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Good One from GMATPrep

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Good One from GMATPrep [#permalink] New post   23 Jan 2006, 00:24
Solve this



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Re: Good One from GMATPrep [#permalink] New post   Updated on: 25 Jan 2006, 03:59
Since OP and OQ are radius:
s^2+t^2 = 4...(1)
From pythagoras, (t-1)^2+(s+root(3))^2 = 8 .... (2)

Solving 1 and 2, s=1, t=root(3) or s=-1, t=-root(3). Since s,t are in the first quadrant, s=1, t=root(3).

Originally posted by watever on 23 Jan 2006, 00:41.
Last edited by watever on 25 Jan 2006, 03:59, edited 1 time in total.
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Re: Good One from GMATPrep [#permalink] New post   23 Jan 2006, 00:52
Bhai wrote:
Solve this

PA=1 OA=sqrt3 so OP=2 10x Pifagorus(not sure if this right spelling).Hence OP=OQ=2 cause they are both radius of this circle PQ=sqrt(2^2+2^2)=sqrt8.So the volume of S would be -sqrt3+sqrt8[/img]
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Re: Good One from GMATPrep [#permalink] New post   23 Jan 2006, 01:36
Another simple way of getting one equation is to compare the slopes of OP and OQ. They should be negative reciprocals.

t/s*-1/root(3) = -1 which gives t=s*root(3). This can be used together with s^2+t^2=4 to give s=1, t=root(3).
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Re: Good One from GMATPrep [#permalink] New post   23 Jan 2006, 03:10
Length OP = SQRT(3+1) = 2
This is a semicircle so OP = OQ = 2

Angle of OP is 60 with -ve side of x axis.

So angle of OQ with +ve x axis = 180 - 90-60 = 30

So Cos(30) = SQRT(3)/2 = s/2 so

s = SQRT(3)
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Re: Good One from GMATPrep [#permalink] New post   23 Jan 2006, 10:52
watever wrote:
Another simple way of getting one equation is to compare the slopes of OP and OQ. They should be negative reciprocals.

t/s*-1/root(3) = -1 which gives t=s*root(3). This can be used together with s^2+t^2=4 to give s=1, t=root(3).


Perfecto watever. s = 1 is the answer
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Re: Good One from GMATPrep [#permalink] New post   23 Jan 2006, 13:02
S=1.

Using distance formula for distance between (0,0) and ( -sqrt(3),1) gives radius = 2.

Also, using pythagoras theorem, we get sqrt(3)s = t. Since radius = 2, distance between (s,t) and (0,0) is s^2 + t^2 = 4. Substituting, t = sqrt(3)s, we get s = 1.
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Re: Good One from GMATPrep [#permalink] New post   23 Jan 2006, 13:12
ps_dahiya wrote:
Length OP = SQRT(3+1) = 2
This is a semicircle so OP = OQ = 2

Angle of OP is 60 with -ve side of x axis.

So angle of OQ with +ve x axis = 180 - 90-60 = 30

So Cos(30) = SQRT(3)/2 = s/2 so

s = SQRT(3)


I did the wrong calculations. Cos(60) is 1/2 and Cos(30) is SQRT(3)/2. But I took totally opposite :wall
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Re: Good One from GMATPrep [#permalink] New post   23 Jan 2006, 13:59
Quote:
Perfecto watever. s = 1 is the answer


Thanks for the encouragement Bhai. You have some relation to Mumbai? ;-)
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Re: Good One from GMATPrep [#permalink] New post   23 Jan 2006, 15:01
watever wrote:
Quote:
Perfecto watever. s = 1 is the answer


Thanks for the encouragement Bhai. You have some relation to Mumbai? ;-)


Well who is not related to Mumbai these days? But not with any Bhai. My bhaigiri is very decent. :stab
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Re: Good One from GMATPrep [#permalink] New post   23 Jan 2006, 23:23
yups got 1 but after 5 min :oops:
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Re: Good One from GMATPrep [#permalink] New post   24 Jan 2006, 13:03
Bhai wrote:
watever wrote:
Quote:
Perfecto watever. s = 1 is the answer


Thanks for the encouragement Bhai. You have some relation to Mumbai? ;-)


Well who is not related to Mumbai these days? But not with any Bhai. My bhaigiri is very decent. :stab


:) :) :) :) :) :)
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Re: Good One from GMATPrep [#permalink] New post   24 Jan 2006, 14:19
watever wrote:
Since OP and OQ are radius:
s^2+t^2 = 4...(1)
From pythagoras, (t-1)^2+(s+root(3))^2 = 8 .... (2)

Solving 1 and 2, s=1, t=root(3) or s=-1, t=-root(3). Since s,t are in the first quadrant, s=-1, t=root(3).


watever: you meant s=1, right? (as opposed to '-1')

also you don't show the substitution of 1 and 2, did you solve for t in equation 1, t=sqrt(4-s^2) into the second equation? I did this and got the same answer but I was wondering if you omitted the work because there was an intuitive solution that you instantly knew.
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Re: Good One from GMATPrep [#permalink] New post   25 Jan 2006, 04:02
Quote:
watever: you meant s=1, right? (as opposed to '-1')

also you don't show the substitution of 1 and 2, did you solve for t in equation 1, t=sqrt(4-s^2) into the second equation? I did this and got the same answer but I was wondering if you omitted the work because there was an intuitive solution that you instantly knew.


Yeah, I meant s=1, I edited the post. Also, I did omit the working for sake of brevity. I however advise my second method be used which used the slopes as one of the equation.

So 1st eqn is s^2 + t^2 = 4 and the second is t/s*-1/root(3) = -1 which gives t=s*root(3). So substition and calculations become highly easy as compared to the first method.



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