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# In the figure above, the point on segment PQ that is twice as far from

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In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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18 Sep 2012, 01:51
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Difficulty:

25% (medium)

Question Stats:

76% (01:39) correct 24% (02:00) wrong based on 1566 sessions

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In the figure above, the point on segment PQ that is twice as far from P as from Q is

(A) (3,1)
(B) (2,1)
(C) (2,-1)
(D) (1.5,0.5)
(E) (1,0)

Practice Questions
Question: 43
Page: 158
Difficulty: 600

Attachment:

Plane.png [ 7.49 KiB | Viewed 43589 times ]

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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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18 Sep 2012, 01:52
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In the figure above, the point on segment PQ that is twice as far from P as from Q is

(A) (3,1)
(B) (2,1)
(C) (2,-1)
(D) (1.5,0.5)
(E) (1,0)

Options A or C cannot be correct answer since these points aren't even on segment PQ. E (1,0) is clearly closer to P, so it's also out, D is right in the middle of the segment, so only option B is left.

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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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26 Dec 2012, 02:06
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Lets see the algebraic solution.

A point P that divides a line AB internally into ratio m1 & m2 is as below.

A(x1,y1)____________m1_____________P(x3,y3)________________________m2__________________________B(x2,y2)

The formula for finding coordinates are as below.

$$x3 = (x1*m2 + x2*m1)/(m1 + m2)$$

$$y3 = (y1*m2 + y2*m1)/(m1 + m2)$$

Back to our question.

P (0,-1 ) & Q (3,2)

We fathom that m1 =2 & m2 = 1

$$x1 = 0 , y1 = -1$$

$$x2 = 3 , y1 = 2$$

Plugging values, we obtain answer as (2,1). Hence B

Once you know the formula its very easy.

P.S. Bunuel's approach is beautifully elegant. However, for me solving algebraically is much faster than figuring the answer choices out.
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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18 Sep 2012, 05:30
1
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Bunuel wrote:

Attachment:
Plane.png
In the figure above, the point on segment PQ that is twice as far from P as from Q is

(A) (3,1)
(B) (2,1)
(C) (2,-1)
(D) (1.5,0.5)
(E) (1,0)

The question ask to divide the line PQ into 2:1 ratio and find the point.
By symmetry, the line segment at x-axis (1,0) will be divided in ratio 1:2.
Similarly, at (2,1) line will be divided in ration 2:1

Hence B
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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18 Sep 2012, 09:32
1
Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

Attachment:
Plane.png
In the figure above, the point on segment PQ that is twice as far from P as from Q is

(A) (3,1)
(B) (2,1)
(C) (2,-1)
(D) (1.5,0.5)
(E) (1,0)

Practice Questions
Question: 43
Page: 158
Difficulty: 600

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By looking at the figure it comes out that option "B" (2,1) is the right choice.
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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26 Oct 2012, 02:25
twice as far from P as from Q

This confused me..I thought teh Qs is asking for the midpoint..

Could somebody explain me what it is asking..

and thanks bunuel for POE method. but how would u solve this algebraically?
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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21 Dec 2012, 14:14
Sachin9 wrote:
twice as far from P as from Q

This confused me..I thought teh Qs is asking for the midpoint..

Could somebody explain me what it is asking..

and thanks bunuel for POE method. but how would u solve this algebraically?

I concur. I think it would be helpful if we can solve this algebraically using the given y-coordinate and point Q, rather than elimination — similar to Bunel's approach to this problem.
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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22 Dec 2012, 00:49
Please suggest a algebraic approach, Bunuel!
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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22 Dec 2012, 04:04
1
Sachin9 wrote:
Please suggest a algebraic approach, Bunuel!

I would never solve this question algebraically, but you can check for the tools for that here: math-coordinate-geometry-87652.html
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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29 Dec 2012, 18:44
The only way to solve this problem is by plugging in answer choices?
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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30 Dec 2012, 00:11
fozzzy wrote:
The only way to solve this problem is by plugging in answer choices?

See this

in-the-figure-above-the-point-on-segment-pq-that-is-twice-a-139117.html#p1161061

If you need further elaboration, let me know.
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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30 Dec 2012, 01:10
I was thinking along those lines by forming a triangle and then solving it. This is a fairly easy question you can still use the answer choices but if its a bit more complicated I would like a fast approach.
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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30 Dec 2012, 01:26
fozzzy wrote:
I was thinking of using similar triangle property, I was thinking along those lines by forming a triangle and then solving it.

