## Significant Figures Rules, Examples | How to Round Significant Figures?

Significant figures help to find and establish the number that is in the form of digits. The significant digits used instead of figures. It is easy to identify significant digits by counting the numbers from the 1st non-zero digit located on the left. Significant figures of any given number are nothing but the significant digits they used to convey the meaning according to its accuracy. For example, 2.238 has four significant digits.

Clearly understand the topics available below.

• Rules to find the number of significant figures
• Rounding off a decimal to the required number of significant figures
• Round off to a special unit.

## Significant Figures Rules

1. All non–zero numbers (1, 2, 3, 4, 5) are always significant.

Example:

• 2154 has four significant figures
• 142.35 has four significant figures

2. All zeros between non-zero numbers are always significant.

Example:

• 305.003 has six significant figures.
• 70.00 has four significant figures.
• 61.04020 has seven significant figures.

3. In a decimal number that lies between 0 and 1, all zeros that are to the right of the decimal point but to the left of a non-zero number are not significant.

Example:

• 0.00365 has only three significant figures.
• 0.006040 has four significant figures.

4. In a whole number if there are zeros to the left of an understood decimal point but to the right of a non-zero digit, the case becomes doubtful.

Example:

• 304000 there is an understood decimal point after the given six digits. There are 3 zeros that present to the left of the understood decimal point but to the right of a non-zero number so the case becomes doubtful.
• It is represented as 3.04 × 10⁵ also it consists of 3 significant figures. Also, it can represent as 3.040 × 10⁵, then the number of significant figures is 4.

5. When a decimal is round off to a given number of decimal places, all the final zeros in a decimal number are significant.

Example:

• If we round off 2.875 to two decimal places, we get 3.90 which has 3 significant figures.

### Rounding Off Decimals to the Required Number of Significant Figures

Rounding off the number correct to three significant figures is explained along with the examples.

(1) 63.314

Given that 63.314. It has 5 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal.
The digit in hundredth place 1 is less than 5. So, the digit in the tenths place remains 3 and the digits 1 and 4 disappear.
Therefore, 63.314 = 63.3 rounded off to 3 significant figures.

(2) 5.3062

Given that 5.3062. It has 5 significant figures. To round off the given number into 3 significant digits, we need to round it off to 2 places after the decimal.
The digit in thousandth place 6 is greater than 5. So, the digit in the hundredth place becomes 1 and the digits 6 and 2 disappear.
Therefore, 5.3062 = 5.31 rounded off to 3 significant figures.

(3) 50.003

Given that 50.003. It has 5 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal.
The digit in hundredth place 0 is less than 5. So, the digit in the tenths place remains 0 and the digits 0 and 3 disappear.
Therefore, 50.003 = 50.0 rounded off to 3 significant figures.

(4) 0.0002489

Given that 0.0002489. It has 4 significant figures. To round off the given number into 3 significant digits, we need to round it off to 6 places after the decimal.
The digit 9 is greater than 5. So, the digit 8 changes to 9, and the digit 9 disappear.
Therefore, 0.0002489 = 0.000249 rounded off to 3 significant figures.

(5) 0.0002477

Given that 0.0002477. It has 4 significant figures. To round off the given number into 3 significant digits, we need to round it off to 6 places after the decimal.
The digit 7 is greater than 5. So, the digit 7 changes to 8, and the digit 7 disappear.
Therefore, 0.0002477 = 0.000248 rounded off to 3 significant figures.

(6) 8.404

Given that 8.404. It has 4 significant figures. To round off the given number into 3 significant digits, we need to round it off to 2 places after the decimal.
The digit 4 is less than 5. So, the digit 0 remains the same 0, and the digit 4 disappears.
Therefore, 8.404 = 8.40 rounded off to 3 significant figures.

(7) 4.888

Given that 4.888. It has 4 significant figures. To round off the given number into 3 significant digits, we need to round it off to 2 places after the decimal.
The digit 8 is greater than 5. So, the digit 8 becomes 9, and the digit 8 disappears.
Therefore, 4.888 = 4.89 rounded off to 3 significant figures.

(8) 6.999

Given that 6.999. It has 4 significant figures. To round off the given number into 3 significant digits, we need to round it off to 0 places after the decimal.
The digit 9 is greater than 5. So, the digit 6 becomes 7, and the digits 9 disappears.
Therefore, 6.999 = 7 rounded off to 3 significant figures.

### Round Off the Following Measurements Examples

(i) 1384.977 kg correct to 6 significant figures.

Given that 1384.977 kg. It has 7 significant figures. To round off the given number into 6 significant digits, we need to round it off to 2 places after the decimal.
The digit 7 is greater than 5. So, the digit 7 becomes 8, and the digits 7 disappears.
Therefore, 1384.977 kg = 1384.98 kg rounded off to 6 significant figures.

(ii) 303.203 g correct to 4 significant figures.

Given that 303.203 g. It has 6 significant figures. To round off the given number into 4 significant digits, we need to round it off to 1 place after the decimal.
The digit 0 is less than 5. So, the digit 2 remains 2, and the digits 0 and 3 disappears.
Therefore, 303.203 g = 303.2 kg rounded off to 4 significant figures.

(iii) 2.0829 mg correct to 2 significant figures.

Given that 2.0829 mg. It has 5 significant figures. To round off the given number into 5 significant digits, we need to round it off to 1 place after the decimal.
The digit 8 is greater than 5. So, the digit 0 becomes 1, and the digits 8, 2, and 9 disappears.
Therefore, 2.0829 mg = 2.1 mg rounded off to 2 significant figures.

(iv) 0.004784 km correct to 1 significant figures.

Given that 0.004784 km. It has 4 significant figures. To round off the given number into 5 significant digits, we need to round it off to 3 places after the decimal.
The digit 7 is greater than 5. So, the digit 4 becomes 5, and the digits 7, 8, and 4 disappear.
Therefore, 0.004784 km = 0.005 km rounded off to 1 significant figures.

### Rounding Off to a Specified Unit Examples

(i) Round off $65537 to the nearest 10 dollars. Given that$ 65537. 65537 is in between 65530 and 65540. 65537 is closer to 65540. Therefore, $65537 to the nearest 10 dollars is$ 65540.

(ii) Round off $208.287 to the nearest 10 cents. Given that$ 208.287. $208.287 is in between 208.20 and 208.30. 208.287 is closer to 208.30. Therefore,$ 208.287 to the nearest 10 cents is $208.30. (iii) Round off 892.58 to the nearest dollar. Given that$ 892.58. $892.58 is in between 892 and 893. 892.58 is closer to 893. Therefore, 892.58 to the nearest dollar is$893.

(iv) Round off 575.085 to the nearest cents.

Given that 575.085. 575.085 is in between 575.080 and 575.90. 575.085 is closer to 575.90. Therefore, 575.085 to the nearest cents is 575.90.

(v) Round off 19.077 cm to the nearest mm.

Given that 19.077 cm. 19.077 cm is in between 19.0 cm and 19.1 cm. 19.077 cm is closer to 19.1 cm. 19.1 cm = 191 mm. Therefore, 19.077 cm to the nearest mm is 191 mm.

(vi) Round off 63.5389 m to the nearest cm.

Given that 63.5389 m. 63.5389 m is closer to 65.54 m. Therefore, 63.5389 m to the nearest cm is 65.54 m.

(vii) 0.00848 kg to nearest g.

Since 1 g = 0.001 kg so to round off 0.00848 kg to the nearest g, we have to round it off to 3 places of decimal. Therefore, 0.00848 kg = 0.008 kg rounded off to the nearest g.

(viii) 18.3373 g to the nearest mg.

Since 1 mg = 0.001 g so to round off 18.3373 g to the nearest mg, we have to round it off to three places of decimal. Therefore, 18.3373 g = 18.337 g rounded off to the nearest mg.

## Parallel and Transversal Lines | Properties of Parallel Lines cut by Transversal

Learn how a pair of angles formed between Parallel and Transversal Lines. When two parallel lines intersect by a Transversal Line, they form different angles. Those angles are given with different names based on their location. Therefore, check what are the angles formed and how to find them, and what names are given to them depending on their location. You can also get all concepts and examples available on Lines and Angles for free on our website.

When the transversal intersects two parallel lines

• Pairs of alternate angles are equal
• Pairs of corresponding angles are equal.
• Interior angles on the same side of the transversal are supplementary.

## Parallel Lines Cut by a Transversal Problems

1. In adjoining figure l ∥ m is cut by the transversal t. If ∠5 = 70, find the measure of ∠7, ∠9, ∠10.

Solution:
Given that ∠5 = 70° ∠5 = ∠7 (Vertically opposite angles)
Therefore, ∠7 = 70°
Now, ∠5 = ∠9 (Corresponding angles)
Therefore, ∠9 = 70°
Also, ∠7 + ∠10 = 180° (Co-interior angles)
70° + ∠10 = 180°

Therefore, ∠10 = 180° – 70° = 110°

2. In the given figure PQ ∥ RS, ∠QAO = 125°, ∠RBO = 40°. Find the measure of ∠AOB.

Solution:

Draw a line AB parallel to PQ and RS passing through O such that PQ ∥ MN and RS ∥ MN
∠QAO + ∠NOA = 180° (Co-interior angles)
Therefore, 125° + ∠NOA = 180°
Therefore, ∠NOA = 180° – 125° = 55°
Also, ∠RBO = ∠NOB (Alternate angles)
Given ∠RBO = 40°
Therefore, ∠NOB = 40°
Then ∠AOB = ∠AON + ∠BON
= 55° + 40° = 95°

3. In the given figure NM ∥ PO ∥ SR and NS ⊥ NM. Also, ∠MNS = 90°. Find the values of ∠a, ∠b and ∠c.

Solution:

b + 45° = 180°
Therefore, ∠b = 180° – 45° (Co-interior angles) = 135°
∠b =∠a (Corresponding angles)
Therefore, ∠a = 135°
Also, 90° + ∠c + 45° = 180°
Therefore, 135° + ∠c = 180°
Therefore, ∠c = 180° – 135° = 45°

The final answer is ∠a = 135°, ∠b = 135°, ∠c = 45°

4. In the given figure, MN ∥ QP, QP ∥ RS, QR ∥ OP. Also, ∠6 = 60°, ∠8 = 55°, then find ∠7, ∠9, ∠10.

Solution:

Since, QR ∥ OP cut by transversal QP
Therefore, ∠8 = ∠10 we know, ∠8 = 55°
Therefore, ∠10 = 55°
Also, QP ∥ XY cut by transversal OP
Therefore, ∠10 = ∠x we know ∠10 = 55°
Therefore,∠x = 55°
Also, ∠x + ∠1 + ∠y = 180°
55° + 60° + ∠y = 180°
115° + ∠y = 180°
∠y = 180° – 115°
Therefore, ∠y = 65°
Now, ∠y + ∠2 = 1800 (Co-interior angles)

65° + ∠7 = 180°
∠7 = 180° – 65°
∠7 = 115°
Since, QP ∥ RS cut by transversal QR
Therefore, ∠8 + ∠9 = 180°
55° + ∠9 = 180°
Therefore, ∠9 = 180° – 55° = 125°

The final answer is ∠7 = 115°, ∠9 = 125°, ∠10 = 55°.

5. In the given figure AB ∥ EF. Also, m : n = 4 : 5 find.

Solution:
Let the common ratio be x
Then m = 4x and n = 5x
Also, ∠n = ∠o (Alternate interior angles)
Since, n = 5x
Therefore, ∠o = 5x [CD ∥ EF cut by transversal t]
Now, ∠o = ∠l (Corresponding angles)
Since, ∠o = 5x
Therefore, ∠l = 5x [AB ∥ CD cut by transversal t]
∠l + ∠m = 180° (Co-interior angles)
5x + 4x = 1800
9x = 180°
x = 180/9
x = 20
Since, m = 4x
Therefore, m = 4 × 20
m = 80°
n = 5x
Therefore, n = 5 × 20
n = 100°
l = 5x
Therefore, l = 5 × 20
l = 100°

Therefore, ∠l = 100°, ∠m = 80°, ∠n = 100°

## Terminating Decimal – Definition, Examples | How to Determine if it is a Terminating Decimal?

Seek help regarding the Terminating Decimals concept from this webpage. Get to know all about Terminating Decimals such as Definitions, How to Determine if it is a Terminating Decimal or not by going through this entire article. We have also listed the steps on how to convert a terminating decimal to a repeating decimal here. Find Solved Examples on Terminating Decimals available here so as to understand the concept clearly.

Also, See:

## Terminating Decimal – Definition

A Terminating Decimal is a decimal that ends. A Decimal Number that contains a finite number of digits next to the decimal point is called a Terminating Decimal. In fact, we can rewrite the terminating decimals as fractions.

Example: 0.52, 0.625, 0.78, etc.

### How to know if it is a Terminating Decimal?

While expressing a fraction in decimal form, when we perform division we observe that the division is complete after few steps i.e. we get a remainder zero. The decimal quotient obtained is known as the terminating decimal. In fact, such decimals will have a finite number of digits after the decimal point.

Terminating Decimals can be expressed as repeating decimals by simply placing zeros that are never-ending. For Example, 12.32 can be written as 12.3200000000…….

### Solved Examples on Terminating Decimals

1. Express 1/4 in decimal form and check if it is terminating decimal?

Solution:

Perform Division Operation and check if the decimal quotient has a finite number of digits or not. If it has a finite number of digits after the decimal point then it is a terminating decimal.

Since 0.25 has a finite number of digits next to the decimal point it is considered a terminating decimal.

2. Express 15/4 in Decimal Form and check if it is terminating decimal?

Solution:

Perform Division Operation and check if the decimal quotient has a finite number of digits or not. If it has a finite number of digits after the decimal point then it is a terminating decimal.

Since 3.75 has a finite number of digits next to the decimal point it is considered as a terminating decimal.

3. Express 23/5 in Decimal Form?

Solution:

Perform Division Operation and check if the decimal quotient has a finite number of digits or not. If it has a finite number of digits after the decimal point then it is a terminating decimal.

Since the decimal quotient, 4.6 has a finite number of decimal places next to the decimal point it is called a terminating decimal.

### FAQs on Terminating Decimal

1. What is a terminating decimal?

Terminating Decimal is a Decimal Number that contains a finite number of digits next to the decimal point.

2. Is 2.3 a terminating decimal?

Yes, 2.3 is a terminating decimal as it has a finite number of digits after the decimal point.

3. How do you find terminating decimals without actual division?

Prime Factorization of the denominator should contain factor 2 or factor 5 or both factors 2 and 5 to tell if it’s a terminating decimal. Any factor other than these gives a non-terminating decimal.

## Engage NY Eureka Math 6th Grade Module 3 Lesson 18 Answer Key

### Eureka Math Grade 6 Module 3 Lesson 18 Example Answer Key

Example 1. The Distance Between Points on an Axis
Consider the points (- 4, 0) and (5, 0).
What do the ordered pairs have in common, and what does that mean about their location in the coordinate plane?
Both of their y-coordinates are zero, so each point lies on the x-axis, the horizontal number line.

How did we find the distance between two numbers on the number line?
We calculated the absolute values of the numbers, which told us how far the numbers were from zero. If the numbers were located on opposite sides of zero, then we added their absolute values together. If the numbers were located on the same side of zero, then we subtracted their absolute values.

Use the same method to find the distance between (- 4, 0) and (5, 0).
|- 4| = 4 and |5| = 5. The numbers are on opposite sides of zero, so the absolute values get combined: 4 + 5 = 9. The distance between (- 4, 0) and (5, 0) is 9 units.

Example 2.
The Length of a Line Segment on an Axis
Consider the line segment with end points (0,- 6) and (0, – 11).
What do the ordered pairs of the end points have in common, and what does that mean about the line segment’s location in the coordinate plane?
The x-coordinates of both end points are zero, so the points lie on the y-axis, the vertical number line. If its end points lie on a vertical number line, then the line segment itself must also lie on the vertical line.

Find the length of the line segment described by finding the distance between its end points (0,- 6) and (0, – 11).
|- 6| = 6 and |- 11| = 11. The numbers are on the same side of zero, which means the longer distance contains the shorter distance, so the absolute values need to be subtracted: 11 – 6 = 5. The distance between (0, – 6) and (0, – 11) is 5 units, so the length of the line segment with end points (0, – 6) and (0, – 11) is 5 units.

Example 3.
Length of a Horizontal or Vertical Line Segment That Does Not Lie on an Axis
Consider the line segment with end points (- 3, 3) and (- 3, – 5).
What do the end points, which are represented by the ordered pairs, have in common? What does that tell us about the location of the line segment on the coordinate plane?
Both end points have x-coordinates of – 3, so the points lie on the vertical line that intersects the x-axis at – 3. This means that the end points of the line segment, and thus the line segment, lie on a vertical line.

Find the length of the line segment by finding the distance between its end points.
The end points are on the same vertical line, so we only need to find the distance between 3 and – 5 on the number line.
|3| = 3 and |- 5| = 5, and the numbers are on opposite sides of zero, so the values must be added: 3 + 5 = 8. So, the distance between (- 3, 3) and (- 3, – 5) is 8 units.

