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# A rectangle was altered by increasing its length by 10 percent and decreasing it...

### Question

A rectangle was altered by increasing its length by 10 percent and decreasing its width by p percent. If these alterations decreased the area of the rectangle by 12 percent, what is the value of p ?

### Options

A)
12
B)
15
C)
20
D)
22

Choice C is correct. Let l and w be the length and width, respectively of the original rectangle. The area of the original rectangle is A = lw. The rectangle is altered by increasing its length by 10 percent and decreasing its width by p percent; thus, the length of the altered rectangle is 1.1l, and the width of the altered rectangle is $$(1 - \frac{p}{100})w$$. The alterations decrease the area by 12 percent, so the area of the altered rectangle is (1 - 0.12)A = 0.88A. The area of the altered rectangle is the product of its length and width, so 0.88A = $$(1.1l)(1 - \frac{p}{100})w$$. Since A = lw, this last equation can be rewritten as 0.88A = (1.1)$$(1 - \frac{p}{100})lw$$ = (1.1)$$(1 - \frac{p}{100})$$A, from which it follows that 0.88 = (1.1), or 0.8 = (1.1)$$(1 - \frac{p}{100})$$, or 0.8 = $$(1 - \frac{p}{100})$$. Therefore,$$\frac{p}{100}$$ = 0.2, and so the value of p is 20.