 silently. In the first column, write down any words and definitions you know. In the second column, write down any words you do not know.

For Tim’s 13th birthday, he received $150 in cash from his mom. His dad took him to the bank to open a savings account. Tim gave the cash to the banker to deposit into the account. The banker credited Tim’s new account $150 and gave Tim a receipt. One week later, Tim deposited another $25 that he had earned as allowance. The next month, Tim’s dad gave him permission to withdraw $35 to buy a new video game. Tim’s dad explained that the bank would charge a $5 fee for each withdrawal from the savings account and that each withdrawal and charge results in a debit to the account.

In the third column, write down any new words and definitions that you learn during the discussion.
Answer:

Example 2:
How Hot, How Cold?
Temperature is commonly measured using one of two scales, Celsius or Fahrenheit. In the United States, the Fahrenheit system continues to be the accepted standard for nonscientific use. All other countries have adopted Celsius as the primary scale in use. The thermometer shows how both scales are related.

a. The boiling point of water is 100°C. Where is loo degrees Celsius located on the thermometer to the right?
Answer:
It is not shown because the greatest temperature shown in Celsius is 50°C.

b. On a vertical number line, describe the position of the integer that represents 100°C.
Answer:
The integer is 100, and it would be located 100 units above zero on the Celsius side of the scale.

c. Write each temperature as an integer.
i. The temperature shown on the thermometer in degrees Fahrenheit:
Answer:
100

ii. The temperature shown on the thermometer in degrees Celsius:
Answer:
38

iii. The freezing point of water in degrees Celsius:
Answer:
0

d. If someone tells you your body temperature is 98. 6°, what scale is being used? How do you know?
Answer:
Since water boils at 100°C, they must be using the Fahrenheit scale.

e. Does the temperature 0 degrees mean the same thing on both scales?
Answer:
No. 0°C corresponds to 32°F, and 0°F corresponds to approximately – 18°C.

Eureka Math Grade 6 Module 3 Lesson 2 Exercise Answer Key

Exercise 1.
Read Example 1 again. With your partner, number the events in the story problem. Write the number above each sentence to show the order of the events.
For Tim’s 13” birthday, he received $150 in cash from his mom. His dad took him to the bank to open a savings account.

Tim gave the cash to the banker to deposit into the account. The banker credited Tim’s new account $150 and gave Tim

a receipt. One week later, Tim deposited another $25 that he had earned as allowance. The next month, Tim’s dad gave

him permission to withdraw $35 to buy a new video game. Tim’s dad explained that the bank would charge a $5 fee for

each withdrawal from the savings account and that each withdrawal and charge results in a debit to the account.
Answer:

Exercise 2.
Write each individual description below as an integer. Model the integer on the number line using an appropriate scale.

Answer:

Exercise 3.
Write each word under the appropriate column, “Positive Number” or “Negative Number.”
Gain Loss Deposit Credit Debit Charge Below Zero Withdraw Owe Receive
Answer:

Exercise 4.
Write an integer to represent each of the following situations:
a. A company loses $345, 000 in 2011.
Answer:
– 345, 000

b. You earned $25 for dog sitting.
Answer:
25

c. Jacob owes his dad $5.
Answer:
– 5

d. The temperature at the sun’s surface is about 5, 500°C
Answer:
5, 500

e. The temperature outside is 4 degrees below zero.
Answer:
– 4

f. A football player lost 10 yards when he was tackled.
Answer:
– 10

Exercise 5.
Describe a situation that can be modeled by the integer – 15. Explain what zero represents in the situation.
Answer:
Answers will vary. I owe my best friend $15. in this situation, 0 represents my owing nothing to my best friend.

Eureka Math Grade 6 Module 3 Lesson 2 Problem Set Answer Key

Question 1.
Express each situation as an integer in the space provided.
a. A gain of 56 points in a game
Answer:
56

b. A fee charged of $2
Answer:
– 2

c. A temperature of 32 degrees below zero
Answer:
– 32

d. A 56-yard loss in a football game
Answer:
– 56

e. The freezing point of water in degrees Celsius
Answer:
0

f. A $12,500 deposit
Answer:
12,500

For Problems 2 – 5, use the thermometer to the right.

Question 2.
Each sentence is stated incorrectly. Rewrite the sentence to correctly describe each situation.
a. The temperature is – 1o degrees Fahrenheit below zero.
Answer:
Correct: The temperature is – 10°F.
OR
The temperature is 10 degrees below zero Fahrenheit.

b. The temperature is – 22 degrees Celsius below zero.
Answer:
Correct: The temperature is – 22°C.
OR
The temperature is 22 degrees below zero Celsius.

Question 3.
Mark the integer on the thermometer that corresponds to the temperature given.
a. 70°F
b. 12°C
c. 110°F
d. – 4°C
Answer:

Question 4.
The boiling point of water is 2 12°F. Can this thermometer be used to record the temperature of a boiling pot of water? Explain.
Answer:
No, it cannot because the highest temperature in Fahrenheit on this thermometer is 120°.

Question 5.
Kaylon shaded the thermometer to represent a temperature of 20 degrees below zero Celsius as shown in the diagram. Is she correct? Why or why not? If necessary, describe how you would fix Kaylon’s shading.
Answer:
She is incorrect because she shaded a temperature of – 20°F. I would fix this by marking a line segment at – 20°C and shade up to that line.

Eureka Math Grade 6 Module 3 Lesson 2 Exit Ticket Answer Key

Question 1.
Write a story problem that includes both integers – 8 and 12.
Answer:
Answers may vary. One boxer gains 12 pounds of muscle to train for a fight. Another boxer loses 8 pounds of fat.

Question 2.
What does zero represent in your story problem?
Answer:
Zero represents no change in the boxer’s weight.

Question 3.
Choose an appropriate scale to graph both integers on the vertical number line. Label the scale.
Answer:
I chose a scale of 1.

Question 4.
Graph both points on the vertical number line.
Answer:

Engage NY Eureka Math 6th Grade Module 2 Mid Module Assessment Answer Key

Eureka Math Grade 6 Module 2 Mid Module Assessment Answer Key

Question 1.
Yasmine is having a birthday party with snacks and activities for her guests. At one table, five people are sharing three-quarters of a pizza. What equal-sized portion of the whole pizza will each of the five people receive?
a. Use a model (e.g., picture, number line, or manipulative materials) to represent the quotient.
Answer:

b. Write a number sentence to represent the situation. Explain your reasoning.
Answer:
Because there are 5 people, we found I out of the 5, which is \(\frac{1}{5}\). I can represent the situation as:
\(\frac{3}{2}\) ÷ 5 = \(\frac{3}{4}\) . \(\frac{1}{5}\) = \(\frac{3}{20}\)

c. If three-quarters of the pizza provided 12 pieces to the table, how many pieces were in the pizza when it was full? Support your answer with models.
Answer:

Question 2.
Yasmine needs to create invitations for the party. She has \(\frac{3}{4}\) of an hour to make the invitations. It takes her \(\frac{1}{12}\) of an hour to make each card. How many invitations can Yasmine create?
a. Use a number line to represent the quotient.
Answer:

b. Draw a model to represent the quotient.
Answer:

c. Compute the quotient without models. Show your work.
Answer:

Question 3.
Yasmine is serving ice cream with the birthday cake at her party. She has purchased 19\(\frac{1}{2}\) pints of ice cream. She will serve \(\frac{3}{4}\) of a pint to each guest.
a. How many guests can be served ice cream?
Answer:

b. Will there be any ice cream left? Justify your answer.
Answer:
My answer, 26, is a whole number, so there will be no ice cream left over. If my answer was 26\(\frac{1}{4}\) or any mixed number, there would be ice cream left over.