### Eureka Math Grade 6 Module 3 Lesson 18 Exercise Answer Key

Exercise

Find the lengths of the line segments whose end points are given below. Explain how you determined that the line segments are horizontal or vertical.
a. (- 3, 4) and (- 3, 9)
Both end points have x-coordinates of – 3, so the points lie on a vertical line that passes through – 3 on the x-axis. |4| = 4 and |9| = 9, and the numbers are on the same side of zero. By subtraction, 9 – 4 = 5, so the length of the line segment with end points (- 3, 4) and (- 3, 9) is 5 units.

b. (2, – 2) and (- 8, – 2)
Both end points have y-coordinates of – 2, so the points lie on a horizontal line that passes through – 2 on the y-axis. |2| = 2 and |- 8| = 8, and the numbers are on opposite sides of zero, so the absolute values must be added. By addition, 8 + 2 = 10, so the length of the line segment with end points (2, – 2) and (- 8, – 2)is 10 units.

c. (- 6, – 6) and (- 6, 1)
Both end points hove x-coordinates of – 6, so the points lie on a vertical line. |- 6| = 6 and |1| = 1, and the numbers are on opposite sides of zero, so the absolute values must be added. By addition, 6 + 1 = 7, so the length of the line segment with end points (- 6, – 6) and (- 6, 1) is 7 units.

d. (- 9, 4) and (- 4, 4)
Both end points have y-coordinates of 4, so the points lie on a horizontal line. |- 9| = 9 and |- 4| = 4, and the numbers are on the same side of zero. By subtraction, 9 – 4 = 5, so the length of the line segment with end points (- 9, 4) and (- 4, 4) is 5 units.

e. (0, – 11) and (0, 8)
Both end points hove x-coordinates of 0, so the points lie on the y-axis. |- 11| = 11 and |8| = 8, and the numbers are on opposite sides of zero, so their absolute values must be added. By addition, 11 + 8 = 19, so the length of the line segment with end points (0, – 11) and (0, 8) is 19 units.

### Eureka Math Grade 6 Module 3 Lesson 18 Problem Set Answer Key

Question 1.
Find the length of the line segment with end points (7, 2) and (- 4, 2), and explain how you arrived at your solution.
11 units. Both points have the same y-coordinate, so I knew they were on the same horizontal line. I found the
distance between the x-coordinates by counting the number of units on a horizontal number line from – 4 to zero and then from zero to 7, and 4 + 7 = 11.
or
I found the distance between the x-coordinates by finding the absolute value of each coordinate. |7| = 7 and |- 4| = 4. The coordinates lie on opposite sides of zero, so I found the length by adding the absolute values together. Therefore, the length of a line segment with end points (7, 2) and (- 4, 2) is 11 units.

Question 2.
Sarah and Jamal were learning partners in math class and were working independently. They each started at the point (- 2, 5) and moved 3 units vertically in the plane. Each student arrived at a different end point. How is this possible? Explain and list the two different end points.
It is possible because Sarah could have counted up and Jamal could have counted down or vice versa. Moving 3 units in either direction vertically would generate the following possible end points: (- 2, 8) or (- 2, 2).

Question 3.
The length of a line segment is 13 units. One end point of the line segment is (- 3,7). Find four points that could be the other end points of the line segment.
(- 3, 20), (- 3, – 6), (- 16, 7) or (10, 7)

### Eureka Math Grade 6 Module 3 Lesson 18 Exit Ticket Answer Key

Question 1.
Determine whether each given pair of end points lies on the same horizontal or vertical line. If so, find the length of the line segment that joins the pair of points. If not, explain how you know the points are not on the same horizontal or vertical line.

a. (0, – 2) and (0, 9)
The end points both have x-coordinates of 0, so they both lie on the y-axis, which is a vertical line. They lie on opposite sides of zero, so their absolute values have to be combined to get the total distance. |- 2| = 2 and |9| = 9, so by addition, 2 + 9 = 11. The length of the line segment with end points (0, – 2) and (0, 9) is 11 units.

b. (11, 4) and (2, 11)
The points do not lie on the same horizontal or vertical line because they do not share a common x- or y-coordinate.

c. (3, – 8) and (3, – 1)
The end points both have x-coordinates of 3, so the points lie on a vertical line that passes through 3 on the x-axis. The y-coordinates lie on the some side of zero. The distance between the points is determined by subtracting their absolute values, |- 8| = 8 and |- 1| = 1. So, by subtraction, 8 – 1 = 7. The length of the line segment with end points (3, – 8) and (3, – 1) is 7 units.

d. (- 4, – 4) and (5, – 4)
The end points have the same y-coordinate of – 4, so they lie on a horizontal line that passes through – 4 on the y-axis. The numbers lie on opposite sides of zero on the number line, so their absolute values must be added to obtain the total distance, |- 4| = 4 and |5| = 5. So, by addition, 4 + 5 = 9. The length of the line segment with endpoints (- 4, – 4) and (5, – 4) is 9 units.

### Eureka Math Grade 6 Module 3 Lesson 18 Opening Exercise Answer Key

Four friends are touring on motorcycles. They come to an intersection of two roads; the road they are on continues straight, and the other is perpendicular to it. The sign at the intersection shows the distances to several towns. Draw a map/diagram of the roads, and use it and the information on the sign to answer the following questions:

What is the distance between Albertsville and Dewey Falls?
Students draw and use their maps to answer. Albertsville is 8 miles to the left, and Dewey Falls is 6 miles to the right. Since the towns are in opposite directions from the intersection, their distances must be combined by addition, 8 + 6 = 14, so the distance between Albertsville and Dewey Falls is 14 miles.

What is the distance between Blossville and Cheyenne?
Blossville and Cheyenne are both straight ahead from the intersection in the direction that they are going. Since they are on the same side of the intersection, Blossville is on the way to Cheyenne, so the distance to Cheyenne includes the 3 miles to Blossville. To find the distance from Blossville to Cheyenne, I have to subtract; 12 – 3 = 9. So, the distance from Blossville to Cheyenne is 9 miles.

On the coordinate plane, what represents the intersection of the two roads?
The intersection is represented by the origin.

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

### Eureka Math Grade 8 Module 2 Lesson 8 Exercise Answer Key

Exercise 1.
The Federal Reserve states that the average household in January of 2013 had $7,122 in credit card debt. About how many times greater is the U.S. national debt, which is$16,755,133,009,522? Rewrite each number to the nearest power of 10 that exceeds it, and then compare.
Household debt=7122<9999<10000=104.
U.S.debt=16 755 133 009 522<99 999 999 999 999<100 000 000 000 000=1014.
$$\frac{10^{14}}{10^{4}}$$ =1014-4=1010. The U.S. national debt is 1010 times greater than the average household’s credit card debt.

Exercise 2.
There are about 3,000,000 students attending school, kindergarten through Grade 12, in New York. Express the number of students as a single-digit integer times a power of 10.
3 000 000=3×106

The average number of students attending a middle school in New York is 8×102. How many times greater is the overall number of K–12 students compared to the average number of middle school students?
$$\frac{3 \times 10^{6}}{8 \times 10^{2}}$$=$$\frac{3}{8}$$×$$\frac{10^{6}}{10^{2}}$$
= $$\frac{3}{8}$$×104
= 0.375×104
There are about 3,750 times more students in K–12 compared to the number of students in middle school.

Exercise 3.
A conservative estimate of the number of stars in the universe is 6×1022. The average human can see about 3,000 stars at night with his naked eye. About how many times more stars are there in the universe compared to the stars a human can actually see?
$$\frac{6 \times 10^{22}}{3 \times 10^{3}}$$)=$$\frac{6}{3}$$×$$\frac{10^{22}}{10^{3}}$$ =2×1022-3=2×1019
There are about 2×1019 times more stars in the universe compared to the number we can actually see.

Exercise 4.
The estimated world population in 2011 was 7×109. Of the total population, 682 million of those people were left-handed. Approximately what percentage of the world population is left-handed according to the 2011 estimation?
682 000 000≈700 000 000=7×108
$$\frac{7 \times 10^{8}}{7 \times 10^{9}}$$=$$\frac{7}{7}$$×$$\frac{10^{8}}{10^{9}}$$
=1×$$\frac{1}{10}$$
=$$\frac{1}{10}$$
About one-tenth of the population is left-handed, which is equal to 10%.

Exercise 5.
The average person takes about 30,000 breaths per day. Express this number as a single-digit integer times a power of 10.
30000=3×104

If the average American lives about 80 years (or about 30,000 days), how many total breaths will a person take in her lifetime?
(3×104 )×(3×104 )=9×108

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

Students practice estimating size of quantities and performing operations on numbers written in the form of a single-digit integer times a power of 10.

Question 1.
The Atlantic Ocean region contains approximately 2×1012 gallons of water. Lake Ontario has approximately 8,000,000,000,000 gallons of water. How many Lake Ontarios would it take to fill the Atlantic Ocean region in terms of gallons of water?
8 000 000 000 000=8×1012
$$\frac{2 \times 10^{16}}{8 \times 10^{12}}$$=$$\frac{2}{8}$$×$$\frac{10^{16}}{10^{12}}$$
=$$\frac{1}{4}$$×104
=0.25×104
=2500
2,500 Lake Ontario’s would be needed to fill the Atlantic Ocean region.

Question 2.
U.S. national forests cover approximately 300,000 square miles. Conservationists want the total square footage of forests to be 300,0002 square miles. When Ivanna used her phone to do the calculation, her screen showed the following:

a. What does the answer on her screen mean? Explain how you know.
The answer means 9×1010. This is because:
(300 000)2=(3×105)2
=32×(105)2
=9×1010

b. Given that the U.S. has approximately 4 million square miles of land, is this a reasonable goal for conservationists? Explain.
4 000 000=4×106. It is unreasonable for conservationists to think the current square mileage of forests could increase that much because that number is greater than the number that represents the total number of square miles in the U.S,
9×1010>4×106.

Question 3.
The average American is responsible for about 20,000 kilograms of carbon emission pollution each year. Express this number as a single-digit integer times a power of 10.
20 000=2×104

Question 4.
The United Kingdom is responsible for about 1×104 kilograms of carbon emission pollution each year. Which country is responsible for greater carbon emission pollution each year? By how much?
2×104>1×104
America is responsible for greater carbon emission pollution each year. America produces twice the amount of the U.K. pollution.

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

Most English-speaking countries use the short-scale naming system, in which a trillion is expressed as 1,000,000,000,000. Some other countries use the long-scale naming system, in which a trillion is expressed as 1,000,000,000,000,000,000,000. Express each number as a single-digit integer times a power of ten. How many times greater is the long-scale naming system than the short-scale?
1 000 000 000 000=1012
1 000 000 000 000 000 000 000=1021
$$\frac{10^{21}}{10^{12}}$$. The long-scale is about 109 times greater than the short-scale.

## Engage NY Eureka Math 5th Grade Module 2 Lesson 11 Answer Key

### Eureka Math Grade 5 Module 2 Lesson 11 Sprint Answer Key

A
Multiply Decimals

3 x 3 = 9, 0.3 x 3 = 0.9, 0.03 x 3 = 0.09, 3 x 2 = 6, 0.3 x 2 = 0.6, 0.03 x 2 = 0.06, 2 x 2 = 4, 0.2 x 2 = 0.4, 0.02 x 2 = 0.04, 5 x 3 = 15, 0.05 x 3 = 0.15, 0.04 x 3 = 0.12, 4 x 3 = 12, 5 x 5 = 25, 0.5 x 5 = 2.5, 0.05 x 5 = 0.25, 7 x 4 = 28, 0.7 x 4 = 2.8, 0.07 x 4 = 0.28, 0.9 x 4 = 3.6.

Explanation:
In the above-given question,
given that,
multiply decimals.
3 x 3 = 9, 0.3 x 3 = 0.9, 0.03 x 3 = 0.09, 3 x 2 = 6, 0.3 x 2 = 0.6, 0.03 x 2 = 0.06, 2 x 2 = 4, 0.2 x 2 = 0.4, 0.02 x 2 = 0.04, 5 x 3 = 15, 0.05 x 3 = 0.15, 0.04 x 3 = 0.12, 4 x 3 = 12, 5 x 5 = 25, 0.5 x 5 = 2.5, 0.05 x 5 = 0.25, 7 x 4 = 28, 0.7 x 4 = 2.8, 0.07 x 4 = 0.28, 0.9 x 4 = 3.6.

8 x 5 = 40, 0.8 x 5 = 4.0, 0.08 x 5 = 0.40, 0.06 x 5 = 0.30, 0.06 = 3 = 0.18, 0.6 x 5 = 3.0, 0.06 x 2 = 0.12, 0.06 x 7 = 0.42, 0.9 x 6 = 5.4, 0.06 x 9 = 0.54, 0.09 x 9 = 0.81, 0.8 x 8 = 0.64, 0.07 x 7 = 0.49, 0.6 x 6 = 3.6, 0.05 x 5 = 0.25, 0.6 x 8 = 4.8, 0.07 x 9 = 0.63, 0.8 x 3 = 2.4, 0.09 x 6 = 0.54, 0.5 x 7 = 3.5, 0.12 x 4 = 0.48, 0.12 x 9 = 1.08.

Explanation:
In the above-given question,
given that,
multiply decimals.
8 x 5 = 40, 0.8 x 5 = 4.0, 0.08 x 5 = 0.40, 0.06 x 5 = 0.30, 0.06 = 3 = 0.18, 0.6 x 5 = 3.0, 0.06 x 2 = 0.12, 0.06 x 7 = 0.42, 0.9 x 6 = 5.4, 0.06 x 9 = 0.54, 0.09 x 9 = 0.81, 0.8 x 8 = 0.64, 0.07 x 7 = 0.49, 0.6 x 6 = 3.6, 0.05 x 5 = 0.25, 0.6 x 8 = 4.8, 0.07 x 9 = 0.63, 0.8 x 3 = 2.4, 0.09 x 6 = 0.54, 0.5 x 7 = 3.5, 0.12 x 4 = 0.48, 0.12 x 9 = 1.08.

Question 1.
3 × 3 = 9.

3 x 3 = 9.

Explanation:
In the above-given question,
given that,
multiply.
3 x 3 = 9

Question 2.

0.3 × 3 = 0.9

0.3 x 3 = 0.9.

Explanation:
In the above-given question,
given that,
multiply.
0.3 x 3 = 0.9

Question 3.
0.03 × 3 = 0.09

0.03 x 3 = 0.09.

Explanation:
In the above-given question,
given that,
multiply.
0.03 x 3 = 0.09

Question 4.
3 × 2 = 6

3 x 2 = 6.

Explanation:
In the above-given question,
given that,
multiply.
3 x 2 = 6.

Question 5.
0.3 × 2 = 0.6

0.3 x 2 = 0.6.

Explanation:
In the above-given question,
given that,
multiply.
0.3 x 2 = 0.6

Question 6.
0.03 × 2 = 0.06

0.03 x 2 = 0.06.

Explanation:
In the above-given question,
given that,
multiply.
0.03 x 2 = 0.06

Question 7.
2 × 2 = 4

2 x 2 = 4.

Explanation:
In the above-given question,
given that,
multiply.
2 x 2 = 4.

Question 8.
0.2 × 2 = 0.4

0.2 x 2 = 0.4.

Explanation:
In the above-given question,
given that,
multiply.
0.2 x 2 = 0.4

Question 9.
0.02 × 2 = 0.04

0.02 x 2 = 0.04.

Explanation:
In the above-given question,
given that,
multiply.
0.02 x 2 = 0.04

Question 10.
5 × 3 = 15

5 x 3 = 15.

Explanation:
In the above-given question,
given that,
multiply.
5 x 3 = 15

Question 11.
0.5 × 3 = 1.5

0.5 x 3 = 1.5.

Explanation:
In the above-given question,
given that,
multiply.
0.5 x 3 = 1.5

Question 12.
0.05 × 3 = 0.15

0.05 x 3 = 0.15.

Explanation:
In the above-given question,
given that,
multiply.
0.05 x 3 = 0.15

Question 13.
0.04 × 3 = 0.12

0.04 x 3 = 0.12.

Explanation:
In the above-given question,
given that,
multiply.
0.04 x 3 = 0.12

Question 14.
0.4 × 3 = 1.2

0.4 x 3 = 1.2.

Explanation:
In the above-given question,
given that,
multiply.
0.4 x 3 = 1.2

Question 15.
4 × 3 = 12

4 x 3 = 12.

Explanation:
In the above-given question,
given that,
multiply.
4 x 3 = 12.

Question 16.
5 × 5 = 25

5 x 5 = 25.

Explanation:
In the above-given question,
given that,
multiply.
5 x 5 = 25

Question 17.
0.5 × 5 = 2.5

3 x 3 = 9.

Explanation:
In the above-given question,
given that,
multiply.
3 x 3 = 9

Question 18.
0.05 × 5 = 025

0.05 x 5 = 0.25.

Explanation:
In the above-given question,
given that,
multiply.
0.05 x 5 = 0.25

Question 19.
7 × 4 = 28

7 x 4 = 28.

Explanation:
In the above-given question,
given that,
multiply.
7 x 4 = 28

Question 20.
0.7 × 4 = 2.8

0.7 x 4 = 2.8.

Explanation:
In the above-given question,
given that,
multiply.
0.7 x 4 = 2.8

Question 21.
0.07 × 4 = 0.28

0.07 x 4 = 0.28.

Explanation:
In the above-given question,
given that,
multiply.
0.07 x 4 = 0.28

Question 22.
0.9 × 4 = 3.6

0.9 x 4 = 3.6.

Explanation:
In the above-given question,
given that,
multiply.
0.9 x 4 = 3.6

Question 23.
8 × 5 = 40

8 x 5 = 40.

Explanation:
In the above-given question,
given that,
multiply.
8 x 5 = 40

Question 24.
0.8 × 5 = 0.40

0.8 x 5 = 4.0.

Explanation:
In the above-given question,
given that,
multiply.
0.8 x 5 = 4.0

Question 25.
0.08 × 5 =  0.40

0.08 x 5 = 0.40.

Explanation:
In the above-given question,
given that,
multiply.
0.08 x 5 = 0.40

Question 26.
0.06 × 5 = 0.30

0.06 x 5 = 0.30.

Explanation:
In the above-given question,
given that,
multiply.
0.06 x 5 = 0.30

Question 27.
0.06 × 3 = 0.18

0.06 x 3 = 0.18.

Explanation:
In the above-given question,
given that,
multiply.
0.06 x 3 = 0.18

Question 28.
0.6 × 5 = 3.0

0.6 x 5 = 3.0.

Explanation:
In the above-given question,
given that,
multiply.
0.6 x 5 = 3.0

Question 29.
0.06 × 2 = 0.12

0.06 x 2 = 0.12.

