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

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. 

Choice A is incorrect and may be the result of confusing the 12 percent decrease in area with the percent decrease in width. Choice B is incorrect because decreasing the width by 15 percent results in a 6.5 percent decrease in area, not a 12 percent decrease. Choice D is incorrect and may be the result of adding the percents given in the question (10 + 12).