Question 4.
L.B. Johnson Middle School held a track and field event during the school year. Miguel took part in a four person shot put team. Shot put is a track and field event where athletes throw (or “put”) a heavy sphere, called a “shot,” as far as possible. To determine a team score, the distances of all team members are added. The team with the greatest score wins first place. The current winning team’s final score at the shot put is 52.08 ft. Miguel’s teammates threw the shot put the following distances: 12.26 ft., 12.82 ft., and 13.75 ft. Exactly how many feet will Miguel need to throw the shot put in order to tie the current first-place score? Show your work.

Answer:

Miguel will need to throw the shot put 13.25 feet to he the current first place score.

Question 5.
The sand pit for the long jump has a width of 2.75 meters and a length of 9.54 meters. Just in case it rains, the principal wants to cover the sand pit with a piece of plastic the night before the event. How many square meters of plastic will the principal need to cover the sand pit?

Answer:

Question 6.
The chess club is selling drinks during the track and field event. The club purchased water, juice boxes, and pouches of lemonade for the event. They spent $138.52 on juice boxes and $75.00 on lemonade. The club purchased three cases of water. Each case of water costs $6.80. What ¡s the total cost of the drinks?
Answer:

Engage NY Eureka Math 6th Grade Module 3 Lesson 1 Answer Key

Eureka Math Grade 6 Module 3 Lesson 1 Exercise Answer Key

Complete the diagrams. Count by ones to label the number lines.

Exercise 1.
Plot your point on both number lines.
Answer:
Answers may vary.

Exercise 2.
Show and explain how to find the opposite of your number on both number lines.
Answer:
In this example, the number chosen was – 4. So – 4 is the first number plotted, and the opposite is 4. Horizontal Number Line: I found my point by starting at zero and counting four units to the left to end on – 4. Then, to find the opposite of my number, I started on zero and counted to the right four units to end on 4. Vertical Number Line: I found my point by starting at zero and counting four units down to end on – 4. I found the opposite of my number by starting at zero and counting four units up to end on 4.

Exercise 3.
Mark the opposite on both number lines.
Answer:
Answers may vary.

Exercise 4.
Choose a group representative to place the opposite number on the class number lines.
Answer:

Exercise 5.
Which group had the opposite of the number on your index card?
Answer:
Answers may vary. Jackie’s group had the opposite of the number on my index card. They had 4.

Eureka Math Grade 6 Module 3 Lesson 1 Problem Set Answer Key

Question 1.
Draw a number line, and create a scale for the number line in order to plot the points – 2, 4, and 6.
a. Graph each point and its opposite on the number line.
Answer:

b. Explain how you found the opposite of each point.
Answer:
To graph each point, I started at zero and moved right or left based on the sign and number (to the right for a positive number and to the left for a negative number). To graph the opposites, I started at zero, but this time I moved in the opposite direction the same number of times.

Question 2.
Carlos uses a vertical number line to graph the points – 4, – 2, 3, and 4. He notices that – 4 is closer to zero than – 2. He is not sure about his diagram. Use what you know about a vertical number line to determine if Carlos made a mistake or not. Support your explanation with a number line diagram.
Answer:
Carlos made a mistake because – 4 is less than – 2, so it should be farther down the number line. Starting at zero, negative numbers decrease as we look farther below zero. So, – 2 lies before – 4 on a number line since – 2 is 2 units below zero and – 4 is 4 units below zero.

Question 3.
Create a scale in order to graph the numbers – 12 through 12 on a number line. What does each tick mark represent?
Answer:
Each tick mark represents 1 unit.

Question 4.
Choose an integer between – 5 and – 10. Label it R on the number line created in Problem 3, and complete the following tasks.
Answers may vary. Refer to the number line above for sample student work. – 6, – 7, – 8, or – 9
a. What is the opposite of R? Label it Q.
Answer:
Answers will vary. 6

b. State a positive integer greater than Q. Label it T.
Answer:
Answers will vary. 11

c. State a negative integer greater than R. Label it S.
Answer:
Answers will vary. – 3

d. State a negative integer less than R. Label it U.
Answer:
Answers will vary. 9

e. State an integer between R and Q. Label it V.
Answer:
Answers will vary. 2

Question 5.
Will the opposite of a positive number always, sometimes, or never be a positive number? Explain your reasoning.
Answer:
The opposite of a positive number will never be a positive number. For two nonzero numbers to be opposites, zero has to be in between both numbers, and the distance from zero to one number has to equal the distance between zero and the other number.

Question 6.
Will the opposite of zero always, sometimes, or never be zero? Explain your reasoning.
Answer:
The opposite of zero will always be zero because zero is its own opposite.

Question 7.
Will the opposite of a number always, sometimes, or never be greater than the number itself? Explain your reasoning. Provide an example to support your reasoning.
Answer:
The opposite of a number will sometimes be greater than the number itself because it depends on the given number. For example, if the number given is – 6, then the opposite is 6, which is greater than – 6. If the number given is 5, then the opposite is – 5, which is not greater than 5. If the number given is 0, then the opposite is 0, which is never greater than itself.

Eureka Math Grade 6 Module 3 Lesson 1 Exit Ticket Answer Key

Question 1.
Draw a number line, and create a scale for the number line in order to plot the points —2, 4, and 6.
a. Graph each point and its opposite on the number line.

Answer:
Answers will vary. One possible answer is a: – 4; b: – 1; C: 1; d: 4.

Question 2.
Below is a list of numbers in order from least to greatest. Use what you know about the number line to complete the list of numbers by filling in the blanks with the missing integers.
– 6, – 5, ________, – 3, – 2, – 1, ________, 1, 2, _______, 4, ________, 6
Answer:
– 6, – 5,   – 4,     – 3, – 2, – 1,    0,    1, 2,    3,    4,     5,      6

Question 3.
Complete the number line scale. Explain and show how to find 2 and the opposite of 2 on a number line.
Answer:
I would start at zero and move 2 units to the left to locate the number – 2 on the number line. So, to locate 2, 1 would start at zero and move 2 units to the right (the opposite direction).

Eureka Math Grade 6 Module 3 Lesson 1 Exploratory Challenge Answer Key

Exploratory Challenge: Constructing the Number Line
Answer:
The purpose of this exercise is to let students construct the number line (positive and negative numbers and zero) using a compass.

Have students draw a line, place a point on the line, and label it 0.

Have students use the compass to locate and label the next point 1, thus creating the scale. Students continue to locate other whole numbers to the right of zero using the same unit measure.

