The area of square region S is what percent greater than

Question

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  D.

rohit8865 · Answer

Let the side of 1st sq =x
then side of 2nd one is = 1.1x
thus thier respective area := x^2 and (1.1x)^2
let area of 2nd sq. be A % greater thus we can calculate as

(1.1x)^2 = x^2 (1+A/100)
so A can be calculated

suff

(2) let area of 2nd sq. be A % greater thus we can calculate as

(22)^2 = 20^2 (1+A/100)
so A can be calculated
(Note:- option 2 is similar to option 1 , as 22 is 10% greater than 20)
suff..

Ans D

michaelkalend · Answer

It is not very clear how you derived on this part:

let area of 2nd sq. be A % greater thus we can calculate as

(1.1x)^2 = x^2 (1+A/100)
so A can be calculated

Kindly elaborate!

amanvermagmat · Answer

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)

longhaul123 · Answer

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

Nunuboy1994 · Answer

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

sadikabid27 · Answer

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

KarishmaB · Answer

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)

push12345 · Answer

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

ScottTargetTestPrep · 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

Princ · Answer

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.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

The area of square region S is what percent greater than

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

The area of square region S is what percent greater than

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards