The area of square region S is what percent greater than

If a number 'x' is A% greater than another number 'y', it means:
x = A% greater than y
x = y + A% of y
x = y + A/100 * y

Taking y common,
x = y (1 + A/100)

we can take values for the sides of the square and check if the percentage remains constant throughout. Therefore statement 1 is sufficient. Statement 2 gives us the values so that's also sufficient. Answer is D

So this is actually a bit trickier than a standard 500 question but that's okay because the directory is not always accurate

St 1

Actually, if you just use 2 simple examples and apply the formula

new-old/old

You will find that the percentage increase in the area of a 1 x 1 and 2 x 2 square is exactly the same

St 2

clearly yes because we can just calculate the area and compare

D

Bunuel, VeritasKarishma, ScottTargetTestPrep please share your approach here. Thanks.

Both regions are squares so their sides will be equal to each other. Area will be side^2.

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

Since side is 10% greater, area will be 21% greater. If you want to see how,

\(SideS = SideT * (1 + 10/100)\)

So \(AreaS = (SideS)^2 = (SideT * 1.1)^2 = 1.21 * SideT^2 = (1 + 21/100) * AreaT^2\)

So AreaS is 21% more than AreaT.

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

You know that you can calculate the area of each here using Area = Side^2 and can find the required percentage.

Hence each statement alone is sufficient.
Answer (D)

As both the figures are squares
So any statement which provides relation between sides of diff squares will allow us to calculate

Both statements dothis So D is the answer

If we know the side length of each square, then we can determine the percent greater that the area of square S is, compared to the square area of square T. For example, if the side length of square T is 10, and that of square S is 11, then the area of square T is 100 and that of square S is 121. So the area of square S is (121 - 100)/100 x 100 = 21 percent greater than that of square T.

Statement One Alone:

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

If we let the side length of square T = 10, then the side length of square S = 11. From the stem analysis, we see that the area of square S is 21 percent greater than the area of square T.

Statement Two Alone:

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

Since we know the side lengths of both squares, we can determine the answer to the question.

Answer: D

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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