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Bunuel
A square of side x has the length of each of its sides doubled. Its new area is what percent of its original area?

A. 100
B. 200
C. 300
D. 400
E. 500
Let side length = 2
Area of small square = 4

Length of sides doubles, new side = 4
New area = 16

New area = what percent OF old?
That's the potential trap. "Of" implies multiplication (division).*

Percent of = \((\frac{Part}{Whole}*100)\)

\((\frac{16}{4}*100)=4*100=400\)
percent

Answer D

*"Percent greater than" implies addition (subtraction): \(\frac{New-Old}{Old}*100\)
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Bunuel
A square of side x has the length of each of its sides doubled. Its new area is what percent of its original area?

A. 100
B. 200
C. 300
D. 400
E. 500

Let x = 1 , So 2x = 2

Initial area is 1 sq units and Increased area is 4 sq units

So, New area is \(\frac{4}{1} *100\) = \(400\) % of original area, Answer must be (D)
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Bunuel
A square of side x has the length of each of its sides doubled. Its new area is what percent of its original area?

A. 100
B. 200
C. 300
D. 400
E. 500

We can let the original side of the square be 2, so the area of the square is 4.

The new (doubled) side is 4, so the new area is16.

The new area is 16/4 x 100 = 400 percent of the original area.

Answer: D
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Bunuel
A square of side x has the length of each of its sides doubled. Its new area is what percent of its original area?

A. 100
B. 200
C. 300
D. 400
E. 500
Let original length of each side be 1 unit ; Increased length is 2

So, Original Area is 1 ; Increased Area is 4

Quote:
new area is what percent of its original area?
\(= \frac{4}{1}*100=400\), Answer must be (D) 400
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