Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 6 Quiz Answer Key.

## Texas Go Math Grade 6 Module 6 Quiz Answer Key

**Texas Go Math Grade 6 Module 6 Ready to Go On? Answer Key**

**6.1 Multiplying Integers**

**Find each product.**

Question 1.

(- 2) (3) ______________

Answer:

First, determine if the product wilt be positive or negative

Since – 2 is negative and 3 is positive (they have opposite signs), the product will be negative.

Then, multiply absolute values of given integers:

|- 2| = 2 and |3| = 3

2 Ã— 3 = 6

The result is 6 or – 6, depending on signs of given integers.

Product is negative so final result is – 6.

Result is – 6.

Find the sign of the product, multiply absolute values of integers and add minus if the sign of product is minus (-).

Question 2.

(- 5) (- 7) _____________

Answer:

First, determine if the product wilt be positive or negative

Since – 5 is negative and – 7 is negative (they have same signs), the product will be positive.

Then, multiply absolute values of given integers:

|- 5| = 5 and |- 7| = 7

5 Ã— 7 = 35

The result is 35 or – 35, depending on signs of given integers.

Product is negative so final result is 35.

Result is 35.

Find the sign of the product, multiply absolute values of integers and add minus if the sign of product is minus (-).

**Go Math Grade 6 Module 6 Answer Key Question 3.**

(8) (- 11) ______________

Answer:

First, determine if the product will be positive or negative

Since 8 is positive and – 11 is negative (they have opposite signs), the product will be negative.

Then, multiply absolute values of given integers:

|8| = 8 and |- 11| = 11

8 Ã— 11 = 88

The result is 88 or – 88, depending on signs of given integers.

Product is negative so final result is – 88

Result is – 88.

Find the sign of the product, multiply absolute values of integers and add minus if the sign of product is minus (-).

Question 4.

(- 3) (2) (- 2) ______________

Answer:

First, calculate the product:

(-3) (2)

Determine if the product will be positive or negative

Since – 3 is negative and 2 is positive (they have opposite signs, the product will be negative.

Then, multiply absolute values of given integers:

|- 3| = 3 and |2| = 2

3 Ã— 2 = 6

The result is 6 or – 6, depending on signs of given integers.

Product is negative so result is – 6.

Now, find the product:

(- 6) (- 2)

Determine if the product will be positive or negative.

Since – 6 is negative and – 2 is negative (they have the same sign), the product will be positive.

Then, multiply absolute values of given integers:

|- 6| = 6 and |- 2| = 2

6 Ã— 2 = 12

The result is 12 or – 12, depending on signs of given integers.

Product is positive so final result is 12.

Therefore:

(- 3) (2) (- 2) = 12

First, calculate the product (- 3)(2) and then multiply it with – 2.

Find the sign of the product, multiply absolute values of integers and add minus if the sign of product is minus (-).

Question 5.

The temperature dropped 3 Â°C every hour for 5 hours.

Write an integer that represents the change Â¡n temperature. ______________

Answer:

Use negative integer to represent how many degrees the temperature dropped each hour.

The expression you get is:

5 (- 3)

First, determine if the product will be positive or negative

Since 5 is positive and – 3 is negative (they have opposite signs), the product will be negative.

Then, multiply absolute values of given integers:

|5| = 5 and |- 3| = 3

5 Ã— 3 = 15

The result is 15 or – 15, depending on signs of given integers.

Product is negative so final result is – 15.

The change in temperature is – 15 degrees.

Write the expression that represents the change in temperature and find the product using rules for multiplying integers.

**6.2 Dividing Integers**

**Find each quotient.**

Question 6.

\(\frac{-63}{7}\)

Answer:

First determine if the quotient will be positive or negative

Since – 63 is negative and 7 is positive (they have opposite signs), the quotient will be negative.

Divide given integers:

\(\frac{-63}{7}\) = – 9

Result is – 9

Find the sign of the quotient (opposite signs of integers then -, the same sign of integers then +) and divide given integers.

Question 7.

\(\frac{-15}{-3}\)

Answer:

First determine if the quotient will be positive or negative

Since – 15 is negative and – 3 is negative (they have same signs), the quotient will be positive.

Divide given integers:

\(\frac{-15}{-3}\) = 5

Result is 5

Find the sign of the quotient (opposite signs of integers then -, the same sign of integers then +) and divide given integers.

**Module 6 Test Answers Go Math Grade 6 Question 8.**

\(\frac{0}{-15}\)

Answer:

First, determine if the quotient wilt be positive or negative

Since dividend is 0 and divisor – 15 is not 0, the quotient is 0.