Using the same process, have students locate the opposites of the whole numbers. Have students label the first point to the left of zero – 1.

Introduce to the class the definition of the opposite of a number.

Sample student work is shown below.

Engage NY Eureka Math 6th Grade Module 2 End of Module Assessment Answer Key

Eureka Math Grade 6 Module 2 End of Module Assessment Answer Key

Question 1.
LB. Johnson Middle School held a track and field event during the school year. The chess club sold various drink and snack items for the participants and the audience. Altogether, they sold 486 items that totaled $2,673.
a. If the chess club sold each item for the same price, calculate the price of each item.
Answer:

Each item’s price is $5.50

b. Explain the value of each digit in your answer to 1(a) using place value terms.
Answer:

Question 2.
The long-jump pit was recently rebuilt to make it level with the runway. Volunteers provided pieces of
wood to frame the pit. Each piece of wood provided measures 6 feet, which is approximately 1.8287
meters.

a. Determine the amount of wood, in meters, needed to rebuild the frame.
Answer:

b. How many boards did the volunteers supply? Round your calculations to the nearest hundredth, and then provide the whole number of boards supplied.
Answer:

Question 3.
Andy runs 436.8 meters in 62.08 seconds.
a. If Andy runs at a constant speed, how far does he run in one second? Give your answer to the nearest tenth of a second.
Answer:

b. Use place value, multiplication with powers of 10, or equivalent fractions to explain what is happening mathematically to the decimal points in the divisor and dividend before dividing.
Answer:

When you write the problem as a fraction, multiply the numerator and denominator by 100. Multiplying each by 100 resulted in both numbers being whole numbers.
436.8 ÷ 62.08 is the same as 43,680 ÷ 6,208.

c. In the following expression, place a decimal point in the divisor and the dividend to create a new problem with the same answer as in 3(a). Then, explain how you know the answer will be the same.
43.68 ÷ 6.208
Answer:

Multiplying or dividing the dividend and divisor by the same power of ten yields the same quotient.

Question 4.
The PTA created a cross-country trail for the meet.
a. The PTA placed a trail marker in the ground every four hundred yards. Every nine hundred yards, the PTA set up a water station. What is the shortest distance a runner will have to run to see both a water station and trail marker at the same location?
Answer:

LCM 2 . 2 . 3 . 3 = 36 hundred
36 hundred yards

b. There are 1,760 yards in one mile. About how many miles will a runner have to run before seeing both a water station and trail marker at the same location? Calculate the answer to the nearest hundredth of a mile.
Answer:

c. The PTA wants to cover the wet areas of the trail with wood chips. They find that one bag of wood
chips covers a 3\(\frac{1}{2}\)-yard section of the trail. If there is a wet section of the trail that is approximately 50\(\frac{1}{4}\) yards long, how many bags of wood chips are needed to cover the wet section of the trail?
Answer:

Question 5.
The Art Club wants to paint a rectangle-shaped mural to celebrate the winners of the track and field meet. They design a checkerboard background for the mural where they will write the winners’ names. The rectangle measures 432 inches in length and 360 inches in width. Apply Euclid’s algorithm to determine the side length of the largest square they can use to fill the checkerboard pattern completely without overlap or gaps.
Answer:

Engage NY Eureka Math 6th Grade Module 2 Lesson 19 Answer Key

Eureka Math Grade 6 Module 2 Lesson 19 Opening Exercise Answer Key

Euclid’s algorithm is used to find the greatest common factor (GCF) of two whole numbers.
1. Divide the larger of the two numbers by the smaller one.
2. If there is a remainder, divide it into the divisor.
3. Continue dividing the last divisor by the last remainder until the remainder is zero.
4. The final divisor is the GCF of the original pair of numbers.

383 ÷ 4 =
Answer:
95.75

432 ÷ 12 =
Answer:
36

403 ÷ 13 =
Answer:
31

Eureka Math Grade 6 Module 2 Lesson 19 Example Answer Key

Example 1:
Euclid’s Algorithm Conceptualized

Answer:
→ Notice that we can use the GCF of 20 to create the largest square tile that covers the rectangle without any overlap or gaps. We used a 20 × 20 tile.
→ But, what if we did not know that? We could start by guessing. What is the biggest square tile that we can guess?
60 × 60

Display the following diagram:

→ It fits, but there are 40 units left over. Do the division problem to prove this.
→ What is the leftover area?
60 × 40
→ What is the largest square tile that we can fit in the leftover area?
40 × 40

Display the following diagram:

→ What is the leftover area?
20 × 40
→ What is the largest square tile that we can fit in the leftover area? 20)40
20 × 20
→ When we divide 40 by 20, there is no remainder. So, we have tiled the entire rectangle.
→ If we had started tiling the whole rectangle with squares, the largest square we could have used would be 20 by 20.

Example 2.
a. Let’s apply Euclid’s algorithm to some of the problems from our last lesson.
i. What is the GCF of 30 and 50?
Answer:
10

ii. Using Euclid’s algorithm, we follow the steps that are listed In the Opening Exercise.
Answer:

When the remainder is zero, the final divisor is the GCF.

b. Apply Euclid’s algorithm to find the GCF (30, 45).
Answer:

15

Example 3.
Larger Numbers
GCF (96, 144)
Answer:

Example 4:
Area Problems
The greatest common factor has many uses. Among them, the GCF lets us find out the maximum size of squares that cover a rectangle. When we solve problems like this, we cannot have any gaps or any overlapping squares. Of course, the maximum size squares is the minimum number of squares needed.

A rectangular computer table measures 30 inches by 50 inches. We need to cover it with square tiles. What is the side length of the largest square tile we can use to completely cover the table without overlap or gaps?
Answer:

a. If we use squares that are 10 by 10, how many do we need?
Answer:
3 . 5, or 15 squares

b. If this were a giant chunk of cheese ¡n a factory, would ¡t change the thinking or the calculations we just did?
Answer:
No

c. How many 10 inch × 10 inch squares of cheese could be cut from the giant 30 inch × 50 inch slab?
Answer:
15

Eureka Math Grade 6 Module 2 Lesson 19 Problem Set Answer Key

Question 1.
Use Euclid’s algorithm to find the greatest common factor of the following pairs of numbers:
a. GCF(12, 78)
Answer:

GCF (12, 78) = 6

b. GCF(18, 176)
Answer:

GCF (18, 176) = 2

Question 2.
Juanita and Samuel are planning a pizza party. They order a rectangular sheet pizza that measures 21 inches by 36 inches. They tell the pizza maker not to cut it because they want to cut it themselves.
a. All pieces of pizza must be square with none left over. What is the side length of the largest square pieces into which Juanita and Samuel can cut the pizza?
Answer:
GCF (21, 36) = 3 They can cut the pizza into 3 inch by 3 inch squares.

b. How many pieces of this size can be cut?
Answer:
7. 12 = 84 Juanita and Samuel can cut 84 pieces.

Question 3.
Shelly and Mickelle are making a quilt. They have a piece of fabric that measures 48 inches by 168 inches.
a. All pieces of fabric must be square with none left over. What is the side length of the largest square pieces into which Shelly and Mickelle can cut the fabric?
Answer:
GCF (48, 168) = 24

b. How many pieces of this size can Shelly and Mickelle cut?
Answer:
2 . 7 = 14 They can cut 14 pieces.

