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Old July 9th, 2009, 09:35 PM
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Exclamation I Need Help With These 2 Algebraic Word Problems

Hi! I'm in 7th grade and I decided to go to summer school so I could get my credit for Algebra early. I came across these two word problems (well, actually 12 but they are similar to these particular 2) that I do not understand.

Could you please show me in steps how to get the answer? Since summer school is so fast paced, we only have a day to grasp this and I want to make sure I get it.


1. A basketball court is 50ft wide. The key is 16ft wide. What is the distance from the sideline to the key?


2. It takes 64in of ribbon to make a bow and wrap the ribbon around the box. The bow takes 30in of ribbon. The width of the box is 12in. What is the height of the box?


Thanks to anyone who can help me!

EDIT!: Wait I Think I figured it out! can you tell me if it is right?

1. The distance is half of the width - half of the key
= 50/2 - 16/2 = 25 - 8 = 17 feet

2. The perimeter of the box is the sum of its sides (that is being wrapped by the ribbon) = 2 x width + 2 x height
34 = 2 x 12 + 2 x height
34 - 24 = 2 x height
height = 10/2
height = 5 inches

is that right?

Last edited by Clarabella; July 9th, 2009 at 09:42 PM. Reason: Solution
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Old July 9th, 2009, 09:44 PM
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Quote:
Originally Posted by Clarabella View Post
Hi! I'm in 7th grade and I decided to go to summer school so I could get my credit for Algebra early. I came across these two word problems (well, actually 12 but they are similar to these particular 2) that I do not understand.

Could you please show me in steps how to get the answer? Since summer school is so fast paced, we only have a day to grasp this and I want to make sure I get it.


1. A basketball court is 50ft wide. The key is 16ft wide. What is the distance from the sideline to the key?


2. It takes 64in of ribbon to make a bow and wrap the ribbon around the box. The bow takes 30in of ribbon. The width of the box is 12in. What is the height of the box?


Thanks to anyone who can help me!

EDIT!: Wait I Think I figured it out! can you tell me if it is right?

1. The distance is half of the width - half of the key
= 50/2 - 16/2 = 25 - 8 = 17 feet

2. The perimeter of the box is the sum of its sides (that is being wrapped by the ribbon) = 2 x width + 2 x height
34 = 2 x 12 + 2 x height
34 - 24 = 2 x height
height = 10/2
height = 5 inches

is that right?
These are both correct.
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