The result of division is:

\(\frac{0}{-15}\) = 0

Result is 0.

Find the sign of the quotient (opposite signs of integers then -, the same sign of integers then +) and divide given integers.

Question 9.

\(\frac{96}{-12}\)

Answer:

First determine if the quotient will be positive or negative

Since 96 is positive and – 12 is negative (they have opposite signs), the quotient will be negative.

Divide given integers:

\(\frac{96}{-12}\) = – 8

Result is – 8.

Find the sign of the quotient (opposite signs of integers then -, the same sign of integers then +) and divide given integers.

Question 10.

An elephant at the zoo lost 24 pounds over 6 months. The elephant lost the same amount of weight each month. Write an integer that represents the change in the elephantâ€™s weight each month.

Answer:

Use negative integer to represent the change in elephantâ€™s weight

The expression you get is:

\(\frac{- 24}{6}\)

First determine if the quotient will be positive or negative

Since – 24 is negative and 6 is positive (they have opposite signs), the quotient wilt be negative.

Divide given integers:

\(\frac{- 24}{6}\) = – 4

The change in weight each month is – 4 pounds.

Write the expression that represents the change in elephantâ€™s weight and find the quotient using rules for dividing integers.

**6.3 Applying Integer Operations**

**Evaluate each expression.**

Question 11.

(- 4) (5) + 8 _____________

Answer:

The expression you need to evaluate is:

(- 4)(5) + 8

First find the product using rules for multiplying integers:

– 20 + 8

Then, use rules for adding integers to find the sum:

– 20 + 8 = – 12

Therefore:

(- 4)(5) + 8 = – 12

The value of the expression is – 12

Use rules for multiplying integers and rules for adding integers to find that value.

Question 12.

(- 3) (-6) – 7 _____________

Answer:

The expression you need to evaluate is:

(- 3) (- 6) – 7

First find the product using rules for multiplying integers:

18 – 7

Then, use rules for subtracting integers to find the difference:

18 – 7 = 11

Therefore:

(- 3) (- 6) – 7 = 11

The value of the expression is 11.

Use rules for multiplying integers and rules for subtracting integers to find the value.

**Go Math Module 6 Review Answer Key Grade 6 Question 13.**

\(\frac{- 27}{9}\) – 11 _____________

Answer:

The expression you need to evaluate is:

\(\frac{-27}{9}\) – 11

First find the quotient using rules for dividing integers:

– 3 – 11

Then, use rules for subtracting integers to find the difference:

– 3 – 11 = – 14

Therefore:

\(\frac{-27}{9}\) – 11 = – 14

The value of the expression is – 14.

Use rules for dividing integers and rules for subtracting integers to find the value.

Question 14.

\(\frac{-24}{-3}\) – (- 2) _____________

Answer:

The expression you need to evaluate is:

\(\frac{-24}{-3}\) – (- 2)

First find the quotient using rules for dividing integers:

8 – (- 2)

Then, use rules for subtracting integers to find the difference:

8 – (- 2) = 10

Therefore:

\(\frac{-24}{-3}\) – (- 2) = 10

The value of the expression is 10.

Use rules for dividing integers and rules for subtracting integers to find the value.

**Essential Question**

Question 15.

Write and solve a real-world problem that can be represented by the expression (- 3) (5) + 10.

Answer:

Jim is at the entrance to the cave. He descends 5 times, each time 3 feet Then, he ascends 10 feet What is his new position relative to the cave entrance

Use negative integer if he descends and positive integer if he ascends.

The expression you get is:

(- 3) (5) + 10

First, find the product using rules for multiplying integers:

– 15 + 10

Then, use rules for adding integers to find the sum

– 15 + 10 = – 5

His new position is 5 feet below the cave entrance.

Jim is at the entrance to the cave. He descends 5 times, each time 3 feet Then, he ascends 10 feet What is his new position relative to the cave entrance?

Jimâ€™s new position is 5 feet below the cave entrance.

Use rules for multiplying integers and rules for adding integers to find his new position.

**Texas Go Math Grade 6 Module 6 Mixed Review Texas Test Prep Answer Key**

**Selected Response**

Question 1.

A diver is at an elevation of – 18 feet relative to sea level. The diver descends to an undersea cave that is 4 times as far from the surface. What is the elevation of the cave?

(A) – 72 feet

(B) – 22 feet

(C) – 18 feet

(D) – 14 feet

Answer:

(A) – 72 feet

Explaination:

The expression that represents elevation of the cave is:

4(- 18)

Calculate that product to find the elevation.