Explanation:
In the above-given question,
given that,
multiply.
0.06 x 2 = 0.12

Question 30.
0.06 × 7 = 0.42

0.06 x 7 = 0.42.

Explanation:
In the above-given question,
given that,
multiply.
0.06 x 7 = 0.42

Question 31.
0.9 × 6 = 0.54

0.9 x 6 = 5.4.

Explanation:
In the above-given question,
given that,
multiply.
0.9 x 6 = 5.4

Question 32.
0.06 × 9 =  0.54

0.06 x 9 = 0.54.

Explanation:
In the above-given question,
given that,
multiply.
0.06 x 9 = 0.54

Question 33.
0.09 × 9 = 0.81

0.09 x 9 = 0.81.

Explanation:
In the above-given question,
given that,
multiply.
0.09 x 9 = 0.81

Question 34.
0.8 × 8 = 6.4

0.8 x 8 = 6.4

Explanation:
In the above-given question,
given that,
multiply.
0.8 x 8 = 6.4

Question 35.
0.07 × 7 = 0.49

0.07 x 7 = 0.49.

Explanation:
In the above-given question,
given that,
multiply.
0.07 x 7 = 0.49

Question 36.
0.6 × 6 = 3.6

0.6 x 6 = 3.6.

Explanation:
In the above-given question,
given that,
multiply.
0.6 x 6 = 3.6

Question 37.
0.05 × 5 = 0.25

0.05 x 5 = 0.25.

Explanation:
In the above-given question,
given that,
multiply.
0.05 x 5 = 0.25

Question 38.
0.6 × 8 = 4.8

0.6 x 8 = 4.8.

Explanation:
In the above-given question,
given that,
multiply.
0.6 x 8 = 4.8

Question 39.
0.07 × 9 = 0.63

0.07 x 9 = 0.63.

Explanation:
In the above-given question,
given that,
multiply.
0.07 x 9 = 0.63

Question 40.
0.8 × 3 = 2.4

0.8 x 3 = 2.4.

Explanation:
In the above-given question,
given that,
multiply.
0.8 x 3 = 2.4

Question 41.
0.09 × 6 = 0.54

0.09 x 6 = 0.54.

Explanation:
In the above-given question,
given that,
multiply.
0.09 x 6 = 0.54

Question 42.
0.5 × 7 = 3.5

0.5 x 7 = 3.5.

Explanation:
In the above-given question,
given that,
multiply.
0.5 x 7 = 3.5

Question 43.
0.12 × 4 = 0.48

0.12 x 4 = 0.48.

Explanation:
In the above-given question,
given that,
multiply.
0.12 x 4 = 0.48

Question 44.
0.12 × 9 = 1.08

0.12 x 9 = 1.08.

Explanation:
In the above-given question,
given that,
multiply.
0.12 x 9 = 1.08

B
Multiply Decimals

2 x 2 = 4, 0.2 x 2 = 0.4, 0.02 x 2 = 0.04, 4 x 2 = 8, 0.4 x 2 = 0.8, 0.04 x 2 = 0.08, 3 x 3 = 9, 0.3 x 3 = 0.9, 0.03 x 3 = 0.09, 4 x 3 = 12, 0.4 x 3 = 1.2, 0.04 x 3 = 0.12, 0.05 x 3 = 0.15, 0.5 x 3 = 1.5, 5 x 3 = 15, 4 x 4 = 16, 0.4 x 4 = 1.6, 0.04 x 4 = 0.16, 8 x 4 = 32, 0.8 x 4 = 3.2, 0.08 x 4 = 0.32, 0.6 x 4 = 2.4.

Explanation:
In the above-given question,
given that,
multiply decimals.
2 x 2 = 4, 0.2 x 2 = 0.4, 0.02 x 2 = 0.04, 4 x 2 = 8, 0.4 x 2 = 0.8, 0.04 x 2 = 0.08, 3 x 3 = 9, 0.3 x 3 = 0.9, 0.03 x 3 = 0.09, 4 x 3 = 12, 0.4 x 3 = 1.2, 0.04 x 3 = 0.12, 0.05 x 3 = 0.15, 0.5 x 3 = 1.5, 5 x 3 = 15, 4 x 4 = 16, 0.4 x 4 = 1.6, 0.04 x 4 = 0.16, 8 x 4 = 32, 0.8 x 4 = 3.2, 0.08 x 4 = 0.32, 0.6 x 4 = 2.4.

6 x 5 = 30, 0.6 x 5 = 3.0, 0.06 x 5 = 0.30, 0.08 x 5 = 0.40, 0.08 x 3 = 0.24, 0.8 x 5 = 4.0, 0.08 x 2 = 0.16, 0.08 x 7 = 0.56, 0.9 x 8 = 7.2, 0.08 x 9 = 0.72, 0.9 x 9 = 8.1, 0.08 x 8 = 0.64, 0.7 x 7 = 4.9, 0.06 x 6 = 0.36, 0.7 x 9 = 6.3, 0.5 x 5 = 2.5, 0.06 x 8 = 0.48, 0.08 x 3 = 0.24, 0.9 x 6 = 5.4, 0.05 x 7 = 0.35, 0.12 x 6 = 0.72, 0.12 x 8 = 0.96.

Explanation:
In the above-given question,
given that,
multiply decimals.
6 x 5 = 30, 0.6 x 5 = 3.0, 0.06 x 5 = 0.30, 0.08 x 5 = 0.40, 0.08 x 3 = 0.24, 0.8 x 5 = 4.0, 0.08 x 2 = 0.16, 0.08 x 7 = 0.56, 0.9 x 8 = 7.2, 0.08 x 9 = 0.72, 0.9 x 9 = 8.1, 0.08 x 8 = 0.64, 0.7 x 7 = 4.9, 0.06 x 6 = 0.36, 0.7 x 9 = 6.3, 0.5 x 5 = 2.5, 0.06 x 8 = 0.48, 0.08 x 3 = 0.24, 0.9 x 6 = 5.4, 0.05 x 7 = 0.35, 0.12 x 6 = 0.72, 0.12 x 8 = 0.96.

Question 1.
2 × 2 = 4

2 x 2 = 4.

Explanation:
In the above-given question,
given that,
multiply.
2 x 2 = 4.

Question 2.
0.2 × 2 = 0.4

0.2 x 2 = 0.4.

Explanation:
In the above-given question,
given that,
multiply.
0.2 x 2 = 0.4.

Question 3.
0.02 × 2 = 0.04

0.02 x 2 = 0.04.

Explanation:
In the above-given question,
given that,
multiply.
0.02 x 2 = 0.04.

Question 4.
4 × 2 = 8

4 x 2 = 8.

Explanation:
In the above-given question,
given that,
multiply.
4 x 2 = 8.

Question 5.
0.4 × 2 = 0.8

0.4 x 2 = 0.8.

Explanation:
In the above-given question,
given that,
multiply.
0.4 x 2 = 0.8.

Question 6.
0.04 × 2 = 0.08.

0.04 x 2 = 0.08.

Explanation:
In the above-given question,
given that,
multiply.
0.04 x 2 = 0.08

Question 7.
3 × 3 = 9

3 x 3 = 9.

Explanation:
In the above-given question,
given that,
multiply.
3 x 3 = 9

Question 8.
0.3 × 3 = 0.9.

0.3 x 3 = 0.9.

Explanation:
In the above-given question,
given that,
multiply.
0.3 x 3 = 0.9

Question 9.
0.03 × 3 = 0.09

0.03 x 3 = 0.09.

Explanation:
In the above-given question,
given that,
multiply.
0.03 x 3 = 0.09.

Question 10.
4 × 3 = 12

4 x 3 = 12.

Explanation:
In the above-given question,
given that,
multiply.
4 x 3 = 12

Question 11.
0.4 × 3 = 1.2

0.4 x 3 = 1.2.

Explanation:
In the above-given question,
given that,
multiply.
0.4 x 3 = 1.2

Question 12.
0.04 × 3 = 0.12

0.04 x 3 = 0.12.

Explanation:
In the above-given question,
given that,
multiply.
0.04 x 3 = 0.12

Question 13.
0.05 × 3 = 0.15

0.05 x 3 = 0.15.

Explanation:
In the above-given question,
given that,
multiply.
0.05 x 3 = 0.15

Question 14.
0.5 × 3 = 1.5

0.5 x 3 = 1.5.

Explanation:
In the above-given question,
given that,
multiply.
0.5 x 3 = 1.5

Question 15.
5 × 3 = 15

5 x 3 = 15.

Explanation:
In the above-given question,
given that,
multiply.
5 x 3 = 15

Question 16.
4 × 4 = 16

4 x 4 = 16.

Explanation:
In the above-given question,
given that,
multiply.
4 x 4 = 16.

Question 17.
0.4 × 4 = 1.6

0.4 x 4 = 1.6.

Explanation:
In the above-given question,
given that,
multiply.
0.4 x 4 = 1.6

Question 18.
0.04 × 4 = 0.16

0.04 x 4 = 0.16.

Explanation:
In the above-given question,
given that,
multiply.
0.04 x 4 = 0.16

Question 19.
8 × 4 = 32

8 x 4 = 32.

Explanation:
In the above-given question,
given that,
multiply.
8 x 4 = 32

Question 20.
0.8 × 4 = 3.2

0.8 x 4 = 3.2.

Explanation:
In the above-given question,
given that,
multiply.
0.8 x 4 = 3.2

Question 21.
0.08 × 4 = 0.32

0.08 x 4 = 0.32.

Explanation:
In the above-given question,
given that,
multiply.
0.08 x 4 = 0.32

Question 22.
0.6 × 4 = 2.4

0.6 x 4 = 2.4.

Explanation:
In the above-given question,
given that,
multiply.
0.6 x 4 = 2.4

Question 23.
6 × 5 = 30

6 x 5 = 30.

Explanation:
In the above-given question,
given that,
multiply.
6 x 5 = 30

Question 24.
0.6 × 5 = 3.0

0.6 x 5 = 3.0.

Explanation:
In the above-given question,
given that,
multiply.
0.6 x 5 = 3.0

Question 25.
0.06 × 5 = 0.30

0.06 x 5 = 0.30.

Explanation:
In the above-given question,
given that,
multiply.
0.06 x 5 = 0.30

Question 26.
0.08 × 5 = 0.40

0.08 x 5 = 0.40.

Explanation:
In the above-given question,
given that,
multiply.
0.08 x 5 = 0.40

Question 27.
0.08 × 3 = 0.24

0.08 x 3 = 0.24.

Explanation:
In the above-given question,
given that,
multiply.
0.08 x 3 = 0.24

Question 28.
0.8 × 5 = 4.0

0.8 x 5 = 4.0

Explanation:
In the above-given question,
given that,
multiply.
0.8 x 5 = 4.0

Question 29.
0.08 × 2 = 0.16

0.08 x 2 = 0.16.

Explanation:
In the above-given question,
given that,
multiply.
0.08 x 2 = 0.16

Question 30.
0.08 × 7 = 0.56

0.08 x 7 = 0.56.

Explanation:
In the above-given question,
given that,
multiply.
0.08 x 7 = 0.56

Question 31.
0.9 × 8 = 7.2

0.08 x 5 = 0.40.

Explanation:
In the above-given question,
given that,
multiply.
0.08 x 5 = 0.40

Question 32.
0.08 × 9 = 0.72

0.08 x 9 = 0.72.

Explanation:
In the above-given question,
given that,
multiply.
0.08 x 9 = 0.72

Question 33.
0.9 × 9 = 8.1

0.9 x 9 = 8.1.

Explanation:
In the above-given question,
given that,
multiply.
0.9 x 9 = 8.1

Question 34.
0.08 × 8 = 0.64

0.08 x 8 = 0.64.

Explanation:
In the above-given question,
given that,
multiply.
0.08 x 8 = 0.64

Question 35.
0.7 × 7 = 4.9

0.7 x 7 = 4.9.

Explanation:
In the above-given question,
given that,
multiply.
0.7 x 7 = 4.9

Question 36.
0.06 × 6 = 0.36

0.06 x 6 = 0.36.

Explanation:
In the above-given question,
given that,
multiply.
0.06 x 6 = 0.36

Question 37.
0.5 × 5 = 2.5

0.5 x 5 = 2.5.

Explanation:
In the above-given question,
given that,
multiply.
0.5 x 5 = 2.5

Question 38.
0.06 × 8 = 0.48

0.06 x 8 = 0.48.

Explanation:
In the above-given question,
given that,
multiply.
0.06 x 8 = 0.48

Question 39.
0.7 × 9 = 6.3

0.7 x 9 = 6.3.

Explanation:
In the above-given question,
given that,
multiply.
0.7 x 9 = 6.3

Question 40.
0.08 × 3 = 0.24

0.08 x 3 = 0.24.

Explanation:
In the above-given question,
given that,
multiply.
0.08 x 3 = 0.24

Question 41.
0.9 × 6 = 5.4

0.9 x 6 = 5.4.

Explanation:
In the above-given question,
given that,
multiply.
0.9 x 6 = 5.4

Question 42.
0.05 × 7 = 0.35

0.05 x 7 = 0.35.

Explanation:
In the above-given question,
given that,
multiply.
0.05 x 7 = 0.35

Question 43.
0.12 × 6 = 0.72

0.12 x 6 = 0.72.

Explanation:
In the above-given question,
given that,
multiply.
0.12 x 6 = 0.72

Question 44.
0.12 × 8 = 0.96

0.12 x 8 = 0.96.

Explanation:
In the above-given question,
given that,
multiply.
0.12 x 8 = 0.96.

### Eureka Math Grade 5 Module 2 Lesson 11 Problem Set Answer Key

Question 1.
Estimate the product. Solve using the standard algorithm. Use the thought bubbles to show your thinking. (Draw an area model on a separate sheet if it helps you.)

a. 1.38 × 32 ≈ ___1.38____ × ___100____ = ___138____ 1.38 × 32 = ___44.16___________

1.38 x 32 = 44.16.

Explanation:
In the above-given question,
given that,
multiply.
1.38 x 32 = 44.16

b. 3.55 × 89 ≈ __3.55_____ × __100_____ = __355_____ 3.55 × 89 = ___315.95___________

3.55 x 89 = 315.95.

Explanation:
In the above-given question,
given that,
multiply.
3.55 x 89 = 315.95.

Question 2.
Solve using the standard algorithm.

a. 5.04 × 8

5.04 x 8 = 40.32.

Explanation:
In the above-given question,
given that,
multiply.
5.04 x 8 = 40.32

b. 147.83 × 67

147.83 x 67 = 40.32.

Explanation:
In the above-given question,
given that,
multiply.
147.83 x 67 = 40.32

c. 83.41 × 504

83.41 x 504 = 42,038

Explanation:
In the above-given question,
given that,
multiply.
83.41 x 504 = 42,038

d. 0.56 × 432

0.56 x 432 = 241.92.

Explanation:
In the above-given question,
given that,
multiply.
0.56 x 432 = 241.92.

Question 3.
Use the whole number product and place value reasoning to place the decimal point in the second product. Explain how you know.
a. If 98 × 768 = 75,264 then 98 × 7.68 = ____752.64__________

98 x 7.68 = 752.64.

Explanation:
In the above-given question,
given that,
multiply.
98 x 7.68 = 752.64

b. If 73 × 1,563 = 114,099 then 73 × 15.63 = _____1140.99_________

15.63 x 73 = 1140.99.

Explanation:
In the above-given question,
given that,
multiply.
15.63 x 73 = 1140.99

c. If 46 × 1,239 = 56,994 then 46 × 123.9 = ___5699.4___________

15.63 x 73 = 5699.4.

Explanation:
In the above-given question,
given that,
multiply.
15.63 x 73 = 5699.4

Question 4.
Jenny buys 22 pens that cost $1.15 each and 15 markers that cost$2.05 each. How much did Jenny spend?

The Jenny spend = $3.2 Explanation: In the above-given question, given that, Jenny buys 22 pens that cost$1.15 each and 15 markers that cost $2.05 each. 1.15 + 2.05 = 3.2 Question 5. A living room measures 24 feet by 15 feet. An adjacent square dining room measures 13 feet on each side. If carpet costs$6.98 per square foot, what is the total cost of putting carpet in both rooms?

The total cost of putting carpet in both rooms = $2603.54 Explanation: In the above-given question, given that, A living room measures 24 feet by 15 feet. An adjacent square dining room measures 13 feet on each side. if carpet costs$6.98 per square foot.
24 x 15 = 360.
13 x 6.98 = 90.74.
360 x 6.98 = 2,512.8.
90.74 + 2512.8 = 2603.54.

### Eureka Math Grade 5 Module 2 Lesson 11 Exit Ticket Answer Key

Use estimation and place value reasoning to find the unknown product. Explain how you know.

Question 1.
If 647 × 63 = 40,761 then 6.47 × 63 = ______405.09________

6.47 x 63 = 405.09.

Explanation:
In the above-given question,
given that,
multiply.
6.47 x 63 = 405.09

Question 2.
Solve using the standard algorithm.
a. 6.13 × 14

6.13 x 14 = 85.82.

Explanation:
In the above-given question,
given that,
multiply.
6.13 x 14 = 85.82

b. 104.35 × 34

104.35 x 34 = 1460.9.

Explanation:
In the above-given question,
given that,
multiply.
104.35 x 34 = 1460.9

### Eureka Math Grade 5 Module 2 Lesson 11 Homework Answer Key

Question 1.
Estimate the product. Solve using the standard algorithm. Use the thought bubbles to show your thinking. (Draw an area model on a separate sheet if it helps you.)
a. 2.42 × 12 ≈ ___2.42____ × __100_____ = ___242____ 2.42 × 12 = ____29.04__________

2.42 x 12 = 29.04.

Explanation:
In the above-given question,
given that,
multiply.
2.42 x 12 = 29.04

b. 4.13 × 37 ≈ __4.13_____ × ___100____ = __413_____ 4.13 × 37 = __152.81____________

4.13 x 37 = 152.81.