Eureka Math Grade 6 Module 2 Lesson 19 Exit Ticket Answer Key

Question 1.
Use Euclid’s algorithm to find the greatest common factor of 45 and 75.
Answer:

GCF (45, 75) = 15

Engage NY Eureka Math 6th Grade Module 2 Lesson 18 Answer Key

Eureka Math Grade 6 Module 2 Lesson 18 Example Answer Key

Find the greatest common factor of 12 and 18.
→ Listing these factor pairs in order helps ensure that no common factors are missed. Start with 1 multiplied by the number.
→ Circle all factors that appear on both lists.
→ Place a triangle around the greatest of these common factors.
GCF (12, 18)

12

Answer:

18

Answer:

Example 2.
Least Common Multiple
Find the least common multiple of 12 and 18.
LCM (12, 18)
Write the first 10 multiples of 12.
Answer:
12, 24, 36, 48, 60, 72, 84, 96, 108, 120

Write the first 10 multiples of 18.
Answer:
18, 36, 54, 72, 90, 108, 126, 144, 162, 180

Circle the multiples that appear on both lists.
Answer:

Put a rectangle around the least of these common multiples.
Answer:

Eureka Math Grade 6 Module 2 Lesson 18 Exercise Answer Key

Exercise 1.
Station 1: Factors and GCF
Choose one of these problems that has not yet been solved. Solve It together on your student page. Then, use your marker to copy your work neatly on the chart paper. Use your marker to cross out your choice so that the next group solves a different problem.
GCF (30, 50)
Answer:
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30                       Factors of 50: 1, 2, 5, 10, 25, 50
Common Factors: 1, 2, 5, 10                                   Greatest Common Factor: 10

GCF (30, 45)
Answer:
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30                        Factors of 45: 1, 3, 5, 9, 15, 45
Common Factors: 1, 3, 5, 15                                     Greatest Common Factor: 15

GCF (45, 60)
Answer:
Factors of 45: 1, 3, 5, 9, 15, 45                        Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common Factors: 1, 3, 5, 15                           Greatest Common Factor: 15

GCF (42, 70)
Answer:
Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42                       Factors of 70: 1, 2, 5, 7, 10, 14, 35, 70
Common Factors: 1, 2, 7, 14                                   Greatest Common Factor: 14

GCF (96, 144)
Answer:
Factors of 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96                       Factors of 144: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
Common Factors: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48                            Greatest Common Factor: 48

Next, choose one of these problems that has not yet been solved:

a. There are 18 girls and 24 boys who want to participate in a Trivia Challenge. If each team must have the same ratio of girls and boys, what is the greatest number of teams that can enter? Find how many boys and girls each team would have.
Answer:
6 teams can enter the Trivia Challenge, each having 3 girls and 4 boys.

b. Ski Club members are preparing Identical welcome kits for new skiers. The Ski Club has 60 hand-warmer packets and 48 foot-warmer packets. Find the greatest number of identical kits they can prepare using all of the hand-warmer and foot-warmer packets. How many hand-warmer packets and foot-warmer packets would each welcome kit have?
Answer:
There would be 12 welcome kits, each having 5 hand-warmer packets and 4 foot-warmer packets.

c. There are 435 representatives and 100 senators serving in the United States Congress. How many Identical groups with the same numbers of representative and senators could be formed from all of Congress if we want the largest groups possible? How many representatives and senators would be In each group?
Answer:
5 identical groups with the same numbers of representatives and senators can be formed, each group with 87 representatives and 20 senators.

d. Is the GCF of a pair of numbers ever equal to one of the numbers? Explain with an example.
Answer:
Yes. Valid examples should show a pair of numbers where the lesser of the two numbers is a factor of the greater number; the greater of the two numbers is a multiple of the lesser number.

e. Is the GCF of a pair of numbers ever greater than both numbers? Explain with an example.
Answer:
No. Factors are, by definition, less than or equal to the number. Therefore, the GCF cannot be greater than both numbers.

Station 2: Multiples and LCM
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper. Use your marker to cross out your choice so that the next group solves a different problem.
LCM (9, 12)
Answer:
Multiples of 9: 9, 18, 27, 36                                                                    Multiples of 12: 12, 24, 36
Least Common Multiple: 36

LCM (8, 18)
Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72                                         Multiples of 18: 18, 36, 54, 72
Least Common Multiple: 72

LCM (4, 30)
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60         Multiples of 30: 30, 60
Least Common Multiple: 60

LCM (12, 30)
Multiples of 12: 12, 24, 36, 48, 60                                                              Multiples of 30: 30 ,60
Least Common Multiple: 60

LCM (20, 50)
Multiples of 20: 20, 40, 60, 80, 100                                                             Multiples of 50: 50, 100
Least Common Multiple: 100

Next, choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on this chart paper and to cross out your choice so that the next group solves a different problem.

a. Hot dogs come packed 10 in a package. Hot dog buns come packed 8 in a package. If we want one hot dog for each bun for a picnic with none left over, what is the least amount of each we need to buy? How many packages of each item would we have to buy?
Answer:
Four packages of hot dogs = 40 hot dogs. Five packages of buns = 40 buns. LCM (8, 10) = 40.

b. Starting at 6:00 a.m., a bus stops at my street corner every 15 minutes. Also starting at 6:00 a.m., a taxi cab comes by every 12 minutes. What is the next time both a bus and a taxi are at the corner at the same time?
Answer:
Both o bus and a taxi arrive at the corner at 7:00 a.m., which is 60 minutes after 6:00 a.m.
LCM (12, 15) = 60.

c. Two gears in a machine are aligned by a mark drawn from the center of one gear to the center of the other. If the first gear has 24 teeth, and the second gear has 40 teeth, how many revolutions of the first gear are needed until the marks line up again?
Answer:
The first gear needs five revolutions. During this time, 120 teeth pass by. The second gear revolves three times. LCM(24, 40) = 120.

d. Is the LCM of a pair of numbers ever equal to one of the numbers? Explain with an example.
Answer:
Yes. Valid examples should show of a pair of numbers where the lesser of the two numbers is a factor of the greater number; the greater of the two numbers is a multiple of the lesser number.

e. Is the LCM of a pair of numbers ever less than both numbers? Explain with an example.
Answer:
No. Multiples are, by definition, equal to or greater than the number.

Station 3: Using Prime Factors to Determine GCF
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.

Next, choose one of these problems that has not yet been solved:

a. Would you rather find all the factors of a number or find all the prime factors of a number? Why?
Answer:
Accept opinions. Students should defend their answer and use accurate mathematical terms in their response.

b. Find the GCF of your original pair of numbers.
Answer:
See answers listed in Exploratory Challenge 1.

c. Is the product of your LCM and GCF less than, greater than, or equal to the product of your numbers?
Answer:
In all cases, GCF (a, b) . LCM (a, b) = a . b.

d. Glenn’s favorite number is very special because it reminds him of the day his daughter, Sarah, was born. The factors of this number do not repeat, and all of the prime numbers are less than 12. What is Glenn’s number? When was Sarah born?
Answer:
2 . 3 . 5 . 7 . 11 = 2,310 Sarah’s birth date is 2/3/10.