First, determine if the product will be positive or negative

Since 4 is positive and – 18 is negative (they have opposite signs), the product will be negative.

Then, multiply absolute values of given integers:

|4| = 4 and |- 18| = 18

4 Ã— 18 = 72

The result is 72 or – 72, depending on signs of given integers.

Product is negative so final result is – 72.

The elevation of the cave is – 72 feet.

**Module 6 Quiz Answers Go Math Grade 6 Question 2.**

The football team lost 4 yards on 2 plays in a row. Which of the following could represent the change in field position?

(A) – 12 yards

(B) – 8 yards

(C) – 6 yards

(D) – 2 yards

Answer:

(B) – 8 yards

Explaination:

Use negative integer because the team lost yards. The expression you get is:

2(- 4)

First, determine if the product will be positive or negative

Since 2 is positive and – 4 is negative (they have opposite signs), the product will be negative.

Then, multiply absolute values of given integers:

|2| = 2 and |- 4| = 4

2 Ã— 4 = 8

The result is 8 or – 8, depending on signs of given integers.

Product is negative so final result is – 8.

The change in field position is – 8 yards.

Question 3.

Clayton climbed down 50 meters. He climbed down in 10-meter intervals. In how many intervals did Clayton make his climb?

(A) 5

(B) 10

(C) 40

(D) 500

Answer:

(A) 5

Explaination:

Divide total distance by distance of one interval to get number of intervals. The expression you get is:

50 Ã· 10

First, determine if the quotient will be positive or negative

Since 50 is positive and 10 is positive (they have the same sign), the quotient will be positive.

Divide given integers:

50 Ã· 10 = 5

He made his climb in 5 intervals.

Question 4.

Which expression results in a negative answer?

(A) a negative number divided by a negative number

(B) a positive number divided by a negative number

(C) a negative number multiplied by a negative number

(D) a positive number multiplied by a positive number

Answer:

(B) a positive number divided by a negative number

Explaination:

The quotient of two integers with opposite signs is negative so the correct answer is B.

The quotient of two integers with the same sign is positive as well as the product of two integers with the same sign so results of expressions A), C) and D) are positive

Question 5.

Clara played a video game before she left the house to go on a walk. She started with 0 points, lost 6 points 3 times, won 4 points, and then lost 2 points. How many points did she have when she left the house to go on the walk?

(A) – 20

(B) 12

(C) – 16

(D) 20

Answer:

(C) – 16

Explaination:

Subtract integer if she lost points and add integer if she won points

The expression you get is:

3(- 6) + 4 – 2

First, find the product using rules for multiplying integers:

– 18 + 4 – 2

Use rules for adding integers:

– 14 – 2

Use rules for subtracting integers to find the final result:

– 14 – 2 = – 16

Clara had – 16 points when she left the house.

Question 6.

Which expression is equal to

(A) \(\frac{-24}{6}\) – 4

(B) \(\frac{-24}{-6}\) + 4

(C) \(\frac{24}{6}\) + 4

(D) \(\frac{-24}{-6}\) – 4

Answer:

(D) \(\frac{-24}{-6}\) – 4

Explaination:

Calculate value of each expression to see which one is equal to 0.

(A) \(\frac{-24}{6}\) – 4

Find the quotient using rules for dividing integers:

– 4 – 4

Use rules for subtracting integers to find the difference:

– 4 – 4 = – 8

(B) \(\frac{-24}{-6}\) + 4

Find the quotient using rules for dividing integers:

4 + 4

Use rules for adding integers to find the sum:

4 + 4 = 8

(C) \(\frac{24}{6}\) + 4

Find the quotient using rules for dividing integers:

4 + 4

Use rules for adding integers to find the sum:

4 + 4 = 8

(D) \(\frac{-24}{-6}\) – 4

Find the quotient using rules for dividing integers:

4 – 4

Use rules for subtracting integers to find the difference:

4 – 4 = 0

Therefore, the correct answer is D.

Gridded Response

**Grade 6 Module 6 Quiz Answer Key Go Math Question 7.**

Rochelle made three $25 withdrawals and then wrote a check for $100. If she started with $200 in her account, find the total amount she has left in her account in dollars.

Answer:

Use positive integer to represent the amount she had at the beginning and use negative integers to represent the amount she withdrew and the value of the check.

The expression you get is:

200 + 3(- 25) + (- 100)

First find the product using rules for multiplying integers:

200 – 75 + (- 100)

Use rules for subtracting integers:

125 – (- 100)

Use rules for adding integers to find the final result:

125 + (- 100) = 25

Rochelle has left $25 in her account