Explanation:
In the above-given question,
given that,
multiply.
4.13 x 37 = 152.81

Question 2.
Solve using the standard algorithm.

a. 2.03 × 13

2.03 x 13 = 26.39.

Explanation:
In the above-given question,
given that,
multiply.
2.03 x 13 = 26.39

b. 53.16 × 34

53.16 x 34 = 1807.44.

Explanation:
In the above-given question,
given that,
multiply.
53.16 x 34 = 1807.44

c. 371.23 × 53

371.23 x 53 = 19675.19.

Explanation:
In the above-given question,
given that,
multiply.
371.23 x 53 = 19675.19

d. 1.57 × 432

1.57 x 432 = 678.24

Explanation:
In the above-given question,
given that,
multiply.
1.57 x 432 = 678.24

Question 3.
Use the whole number product and place value reasoning to place the decimal point in the second product. Explain how you know.
a. If 36 × 134 = 4,824 then 36 × 1.34 = ______48.24________

1.34 x 36 = 48.24.

Explanation:
In the above-given question,
given that,
multiply.
1.34 x 36 = 48.24

b. If 84 × 2,674 = 224,616 then 84 × 26.74 = ____2,246.16__________

84 x 26.74 = 2,246.16.

Explanation:
In the above-given question,
given that,
multiply.
84 x 26.74 = 2,246.16.

c. 19 × 3,211 = 61,009 then 321.1 × 19 = ___6100.9___________

321.1 x 19 = 6100.9.

Explanation:
In the above-given question,
given that,
multiply.
321.1 x 19 = 6100.9

Question 4.
A slice of pizza costs $1.57. How much will 27 slices cost? Answer: The 27 slices cost =$42.39.

Explanation:
In the above-given question,
given that,
A slice of pizza costs $1.57. 27 x 1.57 = 42.39$.

Question 5.
A spool of ribbon holds 6.75 meters. A craft club buys 21 spools.
a. What is the total cost if the ribbon sells for $2 per meter? Answer: The total cost if the ribbon sells for$2 per meter = $13.5. Explanation: In the above-given question, given that, A spoon of ribbon holds 6.75 meters. A craft club buys 21 spools. 6.75 x$2 = $13.5. b. If the club uses 76.54 meters to complete a project, how much ribbon will be left? Answer: The ribbon will be left = 65.21 meters. Explanation: In the above-given question, given that, If the club uses 76.54 meters to complete a project. 6.75 x 21 = 141.75. 141.75 – 76.54 = 65.21. ## Eureka Math Grade 1 Module 1 Lesson 11 Answer Key ## Engage NY Eureka Math 1st Grade Module 1 Lesson 11 Answer Key ### Eureka Math Grade 1 Module 1 Lesson 11 Problem Set Answer Key Question 1. Jill was given a total of 5 flowers for her birthday. Draw more flowers in the vase to show Jill’s birthday flowers. How many flowers did you have to draw? ___ flowers Write a number sentence and a number bond to match the story. Answer: 5 flowers need to draw Question 2. Kate and Nana were baking cookies. They made 2 heart cookies and then made some square cookies. They made 8 cookies altogether. How many square cookies did they make? Draw and count on to show the story. Answer: Write a number sentence and a number bond to match the story. Answer: Show the parts. Write a number bond to match the story. Question 3. Bill has 2 trucks. His friend, James, came over with some more. Together, they had 5 trucks. How many trucks did James bring over? James brought over ______ trucks. Answer: James brought over 3 trucks. Write a number sentence to explain the story. Answer: Question 4. Jane caught 7 fish before she stopped to eat lunch. After lunch, she caught some more. At the end of the day, she had 9 fish. How many fish did she catch after lunch? Jane caught _______ fish after lunch. Answer: Jane caught 2 fish after lunch. Write a number sentence to explain the story. Answer: ### Eureka Math Grade 1 Module 1 Lesson 11 Exit Ticket Answer Key Draw more bears to show that Jen has 8 bears total. I added _____ more bears. Answer: I added 3 more bears. Write a number sentence to show how many bears you drew. Answer: ### Eureka Math Grade 1 Module 1 Lesson 11 Homework Answer Key Question 1. Use the 5-group cards to count on to find the missing number in the number sentences. Answer: Question 2. Match the number sentence to the math story. Draw a picture or use your 5-group cards to solve. a. Scott has 3 cookies. His mom gives him some more. Now, he has 8 cookies. How many cookies did his mom give him? Scott’s mom gave him ________ cookies. b. Kim sees 6 birds in the tree. Some more birds fly in. Kim sees 9 birds in the tree. How many birds flew to the tree? ______ birds flew to the tree. Answer: a. Scott’s mom gave him 5 cookies. b. 3 birds flew to the tree. ## Eureka Math Grade 8 Module 2 Lesson 4 Answer Key ## Engage NY Eureka Math 8th Grade Module 2 Lesson 4 Answer Key ### Eureka Math Grade 8 Module 2 Lesson 4 Exercise Answer Key Exercise 1. Reflect △ABC and Figure D across line L. Label the reflected images. Answer: Exercise 2. Which figure(s) were not moved to a new location on the plane under this transformation? Answer: Point B and line L were not moved to a new location on the plane under this reflection. Exercises 3–5 Exercise 3. Reflect the images across line L. Label the reflected images. Answer: Exercise 4. Answer the questions about the image above. a. Use a protractor to measure the reflected ∠ABC. What do you notice? Answer: The measure of the reflected image of ∠ABC is 66°. b. Use a ruler to measure the length of IJ and the length of the image of IJ after the reflection. What do you notice? Answer: The length of the reflected segment is the same as the original segment, 5 units. Note: This is not something students are expected to know, but it is a preview for what is to come later in this lesson. Exercise 5. Reflect Figure R and △EFG across line L. Label the reflected images. Answer: Exercises 6–9 Use the picture below for Exercises 6–9. Exercise 6. Use the picture to label the unnamed points. Answer: Points are labeled in red above. Exercise 7. What is the measure of ∠JKI? ∠KIJ? ∠ABC? How do you know? Answer: m∠JKI=31°, m∠KIJ=28°, and m∠ABC=150°. Reflections preserve angle measures. Exercise 8. What is the length of segment Reflection(FH)? IJ? How do you know? Answer: |Reflection(FH)|=4 units, and IJ=7 units. Reflections preserve lengths of segments. Exercise 9. What is the location of Reflection(D)? Explain. Answer: Point D and its image are in the same location on the plane. Point D was not moved to another part of the plane because it is on the line of reflection. The image of any point on the line of reflection will remain in the same location as the original point. ### Eureka Math Grade 8 Module 2 Lesson 4 Exit Ticket Answer Key Question 1. Let there be a reflection across line LAB. Reflect △CDE across line LAB. Label the reflected image. Answer: Question 2. Use the diagram above to state the measure of Reflection(∠CDE). Explain. Answer: The measure of Reflection(∠CDE) is 90° because reflections preserve degrees of measures of angles. Question 3. Use the diagram above to state the length of segment Reflection(CE). Explain. Answer: The length of Reflection(CE) is 10 cm because reflections preserve segment lengths. Question 4. Connect point C to its image in the diagram above. What is the relationship between line LAB and the segment that connects point C to its image? Answer: The line of reflection bisects the segment that connects C to its image. ### Eureka Math Grade 8 Module 2 Lesson 4 Problem Set Answer Key Question 1. In the picture below, ∠DEF=56°, ∠ACB=114°, AB=12.6 units, JK=5.32 units, point E is on line L, and point I is off of line L. Let there be a reflection across line L. Reflect and label each of the figures, and answer the questions that follow. Answer: Question 2. What is the measure of Reflection(∠DEF)? Explain. Answer: The measure of Reflection(∠DEF) is 56°. Reflections preserve degrees of angles. Question 3. What is the length of Reflection(JK)? Explain. Answer: The length of Reflection(JK) is 5.32 units. Reflections preserve lengths of segments. Question 4. What is the measure of Reflection(∠ACB)? Answer: The measure of Reflection(∠ACB) is 114°. Question 5. What is the length of Reflection(AB)? Answer: The length of Reflection(AB) is 12.6 units. Question 6. Two figures in the picture were not moved under the reflection. Name the two figures, and explain why they were not moved. Answer: Point E and line L were not moved. All of the points that make up the line of reflection remain in the same location when reflected. Since point E is on the line of reflection, it is not moved. Question 7. Connect points I and I’. Name the point of intersection of the segment with the line of reflection point Q. What do you know about the lengths of segments IQ and QI’? Answer: Segments IQ and QI’ are equal in length. The segment II’ connects point I to its image, I’. The line of reflection will go through the midpoint, or bisect, the segment created when you connect a point to its image. ## Eureka Math Grade 7 Module 1 Lesson 15 Answer Key ## Engage NY Eureka Math 7th Grade Module 1 Lesson 15 Answer Key ### Eureka Math Grade 7 Module 1 Lesson 15 Example Answer Key Example 1. Mother’s 10K Race Sam’s mother has entered a 10K race. Sam and his family want to show their support for their mother, but they need to figure out where they should go along the race course. They also need to determine how long it will take her to run the race so that they will know when to meet her at the finish line. Previously, his mother ran a 5K race with a time of 1$$\frac{1}{2}$$ hours. Assume Sam’s mother will run the same rate as the previous race in order to complete the chart. → Discuss with your partner: Can you find Sam’s mother’s average rate for the entire race based on her previous race time? → 3$$\frac{1}{3}$$ km/h, or $$\frac{10}{3}$$ km/h Create a table that shows how far Sam’s mother has run after each half hour from the start of the race, and graph it on the coordinate plane to the right. Answer: a. What are some specific things you notice about this graph? Answer: It forms a line through the origin; it relates time (in hours) to the distance run (in kilometers); the line through the origin means that the values are proportional. b. What is the connection between the table and the graph? Answer: The time (in hours) is on the horizontal axis, and the distance run (in kilometers) is on the vertical axis; the coordinates of the points on the line are the same as the pairs of numbers in the table. c. What does the point (2, 6$$\frac{2}{3}$$ ) represent in the context of this problem? Answer: After 2 hours, she has run 6 $$\frac{2}{3}$$ km. Discuss the responses with the class and draw a conclusion. → Write an equation that models the data in the chart. Record the student responses so that they can see all of the responses. → D = 3$$\frac{1}{3}$$ H, where D represents the distance, and H represents the hours (or D=$$\frac{10}{3}$$ H). Example 2. Gourmet Cooking After taking a cooking class, you decide to try out your new cooking skills by preparing a meal for your family. You have chosen a recipe that uses gourmet mushrooms as the main ingredient. Using the graph below, complete the table of values and answer the following questions. Answer: a. Is this relationship proportional? How do you know from examining the graph? Answer: Yes, the relationship is proportional because the graph is a line that passes through the origin. b. What is the unit rate for cost per pound? Answer: k = $$\frac{16}{2}$$ = 8. The unit rate is 8. c. Write an equation to model this data. Answer: C = 8w d. What ordered pair represents the unit rate, and what does it mean? Answer: (1, 8) The unit rate is 8, which means that one pound of mushrooms costs$8.00.

e. What does the ordered pair (2, 16) mean in the context of this problem?
(2, 16) This means 2 pounds of mushrooms cost $16.00. f. If you could spend$10.00 on mushrooms, how many pounds could you buy?
C = 8w; C = 10;($$\frac{1}{8}$$ )10 = ($$\frac{1}{8}$$ )8w; 1 $$\frac{1}{4}$$ = w. You can buy 1.25 pounds of mushrooms with $10.00. g. What would be the cost of 30 pounds of mushrooms? Answer: C = 8w; w = 30; C = 8(30); C =$240

### Eureka Math Grade 7 Module 1 Lesson 15 Problem Set Answer Key

Question 1.
Students are responsible for providing snacks and drinks for the Junior Beta Club Induction Reception. Susan and Myra were asked to provide the punch for the 100 students and family members who will attend the event. The chart below will help Susan and Myra determine the proportion of cranberry juice to sparkling water needed to make the punch. Complete the chart, graph the data, and write the equation that models this proportional relationship.

C = $$\frac{4}{5}$$ S, where C represents the number of cups of cranberry juice, and S represents the number of cups of sparkling water.

Question 2.
Jenny is a member of a summer swim team.
a. Using the graph, determine how many calories she burns in one minute.
Jenny burns 100 calories every 15 minutes, so she burns 6 $$\frac{2}{3}$$ calories each minute.

b. Use the graph to determine the equation that models the number of calories Jenny burns within a certain number of minutes.
C = 6 $$\frac{2}{3}$$ t, where C represents the number of calories burned, and t represents the time she swims in minutes.

c. How long will it take her to burn off a 480-calorie smoothie that she had for breakfast?
It will take Jenny 72 minutes of swimming to burn off the smoothie she had for breakfast.

Question 3.
Students in a world geography class want to determine the distances between cities in Europe. The map gives all distances in kilometers. The students want to determine the number of miles between towns so they can compare distances with a unit of measure with which they are already familiar. The graph below shows the relationship between a given number of kilometers and the corresponding number of miles.

a. Find the constant of proportionality, or the rate of miles per kilometer, for this problem, and write the equation that models this relationship.
The constant of proportionality is $$\frac{5}{8}$$ .
The equation that models this situation is M=$$\frac{5}{8}$$ K, where M represents the number of miles, and K represents the number of kilometers.

b. What is the distance in kilometers between towns that are 5 miles apart?
The distance between towns that are 5 miles apart is 8 km.

c. Describe the steps you would take to determine the distance in miles between two towns that are 200 kilometers apart?
Solve the equation M = $$\frac{5}{8}$$ (200). To find the number of miles for 200 km, multiply 200 by $$\frac{5}{8}$$.
200($$\frac{5}{8}$$) = 125. The two towns are 125 miles apart.

Question 4.
During summer vacation, Lydie spent time with her grandmother picking blackberries. They decided to make blackberry jam for their family. Her grandmother said that you must cook the berries until they become juice and then combine the juice with the other ingredients to make the jam.
a. Use the table below to determine the constant of proportionality of cups of juice to cups of blackberries.

k = $$\frac{1}{3}$$ one cup of juice is produced when 3 cups of blackberries are cooked.

b. Write an equation that models the relationship between the number of cups of blackberries and the number of cups of juice.
j = $$\frac{1}{3}$$ b, where j represents the number of cups of juice, and b represents the number of cups of blackberries.

c. How many cups of juice were made from 12 cups of berries? How many cups of berries are needed to make 8 cups of juice?
4 cups of juice are made from 12 cups of berries.
24 cups of berries are needed to make 8 cups of juice.

### Eureka Math Grade 7 Module 1 Lesson 15 Exit Ticket Answer Key

Using the graph and its title:

Question 1.
Describe the relationship that the graph depicts.

The graph shows that in 3 days the water rose to 4 inches. The water has risen at a constant rate. Therefore, the water has risen 1 $$\frac{1}{3}$$ inches per day.

Question 2.
Identify two points on the line, and explain what they mean in the context of the problem.
(6, 8) means that by the 6th day, the water rose 8 inches; (9, 12) means that by the 9th day, the water rose 12 inches.

Question 3.
What is the unit rate?
The unit rate in inches per day is $$\frac{4}{3}$$ .

Question 4.
What point represents the unit rate?
The point that shows the unit rate is (1, 1 $$\frac{1}{3}$$).

Every concept seen in Eureka Math Grade 8 Solution Key is provided with a step-by-step solution in order to make your preparation effective. You can solve the number of questions with the help of Eureka Engage NY Math Book Solutions for Grade 8. Tap the links given below to Download Engage NY Math Grade 8 Answer Key Pdf and start learning the concepts.

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## 7Engage NY Eureka Math 4th Grade Module 2 Lesson 5 Answer Key

### Eureka Math Grade 4 Module 2 Lesson 5 Sprint Answer Key

A
Convert to Kilograms and Grams

Question 1.
2,000 g = 2 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  2,000 g = 2 X 1000 g ÷ 1000 = 2 kg 0 g.

Question 2.
3,000 g = 3 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  3,000 g = 3 X 1000 g ÷ 1000 = 3 kg 0 g.

Question 3.
4,000 g = 4 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  4,000 g = 4 X 1000 g ÷ 1000 = 4 kg 0 g.

Question 4.
9,000 g = 9 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  9,000 g = 9 X 1000 g ÷ 1000 = 9 kg 0 g.

Question 5.
6,000 g = 6 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  6,000 g = 6 X 1000 g ÷ 1000 = 6 kg 0 g.

Question 6.
1,000 g = 1 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  1,000 g = 1 X 1000 g ÷ 1000 = 1 kg 0 g.

Question 7.
8,000 g = 8 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  8,000 g = 8 X 1000 g ÷ 1000 = 8 kg 0 g.

Question 8.
5,000 g = 5 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,000 g = 5 X 1000 g ÷ 1000 = 5 kg 0 g.

Question 9.
7,000 g = 7 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  7,000 g = 7 X 1000 g ÷ 1000 = 7 kg 0 g.

Question 10.
6,100 g = 6 kg 100 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  6,100 g = 6,100 g ÷ 1000 = 6 kg 100 g.

Question 11.
6,110 g = 6 kg 110 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  6,110 g = 6 ,110 g ÷ 1000 = 6 kg 110 g.

Question 12.
6,101 g = 6 kg 101 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  6,101 g = 6,101 g ÷ 1000 = 6 kg 101 g.

Question 13.
6,010 g = 6 kg 10 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  6,010 g = 6,010 g ÷ 1000 = 6 kg 10 g.

Question 14.
6,011 g = 6 kg 11 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  6,011 g = 6,011 g ÷ 1000 = 6 kg 11 g.

Question 15.
6,001 g = 6 kg 1 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  6,001 g = 6,001 g ÷ 1000 = 6 kg 1 g.

Question 16.
8,002 g = 8 kg 2 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  8,002 g = 8,002 g ÷ 1000 = 8 kg 2 g.

