A designer enlarged both the length and the width of a rectangular carpet by 50 percent. The new carpet was too large so the designer was asked to reduce its length and its width by 20 percent. By what percent was the area of the final carpet design greater than the area of the original design?
The answer is 44%
I checked and the answer is definitely 44%. I am guessing you have come here from RSM???
any one cam help me
I can tell you why no math tutor has replied.
#1 Get rid of the all-caps.
#2 Put your subject in the School Subject box instead of all this begging.
And most of all, #3 -- Follow the directions on the Post a New Question page.
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To solve this problem, let's assume that the original length of the rectangular carpet is L and the original width is W.
First, the designer enlarges both the length and width by 50 percent. This means the new length is 1.5L (L + 50% of L) and the new width is 1.5W (W + 50% of W).
Next, the designer is asked to reduce the length and width of the enlarged carpet by 20 percent. This means the final length is 0.8(1.5L) and the final width is 0.8(1.5W).
To find the area of each design, we multiply the length and width. Therefore, the area of the original design is LW, and the area of the final design is 0.8(1.5L) * 0.8(1.5W).
Now, let's calculate the area of the final design:
Area of the final design = 0.8(1.5L) * 0.8(1.5W)
= 0.64(2.25)LW
= 1.44LW
To find the percentage difference in the area, we subtract the area of the original design from the area of the final design and divide by the area of the original design. Then we multiply by 100% to get the percentage:
Percentage difference = [(Area of final design - Area of original design) / Area of original design] * 100%
= [(1.44LW - LW) / LW] * 100%
= (0.44LW / LW) * 100%
= 44%
Therefore, the area of the final carpet design is 44% greater than the area of the original design.