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Is this a GMAT questiongoyalsau wrote:Help Guys,,,,,,,,,
6/7
7/6
5/6
6/5
7/5
this question is drawn from CAT administered in India. The CAT exam content is broad and not being limited to the topics embraced in GMAT quantitative curriculum. Having said this, I believe this is not GMAT question, and it tests knowledge of 'similar & congruent triangle' properties.prachich1987 wrote:Is this a GMAT questiongoyalsau wrote:Help Guys,,,,,,,,,
6/7
7/6
5/6
6/5
7/5
I tried hard But couldn't do
someone plz help
The above explanation is valid provided the triangles are similar which is not proved.frank1 wrote:I think answer is 6/5
Permimeter of Triangle 1 =10+12+8=30
perimeter of second is 10+15(found)+25/2(found)=50/2..I didn't understand this calculation The third side would be 12.5.So So perimeter is 37.5
that means.
30/(50/2)
(30x 2)/50= 6/5
Now if i separate the figure it will be more clear
thanks
I am not clear on the aboveRahul@gurome wrote:∆PQS and ∆PQR are similar triangles.
Thus,Perimeter of ∆QSR = (QS + SR + QR) = (10 + 10 + 15) = 35
- 1. PQ/QS = PR/PQ = 12/8 => PR = (12/8)PQ = (12/8)*12 = 18
2. PQ/QS = PR/QR = 12/10 => QR = (10/12)PR = (10/12)*18 = 15
3. SR = PR - PS = 18 - 8 = 10
Perimeter of ∆PQS = (PQ + QS + PS) = (12 + 10 + 18) = 30
Required ratio = 30/35 = 6/7
The correct answer is A.
Edit: (Proof of the similarity)
In ∆PQS and ∆PQR,Thus they are similar by angle-angle similarity.
- 1. angle PQS = angle PRQ (Given)
2. angle QPS = angle QPR (Common angle)
The Source of this question is www.time4education.com.Night reader wrote: this question is drawn from CAT administered in India. The CAT exam content is broad and not being limited to the topics embraced in GMAT quantitative curriculum. Having said this, I believe this is not GMAT question, and it tests knowledge of 'similar & congruent triangle' properties.
∆PRQ is essentially same as ∆PQR.prachich1987 wrote:I am not clear on the above
how can ∆PQS and ∆PQR be similar
∆PQS can be similar to ∆PRQ but not to ∆PQR
Prachi I will suggest you to Draw Three different Triangles on your scratch paper,, I m sure it will help in understanding.....prachich1987 wrote: I am not clear on the above
how can ∆PQS and ∆PQR be similar
∆PQS can be similar to ∆PRQ but not to ∆PQR
As far as I know the GMAT Quant is intellectual domain of GMAC and the institute mentioned by you is not affiliated with GMAC, correct? When it comes to preparation, I agree with you that GMAT is like marathon in terms of testing one's endurance. I might try to solve matrix by Gauss, or prove (disprove) theorem in Geo, yet GMAT is quite different story...goyalsau wrote:The Source of this question is www.time4education.com.Night reader wrote: this question is drawn from CAT administered in India. The CAT exam content is broad and not being limited to the topics embraced in GMAT quantitative curriculum. Having said this, I believe this is not GMAT question, and it tests knowledge of 'similar & congruent triangle' properties.
I am a member of the institute and they have uploaded the Gmat Quant material on the website. And it is Gmat Quant Section... Night reader Its always good to be smart with the syllabus,,,But saying that question is out of syllabus is not a great response,.......... If you have seen the interview of jyoti with Eric,,,, They said couple of times that preparing for Gmat is like marathon training........
Marathon is of 26 miles and people tend to run 52 miles before the race so when it comes to the real test, They can give there best. Now its up to the person ,, How much preparation one needs to do.........???????
You'll need to flip the triangle.prachich1987 wrote:I am not clear on the aboveRahul@gurome wrote:∆PQS and ∆PQR are similar triangles.
Thus,Perimeter of ∆QSR = (QS + SR + QR) = (10 + 10 + 15) = 35
- 1. PQ/QS = PR/PQ = 12/8 => PR = (12/8)PQ = (12/8)*12 = 18
2. PQ/QS = PR/QR = 12/10 => QR = (10/12)PR = (10/12)*18 = 15
3. SR = PR - PS = 18 - 8 = 10
Perimeter of ∆PQS = (PQ + QS + PS) = (12 + 10 + 18) = 30
Required ratio = 30/35 = 6/7
The correct answer is A.
Edit: (Proof of the similarity)
In ∆PQS and ∆PQR,Thus they are similar by angle-angle similarity.
- 1. angle PQS = angle PRQ (Given)
2. angle QPS = angle QPR (Common angle)
how can ∆PQS and ∆PQR be similar
∆PQS can be similar to ∆PRQ but not to ∆PQR
Rahul@gurome wrote:∆PQS and ∆PQR are similar triangles.
Thus,Perimeter of ∆QSR = (QS + SR + QR) = (10 + 10 + 15) = 35
- 1. PQ/QS = PR/PQ = 12/8 => PR = (12/8)PQ = (12/8)*12 = 18
2. PQ/QS = PR/QR = 12/10 => QR = (10/12)PR = (10/12)*18 = 15
3. SR = PR - PS = 18 - 8 = 10
Perimeter of ∆PQS = (PQ + QS + PS) = (12 + 10 + 18) = 30
Required ratio = 30/35 = 6/7
The correct answer is A.
Edit: (Proof of the similarity)
In ∆PQS and ∆PQR,Thus they are similar by angle-angle similarity.
- 1. angle PQS = angle PRQ (Given)
2. angle QPS = angle QPR (Common angle)
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