Question 17.
8,020 g = 8 kg 20 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  8,002 g = 8,002 g ÷ 1000 = 8 kg 2 g.

Question 18.
8,200 g = 8 kg 200 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  8,200 g = 8,200 g ÷ 1000 = 8 kg 200 g.

Question 19.
8,022 g = 8 kg 22 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  8,022 g = 8,022 g ÷ 1000 = 8 kg 22 g.

Question 20.
8,220 g = 8 kg  220 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  8,220 g = 8,220 g ÷ 1000 = 8 kg 220 g.

Question 21.
8,222 g = 8 kg 222 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  8,222 g = 8,222 g ÷ 1000 = 8 kg 222 g.

Question 22.
7,256 g = 7 kg 256 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  7,256 g = 7,256 g ÷ 1000 = 7 kg 256 g.

Question 23.
3,800 g = 3 kg 800 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  3,800 g = 3,800 g ÷ 1000 = 3 kg 800 g.

Question 24.
4,770 g = 4 kg 770 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  4,770 g = 4,770 g ÷ 1000 = 4 kg 770 g.

Question 25.
4,807 g = 4 kg 807 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  4,807 g = 4,807 g ÷ 1000 = 4 kg 807 g.

Question 26.
5,065 g = 5 kg 065 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,065 g = 5,065 g ÷ 1000 = 5 kg 065 g.

Question 27.
5,040 g = 5 kg 40 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,040 g = 5,040 g ÷ 1000 = 5 kg 40 g.

Question 28.
6,007 g = 6 kg 7 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  6,007 g = 6,007 g ÷ 1000 = 6 kg 7 g.

Question 29.
2,003 g = 2 kg 3 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  2,003 g = 2,003 g ÷ 1000 = 2 kg 3 g.

Question 30.
1,090 g = 1 kg 90 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  1,090 g = 1,090 g ÷ 1000 = 1 kg 90 g.

Question 31.
1,055 g = 1 kg 55 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  1,055 g = 1,055 g ÷ 1000 = 1 kg 55 g.

Question 32.
9,404 g = 9 kg 404 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  9,404 g = 9,404 g ÷ 1000 = 9 kg 404 g.

Question 33.
9,330 g = 9 kg 330 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 9,330 g = 9,330 g ÷ 1000 = 9 kg 330 g.

Question 34.
3,400 g = 3 kg 400 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  3,400 g = 3,400 g ÷ 1000 = 3 kg 400 g.

Question 35.
4,000 g + 2,000 g = 6 kg 0 g,

Explanation:
Given 4,000 g + 2,000 g = 6,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 4,000 g + 2,000 g = 6,000 g = 6,000 g ÷ 1000 = 6 kg 0 g.

Question 36.
5,000 g + 3,000 g = 8 kg 0 g,

Explanation:
Given 5,000 g + 3,000 g = 8,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 5,000 g + 3,000 g = 8,000 g = 8,000 g ÷ 1000 = 8 kg 0 g.

Question 37.
4,000 g + 4,000 g = 8 kg 0 g,

Explanation:
Given 4,000 g + 4,000 g = 8,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  4,000 g + 4,000 g = 8,000 g = 8,000 g ÷ 1000 = 8 kg 0 g.

Question 38.
8 × 7,000 g = 56 kg 0 g,

Explanation:
Given 8 X 7,000 g = 56,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 8 × 7,000 g = 56,000 g = 56,000 g ÷ 1000 = 56 kg 0 g.

Question 39.
49,000 g ÷ 7 = 7 kg 0 g,

Explanation:
Given 49,000 g ÷ 7 = 7,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 49,000 g ÷ 7 = 7,000 g = 7,000 g ÷ 1000 = 7 kg 0 g.

Question 40.
16,000 g × 5 = 80 kg 0 g,

Explanation:
Given 16,000 g X 5 = 80,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 16,000 g X 5 = 80,000 g = 80,000 g ÷ 1000 = 80 kg 0 g.

Question 41.
63,000 g ÷ 7 = 9 kg 0 g,

Explanation:
Given 63,000 g ÷ 7 = 9,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 63,000 g ÷ 7 = 9,000 g = 9,000 g ÷ 1000 = 9 kg 0 g.

Question 42.
17 × 4,000 g = 68 kg 0 g,

Explanation:
Given 17 X 4,000 g  = 68,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 17 X 4,000 g = 68,000 g = 68,000 g ÷ 1000 = 68 kg 0 g.

Question 43.
13,000 g × 5 = 65 kg 0 g,

Explanation:
Given 13,000 g X 5 = 65,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 13,000 g X 5 = 65,000 g = 65,000 g ÷ 1000 = 65 kg 0 g.

Question 44.
84,000 g ÷ 7 = 12 kg 0 g,

Explanation:
Given 84,000 g ÷ 7 = 12,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 84,000 g ÷ 7 = 12,000 g = 12,000 g ÷ 1000 = 12 kg 0 g.

B
Convert to Kilograms and Grams

Question 1.
1,000 g = 1 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  1,000 g = 1 X 1000 g ÷ 1000 = 1 kg 0 g.

Question 2.
2,000 g = 2 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  2,000 g = 2 X 1000 g ÷ 1000 = 2 kg 0 g.

Question 3.
3,000 g = 3 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  3,000 g = 3 X 1000 g ÷ 1000 = 3 kg 0 g.

Question 4.
8,000 g = 8 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  8,000 g = 8 X 1000 g ÷ 1000 = 8 kg 0 g.

Question 5.
6,000 g = 6 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  6,000 g = 6 X 1000 g ÷ 1000 = 6 kg 0 g.

Question 6.
9,000 g = 9 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  9,000 g = 9 X 1000 g ÷ 1000 = 9 kg 0 g.

Question 7.
4,000 g = 4 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  4,000 g = 4 X 1000 g ÷ 1000 = 4 kg 0 g.

Question 8.
7,000 g = 7 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  7,000 g = 7 X 1000 g ÷ 1000 = 7 kg 0 g.

Question 9.
5,000 g = 5 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,000 g = 5 X 1000 g ÷ 1000 = 5 kg 0 g.

Question 10.
5,100 g = 5 kg 100 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,100 g = 5,100 g ÷ 1000 = 5 kg 100 g.

Question 11.
5,110 g = 5 kg 110 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,110 g = 5 ,110 g ÷ 1000 = 5 kg 110 g.

Question 12.
5,101 g = 5 kg 101 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,101 g = 5,101 g ÷ 1000 = 5 kg 101 g.

Question 13.
5,010 g = 5 kg 10 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,010 g = 5,010 g ÷ 1000 = 5 kg 10 g.

Question 14.
5,011 g = 5 kg 11 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,011 g = 5,011 g ÷ 1000 = 5 kg 11 g.

Question 15.
5,001 g = 5 kg 1 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,001 g = 5,001 g ÷ 1000 = 5 kg 1 g.

Question 16.
7,002 g = 7 kg 2 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  7,002 g = 7,002 g ÷ 1000 = 7 kg 2 g.

Question 17.
7,020 g = 7kg 20 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  7,020 g = 7,020 g ÷ 1000 = 7 kg 20 g.

Question 18.
7,200 g = 7 kg 200 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  7,200 g = 7,200 g ÷ 1000 = 7 kg 200 g.

Question 19.
7,022 g = 7 kg 22 g,
7,022 g = 7 kg 22 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  7,022 g = 7,022 g ÷ 1000 = 7 kg 22 g.

Question 20.
7,220 g = 7 kg 220 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  7,220 g = 7,220 g ÷ 1000 = 7 kg 220 g.

Question 21.
7,222 g = 7 kg 222 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  7,222 g = 7,222 g ÷ 1000 = 7 kg 222 g.

Question 22.
4,378 g = 4 kg 378 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  4,378 g = 4,378 g ÷ 1000 = 4 kg 378 g.

Question 23.
2,700 g = 2 kg 700 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  2,700 g = 2,700 g ÷ 1000 = 2 kg 700 g.

Question 24.
3,660 g = 3 kg 660 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  3,660 g = 3,660 g ÷ 1000 = 3 kg 660 g.

Question 25.
3,706 g = 3 kg 706 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  3,706 g = 3,706 g ÷ 1000 = 3 kg 706 g.

Question 26.
4,095 g = 4 kg 95 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  4,095 g = 4,095 g ÷ 1000 = 4 kg 95 g.

Question 27.
4,030 g = 4 kg 30 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  4,030 g = 4,030 g ÷ 1000 = 4 kg 30 g.

Question 28.
5,006 g = 5 kg 6 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  5,006 g = 5,006 g ÷ 1000 = 5 kg 6 g.

Question 29.
3,004 g = 3 kg 4 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  3,004 g = 3,004 g ÷ 1000 = 3 kg 4 g.

Question 30.
2,010 g = 2 kg 10 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  2,010 g = 2,010 g ÷ 1000 = 2 kg 10 g.

Question 31.
2,075 g = 2 kg 75 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  2,075 g = 2,075 g ÷ 1000 = 2 kg 75 g.

Question 32.
1,504 g = 1 kg 504 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  1,504 g = 1,504 g ÷ 1000 = 1 kg 504 g.

Question 33.
1,440 g = 1 kg 440 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  1,440 g = 1,440 g ÷ 1000 = 1 kg 440 g.

Question 34.
4,500 g = 4 kg 500 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  4,500 g = 4,500 g ÷ 1000 = 4 kg 500 g.

Question 35.
3,000 g + 2,000 g = 5 kg 0 g,

Explanation:
Given 3,000 g + 2,000 g = 5,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So  3,000 g + 2,000 g = 5,000 g = 5,000 g ÷ 1000 = 5 kg 0 g.

Question 36.
4,000 g + 3,000 g = 7 kg 0 g,

Explanation:
Given 4,000 g + 3,000 g = 7,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 4,000 g + 3,000 g = 7,000 g = 7,000 g ÷ 1000 = 7 kg 0 g.

Question 37.
5,000 g + 4,000 g = 9 kg 0 g,

Explanation:
Given 5,000 g + 4,000 g = 9,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 5,000 g + 4,000 g = 9,000 g = 9,000 g ÷ 1000 = 9 kg 0 g.

Question 38.
9 × 8,000 g = 72 kg 0 g,

Explanation:
Given 9 X 8,000 g = 72,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 9 × 8,000 g = 72,000 g = 72,000 g ÷ 1000 = 72 kg 0 g.

Question 39.
64,000 g ÷ 8 = 8 kg 0 g,

Explanation:
Given 64,000 g ÷ 8 = 8,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 64,000 g ÷ 8 = 8,000 g = 8,000 g ÷ 1000 = 8 kg 0 g.

Question 40.
17,000 g × 5 = 85 kg 0 g,

Explanation:
Given  17,000 g X 5  = 85,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 17,000 g X 5 = 85,000 g = 85,000 g ÷ 1000 = 85 kg 0 g.

Question 41.
54,000 g ÷ 6 = 9 kg 0 g,

Explanation:
Given 54,000 g ÷ 6 = 9,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 54,000 g ÷ 6 = 9,000 g = 9,000 g ÷ 1000 = 9 kg 0 g.

Question 42.
18,000 g × 4 = 72 kg 0 g,

Explanation:
Given  18,000 g X 4  = 72,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 18,000 g X 4 = 72,000 g = 72,000 g ÷ 1000 = 72 kg 0 g.

Question 43.
14 × 5,000 g = 70 kg 0 g,

Explanation:
Given  14 X 5,000 g  = 70,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 14 X 5,000 g = 70,000 g = 70,000 g ÷ 1000 = 70 kg 0 g.

Question 44.
96,000 g ÷ 8 = 12 kg 0 g,

Explanation:
Given 96,000 g ÷ 8 = 12,000 g
Converted the conversion table as shown above as
We know 1000 g or 103  = 1 kg,
So 96,000 g ÷ 8 = 12,000 g = 12,000 g ÷ 1000 = 12 kg 0 g.

### Eureka Math Grade 4 Module 2 Lesson 5 Problem Set Answer Key

Model each problem with a tape diagram.
Solve and answer with a statement.

Question 1.
The potatoes Beth bought weighed 3 kilograms 420 grams.
Her onions weighed 1,050 grams less than the potatoes.
How much did the potatoes and onions weigh together?

The potatoes and onions wiegh together 5,790 g or 5 kg 790 g,

Statement: Both the potatoes and onions wiegh together are
five thousand seven hundred ninety grams or five kilograms and
seven hundred ninety grams,

Explanation:
Given the potatoes Beth bought weighed 3 kilograms 420 grams.
Her onions weighed 1,050 grams less than the potatoes.
So onions wiegh is 3 kg 420 g – 1,050 g =
3 X 1000 g + 420 g = 3,420 g,
now 3,420 g – 1,050 g = 2,370 g,
So the potatoes and onions weigh together are
3,420 g + 2,370 g = 5,790 g or 5,790 ÷ 1,000 = 5 kg 790 g,
Therefore, the potatoes and onions weigh together
5,790 g or 5 kg 790 g,
Modeled  problem with a tape diagram as shown above and
statement is both the potatoes and onions wiegh together are
five thousand seven hundred ninety grams or five kilograms and
seven hundred ninety grams.

Question 2.
Adele let out 18 meters 46 centimeters of string to fly her kite.
She then let out 13 meters 78 centimeters more before
reeling back in 590 centimeters. How long was her string
after reeling it in?

Adele string is 26,34 centimeters or 26 meters 34 centimeters long
after reeling it in.

Statement: Adele string is twenty six thousand thirty centimeters or
twenty six meters and thirty centimeters long after reeling it in,

Explanation:
Given Adele let out 18 meters 46 centimeters of string to fly her kite.
She then let out 13 meters 78 centimeters more before
reeling back in 590 centimeters. So Adele string length after
reeling it in is 18 m 46 cm + 13 m 78 cm – 590 cm =
as 18 m 46 cm = 18 x 100 cm + 46 cm = 18,00 cm + 46 cm =
18,46 cm and 13 m 78 cm = 13  X 100 cm + 78 cm =
13,00 cm + 78 cm = 13,78 cm.
Now (18,46 cm + 13,78 cm) – 590 cm = 32,24 cm – 590 cm =
26,34 cm or  26,34 cm ÷ 100 = 26 m 34 cm,
Therefore, Adele string is 26,34 centimeters or
26 meters 34 centimeters long after reeling it in.
Modeled  problem with a tape diagram as shown above and
statement is Adele string is twenty six thousand thirty centimeters or
twenty six meters and thirty centimeters long after reeling it in.

Question 3.
Shyan’s barrel contained 6 liters 775 milliliters of paint.
She poured in 1 liter 118 milliliters more.
The first day, Shyan used 2 liters 125 milliliters of the paint.
At the end of the second day, there were 1,769 milliliters of paint remaining in the barrel.
How much paint did Shyan use on the second day?

On the second day Shyan uses 3,999 milliliters or
3 liters 999 milliliters of the paint,

Statement : Shyan on the second day uses three thousand
nine hundred ninety nine milliliters or 3 liters and nine
hundred ninety nine milliliters of the paint,

Explanation:
Given Shyan’s barrel contained 6 liters 775 milliliters of paint.
She poured in 1 liter 118 milliliters more.
The first day, Shyan used 2 liters 125 milliliters of the paint.
At the end of the second day, there were 1,769 milliliters of paint remaining in the barrel. Total amount of the paint is
6 L 775 mL + 1 L 118 mL = 7 L 893 mL,
Paint left after first day is 7 L 893 mL – 2 L 125 mL =
7 L 893 mL = 7 X 1000 mL + 893 mL = 7,893 mL,
2 L 125 mL = 2 X 1000 mL + 125 mL = 2,125 mL,
So 7,893 mL – 2,125 mL = 5,768 mL paint is left after first day,
Now Second day paint used is 5,768 mL – 1,769 mL = 3,999 mL
or 3,999 mL ÷ 1000 = 3 L 999 mL paint has been
used on the second day by Shyan.
Modeled  problem with a tape diagram as shown above and
statement Shyan on the second day uses three thousand
nine hundred ninety nine milliliters or 3 liters and nine
hundred ninety nine milliliters of the paint.

Question 4.
On Thursday, the pizzeria used 2 kilograms 180 grams
less flour than they used on Friday. On Friday, they used
12 kilograms 240 grams. On Saturday, they used 1,888 grams
more than on Friday. What was the total amount of flour used
over the three days?

The total amount of flour used over three days is
36,428 grams or 36 kilograms 428 grams,

Statement : The total amount of flour used over three days is
thirty six thousand four hundred twenty eight grams or thirty six
kilograms and four hundred twenty eight grams,

Explanation:
Given On Thursday, the pizzeria used less 2 kilograms 180 grams
less flour than they used on Friday. On Friday, they used
12 kilograms 240 grams. On Saturday, they used 1,888 grams
more than on Friday. So on Thursday they used
12 kg 240 g – 2 kg 180 g =
12 X 1000 g + 240 g = 12,240 g,
2 X 1000 + 180 g = 2,180g,
12,240 g – 2,180 g = 10,060 g on Thursday,
On Saturday they used 1,888 g + 12,240 g = 14,128 g,
Now the total amount of flour used over the three days is
10,060 g + 12,240 g + 14,128 g = 36,428 g or 36,428 g ÷ 1000 =
36 kg 428 g. Therefore the total amount of flour used over three days is
36,428 grams or 36 kilograms 428 grams,
Modeled problem with a tape diagram as shown above and
statement is the total amount of flour used over three days is
thirty six thousand four hundred twenty eight grams or thirty six
kilograms and four hundred twenty eight grams.

Question 5.
The gas tank in Zachary’s car has a capacity of 60 liters.
He adds 23 liters 825 milliliters of gas to the tank,
which already has 2,050 milliliters of gas.
How much more gas can Zachary add to the gas tank?

