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 80 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?	Answers.
A) 75% B) 80% C) 100% D) 150% E) 159.2%
Let length of each side of square A be 10   Area of A = 10^2 = 100   Since , length of each side of square A is increased by 100 percent to make square B   length of each side of square B = 2*10= 
<gadget id="calculator">2*10</gadget>
<output>20</output>
20   Area of B= 20^2 = 400   Since , length of the side of square B is increased by 80 percent to make square C   length of each side of square C= 1.8*20= 
<gadget id="calculator">1.8*20</gadget>
<output>36</output>
36   Area of C= 36^2 = 1296   Difference in areas of C and cummulative areas of A and B = 1296 -(400+100)= 
<gadget id="calculator">1296 -(400+100)</gadget>
<output>796</output>
796   percent is the area of square C greater than the sum of the areas of squares A and B= (796/500) * 100= 
<gadget id="calculator">(796/500) * 100</gadget>
<output>796/5 = around 159.2</output>
159.2%   Answer 159.2%
<result>E</result>
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 80 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?