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Director  B
Status: I don't stop when I'm Tired,I stop when I'm done
Joined: 11 May 2014
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In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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6
39 00:00

Difficulty:   25% (medium)

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

### HideShow timer Statistics OG Q 2017 New Question(Book Question: 184)

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Math Expert V
Joined: 02 Sep 2009
Posts: 62637
Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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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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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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Senior Manager  Joined: 02 Mar 2012
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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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
Intern  Joined: 06 Jul 2015
Posts: 18
Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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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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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.
Intern  B
Joined: 06 Oct 2015
Posts: 45
Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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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.
Intern  B
Joined: 06 Oct 2015
Posts: 45
Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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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
Intern  B
Joined: 21 Sep 2016
Posts: 35
Location: India
GMAT 1: 430 Q38 V13 GPA: 3.55
Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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@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?
Intern  B
Joined: 24 Apr 2018
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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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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Guys one question. if we know that it is not a right angle , why did u used phythagorean theorem ?
VP  D
Joined: 14 Feb 2017
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Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
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In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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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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_________________
Here's how I went from 430 to 710, and how you can do it yourself:

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  D
Joined: 14 Feb 2017
Posts: 1367
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31
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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But this is the construction if we randomly split each side.
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_________________
Here's how I went from 430 to 710, and how you can do it yourself:
Intern  B
Joined: 21 Apr 2019
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Re: In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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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
Intern  B
Joined: 01 Oct 2019
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In the quilting pattern shown above, a small square has its vertices o  [#permalink]

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