Station 4: Applying Factors to the Distributive Property

Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.

Find the GCF from the two numbers, and rewrite the sum using the distributive property.

1. 12 + 18 =
Answer:
6(2) + 6(3) = 6(2 + 3) = 6(5) = 30

2. 42 + 14 =
Answer:
7(6) + 7(2) = 7(6 + 2) = 7(8) = 56

3. 36 + 27 =
Answer:
9(4) + 9(3) = 9(4 + 3) = 9(7) = 63

4. 16 + 72 =
Answer:
8(2) + 8(9) = 8(2 + 9) = 8(11) = 88

5. 44 + 33 =
Answer:
11(4) + 11(3) = 11(4 + 3) = 11(7) = 77

Next, add another example to one of these two statements applying factors to the distributive property.

Choose any numbers for n, a, and b.

n(a) + n(b) = n(a + b)
Answer:
Accept all mathematically correct responses.

n(a) – n(b) = n(a – b)
Answer:
The distributive property holds for addition as well as subtraction. Accept all mathematically correct responses.

Eureka Math Grade 6 Module 2 Lesson 18 Problem Set Answer Key

Station 1: Factors and GCF
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.
Find the greatest common factor of one of these pairs: 30, 50; 30, 45; 45, 60; 42, 70; 96, 144.
Answer:

Next, choose one of these problems that has not yet been solved:
a. There are 18 girls and 24 boys who want to participate in a Trivia Challenge. If each team must have the same ratio of girls and boys, what is the greatest number of teams that can enter? Find how many boys and girls each team would have.
Answer:

b. Ski Club members are preparing identical welcome kits for new skiers. The Ski Club has 60 hand-warmer packets and 48 foot-warmer packets. Find the greatest number of identical kits they can prepare using all of the hand-warmer and foot-warmer packets. How many hand-warmer packets and foot-warmer packets would each welcome kit have?
Answer:

c. There are 435 representatives and loo senators serving in the United States Congress. How many identical groups with the same numbers of representatives and senators could be formed from all of Congress if we want the largest groups possible? How many representatives and senators would be in each group?
Answer:

d. Is the GCF of a pair of numbers ever equal to one of the numbers? Explain with an example.
Answer:

e. Is the GCF of a pair of numbers ever greater than both numbers? Explain with an example.
Answer:

Station 2: Multiples and LCM
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.
Find the least common multiple of one of these pairs: 9, 12; 8, 18; 4, 30; 12, 30; 20, 50.
Answer:

Next, choose one of these problems that has not yet been solved:
a. Hot dogs come packed 10 in a package. Hot dog buns come packed 8 in a package. If we want one hot dog for each bun for a picnic, with none left over, what is the least amount of each we need to buy? How many packages of each item would we have to buy?
Answer:

b. Starting at 6:00 a.m., a bus stops at my street corner every 15 minutes. Also starting at 6:00 a.m., a taxi cab comes by every 12 minutes. What is the next time both a bus and a taxi are at the corner at the same time?
Answer:

c. Two gears in a machine are aligned by a mark drawn from the center of one gear to the center of the other. If the first gear has 24 teeth, and the second gear has 40 teeth, how many revolutions of the first gear are needed until the marks line up again?
Answer:

d. Is the LCM of a pair of numbers ever equal to one of the numbers? Explain with an example.
Answer:

e. Is the LCM of a pair of numbers ever less than both numbers? Explain with an example.
Answer:

Solve it together on your student page. Then, use your marker to copy your work neatly on this chart paper and to cross out your choice so that the next group solves a different problem.

Station 3: Using Prime Factors to Determine GCF

Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.

Use prime factors to find the greatest common factor of one of the following pairs of numbers:
30, 50 30, 45 45, 60 42, 70 96, 144
Answer:

Next choose one of these problems that has not yet been solved:
a. Would you rather find all the factors of a number or find all the prime factors of a number? Why?
Answer:

b. Find the GCF of your original pair of numbers.
Answer:

c. Is the product of your LCM and GCF less than, greater than, or equal to the product of your numbers?
Answer:

d. Glenn’s favorite number is very special because it reminds him of the day his daughter, Sarah, was born. The factors of this number do not repeat, and all of the prime numbers are less than 12. What is Glenn’s number? When was Sarah born?
Answer:

Station 4: Applying Factors to the Distributive Property
Study these examples of how factors apply to the distributive property.

8 + 12 = 4(2) + 4(3) = 4(2 + 3) = 20
4(2) + 4(3) = 4(5) = 20

15 + 25 = 5(3) + 5(5) = 5(3 + 5) = 40
5(3) + 5(5) = 5(8) = 40

36 – 24 = 4(9) – 4(6) = 4(9 – 6) = 12
4(9) – 4(6) = 4(3) = 12

Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.

Find the GCF from the two numbers, and rewrite the sum using the distributive property.
Question 1.
12 + 18 =
Answer:

Question 2.
42 + 14 =
Answer:

Question 3.
36 + 27 =
Answer:

Question 4.
16 + 72 =
Answer:

Question 5.
44 + 33 =
Answer:

Next, add another example to one of these two statements applying factors to the distributive property.
Choose any numbers for n, a, and b.
n(a) + n(b) = n(a + b)
Answer:

n(a) – n(b) = n(a – b)
Answer:

Eureka Math Grade 6 Module 2 Lesson 18 Exit Ticket Answer Key

Question 1.
Find the LCM and GCF of 12 and 15.
Answer:
LCM: 60; GCF: 3

Question 2.
Write two numbers, neither of which is 8, whose GCF is 8.
Answer:
Answers will vary (e.g., 16 and 24, or 24 and 32).

Question 3.
Write two numbers, neither of which is 28, whose LCM is 28.
Answer:
Answers will vary (e.g., 4 and 14, or 4 and 7).

Rate each of the stations you visited today. Use this scale:
3 – Easy – I’ve got It, I don’t need any help.
2 – Medium – I need more practice, but I understand some of It.
1 – Hard – I’m not getting this yet.

Complete the following chart:

Engage NY Eureka Math 6th Grade Module 2 Lesson 17 Answer Key

Eureka Math Grade 6 Module 2 Lesson 17 Opening Exercise Answer Key

Below is a list of 10 numbers. Place each number in the circle(s) that is a factor of the number. Some numbers can be placed in more than one circle. For example, if 32 were on the list, it would be placed in the circles with 2, 4, and 8 because they are all factors of 32.
24; 36; 80; 115; 214; 360; 975; 4,678; 29,785; 414, 940

Answer:

Discussion:

Divisibility rule for 2: If and only if its unit digit is 0, 2, 4, 6, or 8

Divisibility rule for 4: If and only if its last two digits are a number divisible by 4

Divisibility rule for 5: If and only if its unit digit is 0 or 5

Divisibility rule for 8: If and only if its last three digits are a number divisible by 8

Divisibility rule for 10: If and only if its unit digit is 0

Decimal numbers with fraction parts do not follow the divisibility tests:
Students learn two new divisibility rules today. The rules are used to determine if numbers are divisible by 3 or 9. Start with students who already know the factors of 3 and 9, so they can see that the rule works.