More 34,125 milliliters or 34 L 125 mL gas can Zachary

Statement : More thirty four thousand one hundred twenty five milliliters or thirty four liters and one hundred twenty five milliliters gas can
Zachary add to the gas tank,

Explanation:
Given the gas tank in Zachary’s car has a capacity of 60 liters.
He adds 23 liters 825 milliliters of gas to the tank,
which already has 2,050 milliliters of gas.
So  the gas tank has  23 L 825 mL + 2,050 mL =
as 23 L 825 mL = 23 X 1000 mL + 825 mL =
23,000 mL + 825 mL = 23,825 mL gas,
Now 23,825 mL + 2,050 mL = 25,875 mL
More gas can Zachary add to the gas tank is 60 L – 25,875 mL =
as 60 L = 60 X 1,000 mL = 60,000 mL,
So 60,000 – 25,875 mL = 34,125 mL or 34,125 mL ÷ 1000 =
34 L 125 mL, therefore, more 34,125 milliliters or
34 L 125 mL gas can Zachary add to the gas tank.
Modeled problem with a tape diagram as shown above and
statement is more thirty four thousand one hundred
twenty five milliliters or thirty four liters and one hundred
twenty five milliliters gas can Zachary add to the gas tank.

Question 6.
A giraffe is 5 meters 20 centimeters tall. An elephant is
1 meter 77 centimeters shorter than the giraffe.
A rhinoceros is 1 meter 58 centimeters shorter than
the elephant. How tall is the rhinoceros?

The rhinoceros is 185 centimeters tall or 1 meter 85 centimeters tall,

Statement: The rhinoceros is one hundred eighty five
centimeters tall or one meter eighty five centimeters tall,

Explanation:
Given A giraffe is 5 meters 20 centimeters tall. An elephant is
1 meter 77 centimeters shorter than the giraffe and
A rhinoceros is 1 meter 58 centimeters shorter than
the elephant. So height of rhinoceros is first we see
height of elephant as 5 m 20 cm – 1 m 77 cm =
as 5 X 100 cm + 20 cm = 5,20 cm and
1 m 77 cm = 1 X 100 cm +77 cm = 1,77 cm,
so 520 cm – 177 cm = 343 cm, Now rhinoceros is
343 cm – 1 m 58 cm =
as 1 m 58 cm = 1 X 100 cm + 58 cm = 158 cm
so 343 cm – 158 cm = 185 cm or 185 cm ÷ 100 = 1 m 85 cm,
therefore the rhinoceros is 185 centimeters tall or
1 meter 85 centimeters tall, modeled problem with a
tape diagram as shown above and statement is the
rhinoceros is one hundred eighty five centimeters tall or
one meter eighty five centimeters tall.

### Eureka Math Grade 4 Module 2 Lesson 5 Exit Ticket Answer Key

Model each problem with a tape diagram.
Solve and answer with a statement.

Question 1.
Jeff places a pineapple with a mass of 890 grams on a
balance scale. He balances the scale by placing two oranges,
an apple, and a lemon on the other side. Each orange weighs
280 grams. The lemon weighs 195 grams less than each orange.
What is the mass of the apple?

The mass of the apple is 245 grams,

Statement : The mass of the apple is two hundred forty five grams,

Explanation:
Given Jeff places a pineapple with a mass of 890 grams on a
balance scale. He balances the scale by placing two oranges,
an apple, and a lemon on the other side. Each orange weighs
280 grams. The lemon weighs 195 grams less than each orange.
So the lemon weighs 280 g – 195 g =  85 grams,
Now weigh of two oranges is 280 g + 280 g = 560 grams,
given  2 oranges + lemon + apple = 890 grams, therefore weigh of
apple = 890 g – (560g + 85 g) = 890 g – 645 g = 245 g,
therefore, the mass of the apple is 245 g, modeled problem
with a tape diagram as shown above and statement is mass
of apple is two hundred fortyy five grams.

Question 2.
Brian is 1 meter 87 centimeters tall. Bonnie is 58 centimeters
shorter than Brian. Betina is 26 centimeters taller than Bonnie.
How tall is Betina?
Betina is 155 cm or 1 m 55 cm tall,

Statement: Betina is one hundred fifty five centimeters or
one meter fifty five centimeters tall,

Explanation:
Given Brian is 1 meter 87 centimeters tall.
Bonnie is 58 centimeters shorter than Brian means
Bonnie is 1 m 87 cm – 58 = as 1 m 87 cm =
1 X 100 cm + 87 cm = 187 cm, 187 cm – 58 cm = 129 cm,
Bonnie is 129 cm and Betina is 26 centimeters taller than
Bonnie means Betina is 129 cm + 26 cm = 155 cm or
155  cm ÷ 100 = 1 m 55 cm,
So Betina is 155 cm or 1 m 55 cm tall, modeled problem
with a tape diagram as shown above and statement is
Betina is one hundred fifty five centimeters or
one meter fifty five centimeters tall.

### Eureka Math Grade 4 Module 2 Lesson 5 Homework Answer Key

Model each problem with a tape diagram. Solve and answer with a statement.

Question 1.
The capacity of Jose’s vase is 2,419 milliliters of water.
He poured 1 liter 299 milliliters of water into the empty vase.
Then, he added 398 milliliters. How much more water will
the vase hold?

Jose’s vase can hold more 722 milliliters of water,

Statement : Jose’s vase can hold more seven hundred twenty
two milliliters of water,

Explanation:
Given the capacity of Jose’s vase is 2,419 milliliters of water.
He poured 1 liter 299 milliliters of water into the empty vase.
Then, he added 398 milliliters. So now water in the vase is
1 L 299 mL + 398 mL as 1 L 299 mL = 1 X 1000 mL + 299 mL =
1,299 mL +  398 mL = 1,697 mL of water,
Now we can pour more is 2,419 mL – 1,697 mL = 722 mL more,
therefore, Jose’s vase can hold more 722 milliliters of water,
modeled problem with a tape diagram as shown above and
statement is Jose’s vase can hold more seven hundred
twenty two milliliters of water.

Question 2.
Eric biked 1 kilometer 125 meters on Monday.
On Tuesday, he biked 375 meters less than on Monday.
How far did he bike both days?

Eric biked both days as 1,875 meters or 1 kilometer 875 meters,

Statement : Eric biked both days as one thousand eight
hundred seventy five meters or one kilometer eight
hundred seventy five meters,

Explanation:
Given Eric biked 1 kilometer 125 meters on Monday.
On Tuesday, he biked 375 meters less than on Monday.
So on Tuesday he biked 1 k 125 m – 375 m =
as 1 X 1000 m + 125 m = 1,125 m ,
1125 m – 375 m = 750 m,
therefore Eric biked on Monday and Tuesday is
1,125 m + 750 m = 1,875 m or 1,875 m ÷ 1,000 = 1 k 875 m,
therefore, Eric biked both days as 1,875 meters or
1 kilometer 875 meters, modeled problem with a tape diagram
as shown above and statement is Eric biked both days as one thousand eight hundred seventy five meters or one kilometer eight
hundred seventy five meters.

Question 3.
Zachary weighs 37 kilograms 95 grams. Gabe weighs
4,650 grams less than Zachary. Harry weighs
2,905 grams less than Gabe. How much does Harry weigh?

Harry weigh 29,540 grams or 29 kilogram 540 grams,

Statement : Harry weigh’s twenty nine thousand five hundred
forty grams or twenty nine kilograms five hundred
forty grams,

Explanation:
Given Zachary weighs 37 kilograms 95 grams. Gabe weighs
4,650 grams less than Zachary and Harry weighs
2,905 grams less than Gabe. So Gabe weigh’s
37 kg 95 g – 4,650 g as 37 kg 95 g =
37 X 1000 g + 95 g = 37,095 g so Gabe is
37,095 g – 4,650 g = 32,445 g,
Now Harry weigh’s 32,445 g – 2,905 g = 29,540 grams or
29,540 g ÷ 1,000 = 29 kilograms 540 grams,
therefore, Harry weigh 29,540 grams or 29 kilogram 540 grams,
modeled problem with a tape diagram as shown above and
statement is Harry weigh’s twenty nine thousand five hundred
forty grams or twenty nine kilograms five hundred
forty grams.

Question 4.
A Springer Spaniel weighs 20 kilograms 490 grams.
A Cocker Spaniel weighs 7,590 grams less than a
Springer Spaniel. A Newfoundland weighs
52 kilograms 656 grams more than a Cocker Spaniel.
What is the difference, in grams, between the weights of the Newfoundland and the Springer Spaniel?

The difference in grams, between the weights of the
Newfoundland and the Springer Spaniel is 45,066 grams,
Statement: The difference in grams, between the weights of the
Newfoundland and the Springer Spaniel is forty five thousand
sixty six grams,

Explanation:
Given a Springer Spaniel weighs 20 kilograms 490 grams.
A Cocker Spaniel weighs 7,590 grams less than a
Springer Spaniel. A Newfoundland weighs
52 kilograms 656 grams more than a Cocker Spaniel.
So Cocker spaniel weigh’s 20 kg 490 g – 7,590 g
as 20 kg 490 g = 20 X 1000 g + 490 g = 20,490 g,
So 20,490 g – 7,590 g = 12,900 g,
Now Newfoundland weighs 52 kg 656 g +12,900 g as
52 kg 656 g = 52 X 1000 g + 656 g = 52,656 g,
52,656 g + 12,900 g = 65,556 g,
Now the difference in grams, between the weights of the
Newfoundland and the Springer Spaniel is
65,556 g – 20,490 g = 45,066 grams,
modeled problem with a tape diagram as shown above and
statement is the difference in grams, between the weights of the
Newfoundland and the Springer Spaniel is forty five thousand
sixty six grams.

Question 5.
Marsha has three rugs. The first rug is 2 meters 87 centimeters long.
The second rug has a length 98 centimeters
less than the first. The third rug is 111 centimeters
longer than the second rug. What is the difference in
centimeters between the length of the first rug and the third rug?

The difference in centimeters between the length of
the first rug and the third rug is 13 cm,

Statement : The difference in centimeters between the length of
the first rug and the third rug is thirteen centimeters,

Explanation:
Given Marsha has three rugs. The first rug is
2 meters 87 centimeters long.
The second rug has a length 98 centimeters
less than the first. The third rug is 111 centimeters
longer than the second rug so the second rug is
2 m 87 cm – 98 cm as 2 m 87 cm = 2 X 100 cm + 87 cm = 287 cm,
Now 287 cm – 98 cm = 189 cm and the third rug is
189 cm + 111 cm = 300 cm,
first rug is 287 cm, second rug is 189 cm and third rug is 300 cm,
Now the difference in centimeters between the length of
the first rug and the third rug is 300 cm – 287 cm = 13 cm,
modeled problem with a tape diagram as shown above and
statement is the difference in centimeters between the length of
the first rug and the third rug is thirteen centimeters.

Question 6.
One barrel held 60 liters 868 milliliters of sap.
A second barrel held 20,089 milliliters more sap than the first.
A third barrel held 40 liters 82 milliliters less sap than the second.
If the sap from the three barrels was poured into a
larger container, how much sap would there be in all ?

182,700 milliliters or 182 liters 700 milliliters of sap
would be there in all,

Statement: one lakh eighty two thousand seven hundred
milliliters or one hundred eighty two liters and
seven hundred milliliters of sap would be there in all,

Explanation:
Given one barrel held 60 liters 868 milliliters of sap.
A second barrel held 20,089 milliliters more sap than the first.
A third barrel held 40 liters 82 milliliters less sap than the second.
Second barrel holds 60 L 868 mL + 20,089 mL as
60 L 868 mL = 60 X 1000 mL + 868 mL = 60,868 mL,
60,868 mL + 20,089 mL = 80,957 mL,
Third barrel holds 80,957 mL – 40 L 82 mL as
40 L 82 mL = 40 X 1000 mL + 82 mL = 40,082 mL,
so 80,957 mL – 40,082 mL = 40,875 mL,
First barrel holds 60,868 mL,
Second barrel holds 80,957 mL,
Third barrel holds 40,875 mL,
Now if the sap from the three barrels was poured into a
larger container it will contain
60,868 mL + 80,957 mL + 40,875 mL = 182,700 mL,
modeled problem with a tape diagram as shown above and
statement is one lakh eighty two thousand seven hundred
milliliters or one hundred eighty two liters and
seven hundred milliliters of sap would be there in all.

## Engage NY Eureka Math 4th Grade Module 1 Lesson 1 Answer Key

### Eureka Math Grade 4 Module 1 Lesson 1 Sprint Answer Key

A
Multiply and Divide by 10

Question 1.
2 × 10 =
2 × 10 = 20 ones = 2 tens
Explanation:
10 times of 2 ones is 20 ones that is 2 tens.

Question 2.
3 × 10 =
3 × 10 = 30 ones = 3 tens
Explanation:
10 times of 3 ones is 30 ones that is 3 tens.

Question 3.
4 × 10 =
4 × 10 = 40 ones = 4 tens
Explanation:
10 times of 4 ones is 40 ones that is 4 tens.

Question 4.
5 × 10 =
5 × 10 = 50 ones = 5 tens
Explanation:
10 times of 5 ones is 50 ones that is 5 tens.

Question 5.
1 × 10 =
1 × 10 = 10 ones = 1 ten
Explanation:
10 times of 1 ones is 10 ones that is 1 tens.

Question 6.
20 ÷ 10 =
20 ÷ 10 = 2 ones
2 ones × 10 = 20 ones = 2 tens
Explanation:
10 times of 2 ones is 20 ones that is 2 tens.
20 ÷ 10 = 2 ones.

Question 7.
30 ÷ 10 =
30 ÷ 10 = 3 ones
3 ones × 10 = 30 ones
Explanation:
10 times of 3 ones is 30 ones that is 3 tens.
30 ÷ 10 = 3 ones.

Question 8.
50 ÷ 10 =
50 ÷ 10 = 5 ones
5 ones × 10 = 50 ones = 5 tens
Explanation:
10 times of 5 ones is 50 ones that is 5 tens.
50 ÷ 10 = 5 ones.

Question 9.
10 ÷ 10 =
10 ÷ 10 = 1 ones
1 ones × 10 = 10 ones = 1 tens
Explanation:
10 times of 1 ones is 10 ones that is 1 tens.
10 ÷ 10 = 1 ones.

Question 10.
40 ÷ 10 =
40 ÷ 10 = 4 ones
4 ones × 10 = 40 ones = 4 tens
Explanation:
10 times of 4 ones is 40 ones that is 4 tens.
40 ÷ 10 = 4 ones.

Question 11.
6 × 10 =
6 × 10 = 60
Explanation:
10 times of 6 ones is 60 ones that is 6 tens.

Question 12.
7 × 10 =
7 × 10 = 70
Explanation:
10 times of 7 ones is 70 ones that is 7 tens.

Question 13.
8 × 10 =
8 × 10 = 80
Explanation:
10 times of 8 ones is 80 ones that is 8 tens.

Question 14.
9 × 10 =
9 × 10 = 90
Explanation:
10 times of 9 ones is 90 ones that is 9 tens.

Question 15.
10 × 10 =
10 × 10 = 100
Explanation:
10 times of 10 ones is 100 ones that is 10 tens.

Question 16.
80 ÷ 10 =
80 ÷ 10 = 8 ones
8 ones × 10 = 80 ones = 8 tens
Explanation:
10 times of 8 ones is 80 ones that is 8 tens.
80 ÷ 10 = 8 ones.

Question 17.
70 ÷ 10 =
70 ÷ 10 = 7 ones
7 ones × 10 = 70 ones = 7 tens
Explanation:
10 times of 7 ones is 70 ones that is 7 tens.
70 ÷ 10 = 7 ones.

Question 18.
90 ÷ 10 =
90 ÷ 10 = 9 ones
9 ones × 10 = 90 ones = 9 tens
Explanation:
10 times of 9 ones is 90 ones that is 9 tens.
90 ÷ 10 = 9 ones.

Question 19.
60 ÷ 10 =
60 ÷ 10 = 6 ones
6 ones × 10 = 60 ones = 6 tens
Explanation:
10 times of 6 ones is 60 ones that is 6 tens.
60 ÷ 10 = 6 ones.

Question 20.
100 ÷ 10 =
100 ÷ 10 = 10 ones
10 ones × 10 = 100 ones = 10 tens
Explanation:
10 times of 10 ones is 100 ones that is 10 tens.
100 ÷ 10 = 10 ones = 1 tens.

Question 21.
__ × 10 = 50
5 × 10 = 50
Explanation:
10 times of 5 ones is 50 ones that is 5 tens.

Question 22.
__ × 10 = 10
1 × 10 = 10
Explanation:
10 times of 1 ones is 10 ones that is 1 tens.

Question 23.
__ × 10 = 100
10 × 10 = 100
Explanation:
10 times of 10 ones is 100 ones that is 10 tens.

Question 24.
__ × 10 = 20
2 × 10 = 20
Explanation:
10 times of 2 ones is 20 ones that is 2 tens.

Question 25.
__ × 10 = 30
3 × 10 = 30
Explanation:
10 times of 3 ones is 30 ones that is 3 tens.

Question 26.
100 ÷ 10 =
100 ÷ 10 = 10
10 ones × 10 = 100 ones = 10 tens
Explanation:
10 times of 10 ones is 100 ones that is 10 tens.
100 ÷ 10 = 10 ones = 1 tens.

Question 27.
50 ÷ 10 =
50 ÷ 10 = 5
5 ones × 10 = 50 ones = 5 tens
Explanation:
10 times of 5 ones is 50 ones that is 5 tens.
50 ÷ 10 = 5 ones.

Question 28.
10 ÷ 10 =
10 ÷ 10 = 1
1 ones × 10 = 10 ones = 1 tens.
Explanation:
10 times of 1 ones is 10 ones that is 1 tens.
10 ÷ 10 = 1 ones.

Question 29.
20 ÷ 10 =
20 ÷ 10 = 2 ones
2 ones × 10 = 20 ones = 2 tens
Explanation:
10 times of 2 ones is 20 ones that is 2 tens.
20 ÷ 10 = 2 ones.

