Common Core: 7th Grade Math : Ratios & Proportional Relationships

Study concepts, example questions & explanations for Common Core: 7th Grade Math

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Example Questions

Example Question #51 : Ratios & Proportional Relationships

A baker can decorate  of a wedding cake in  of an hour. If the baker continues this rate, how much of the wedding cake can he decorate per hour? 

Possible Answers:

Correct answer:

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have the portion of the cake decorated, , divided by hours, :

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve. 

Therefore:

The baker can decorate  of the wedding cake per hour. 

Example Question #51 : Grade 7

A baker can decorate  of a wedding cake in  of an hour. If the baker continues this rate, how much of the wedding cake can he decorate per hour? 

Possible Answers:

Correct answer:

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have the portion of the cake decorated, , divided by hours, :

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve. 

Therefore:

The baker can decorate  of the wedding cake per hour. 

Example Question #52 : Ratios & Proportional Relationships

A painter can paint  of a house in  of an hour. If he continues this rate, how much of the house can he paint per hour? 

Possible Answers:

Correct answer:

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have portion of the house painted, , divided by hours, :

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve. 

Therefore:

The painter can paint  of a house per hour. 

Example Question #53 : Ratios & Proportional Relationships

A painter can paint  of a house in  of an hour. If he continues this rate, how much of the house can he paint per hour? 

Possible Answers:

Correct answer:

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have portion of the house painted, , divided by hours, :

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve. 

Therefore:

The painter can paint  of a house per hour. 

Example Question #54 : Ratios & Proportional Relationships

A painter can paint  of a house in  of an hour. If he continues this rate, how much of the house can he paint per hour? 

Possible Answers:

Correct answer:

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have portion of the house painted, , divided by hours, :

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve. 

Therefore:

The painter can paint  of a house per hour. 

Example Question #55 : Ratios & Proportional Relationships

A painter can paint  of a house in  of an hour. If he continues this rate, how much of the house can he paint per hour? 

Possible Answers:

Correct answer:

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have portion of the house painted, , divided by hours, :

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve. 

Therefore:

The painter can paint  of a house per hour. 

Example Question #56 : Ratios & Proportional Relationships

A painter can paint  of a house in  of an hour. If he continues this rate, how much of the house can he paint per hour? 

Possible Answers:

Correct answer:

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have portion of the house painted, , divided by hours, :

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve. 

Therefore:

The painter can paint  of a house per hour. 

Example Question #57 : Ratios & Proportional Relationships

A landscaper can mow  of a yard in  of an hour. If he continues at this rate, how many yards can the landscaper mow per hour? 

Possible Answers:

Correct answer:

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have yards, , divided by hours, :

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve. 

Therefore:

The landscaper can mow  yards per hour. 

Example Question #58 : Ratios & Proportional Relationships

A landscaper can mow  of a yard in  of an hour. If he continues at this rate, how many yards can the landscaper mow per hour? 

Possible Answers:

Correct answer:

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have yards, , divided by hours, :

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve. 

Therefore:

The landscaper can mow  yards per hour. 

Example Question #59 : Ratios & Proportional Relationships

A landscaper can mow  of a yard in  of an hour. If he continues at this rate, how many yards can the landscaper mow per hour? 

Possible Answers:

Correct answer:

Explanation:

The phrase "per hour" gives us a clue that we are going to divide. In this problem, we can replace the word "per" with a division sign; therefore, we will have yards, , divided by hours, :

Remember that when we divide fractions, we can simply multiply by the reciprocal of the denominator to solve. 

Therefore:

The landscaper can mow  yards per hour. 

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