All Common Core: 7th Grade Math Resources
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?
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?
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?
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?
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 #51 : Grade 7
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?
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?
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?
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?
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?
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 #51 : Compute Unit Rates Associated With Ratios Of Fractions: Ccss.Math.Content.7.Rp.A.1
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?
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.