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# In the quilting pattern shown above, a small square has its vertices o

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Director
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In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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01 Jul 2016, 04:43
6
39
00:00

Difficulty:

25% (medium)

Question Stats:

78% (01:32) correct 22% (01:40) wrong based on 933 sessions

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OG Q 2017 New Question(Book Question: 184)

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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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21 May 2017, 19:17
6
pafrompa wrote:
GMAT198917 wrote:
For statement (1) why can't we assume this is a common triangle?

We know the hypotenuse is 10 and that it is a right triangle. With this information the side lengths of the large triangle should be 6 & 8.

Could someone please elaborate on this a bit? I also put choice one, because I assumed that the triangles formed in the corners were 30,60,90 triangles. Therefore, I ended up picking A because I had just assumed that you can back into x,x\sqrt{3},2x where 2x=10.

You are saying yourself that you assumed that it must be 30-60-90 triangle. Why? Do you have any reasons supporting this claim? No.

So, you assumed with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if $$a^2+b^2=10^2$$ DOES NOT mean that $$a=6$$ and $$b=8$$, certainly this is one of the possibilities but definitely not the only one. In fact $$a^2+b^2=10^2$$ has infinitely many solutions for $$a$$ and $$b$$ and only one of them is $$a=6$$ and $$b=10$$.

Hope it's clear.
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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01 Jul 2016, 07:53
5
2
1 is clearly insufficient.
2. the ratio is given as 1:2 again insufficient.
combining both
using Pythagoras theorem.
10^2=x^2+(2x)^2
100= 5x^2
this should give us x followed by the length of the side of the larger square.
Therefore C
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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01 Jul 2016, 07:10
C

1)side of small sq is given but no info about large sq. Insuff
2)side of small sq not given and info about large sq is there. insuff

suppose side of large sq is 3x so its divided into x and 2x and side of small square is given(which acts as hypotenuse of all4 triangles)

apply pythagos. theorem and get value of x
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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02 Jul 2016, 11:25
1
There is one unknown, that is the side of the larger square.
We need something that gives us the side of the bigger square that is in either a variable form which could be determined or in a direct value form.
Statement 1 does not have any relation to side of bigger square, hence insufficient.
Statement 2 gives a variable relation as, if side of bigger square is 3x, then the triangles whose hypotenuse is given by Statement 1 has sides 2x and x which gives an equation to determine x (4x^2 + X^2 = 10^2).
Therefore both statement are required together to arrive at a definite answer.
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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17 Feb 2017, 04:30
For statement (1) why can't we assume this is a common triangle?

We know the hypotenuse is 10 and that it is a right triangle. With this information the side lengths of the large triangle should be 6 & 8.
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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21 May 2017, 13:29
GMAT198917 wrote:
For statement (1) why can't we assume this is a common triangle?

We know the hypotenuse is 10 and that it is a right triangle. With this information the side lengths of the large triangle should be 6 & 8.

Could someone please elaborate on this a bit? I also put choice one, because I assumed that the triangles formed in the corners were 30,60,90 triangles. Therefore, I ended up picking A because I had just assumed that you can back into x,x\sqrt{3},2x where 2x=10.
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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22 May 2017, 16:15
Bunuel wrote:
pafrompa wrote:
GMAT198917 wrote:
For statement (1) why can't we assume this is a common triangle?

We know the hypotenuse is 10 and that it is a right triangle. With this information the side lengths of the large triangle should be 6 & 8.

Could someone please elaborate on this a bit? I also put choice one, because I assumed that the triangles formed in the corners were 30,60,90 triangles. Therefore, I ended up picking A because I had just assumed that you can back into x,x\sqrt{3},2x where 2x=10.

You are saying yourself that you assumed that it must be 30-60-90 triangle. Why? Do you have any reasons supporting this claim? No.

So, you assumed with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if $$a^2+b^2=10^2$$ DOES NOT mean that $$a=6$$ and $$b=8$$, certainly this is one of the possibilities but definitely not the only one. In fact $$a^2+b^2=10^2$$ has infinitely many solutions for $$a$$ and $$b$$ and only one of them is $$a=6$$ and $$b=10$$.

Hope it's clear.

Clear as day, thank you. I guess sometimes I force myself to try to see things that aren't there with DS. Thanks Bunuel
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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08 Apr 2019, 21:34
@Bunnel,

Why can't we have answer as 'A'
My Intake:

st1: Side length of smaller square is 10 cm.
Can't we say area of smaller square=2*area of larger square
10*10=2 S^2
S^2=50
S=root 50

Is this possible?
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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12 May 2019, 09:21
1
Saumya2403 wrote:
@Bunnel,
Can't we say area of smaller square=2*area of larger square

How?
If it is mentioned in the question then only you can consider it, by default you are not allowed to assume anything.
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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13 May 2019, 04:46
Guys one question. if we know that it is not a right angle , why did u used phythagorean theorem ?
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In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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Updated on: 28 Aug 2019, 17:05
I'm a bit confused.

Bunuel
Are we to assume that the ratio of each side is in line with the dimensions formed by the verticie?

This is the assumed construction
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Originally posted by dcummins on 28 Aug 2019, 17:03.
Last edited by dcummins on 28 Aug 2019, 17:05, edited 1 time in total.
VP
Joined: 14 Feb 2017
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GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
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GMAT 6: 600 Q38 V35
GMAT 7: 710 Q47 V41
GPA: 3
WE: Management Consulting (Consulting)
Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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28 Aug 2019, 17:04
But this is the construction if we randomly split each side.
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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08 Dec 2019, 22:26
1. We have one number - 10 as the hypotenuse and we know that its a square. Nothing can be done with this info

2. Now this statement gives the ratio of the sides. We have a ratio with respect to the larger sqaure and no other info on the larger square.

combining, we'd get the hypotenuse and the side ratios, hence sufficient to figure out the side of the larger square
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In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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10 Jan 2020, 01:38
dcummins wrote:
But this is the construction if we randomly split each side.

I liked the question you asked. Because I had read somewhere that the DS questions do not have the figure drawn to scale. It got me thinking about the possibilities. Then I realized that since the figure inside is a square, we should get consistent lengths of each of the sides which will not be the case if we interpret the statements in the way shown in your figure. So the only possibility is one side being x and the other one being 2x.
In the quilting pattern shown above, a small square has its vertices o   [#permalink] 10 Jan 2020, 01:38
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