The length of each side of square A is increased by 100 percent to make square B.
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 25 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?
Answer/Solution
45%
Steps/Work
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 = 20
Area of B = 20^2 = 400
Since , length of the side of square B is increased by 25 percent to make square C
length of each side of square C= 1.25*20 = 25
Area of C= 25^2 = 625
Difference in areas of C and cummulative areas of A and B = 625 -(400+100) = 225
percent is the area of square C greater than the sum of the areas of squares A and B = (225/500) * 100 % = 45%
Answer D
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 = 20
Area of B = 20^2 = 400
Since , length of the side of square B is increased by 25 percent to make square C
length of each side of square C= 1.25*20 = 25
Area of C= 25^2 = 625
Difference in areas of C and cummulative areas of A and B = 625 -(400+100) = 225
percent is the area of square C greater than the sum of the areas of squares A and B = (225/500) * 100 % = 45%
Answer D