→ What do the numbers 12, 18, 30, 66, and 93 all have in common?
Each is divisible by 3.

→ Calculate the sum of the digits for each given number. For example, the sum of the digits in the number 12 is 3 because 1 + 2 = 3.

Give students time to find the sums. Record the sums on the board.
→ What do these sums have in common?
They are divisible by 3.

→ When the sum of a number’s digits is divisible by 3, the entire number is divisible by 3. Now let’s examine a different set of numbers: 27, 36, 54, 72, and 99. What do these numbers have in common?
They are divisible by 9.

→ Calculate the sum of the digits for each given number.
Provide time for students to find the sums. Record the sums on the board.

→ What do all the sums have in common?
They are divisible by 9.

→ When the sum of the digits is divisible by 3 and 9, the entire number is divisible by 9. Let’s use this knowledge to determine if a large number is divisible by 3, 9, or both. The number 765 is divisible by both 3 and 9.

→ We can use what we know about the distributive property to prove that 765 is divisible by 3 and by 9.
→ Let’s begin by expanding 765.

Display the following progression.
→ We can represent 765 as:
7 × 100 + 6 × 10 + 5 × 1

→ We can further decompose the numbers that are easily seen as divisible by 3 and 9.
→ Let’s decompose 100 to 99 + 1. Why would we do this?
Because we already know that 99 is divisible by 3 and 9.
7(99 + 1) + 6 × 10 + 5

→ We can also decompose 10 to 9 + 1. Why would we do this?
Because we already know that 9 is divisible by 3 and 9.
7(99 + 1) + 6(9+ 1) + 5

→ Let’s use the distributive property further to distribute the factor 7 in the expression. We can represent 7
times the quantity (99 + 1) as 7(99) + 7 × 1, or 7(99) + 7.
7(99) + 7+ 6(9 + 1) + 5

→ We can distribute the factor 6 in the same fashion. How can we distribute the factor 6 in the expression?
We can represent 6 times the quantity (9 + 1) as 6(9) + 6 × 1, or 6(9) + 6.
7(99) + 7 + 6(9) + 6 + 5

→ Since we know that 9 is divisible by both 3 and 9, let’s factor the 9 out of the expression. We can use the commutative and associative properties to easily see this.
7(99) + 7 + 6(9) + 6 + 5
7(9 × 11) + 6(9 × 1) + 7 + 6 + 5
9(7 × 11 + 6) + 7 + 6 + 5

→ Let’s investigate our current expression. Obviously the product of 9(7 × 11 + 6) is divisible by 9 since the 9 is already factored out.

→ What about the sum of 7 + 6 + 5? What is the sum? Is this sum divisible by 3 and 9?
7 + 6 + 5 = 18.
This sum is divisible by 3 and 9.

→ Are 3 and 9 both factors of 18?
Yes

→ What do you notice about the addends 7 + 6 + 5?
They are the digits of our original number 765.

Let’s look once more at the expression:

→ Since the first term is divisible by 3 and 9, the number 765 is divisible by 3 and 9 if and only if 7 + 6 + 5 is also divisible by 3 and 9.
→ This process can be used for any decimal whole number!
Introduce the divisibility rules for 3 and 9. Have students record the rules in their student materials.

Divisibility rule for 3: If the sum of the digits is divisible by 3, then the number is divisible by 3.

Divisibility rule for 9: If the sum of the digits is divisible by 9, then the number is divisible by 9.

Eureka Math Grade 6 Module 2 Lesson 17 Example Answer Key

Example 1.
This example shows how to apply the two new divisibility rules we just discussed.
Explain why 378 is divisible by 3 and 9.

a. Expand 378.
Answer:
300 + 70 + 8
3 × 100 + 7 × 1o + 8

b. Decompose the expression to factor by 9.
Answer:
3(99 + 1) + 7(9 + 1) + 8
3(99) + 3 + 7(9) + 7 + 8

c. Factor the 9.
Answer:
3(9 × 11) + 3 + 7(9 × 1) + 7 + 8
9(3 × 11 + 7) + 3 + 7 + 8

d. What is the sum of the three digits?
Answer:
3 + 7 + 8 = 18; the sum of the three digits is 18.

e. Is 18 divisible by 9?
Answer:
Yes

f. Is the entire number 378 divisible by 9? Why or why not?
Answer:
The number 378 is divisible by 9 because the sum of the digits is divisible by 9.

g. Is the number 378 divisible by 3? Why or why not?
Answer:
Three is a factor of 378 because if 9 is a factor of 378, then 3 will also be a factor. OR
The number 378 is divisible by 3 because the sum of the digits is divisible by 3.

Example 2.
Is 3,822 divisible by 3 or 9? Why or why not?
Answer:
The number 3,822 is divisible by 3 but not by 9 because the sum of the digits is 3 + 8 + 2 + 2 = 15, and 15 is divisible by 3 but not by 9.

Eureka Math Grade 6 Module 2 Lesson 17 Exercise Answer Key

Circle ALL the numbers that are factors of the given number. Complete any necessary work in the space provided.

Exercise 1.
2,838 is divisible by

Answer:

Explain your reasoning for your choice(s).
Answer:
The number 2,838 is divisible by 3 because 3 is a factor of 2,838. I know this because the sum of the digits is 21, which is divisible by 3. The number 2,838 is not divisible by 9 because 21 is not divisible by 9, and 2,838 is not divisible by 4 because the last two digits (38) are not divisible by 4.

Exercise 2.
34,515 is divisible by

Answer:

Explain your reasoning for your choice (s).
Answer:
The number 34,515 is divisible by 3 and 9 because both 3 and 9 are factors of 34,515. I know this because the sum of the digits is 18, and 18 is divisible by both 3 and 9. The number 34,515 is also divisible by 5 because the unit digit is a 5.

Exercise 3.
10,534,341 is divisible by

Answer:

Explain your reasoning for your choice(s).
Answer:
The number 10,534,341 is divisible by 3 but not 9 because 3 is afactor of 10,534,341, but 9 is not. I know this because the sum of the digits is 21, which is divisible by 3 but not 9. The number 10, 534, 341 is not divisible by 2 because it does not end with 0, 2, 4, 6, or 8.

Exercise 4.
4,320 is divisible by

Answer:

Explain your reasoning for your choice(s).
Answer:
The number 4,320 is divisible by 3 and 9 because 3 and 9 are factors of 4,320. I know this because the sum of the digits is 9, which is divisible by 3 and 9. The number 4,320 is also divisible by 10 because 10 is a factor of 4,320. I know this because the unit digit is 0.

Exercise 5.
6,240 is divisible by

Answer:

Explain your reasoning for your choice(s).
Answer:
The number 6,240 is divisible by 3 but not divisible by 9 because 3 is a factor of 6,240, but 9 is not. I know this because the sum of the digits is 12, which is divisible by 3 but not divisible by 9. The number 6,240 is divisible by 8 because the last three digits (240) is divisible by 8.