Question 30.
30 ÷ 10 =
30 ÷ 10 = 3 ones
3 ones × 10 = 30 ones = 3 tens
Explanation:
10 times of 3 ones is 30 ones that is 3 tens.
30 ÷ 10 = 3 ones.

Question 31.
__ × 10 = 60
6 × 10 = 60
60 ÷ 10 = 6 ones.
Explanation:
10 times of 6 ones is 60 ones that is 6 tens.

Question 32.
__ × 10 = 70
7 × 10 = 70
70 ÷ 10 = 7 ones.
Explanation:
10 times of 7 ones is 70 ones that is 7 tens.

Question 33.
__ × 10 = 90
9 × 10 = 90
90 ÷ 10 = 9 ones.
Explanation:
10 times of 9 ones is 90 ones that is 9 tens.

Question 34.
__ × 10 = 80
8
× 10 = 80
80 ÷ 10 = 8 ones.
Explanation:
10 times of 8 ones is 80 ones that is 8 tens.

Question 35.
70 ÷ 10 =
70 ÷ 10 = 7
7 ones × 10 = 70 ones = 7 tens.
Explanation:
10 times of 7 ones is 70 ones that is 7 tens.
70 ÷ 10 = 7 ones.

Question 36.
90 ÷ 10 =
90 ÷ 10 = 9
9 ones × 10 = 90 ones = 9 tens.
Explanation:
10 times of 9 ones is 90 ones that is 9 tens.
90 ÷ 10 = 9 ones.

Question 37.
60 ÷ 10 =
60 ÷ 10 = 6
6 ones × 10 = 60 ones = 6 tens.
Explanation:
10 times of 6 ones is 60 ones that is 6 tens.
60 ÷ 10 = 6 ones.

Question 38.
80 ÷ 10 =
80 ÷ 10 = 8
8 ones × 10 = 80 ones = 8 tens.
Explanation:
10 times of 8 ones is 80 ones that is 8 tens.
80 ÷ 10 = 8 ones.

Question 39.
11 × 10 =
11 × 10 = 110
Explanation:
10 times of 11 ones is 110 ones that is 11 tens.

Question 40.
110 ÷ 10 =
110 ÷ 10 = 11
11 ones × 10 = 110 ones = 11 tens.
Explanation:
10 times of 11 ones is 110 ones that is 11 tens.
110 ÷ 10 = 11 ones.

Question 41.
30 ÷ 10 =
30 ÷ 10 = 3
3 ones × 10 = 30 ones.
Explanation:
10 times of 3 ones is 30 ones that is 3 tens.
130 ÷ 10 = 13 ones.

Question 42.
120 ÷ 10 =
120 ÷ 10 = 12
10 ones × 10 = 120 ones = 12 tens.
Explanation:
10 times of 12 ones is 120 ones that is 12 tens.
120 ÷ 10 = 12 ones.

Question 43.
14 × 10 =
14 × 10 = 140
Explanation:
10 times of 14 ones is 140 ones that is 14 tens.

Question 44.
140 ÷ 10 =
140 ÷ 10 = 14
14 ones × 10 = 140 ones = 14 tens.
Explanation:
10 times of 14 ones is 140 ones that is 14 tens.
140 ÷ 10 = 14 ones.

B
Multiply and Divide by 10

Question 1.
1 × 10 =
1 ones × 10 = 10 ones = 1 tens.
Explanation:
10 times of 1 ones is 10 ones that is 1 tens.

Question 2.
2 × 10 =
2 × 10 = 20 ones = 2 tens.
Explanation:
10 times of 2 ones is 20 ones that is 2 tens.

Question 3.
3 × 10 =
3 × 10 = 30
Explanation:
10 times of 3 ones is 30 ones that is 3 tens.

Question 4.
4 × 10 =
4 ones × 10 = 40 ones = 4 tens.
Explanation:
10 times of 4 ones is 40 ones that is 4 tens.

Question 5.
5 × 10 =
5 ones × 10 = 50 ones = 5 tens
Explanation:
10 times of 5 ones is 50 ones that is 5 tens.

Question 6.
30 ÷ 10 =
30 ÷ 10 = 3 ones
3 ones × 10 = 30 ones = 3 tens.
Explanation:
10 times of 3 ones is 30 ones that is 3 tens.
30 ÷ 10 = 3.

Question 7.
20 ÷ 10 =
20 ÷ 10 = 2 ones.
2 ones × 10 = 20 ones = 2 tens.
Explanation:
10 times of 2 ones is 20 ones that is 2 tens.
20 ÷ 10 = 2 ones.

Question 8.
40 ÷ 10 =
40 ÷ 10 = 4 ones
4 ones × 10 = 40 ones = 4 tens.
Explanation:
10 times of 4 ones is 40 ones that is 4 tens.
40 ÷ 10 = 4 ones.

Question 9.
10 ÷ 10 =
10 ÷ 10 = 1 ones.
1 ones × 10 = 10 ones = 1 tens.
Explanation:
10 times of 1 ones is 10 ones that is 1 tens.
10 ÷ 10 = 1 ones.

Question 10.
50 ÷ 10 =
50  ÷ 10 = 5 ones .
5 ones × 10 = 50 ones = 5 tens.
Explanation:
10 times of 5 ones is 50 ones that is 5 tens.
50 ÷ 10 = 5 ones.

Question 11.
10 × 10 =
10 × 10 = 100 ones = 10 tens.
Explanation:
10 times of 10 ones is 100 ones that is 10 tens.

Question 12.
6 × 10 =
6 ones × 10 = 60 ones = 6 tens
Explanation:
10 times of 6 ones is 60 ones that is 6 tens.

Question 13.
7 × 10 =
7 × 10 = 70 ones = 7 tens
Explanation:
10 times of 7 ones is 70 ones that is 7 tens.

Question 14.
8 × 10 =
8 ones × 10 = 80 ones = 8 tens.
Explanation:
10 times of 8 ones is 80 ones that is 8 tens.

Question 15.
9 × 10 =
9 × 10 = 90
Explanation:
10 times of 9 ones is 90 ones that is 9 tens.

Question 16.
70 ÷ 10 =
70 ÷ 10 = 7 ones.
7 ones × 10 = 70 ones = 7 tens.
Explanation:
10 times of 7 ones is 70 ones that is 7 tens.
70 ÷ 10 = 7.

Question 17.
60 ÷ 10 =
60 ÷ 10 = 6 ones.
6 ones × 10 = 60 ones = 6 tens.
Explanation:
10 times of 6 ones is 60 ones that is 6 tens.
60 ÷ 10 = 6 ones.

Question 18.
80 ÷ 10 =
80 ÷ 10 = 8 ones.
8 ones × 10 = 80 ones = 8 tens.
Explanation:
10 times of 8 ones is 80 ones that is 8 tens.
80 ÷ 10 = 8 ones.

Question 19.
100 ÷ 10 =
100 ÷ 10 = 10 ones = 1 tens
10 ones × 10 = 100 ones = 10 tens.
Explanation:
10 times of 10 ones is 100 ones that is 10 tens.
100 ÷ 10 = 10 ones = 1 tens.

Question 20.
90 ÷ 10 =
90 ÷ 10 = 9 ones.
9 ones × 10 = 90 ones = 9 tens.
Explanation:
10 times of 9 ones is 90 ones that is 9 tens.
90 ÷ 10 = 9 ones.

Question 21.
__ × 10 = 10
1 × 10 = 10
10 ÷ 10 = 1 ones.
Explanation:
10 times of 1 ones is 10 ones that is 1 tens.

Question 22.
__ × 10 = 50
5 × 10 = 50
50 ÷ 10 = 5 ones.
Explanation:
10 times of 5 ones is 50 ones that is 5 tens.

Question 23.
__ × 10 = 20
2 × 10 = 20
20 ÷ 10 = 2 ones.
Explanation:
10 times of 2 ones is 20 ones that is 2 tens.

Question 24.
__ × 10 = 100
10 × 10 = 100
100 ÷ 10 = 10 ones = 1 ten.
Explanation:
10 times of 10 ones is 100 ones that is 10 tens.

Question 25.
__ × 10 = 30
3 × 10 = 30
30 ÷ 10 = 3 ones.
Explanation:
10 times of 3 ones is 30 ones that is 3 tens.

Question 26.
20 ÷ 10 =
20 ÷ 10 = 2 ones.
2 ones × 10 = 20 ones = 2 tens.
Explanation:
10 times of 2 ones is 20 ones that is 2 tens.
20 ÷ 10 = 2 ones.

Question 27.
10 ÷ 10 =
10 ÷ 10 = 1 ones
1 ones × 10 = 10 ones = 1 tens.
Explanation:
10 times of 1 ones is 10 ones that is 1 tens.
10 ÷ 10 = 1 ones.

Question 28.
100 ÷ 10 =
100 ÷ 10 = 10 ones
10 ones × 10 = 100 ones = 10 tens.
Explanation:
10 times of 10 ones is 100 ones that is 10 tens.
100 ÷ 10 = 10 ones = 1 ten.

Question 29.
50 ÷ 10 =
50 ÷ 10 = 5 ones.
5 ones × 10 = 50 ones = 5 tens.
Explanation:
10 times of 5 ones is 50 ones that is 5 tens.
50 ÷ 10 = 5.

Question 30.
30 ÷ 10 =
30 ÷ 10 = 3
3 ones × 10 = 30 ones = 3 tens.
Explanation:
10 times of 3 ones is 30 ones that is 3 tens.
30 ÷ 10 = 3.

Question 31.
__ × 10 = 30
3 × 10 = 30
30 ÷ 10 = 3 ones.
Explanation:
10 times of 3 ones is 30 ones that is 3 tens.

Question 32.
__ × 10 = 40
4 × 10 = 40
40 ÷ 10 = 4 ones.
Explanation:
10 times of 4 ones is 40 ones that is 4 tens.

Question 33.
__ × 10 = 90
9 × 10 = 90
90 ÷ 10 = 9 ones.
Explanation:
10 times of 9 ones is 90 ones that is 9 tens.

Question 34.
__ × 10 = 70
7 × 10 = 70
70 ÷ 10 = 7 ones.
Explanation:
10 times of 7 ones is 70 ones that is 7 tens.

Question 35.
80 ÷ 10 =
80 ÷ 10 = 8
8 ones × 10 = 80 ones = 8 tens.
Explanation:
10 times of 8 ones is 80 ones that is 8 tens.
80 ÷ 10 = 8 ones.

Question 36.
90 ÷ 10 =
90 ÷ 10 = 9 ones.
9 ones × 10 = 90 ones = 9 tens.
Explanation:
10 times of 90 ones is 90 ones that is 9 tens.
90 ÷ 10 = 9 ones.

Question 37.
60 ÷ 10 =
60 ÷ 10 = 6 ones.
6 ones × 10 = 60 ones = 6 tens.
Explanation:
10 times of 6 ones is 60 ones that is 6 tens.
60 ÷ 10 = 6.

Question 38.
70 ÷ 10 =
70 ÷ 10 = 7
7 ones × 10 = 70 ones = 7 tens.
Explanation:
10 times of 7 ones is 70 ones that is 7 tens.
70 ÷ 10 = 7.

Question 39.
11 × 10 =
11 × 10 = 110
Explanation:
10 times of 11 ones is 110 ones that is 11 tens.

Question 40.
110 ÷ 10 =
110 ÷ 10 = 11
11 ones × 10 = 110 ones = 11 tens.
Explanation:
10 times of 11 ones is 110 ones that is 11 tens.
110 ÷ 10 = 11.

Question 41.
12 × 10 =
12 × 10 = 120
Explanation:
10 times of 12 ones is 120 ones that is 12 tens.

Question 42.
120 ÷ 10 =
120 ÷ 10 = 12
12 ones × 10 = 120 ones = 12 tens.
Explanation:
10 times of 12 ones is 120 ones that is 12 tens.
120 ÷ 10 = 12.

Question 43.
13 × 10 =
13 × 10 = 130
Explanation:
10 times of 13 ones is 130 ones that is 13 tens.

Question 44.
130 ÷ 10 =
130 ÷ 10 = 13
13 ones × 10 = 130 ones = 13 tens.
Explanation:
10 times of 13 ones is 130 ones that is 13 tens.
130 ÷ 10 = 13.

### Eureka Math Grade 4 Module 1 Lesson 1 Problem Set Answer Key

Question 1.
Label the place value charts. Fill in the blanks to make the following equations true. Draw disks in the place value chart to show how you got your answer, using arrows to show any bundling.
a. 10 × 3 ones = ________ ones = __________

b. 10 × 2 tens =_________ tens = _________

c. 4 hundreds × 10 = _________ hundreds = _________

Question 2.
Complete the following statements using your knowledge of place value:
a. 10 times as many as 1 ten is ________tens.
10 times as many as 1 ten is 10 tens.
b. 10 times as many as _________ tens is 30 tens or ________ hundreds.
10 times as many as 3 tens is 30 tens or 3 hundreds.
c. _____________________________ as 9 hundreds is 9 thousands.
__10 times_____ as 9 hundreds is 9 thousands
d. _________ thousands is the same as 20 hundreds.
__2__ thousands is the same as 20 hundreds.

Use pictures, numbers, or words to explain how you got your answer for Part (d).

Question 3.
Matthew has 30 stamps in his collection. Matthew’s father has 10 times as many stamps as Matthew. How many stamps does Matthew’s father have? Use numbers or words to explain how you got your answer.
Number of stamps in Matthew collection are = 30
30 ones = 3 tens
Matthew’s father have 10 times as many as Matthew = 30 × 10 = 300 ones or 30 tens. or 3 hundred stamps.

Question 4.
Jane saved $800. Her sister has 10 times as much money. How much money does Jane’s sister have? Use numbers or words to explain how you got your answer. Answer: Money saved by Jane =$800
Her sister has 10 times as much money as Jane = 800 × 10
800 × 10 = $8000 Jane’s sister have$8000.

Question 5.
Fill in the blanks to make the statements true.
a. 2 times as much as 4 is _______.
Answer: 2 times as much as 4 is __8__.
b. 10 times as much as 4 is _______.
Answer: 10 times as much as 4 is __40_____.
c. 500 is 10 times as much as _______.
Answer:  500 is 10 times as much as __50_____.
d. 6,000 is ________________________________ as 600.
Answer:  6,000 is ___10 times __ as 600.

Question 6.
Sarah is 9 years old. Sarah’s grandfather is 90 years old. Sarah’s grandfather is how many times as old as Sarah?

Sarah’s grandfather is _______ times as old as Sarah.
Sarah age is = 9 years old.
Sarah’s grandfather age is = 90 years old.
90 ÷ 9 = 10.
9 × 10 = 90.
Sarah’s grandfather is 10 times as old as Sarah.

### Eureka Math Grade 4 Module 1 Lesson 1 Exit Ticket Answer Key

Use the disks in the place value chart below to complete the following problems:

Question 1.
Label the place value chart.

Question 2.
Tell about the movement of the disks in the place value chart by filling in the blanks to make the following equation match the drawing in the place value chart:
_____________ × 10 = ___________________ = ________________

Question 3.
10 times as many as 4 hundred is  40 hundred (4000) = 4 thousands

### Eureka Math Grade 4 Module 1 Lesson 1 Homework Answer Key

Question 1.
Label the place value charts. Fill in the blanks to make the following equations true. Draw disks in the place value chart to show how you got your answer, using arrows to show any regrouping.
a. 10 × 4 ones = ________ ones = __________

b. 10 × 2 tens =_________ tens = _________

c. 5 hundreds × 10 = _________ hundreds = _________

Question 2.
Complete the following statements using your knowledge of place value:
a. 10 times as many as 1 hundred is ______ hundreds or ________ thousand.
Answer: 10 times as many as 1 hundred is _10__ hundreds or __1__ thousand.
b. 10 times as many as _________ hundreds is 60 hundreds or ________ thousands.
Answer:  10 times as many as ___6__hundreds is 60 hundreds or __6_ thousands.
c. _____________________________ as 8 hundreds is 8 thousands.
Answer:  __10 times as many_ as 8 hundreds is 8 thousands.
d. _________ hundreds is the same as 4 thousands.
Answer:  __40___ hundreds is the same as 4 thousands.

Use pictures, numbers, or words to explain how you got your answer for Part (d).

Question 3.
Katrina has 60 GB of storage on her tablet. Katrina’s father has 10 times as much storage on his computer. How much storage does Katrina’s father have? Use numbers or words to explain how you got your answer.
Total storage Katrina tablet has = 60 GB
Katrina’s fathers computer storage has 10 times as much = 65 GB × 10
65 × 10 = 650 ones = 65 tens.

Question 4.
Katrina saved $200 to purchase her tablet. Her father spent 10 times as much money to buy his new computer. How much did her father’s computer cost? Use numbers or words to explain how you got your answer. Answer: Money Katrina saved on purchasing her tablet =$200.
Money saved by her father on buying new computer is =10 times as much as Katrina
$200 × 10 =$2000.
2 hundreds × 10 =20 hundreds = 2 thousands.

Question 5.
Fill in the blanks to make the statements true.
a. 4 times as much as 3 is _______.
Answer: 4 times as much as 3 is _12__.
b. 10 times as much as 9 is _______.
Answer:  10 times as much as 9 is __90_____.
c. 700 is 10 times as much as _______.
Answer:  700 is 10 times as much as __70_____.
d. 8,000 is ________________________________ as 800.
Answer:  8,000 is __10 times as much________ as 800.

Question 6.
Tomas’s grandfather is 100 years old. Tomas’s grandfather is 10 times as old as Tomas. How old is Tomas?
Tomas’s grandfather is = 100 years old
Tomas’s grandfather is = 10 times old as Tomas
Tomas age = 100 ÷ 10 = 10 years old.
Thomas is 10 years old

Students of Grade 3 Module 3 can get a strong foundation on mathematics concepts by referring to the Eureka Math Book. It was developed by highly professional mathematics educators and the solutions prepared by them are in a concise manner for easy grasping. To achieve high scores in Grade 3, students need to solve all questions and exercises included in Eureka’s Grade 3 Textbook.

