VJesus12 wrote:The length of each side of square A is increased by 100 percent to make square B. If the length of the side of square B is increased by 50 percent to make square C, by what percent is the area of square C greater than the sum of the areas of squares A and B?
A. 75%
B. 80%
C. 100%
D. 150%
E. 180%
In any square:
Area = s².
Square A:
Let s = 1, implying that the area of A = 1² = 1.
Square B:
If a value increases by 100%, it DOUBLES.
Thus, each side of B is twice that of A.
Since each side of B = 2*1 = 2, the area of B = 2² = 4.
Square C:
Here, s = 50% more than each side of B = 2 + (1/2)(2) = 3.
Thus, the area of C = 3² = 9.
The area of square C is what percent greater than the sum of the areas of squares A and B?
Since C=9 and A+B = 1+4 = 5, we get:
Percent increase from 5 to 9 = Difference/Smaller * 100 = (9-5)/5 * 100 = 80.
The correct answer is
B.
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