Eureka Math Grade 6 Module 2 Lesson 17 Problem Set Answer Key

Question 1.
Is 32, 643 divisible by both 3 and 9? Why or why not?
Answer:
The number 32,643 is divisible by both 3 and 9 because the sum of the digits is 18, which is divisible by 3 and 9.

Question 2.
Circle all the factors of 424,380 from the list below.
Answer:

Question 3.
Circle all the factors of 322,875 from the list below.
Answer:

Question 4.
Write a 3-digit number that ¡s divisible by both 3 and 4. Explain how you know this number is divisible by 3 and 4.
Answer:
Answers will vary. Possible student response: The sum of the digits is divisible by 3, and that’s how I know the number is divisible by 3. The last 2 digits are divisible by 4, so the entire number is divisible by 4.

Question 5.
Write a 4-digit number that Is divisible by both 5 and 9. Explain how you know this number Is divisible by 5 and 9.
Answer:
Answers will vary. Possible student response: The number ends with a 5 or 0, so the entire number is divisible by 5. The sum of the digits is divisible by 9, so the entire number is divisible by 9.

Eureka Math Grade 6 Module 2 Lesson 17 Exit Ticket Answer Key

Question 1.
Is 26, 341 divisible by 3? If it is, write the number as the product of 3 and another factor. If not, explain.
Answer:
The number 26,341 is not divisible by 3 because the sum of the digits is 16, which is not divisible by 3.

Question 2.
Is 8,397 divisible by 9? If it is, write the number as the product of 9 and another factor. If not, explain.
Answer:
The number 8,397 is divisible by 9 because the sum of the digits is 27, which is divisible by 9. Nine is a factor of 8,397 because 9 × 933 = 8,397.

Question 3.
Explain why 186,426 is divisible by both 3 and 9.
Answer:
The number 186, 426 is divisible by both 3 and 9 because the sum of the digits is 27, which is divisible by both 3 and 9.

Engage NY Eureka Math 6th Grade Module 2 Lesson 16 Answer Key

Eureka Math Grade 6 Module 2 Lesson 16 Opening Exercise Answer Key

a. What is an even number?
Answer:
Possible student responses:
An integer that can be evenly divided by 2
A number whose unit digit is 0, 2, 4, 6, or 8
All the multiples of 2

b. List some examples of even numbers.
Answer:
Answers will vary.

c. What is an odd number?
Answer:
Possible student responses:
An integer that CANNOT be evenly divided by 2
A number whose unit digit is 1, 3, 5, 7, or 9
All the numbers that are NOT multiples of 2

d. List some examples of odd numbers.
Answer:
Answers will vary.

What happens when we add two even numbers? Do we always get an even number?
Answer:
Before holding a discussion about the process to answer the following questions, have students write or share their predictions.

Eureka Math Grade 6 Module 2 Lesson 16 Exercise Answer Key

Exercise 1.
Why is the sum of two even numbers even?
a. Think of the problem 12 + 14. Draw dots to represent each number.
Answer:

b. Circle pairs of dots to determine if any of the dots are left over.
Answer:

c. Is this true every time two even numbers are added together? Why or why not?
Answer:
Since 12 is represented by 6 sets of two dots, and 14 is represented by 7 sets of two dots, the sum is 13 sets of two dots. This is true every time two even numbers are added together because even numbers never have dots left over when we are circling pairs. Therefore, the answer is always even.

Exercise 2.
Why is the sum of two odd numbers even?
a. Think of the problem 11 + 15. Draw dots to represent each number.
Answer:

b. Circle pairs of dots to determine if any of the dots are left over.
Answer:

When we circle groups of two dots, there is one dot remaining in each representation because each addend is an odd number. When we look at the sum, however, the two remaining dots can form a pair, leaving us with a sum that is represented by groups of two dots. The sum is, therefore, even. Since each addend is odd, there is one dot for each addend that does not have a pair. However, these two dots can be paired together, which means there are no dots without a pair, making the sum an even number.

c. Is this true every time two odd numbers are added together? Why or why not?
Answer:
This is true every time two odd numbers are added together because every odd number has one dot remaining when we circle pairs of dots. Since each number has one dot remaining, these dots can be combined to make another pair. Therefore, no dots remain, resulting in an even sum.

Exercise 3.
Why is the sum of an even number and an odd number odd?
a. Think of the problem 14 + 11. Draw dots to represent each number.
Answer:
Students draw dots to represent each number. After circling pairs of dots, there is one dot left for the number 11, and the number 14 has no dots remaining. Since there is one dot left over, the sum is odd because not every dot has a pair.

b. Circle pairs of dots to determine if any of the dots are left over.
Answer:
Students draw dots to represent each number. After circling pairs of dots, there is one dot left for the number 11, and the number 14 has no dots remaining. Since there is one dot left over, the sum is odd because not every dot has a pair.

C. Is this true every time an even number and an odd number are added together? Why or why not?
Answer:
This is always true when an even number and an odd number are added together because only the odd number will have a dot remaining after we circle pairs of dots. Since this dot does not have a pair, the sum is odd.

d. What if the first addend is odd and the second is even? Is the sum still odd? Why or why not? For example, If we had 11 + 14, would the sum be odd?
Answer:
The sum is still odd for two reasons. First, the commutative property states that changing the order of an addition problem does not change the answer. Because an even number plus an odd number is odd, then an odd number plus an even number is also odd. Second, it does not matter which addend is odd; there is still one dot remaining, making the sum odd.

Let’s sum it up:
Answer:
→ “Even” + “even” = “even”
→ “Odd” + “odd” = “even”
→ “Odd” + “even” = “odd”

Exploratory Challenge/Exercises 4-6

Exercise 4.
The product of two even numbers is even.
Answer:
Answers will vary, but one example answer is provided.

Using the problem 6 × 14, students know that this is equivalent to six groups of fourteen, or
14 + 14 + 14 + 14 + 14 + 14. Students also know that the sum of two even numbers is even; therefore, when adding the addends two at a time, the sum is always even. This means the sum of six even numbers is even, making the product even since It is equivalent to the sum.

Using the problem 6 × 14, students can use the dots from previous examples.

From here, students can circle dots and see that there are no dots remaining, so the answer must be even.

Exercise 5.
The product of two odd numbers is odd.
Answer:
Answers will vary, but an example answer is provided.
Using the problem 5 × 15, students know that this is equivalent to five groups of fifteen, or
15 + 15 + 15 + 15 + 15. Students also know that the sum of two odd numbers is even, and the sum of an odd and even number is odd. When adding two of the addends together at a time, the answer rotates between even and odd. When the final two numbers are added together, one is even and the other odd. Therefore, the sum is odd, which makes the product odd since it is equivalent to the sum.

Using the problem 5 × 15, students may also use the dot method.

After students circle the pairs of dots, one dot from each set of 15 remains, for a total of 5 dots. Students can group these together and circle more pairs, as shown below.

Since there is still one dot remaining, the product of two odd numbers is odd.

Exercise 6.
The product of an even number and an odd number is even.
Answer:
Answers will vary, but one example is provided.
Using the problem 6 × 7, students know that this is equivalent to the sum of six sevens, or 7 + 7 + 7 + 7 + 7 + 7.
Students also know that the sum of two odd numbers is even, and the sum of two even numbers is even. Therefore, when adding two addends at a time, the result is an even number. The sum of these even numbers is also even, which means the total sum is even. This also Implies the product is even since the sum and product are equivalent.

Using the problem 6 × 7, students may also use the dot method.

After students circle the pairs of dots, one dot from each set of 7 remains, for a total of 6 dots. Students can group these together and circle more pairs, as shown below.

Since there are no dots remaining, the product of an even number and an odd number is even.

Eureka Math Grade 6 Module 2 Lesson 16 Problem Set Answer Key

Without solving, tell whether each sum or product Is even or odd. Explain your reasoning.

Question 1.
346 + 721
Answer:
The sum is odd because the sum of an even and an odd number is odd.

Question 2.
4,690 × 141
Answer:
The product is even because the product of an even and an odd number is even.

Question 3.
1,462,891 × 745,629
Answer:
The product is odd because the product of two odd numbers is odd.

Question 4.
425,922 + 32,481,064
Answer:
The sum is even because the sum of two even numbers is even.

Question 5.
32 + 45 + 67 + 91 + 34 + 56
Answer:
The first two addends are odd because an even and an odd is odd.
Odd number +67 is even because the sum of two odd numbers is even.
Even number +91 is odd because the sum of an even and an odd number is odd.
Odd number +34 is odd because the sum of an odd and an even number is odd.
Odd number +56 is odd because the sum of an odd andan even number is odd.
Therefore, the final sum is odd.

Eureka Math Grade 6 Module 2 Lesson 16 Exit Ticket Answer Key

Determine whether each sum or product is even or odd. Explain your reasoning.

Question 1.
56,426+17,895
Answer:
The sum is odd because the sum of an even number and an odd number is odd.

Question 2.
317,362 × 129,324
Answer:
The product is even because the product of two even numbers is even.

Question 3.
104,81 + 4,569
Answer:
The sum is even because the sum of two odd numbers is even.

Question 4.
32,457 × 12,781
Answer:
The product is odd because the product of two odd numbers is odd.

Question 5.
Show or explain why 12 + 13 + 14 + 15 + 16 results in an even sum.
Answer:
12 + 13 is odd because even + odd is odd.
Odd number +14 is odd because odd + even is odd.
Odd number +15 is even because odd + odd is even.
Even number +16 is even because even + even is even.
OR
Students may group even numbers together, 12 + 14 + 16, which results in an even number. Then, when students combine the two odd numbers, 13 + 15, the result is another even number. We know that the sum of two evens results in another even number.

Engage NY Eureka Math 8th Grade Module 2 Lesson 5 Answer Key

Eureka Math Grade 8 Module 2 Lesson 5 Exercise Answer Key

Exercise 1.
Let there be a rotation of d degrees around center O. Let P be a point other than O. Select d so that d≥0. Find P’ (i.e., the rotation of point P) using a transparency.

Answer:
Verify that students have rotated around center O in the counterclockwise direction.

Exercise 2.
Let there be a rotation of d degrees around center O. Let P be a point other than O. Select d so that d<0. Find P’ (i.e., the rotation of point P) using a transparency.

Answer:
Verify that students have rotated around center O in the clockwise direction.

Exercise 3.
Which direction did the point P rotate when d≥0?
Answer:
It rotated counterclockwise, or to the left of the original point.

Exercise 4.
Which direction did the point P rotate when d<0?
Answer:
It rotated clockwise, or to the right of the original point.

Exercises 5–6

Exercise 5.
Let L be a line, \(\overrightarrow{A B}\) be a ray, \(\overline{\boldsymbol{C D}}\) be a segment, and ∠EFG be an angle, as shown. Let there be a rotation of d degrees around point O. Find the images of all figures when d≥0.

Answer:

Verify that students have rotated around center O in the counterclockwise direction.

Exercise 6.
Let \(\overline{A B}\) be a segment of length 4 units and ∠CDE be an angle of size 45°. Let there be a rotation by d degrees, where d<0, about O. Find the images of the given figures. Answer the questions that follow.

Answer:

Verify that students have rotated around center O in the clockwise direction.

a. What is the length of the rotated segment Rotation(AB)?
Answer:
The length of the rotated segment is 4 units.

b. What is the degree of the rotated angle Rotation (∠CDE)?
Answer:
The degree of the rotated angle is 45°.

Exercises 7–8

Exercise 7.
Let L₁ and L₂ be parallel lines. Let there be a rotation by d degrees, where -360<d<360, about O.
Is (L₁ )’∥(L₂)’?

Answer:

Verify that students have rotated around center O in either direction. Students should respond that (L₁)’ || (L₂)’.

Exercise 8.
Let L be a line and O be the center of rotation. Let there be a rotation by d degrees, where d≠180 about O. Are the lines L and L’ parallel?

Answer:

Verify that students have rotated around center O in either direction any degree other than 180. Students should respond that L and L’ are not parallel.

Eureka Math Grade 8 Module 2 Lesson 5 Exit Ticket Answer Key

Question 1.
Given the figure H, let there be a rotation by d degrees, where d≥0, about O. Let Rotation(H) be H’.

Answer:

Sample rotation shown above. Verify that the figure H’ has been rotated counterclockwise with center O.

Question 2.
Using the drawing above, let Rotation₁ be the rotation d degrees with d<0, about O. Let Rotation_1 (H) be H”.
Answer:
Sample rotation shown above. Verify that the figure H” has been rotated clockwise with center O.

Eureka Math Grade 8 Module 2 Lesson 5 Problem Set Answer Key

Question 1.
Let there be a rotation by -90° around the center O.

Answer:
Rotated figures are shown in red.

Question 2.
Explain why a rotation of 90 degrees around any point O never maps a line to a line parallel to itself.
Answer:
A 90-degree rotation around point O will move a given line L to L’. Parallel lines never intersect, so it is obvious that a 90-degree rotation in either direction does not make lines L and L’ parallel. Additionally, we know that there exists just one line parallel to the given line L that goes through a point not on L. If we let P be a point not on L, the line L’ must go through it in order to be parallel to L. L’ does not go through point P; therefore, L and L’ are not parallel lines. Assume we rotate line L first and then place a point P on line L’ to get the desired effect (a line through P). This contradicts our definition of parallel (i.e., parallel lines never intersect); so, again, we know that line L is not parallel to L’.

Question 3.
A segment of length 94 cm has been rotated d degrees around a center O. What is the length of the rotated segment? How do you know?
Answer:
The rotated segment will be 94 cm in length. (Rotation 2) states that rotations preserve lengths of segments, so the length of the rotated segment will remain the same as the original.

Question 4.
An angle of size 124° has been rotated d degrees around a center O. What is the size of the rotated angle? How do you know?
Answer:
The rotated angle will be 124°. (Rotation 3) states that rotations preserve the degrees of angles, so the rotated angle will be the same size as the original.