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# The area of square region S is what percent greater than

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Intern
Joined: 08 Feb 2017
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The area of square region S is what percent greater than  [#permalink]

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15 Apr 2017, 17:53
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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.
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Re: The area of square region S is what percent greater than  [#permalink]

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15 Apr 2017, 19:12
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
D.

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
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Re: The area of square region S is what percent greater than  [#permalink]

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23 May 2017, 01:03
rohit8865 wrote:
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
D.

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

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!
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Re: The area of square region S is what percent greater than  [#permalink]

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23 May 2017, 01:30
1
michaelkalend wrote:
rohit8865 wrote:
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
D.

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

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!

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)
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Re: The area of square region S is what percent greater than  [#permalink]

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29 Jul 2017, 21:32
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
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The area of square region S is what percent greater than  [#permalink]

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05 Sep 2017, 18:00
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
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
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Re: The area of square region S is what percent greater than  [#permalink]

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04 Oct 2018, 04:03
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Re: The area of square region S is what percent greater than  [#permalink]

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04 Oct 2018, 05:42
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
D.

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.
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Re: The area of square region S is what percent greater than  [#permalink]

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04 Oct 2018, 21:33
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
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Re: The area of square region S is what percent greater than  [#permalink]

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07 Oct 2018, 19:23
1
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.

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Re: The area of square region S is what percent greater than  [#permalink]

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07 Oct 2018, 21:34
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
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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Re: The area of square region S is what percent greater than   [#permalink] 07 Oct 2018, 21:34
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