Answer :
Based on the given information, it appears that there is no combination of length and width that satisfies both the condition of the area being 120 square meters and the length being less than twice the width.
The problem states that the room of a patient is 120 square meters and the length is less than twice the width. To find the length and width of the room, we can set up an equation and solve for the variables.
Let's denote the width of the room as "w" and the length of the room as "l". We know that the area of a rectangle is given by the formula A = length × width.
From the problem, we are given that the area of the room is 120 square meters, so we can write the equation as:
120 = l × w
Additionally, we are told that the length is less than twice the width. In other words, l < 2w.
Now we have a system of equations:
120 = l × w
l < 2w
To find the length and width, we can solve this system of equations.
Let's consider some possible solutions:
Assume the width is 1 meter. If we substitute w = 1 into the equation l < 2w, we get l < 2. This means the length can be any value less than 2. If we choose l = 1.5, then the area of the room is 1.5 × 1 = 1.5 square meters, which is less than 120. Therefore, this solution doesn't work.
Assume the width is 2 meters. If we substitute w = 2 into the equation l < 2w, we get l < 4. Again, the length can be any value less than 4. Let's choose l = 3. Now the area of the room is 3 × 2 = 6 square meters, which is still less than 120. So, this solution also doesn't work
Assume the width is 3 meters. If we substitute w = 3 into the equation l < 2w, we get l < 6. Again, the length can be any value less than 6. Let's choose l = 5. Now the area of the room is 5 × 3 = 15 square meters, which is still less than 120. This solution doesn't work either.
As we can see, none of the solutions we've tried so far satisfy the condition that the area of the room is 120 square meters. Therefore, there doesn't seem to be a valid solution to this problem.
In conclusion, based on the given information, it appears that there is no combination of length and width that satisfies both the condition of the area being 120 square meters and the length being less than twice the width.
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