## Engage NY Eureka Math 3rd Grade Module 3 Lesson 2 Answer Key

So teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Eureka’s primary school Grade 3 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that.

### Eureka Math Grade 3 Module 3 Lesson 2 Sprint Answer Key

Use the Commutative Property to Multiply

2 x 2 = 4, 3 x 2 = 6, 2 x 3 = 6, 2 x 4 = 8, 4 x 2= 8, 2 x 5 = 10, 5 x 2 = 10, 2 x 6 = 12, 6 x 2 = 12, 2 x 7 = 14, 7 x 2 = 14, 2 x 8 = 16, 8 x 2 = 16, 2 x 9 = 18, 9 x 2 = 18, 2 x 10 = 20, 10 x 2 = 20, 5 x 3 = 15, 3 x 5 = 15, 4 x 5 = 20, 5 x 5 =25,
5 x 6 = 30, 6 x 5 = 30, 5 x 7 = 35, 7 x 5 = 35, 5 x 8 = 40, 8 x 5 = 40, 5 x 9 = 45, 9 x 5 = 45, 5 x 10 = 50, 10 x 5 = 50, 3 x 3 = 9, 3 x 4 = 12, 4 x 3 = 12, 3 x 6 = 18, 6 x 3 = 18, 3 x 7 = 21, 7 x 3 = 21,  3 x 8 = 24, 8 x 3 = 24, 3 x 9 = 27, 9 x 3 = 27, 4 x 4 =16.

Explanation:
In the above-given question,
given that,
the commutative property.
2 x 2 = 4, 3 x 2 = 6, 2 x 3 = 6, 2 x 4 = 8, 4 x 2= 8, 2 x 5 = 10, 5 x 2 = 10, 2 x 6 = 12, 6 x 2 = 12, 2 x 7 = 14, 7 x 2 = 14, 2 x 8 = 16, 8 x 2 = 16, 2 x 9 = 18, 9 x 2 = 18, 2 x 10 = 20, 10 x 2 = 20, 5 x 3 = 15, 3 x 5 = 15, 4 x 5 = 20, 5 x 5 =25,
5 x 6 = 30, 6 x 5 = 30, 5 x 7 = 35, 7 x 5 = 35, 5 x 8 = 40, 8 x 5 = 40, 5 x 9 = 45, 9 x 5 = 45, 5 x 10 = 50, 10 x 5 = 50, 3 x 3 = 9, 3 x 4 = 12, 4 x 3 = 12, 3 x 6 = 18, 6 x 3 = 18, 3 x 7 = 21, 7 x 3 = 21,  3 x 8 = 24, 8 x 3 = 24, 3 x 9 = 27, 9 x 3 = 27, 4 x 4 =16.

Use the Commutative Property to Multiply

5 x 2 = 10, 2 x 5 = 10, 5 x 3 = 15, 3 x 5 = 15, 5 x4 = 20, 4 x 5 = 20, 5 x 5 = 25, 5 x 6 = 30, 6 x 5 = 30, 5 x 7 = 35, 7 x 5 = 35, 5 x 8 = 40, 8 x 5 = 40, 5 x 9 = 45, 9 x 5 = 45, 5 x 10 = 50, 10 x 5 = 50, 2 x 2 = 4, 2 x 3 = 6, 3 x 2 = 6, 2 x 4 = 8, 4 x 2 = 8, 6 x 2 = 12, 2 x 6 = 12, 2 x 7 = 14,7 x 2 = 14, 2 x 8 = 16, 8 x 2 = 16, 2 x 9 = 18, 9 x 2 = 18, 2 x 10 = 20, 10 x 2 = 20, 3 x 3 = 9, 3 x 4 = 12, 4 x 3 = 12, 3 x 6 = 18, 6 x 3 = 18, 3 x 7 = 21, 7 x 3 = 21, 3 x 8 = 24, 8 x 3 = 24, 3 x 9 = 27, 9 x 3 = 27, 4 x 4 = 16.

Explanation:
In the above-given question,
given that,
the figure shows the commutative property.
5 x 2 = 10, 2 x 5 = 10, 5 x 3 = 15, 3 x 5 = 15, 5 x4 = 20, 4 x 5 = 20, 5 x 5 = 25, 5 x 6 = 30, 6 x 5 = 30, 5 x 7 = 35, 7 x 5 = 35, 5 x 8 = 40, 8 x 5 = 40, 5 x 9 = 45, 9 x 5 = 45, 5 x 10 = 50, 10 x 5 = 50, 2 x 2 = 4, 2 x 3 = 6, 3 x 2 = 6, 2 x 4 = 8, 4 x 2 = 8, 6 x 2 = 12, 2 x 6 = 12, 2 x 7 = 14,7 x 2 = 14, 2 x 8 = 16, 8 x 2 = 16, 2 x 9 = 18, 9 x 2 = 18, 2 x 10 = 20, 10 x 2 = 20, 3 x 3 = 9, 3 x 4 = 12, 4 x 3 = 12, 3 x 6 = 18, 6 x 3 = 18, 3 x 7 = 21, 7 x 3 = 21, 3 x 8 = 24, 8 x 3 = 24, 3 x 9 = 27, 9 x 3 = 27, 4 x 4 = 16.

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

Question 1.
Each has a value of 7.

Unit form: 5 ____cubes___
Facts: 5 × ___7___ = ___7___ × 5
Total = ___35___

The unit form is 5 cubes.
5 x 7 = 35.

Explanation:
In the above-given question,
given that,
Each cube has a value of 7.
the unit form is 5 cubes.
facts are 5 x 7 = 7 x 5.
total = 35.

Unit form: 6 sevens = __5____ sevens + ____1__ seven
= 35 + __7____
= ___42___
Facts: __6____ × ___7___ = ____42__
___7___ × __6____ = __42_____

The unit form is 6 cubes.
6 x 7 = 42.

Explanation:
In the above-given question,
given that,
Each cube has a value of 7.
the unit form is 6 cubes.
facts are 6 x 7 = 7 x 6.
total = 42.

Question 2.
a. Each dot has a value of 8

Unit form: 5 __dots____
Facts: 5 × ___8___ = __8____× 5
Total = ___40___

The unit form is 5 dots.
5 x 8 = 40.

Explanation:
In the above-given question,
given that,
Each dot has a value of 8.
the unit form is 5 dots.
facts are 5 x 8 = 8 x 5.
total = 40.

b. Use the fact above to find 8 × 6. Show your work using pictures, numbers, or words.

The unit form is 6 dots.
6 x 8 = 48.

Explanation:
In the above-given question,
given that,
Each dot has a value of 8.
the unit form is 6 dots.
facts are 6 x 8 = 8 x 6.
total = 48.

Question 3.
An author writes 9 pages of her book each week. How many pages does she write in 7 weeks? Use a fives fact to solve.

The author writes in 7 weeks is 7 x 9 = 63.

Explanation:
In the above-given question,
given that,
An author writes 9 pages of her book each week.
the author writes in 7 weeks is 7 x 9 = 63.
facts are 7 x 9 = 9 x 7.
total = 63.

Question 4.
Mrs. Gonzalez buys a total of 32 crayons for her classroom. Each pack contains 8 crayons. How many packs of crayons does Mrs. Gonzalez buy?

The number of crayons of packs = 4.

Explanation:
In the above-given question,
given that,
Mrs. Gonzalez buys a total of 32 crayons for her classroom.
Each pack contains 8 crayons.
8 x 4 = 32.
so the number of packs of crayons = 4.

Question 5.
Hannah has $500. She buys a camera for$435 and 4 other items for $9 each. Now Hannah wants to buy speakers for$50. Does she have enough money to buy the speakers? Explain

No, she has not enough money to buy the speakers.

Explanation:
In the above-given question,
given that,
Hannah has $500. she buys a camera for$435 and 4 other items for $9 each. 435 + 9 + 9 + 9 + 9. 435 + 36. 471. 471 + 50 = 521. so she does not have enough money to buy the speakers. ### Eureka Math Grade 3 Module 3 Lesson 2 Problem Set Answer Key Use a fives fact to help you solve 7 × 6. Show your work using pictures, numbers, or words. Answer: The unit form is 6 cubes. 6 x 7 = 42. Explanation: In the above-given question, given that, Each cube has a value of 7. the unit form is 6 cubes. facts are 6 x 7 = 7 x 6. total = 42. ### Eureka Math Grade 3 Module 3 Lesson 2 Homework Answer Key Question 1. Each has a value of 9. Unit form: ___5 cubes_____ Facts: 5 × ___9___ = ___9___ × 5 Total = ___45___ Answer: The unit form is 5 cubes. 5 x 9 = 45. Explanation: In the above-given question, given that, Each cube has a value of 9. the unit form is 5 cubes. facts are 5 x 9 = 9 x 5. total = 45. Unit form: 6 nines = ___5___ nines + ___1___ nine = 45 + __9____ = ___54___ Facts: ___6___ × ___9___ = __54____ ___9___ × __6____ = ___54____ Answer: The unit form is 6 cubes. 6 x 9 = 54. Explanation: In the above-given question, given that, Each cube has a value of 9. the unit form is 6 cubes. facts are 6 x 9 = 9 x 6. total = 54. Question 2. There are 6 blades on each windmill. How many total blades are on 7 windmills? Use a fives fact to solve. Answer: The total blades on 7 windmills = 42. 7 x 6 = 42. Explanation: In the above-given question, given that, There are 6 blades on each windmill. the total blades on 7 windmills. facts are 7 x 6 = 6 x 7. total = 42. Question 3. Juanita organizes her magazines into 3 equal piles. She has a total of 18 magazines. How many magazines are in each pile? Answer: The number of magazines in each pile = 6. Explanation: In the above-given question, given that, Juanita organizes her magazines into 3 equal piles. She has a total of 18 magazines. 6 x 3 = 18. so the number of magazines in each pile = 6. Question 4. Markuo spends$27 on some plants. Each plant costs $9. How many plants does he buy? Answer: The number of plants he buys = 3. Explanation: In the above-given question, given that, Markuo spends$27 on some plants.
Each plant costs \$9.
9 x 3 = 27.
so the number of plants he buys = 3.

Students of Grade 3 Module 3 can get a strong foundation on mathematics concepts by referring to the Eureka Math Book. It was developed by highly professional mathematics educators and the solutions prepared by them are in a concise manner for easy grasping. To achieve high scores in Grade 3, students need to solve all questions and exercises included in Eureka’s Grade 3 Textbook.

## Engage NY Eureka Math 3rd Grade Module 3 Lesson 4 Answer Key

So teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Eureka’s primary school Grade 3 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that.

### Eureka Math Grade 3 Module 3 Lesson 4 Problem Set Answer Key

Question 1.

The missing numbers are 12, 24, 42, and 54.

Explanation:
In the above-given question,
given that,
the numbers are 6, 18, 30, 36, 48, and 60.
the missing numbers are 12, 24, 42, and 54.

Question 2.
Count by six to fill in the blanks below.

6, __12_____, ___18____, ___24____

The missing numbers are 12, 18, and 24.

Explanation:
In the above-given question,
given that,
the table of 6.
count by six.
so the missing numbers are 12, 18, and 24.

Complete the multiplication equation that represents the final number in your count-by.

6 × ___4____ = ___24____

6 x 4 = 24.

Explanation:
In the above-given question,
given that,
the table of 6 x 4 = 24.
so the missing number is 6 x 4 = 24.

Complete the division equation that represents your count-by.

____24___ ÷ 6 = ___4____

24 / 6 = 4.

Explanation:
In the above-given question,
given that,
24 / 6 = 4.
6 x 4 = 24.

Question 3.
Count by six to fill in the blanks below.

6     , ___12___, ___18___, ___24___, ___30___, __36___, __42___

The missing numbers are 12, 18, 24, 30, 36, and 42.

Explanation:
In the above-given question,
given that,
the numbers are 6, 12, 18, 24, 30, 36, and 42.
so the missing numbers are 12, 18, 24, 30, 36, and 42.

Complete the multiplication equation that represents the final number in your count-by.

6 × ___7____ = ___42____

6 x 7 = 42.

Explanation:
In the above-given question,
given that,
the numbers are 6 x 7 = 42.
6 x 7 = 42.

Complete the division equation that represents your count-by.

___42____ ÷ 6 = ___7____

42 / 6 = 7.

Explanation:
In the above-given question,
given that,
42 / 6 = 7.
6 x 7 = 42.

Question 4.
Mrs. Byrne’s class skip-counts by six for a group counting activity. When she points up, they count up by six, and when she points down, they count down by six. The arrows show when she changes direction.

a. Fill in the blanks below to show the group counting answers.
↑ 0, 6, __12___, 18, __24___↓ ___18__, 12 ↑ __18___, 24, 30, __36___ ↓ 30, 24, ___18__ ↑ 24, __30___, 36, __42___, 48

The missing numbers are 12, 24, 18, 18, 36, 18, 30, and 42.

Explanation:
In the above-given question,
given that,
The numbers are 0, 6, 12, 18, 24, 18, 12, 18, 24, 30, 36, 30, 24, 18, 24, 30, 36, 42, and 48.
the missing numbers are 12, 24, 18, 36, 18, 30, and 42.

b. Mrs. Byrne says the last number that the class counts is the product of 6 and another number. Write a multiplication sentence and a division sentence to show she’s right.
6 × ____8___ = 48
48 ÷ 6 = ___8____

6 x 8 = 48.
48 / 6 = 8.

Explanation:
In the above-given question,
given that,
Mrs. Byme says the last number that the class counts is the product of 6 and another number.
6 x 8 = 48.
48 / 6 = 8.

Question 5.
Julie counts by six to solve 6 × 7. She says the answer is 36. Is she right? Explain your answer.

No, she was not correct.
6 x 7 = 42.

Explanation:
In the above-given question,
given that,
Julie counts by six to solve 6 x 7.
she says the answer is 36.
no she was not correct.
6 x 7 = 42.

### Eureka Math Grade 3 Module 3 Lesson 4 Exit Ticket Answer Key

Question 1.
Sylvia solves 6 × 9 by adding 48 + 6. Show how Sylvia breaks apart and bonds her numbers to complete the ten. Then, solve.

6 x 9 = 54.
48 + 6 = 54.

Explanation:
In the above-given question,
given that,
Sylvia solves 6 x 9 by adding 48 + 6.
10 + 10 + 10 + 10 + 10 + 4 = 54.
6 x 9 = 54.
48 + 6 = 54.

Question 2.
Skip-count by six to solve the following:
a. 8 × 6 = ___48___

8 x 6 = 48.

Explanation:
In the above-given question,
given that,
6 x 8 = 48.
8 x 6 = 48.

b. 54 ÷ 6 = ____9__

54 / 6 = 9.

Explanation:
In the above-given question,
given that,
6 x 9 = 54.
54 / 6 = 9.

### Eureka Math Grade 3 Module 3 Lesson 4 Homework Answer Key

Question 1.
Use number bonds to help you skip-count by six by either making a ten or adding to the ones.

10 + 8 = 18, 30 + 6 = 36, 10 + 20 = 30, 30 + 12 = 42, 40 + 8 = 48, 40 + 14 = 54, 50 + 10 = 60.

Explanation:
In the above-given question,
given that,
the numbers are 10 + 8 = 18.
30 + 6 = 36.
10 + 20 = 30.
30 + 12 = 42.
40 + 8 = 48.
40 + 14 = 54.
50 + 10 = 60.

Question 2.
Count by six to fill in the blanks below.
6, ___12____, ___18____, ___24____, ___30____

The missing numbers are 6, 12, 18, 24, and 30.

Explanation:
In the above-given question,
given that,
the tables of 6.
6, 12, 18, 24, and 30.
so the missing numbers are 6, 12, 18, 24, and 30.

Complete the multiplication equation that represents the final number in your count-by.

6 × ___5____ = ___30____

6 x 5 = 30.

Explanation:
In the above-given question,
given that,
the tables of 6.
6 x 5 = 30.

Complete the division equation that represents your count-by.

__30_____ ÷ 6 = ___5____

30 / 6 = 5.

Explanation:
In the above-given question,
given that,
6 x 5 = 30.
30 / 6 = 5.

Question 3.
Count by six to fill in the blanks below.
6, __12____, __18____, ___24___, __30____, __36____

The missing numbers are 12, 18, 24, 30, and 36.

Explanation:
In the above-given question,
given that,
the numbers are 6, 12, 18, 24, 30, and 36.
so the missing numbers are 12, 18, 24, 30, and 36.

Complete the multiplication equation that represents the final number in your count-by.

6 × ___6____ = ___36____

6 x 6 = 36.

Explanation:
In the above-given question,
given that,
the table of 6.
6 x 6 = 36.

Complete the division equation that represents your count-by.

____36___ ÷ 6 = ___6____

36 / 6 = 6.

Explanation:
In the above-given question,
given that,
6 x 6 = 36.
36 / 6 = 6.

Question 4.
Count by six to solve 48 ÷ 6. Show your work below.

48 / 6 = 8.

Explanation:
In the above-given question,
given that,
6 x 8 = 48.
48 / 6 = 8.

Practice using the Eureka Math 3rd Grade Module 3 Answers to track your progress and identify areas of need. For your comfort, we have organized the Topicwise Eureka Math Grade 3 Module 3 Solutions in an efficient manner. Use whichever topic you would like to prepare by simply clicking on the quick links available. EngageNY Math Grade 3 Module 3 Answer Key helps you gain a complete body of knowledge, not just a discrete set of skills.

Eureka Math Grade 3 Module 3 Multiplication and Division with Units of 0, 1, 6–9, and Multiples of 10

Eureka Math Grade 3 Module 3 Topic A The Properties of Multiplication and Division

Eureka Math 3rd Grade Module 3 Topic B Multiplication and Division Using Units of 6 and 7

Engage NY Math 3rd Grade Module 3 Topic C Multiplication and Division Using Units up to 8