SOLUTION: The floor of a conference hall can be covered completely with tiles.Its length is 10m longer than its width.The area of the floor is grater than 600m² How would you represent t

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Question 1170073: The floor of a conference hall can be covered completely with tiles.Its length is 10m longer than its width.The area of the floor is grater than 600m²
How would you represent the width of the floor?how about its length?
What is mathematical sentence would represent the given situation?
What are the possible dimension of the floor?
What are the possible areas of the floor?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.

Hello,

        my post is of two parts.

Part 1 

    The last question in your post is "What are the possible areas of the floor?"

    But the condition just said "The area of the floor is grater than 600m²".

    Then WHY do you ask AGAIN about it ?


    In Math problems, written professionally, it is not accepted to repeat the same thought twice.


Part 2

Yesterday late in the evening I solved 3 (three) TWIN problems to it.

Their links are

https://www.algebra.com/algebra/homework/coordinate/word/Linear_Equations_And_Systems_Word_Problems.faq.question.1169994.html

https://www.algebra.com/algebra/homework/coordinate/word/Linear_Equations_And_Systems_Word_Problems.faq.question.1169993.html

https://www.algebra.com/algebra/homework/coordinate/word/Linear_Equations_And_Systems_Word_Problems.faq.question.1169995.html


It is more than enough to learn the subject.


Happy learning (!)



Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

W = width
W+10 = length, since it is 10 meters longer than the width

Area = Length*Width
Area = (W+10)*W
Area = W^2+10W

The area is greater than 600 square meters, so,

Area > 600
W^2+10W > 600
W^2+10W-600 > 0
(W+30)(W-20) > 0

There are two ways to have the factors (W+30) and (W-20) multiply to a positive value

Case 1: Both factors (W+30) and (W-20) are positive
Case 2: Both (W+30) and (W-20) are negative


Let's focus on case 1
If W+30 > 0, then W > -30
If W-20 > 0, then W > 20
If both are the case, then W > 20 which is what both intervals have in common.

Now move onto case 2
If W+30 < 0, then W < -30
If W-20 < 0, then W < 20
Those two intervals overlap over W < -30 since something like W = -35 is in both intervals.
However, we can't have negative widths. So case 2 in its entirety is going to be ignored.


Therefore, the only possible values of W are ones such that W > 20.
In words: the width must be larger than 20.

Let's say the width was W = 25 meters
This makes the length to be W+10 = 25+10 = 35 meters
The resulting area is length*width = 35*25 = 875 square meters, which exceeds 600 square meters.

One possible dimension of the floor is 25 meters by 35 meters, which has an area of 875 square meters.

There are infinitely more possible answers since all that matters is W > 20.
So you could have W = 30, W = 40, etc
There is no upper bound. Though in a realistic sense, you'll run out of room eventually. At the same time, we can still pick infinitely many values between two fixed points. For example, there are infinitely many numbers between W = 30 and W = 40 if we include all real numbers, and not just whole numbers.

If your teacher made a restriction like "W is whole number between 21 and 30", then we would have finitely many values of W and therefore finitely many answers. However, this isn't the case here.


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