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 60 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?
Choose the correct answer.
A) 75%  B) 80%  C) 104.8%  D) 150.5%  E) 180%
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 60 percent to make square 104.8%   length of each side of square 104.8%= 1.6*20= 
<gadget id="calculator">1.6*20</gadget>
<output>32</output>
32   Area of 104.8%= 32^2 = 1024   Difference in areas of 104.8% and cummulative areas of A and B = 1024 -(400+100)= 
<gadget id="calculator">1024 -(400+100)</gadget>
<output>524</output>
524   percent is the area of square 104.8% greater than the sum of the areas of squares A and B= (524/500) * 100= 
<gadget id="calculator">(524/500) * 100</gadget>
<output>524/5 = around 104.8</output>
104.8%   Answer 104.8%
<result>C</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 60 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?