kimbercs wrote:

The area of square region S is what percent greater than the area of square region T?

1) The length of a side of square S is 10% greater than the length of a side of square T.

2) The length of a side of square S is 22 and the length of a side of square T is 20.

OA

OA: D

We have to find \(\frac{Area_{S}-Area_{T}}{Area_{T}}*100\)

1) The length of a side of square S is 10% greater than the length of a side of square T.

Let length of side of square T be \(a\).

\(Area_{T}=a^2\)

Let length of side of square S be \(a+\frac{10}{100}a=\frac{11}{10}a\).

\(Area_{S} =(\frac{11}{10}a)^2=\frac{121}{100}a^2\)

\(\frac{Area_{S}-Area_{T}}{Area_{T}}*100=\frac{\frac{121}{100}a^2-a^2}{a^2}*100=\frac{\frac{121}{100}-1}{1}*100=\frac{121-100}{100}*100=21\)%

Statement \(1\) alone is sufficient.

2) The length of a side of square S is \(22\) and the length of a side of square T is \(20\).

\(Area_{S} = 22^2 =484\)

\(Area_{T} = 20^2 =400\)

\(\frac{Area_{S}-Area_{T}}{Area_{T}}*100=\frac{484-400}{400}*100=\frac{84}{400}*100=21\)%

Statement \(2\) alone is sufficient.

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