Gmat DS
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I'm missing some fundamentals here, angle prs is directly horizontally across from the 90 degree angle right and angle pqr is directly above it. Where is angle PRQ? I was under the impression that prq and pqr where the same angle obviously not so... What needs to be learned here
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From the figure:
pqr + qps =90 =>pqr+ (qpr+rps) =90 ....(a)
and, rps + prs =90...(b)
We need to find |prs-pqr|
stm 1: qpr=30
from (a), pqr + 30 + rps =90, or rps = 60-pqr
from (b), rps =90 -prs
So, 60-pqr=90 -prs, hence we know the value of |prs-pqr|. Sufficient
stm 2: pqr + (180-prs) = 150, again this is sufficient to calculate |prs-pqr|.
Hence answer is D
pqr + qps =90 =>pqr+ (qpr+rps) =90 ....(a)
and, rps + prs =90...(b)
We need to find |prs-pqr|
stm 1: qpr=30
from (a), pqr + 30 + rps =90, or rps = 60-pqr
from (b), rps =90 -prs
So, 60-pqr=90 -prs, hence we know the value of |prs-pqr|. Sufficient
stm 2: pqr + (180-prs) = 150, again this is sufficient to calculate |prs-pqr|.
Hence answer is D
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Sorry, I should have explicity stated that pqr refers to the angle that is formed by line segments PQ and QR. Similarly, qps is the angle formed by line segments QP and PS.vladmire wrote:Can anyone elaborate just alittle more. I still don't know pqr or qps
Hope this helps.
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You are asked to find : PRS - PQR
In any triangle : Exterior angle = sum of interior opposite angles
So in PQR : PRS is the exterior angle.
From 1: PRS = PQR + QPR => PRS -PQR = PRQ = 30 Sufficient
From 2: PQR + PRQ = 150.
PRQ = 180- PRS
PQR + 180 - PRS = 150 => PRS - PQR = 30
Hope this helps
- pradeep
In any triangle : Exterior angle = sum of interior opposite angles
So in PQR : PRS is the exterior angle.
From 1: PRS = PQR + QPR => PRS -PQR = PRQ = 30 Sufficient
From 2: PQR + PRQ = 150.
PRQ = 180- PRS
PQR + 180 - PRS = 150 => PRS - PQR = 30
Hope this helps
- pradeep
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GMATPREP1 - 550 (Oct 08)
MGMAT FREE CAT - 600 (Dec 08)
MGMAT CAT1 - 670 (Jan 09)
MGMAT CAT2 - 550 (Jan 09)
MGMAT CAT3 - 640 ( Feb 09)
MGMAT CAT4 - 660 ( Feb 09)