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 40 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? Pick.
A) 75.00%  B) 56.80%  C) 110%  D) 150%  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 56.80%   length of each side of square 56.80% = 2*10= 
<gadget id="calculator">2*10</gadget>
<output>20</output>
20   Area of 56.80%= 20^2 = 400   Since , length of the side of square 56.80% is increased by 40 percent to make square C   length of each side of square C= 1.4*20= 
<gadget id="calculator">1.4*20</gadget>
<output>28</output>
28   Area of C= 28^2 = 784   Difference in areas of C and cummulative areas of A and 56.80% = 784 -(400+100)= 
<gadget id="calculator">784 -(400+100)</gadget>
<output>284</output>
284   percent is the area of square C greater than the sum of the areas of squares A and 56.80%= (284/500) * 100= 
<gadget id="calculator">(284/500) * 100</gadget>
<output>284/5 = around 56.8</output>
56.80%   Answer 56.80%
<result>B</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 40 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?