You can solve it easily using similar triangles too but you've picked the wrong triangles

Take the triangle with vertices [ (-1, 0) , (1,1) , (0,0) ]

and the triangle with vertices [ (3, 2) , (1,1) , (3,0) ]

These two triangles are similar since their lengths are in the ration 1:2

and then you can proceed.
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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19 Jan 2013, 03:24
1
eaakbari wrote:
Lets see the algebraic solution.

A point P that divides a line AB internally into ratio m1 & m2 is as below.

A(x1,y1)____________m1_____________P(x3,y3)________________________m2__________________________B(x2,y2)

The formula for finding coordinates are as below.

$$x3 = (x1*m2 + x2*m1)/(m1 + m2)$$

$$y3 = (y1*m2 + y2*m1)/(m1 + m2)$$

Back to our question.

P (0,-1 ) & Q (3,2)

We fathom that m1 =2 & m2 = 1

$$x1 = 0 , y1 = -1$$

$$x2 = 3 , y1 = 2$$

Plugging values, we obtain answer as (2,1). Hence B

Once you know the formula its very easy.

P.S. Bunuel's approach is beautifully elegant. However, for me solving algebraically is much faster than figuring the answer choices out.

This is called section formula. It will be helpful. Thanks
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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19 Jan 2013, 09:25
http://www.teacherschoice.com.au/Maths_ ... Geom_3.htm

go through the page..

ans is (2,1)
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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20 Jan 2013, 01:11
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I think with questions like these, the test writers are testing whether you'd quickly jump to using an algebraic approach, which in this case is much more time consuming, as compared to making the answer choices a part of your toolbox for finding the correct answer..
The question itself tells us we need to split the line into 3 equal parts with the asked coordinate being 2 parts away from P
A quick glance at the graph gives us the slope of 1, which easily shows us which points will cover the three segments
P(0,-1) --> (1,0) --> (2,1) --> Q(3,2)
Thus (2,1) being twice as far from P as from Q

True, an algebraic approach might be required for more complex problems where the slope isn't easily determined or the line segment might be split into a different ratio, but this question isn't testing that
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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06 Aug 2013, 00:41
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Coordinates of point P = (0,-1)
Coordinates of point Q = (3,2)
Point required to be found is on segment PQ that is twice as far from P as from Q
So, adding (2,2) to point P (0,-1)
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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10 Aug 2014, 22:59
We can solve this question by applying distance formula also.

distance between two points

sqrt((x2-x1) +(y2-y1))^2

Here we take options one by one

P(0,-1) q(3,2)

when we take option (2,1)

we get distance sqrt(8) from point p sqrt (2) from point q.

so this point is twice as far as from point q.
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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27 May 2016, 06:23
2
Bunuel wrote:
Attachment:
Plane.png
In the figure above, the point on segment PQ that is twice as far from P as from Q is

(A) (3,1)
(B) (2,1)
(C) (2,-1)
(D) (1.5,0.5)
(E) (1,0)

Solution:

This is an interesting problem because, although we usually prefer an algebraic approach to solving GMAT questions, the quickest way to solve this problem is to analyze each answer choice in relation to the graph. We need to determine which answer choice gives us a point on the graph that is twice as far from P as from Q. This means that it is closer to Q than to P.

We are given that point Q is at (3,2) and point P is at (0,-1). Let’s start with answer choice A.

A) (3,1)

Looking at the graph we see that (3,1) is not even on line segment PQ. Answer choice A is not correct.

B) (2,1)

Looking at the graph we see that (2,1) is on line segment PQ and it is closer to Q than it is to P. We could use the distance formula to determine the actual distances, but let's wait to see if this is necessary. Let’s test the other answer choices to be certain that answer choice B is correct.

C) (2,-1)

Looking at the graph we see that (2,-1) is not even on line segment PQ. Answer choice C is not correct.

D) (1.5,0.5)

Looking at the graph we see that (1.5,0.5) is on line segment PQ; however, it appears to be about halfway between P and Q. Answer choice D is not correct.

E) (1,0)

Looking at the graph we see that (1,0) is on line segment PQ; however, it is closer to P than to Q. Answer choice E is not correct.
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Re: In the figure above, the point on segment PQ that is twice as far from  [#permalink]

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21 Nov 2016, 11:01
I felt the question is confusing. Shouldn't it be "In the figure above, the point on segment PQ that is half as far from P as from Q is"?
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Re: In the figure above, the point on segment PQ that is twice as far from   [#permalink] 21 Nov 2016, 11:01

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