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Geometry Problem

by knight247 » Sun Jul 03, 2011 9:40 am
In the figure above, Angle BAC = 90° and (p-q)/(p+q)=1/9. Find the value of p.
(A) 100
(B) 120
(C) 135
(D) 150
(E) 180

Detailed explanations would be appreciated. I do not have an OA
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Last edited by knight247 on Sun Jul 03, 2011 11:44 am, edited 1 time in total.
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by Frankenstein » Sun Jul 03, 2011 9:49 am
Hi,
(p + q)/(p-q) = 1/9 ?
How is this possible. How can difference between 2 positive quantities be 9 times sum of them.I guess it is reverse
p+q/(p-q) = 9 => p+q = 9p-9q.
So, 8p = 10q =>q=4p/5
p+q = 270.
So, p+(4p/5) = 270. So, p = 150
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by MBA.Aspirant » Sun Jul 03, 2011 10:01 am
I solved it and got 150

The ratio however should be p-q/p+q = 1/9

you can backsolve this one

from the figure we know p+q = 360 - 90 = 270

Plugin B) 120

120-q/120+q = 1/9

q= 96

120 + 96 = 216 < 270

Plugin D) 150

150 -q /150+q = 1/9

q = 120

150 + 120 = 270
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by knight247 » Sun Jul 03, 2011 11:44 am
Hey Frank,
Sorry i made a mistake while typing...Its (p-q)/(p+q)=1/9
i've edited the post also. Hope u can gimme ur opinion
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by edvhou812 » Sun Jul 03, 2011 2:02 pm
p+q = 360-90 = 270

p-q = 1/9(270)
p-q = 30

(p+q = 270) - (p+q = 30) => p = 120

270-120 = 150

q = 150.
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by smackmartine » Sun Jul 03, 2011 2:12 pm
IMO D
9(p-q)=(p+q)
8p=10q
also, p+q+90 = 360
p+q=270
p+0.8p=270
p=2700/18
p=300/2=150
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