Hey guys,
I am an engineer by profession and would like to discuss the toughest questions on the GMAT.The buzz is that under the Pearsonvue the standard of quant questions has increased exponentially , so lets take no chances.Starting tommorrow i will post some of the tough questions i have come across during my preparation for the past couple of months.
_Kiran..
Discuss the toughest problem solving questions.
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The points P, Q, R lie on a line in that order with PQ=9, QR=21. Let O be a point not on PR such that PO=RO and the distances PO and QO are integral.
Then sum of all possible perimeters of triangle PRO is
(a) 320 (b) 350 (c) 380 (d) 410
The answer will be posted tomm.
Then sum of all possible perimeters of triangle PRO is
(a) 320 (b) 350 (c) 380 (d) 410
The answer will be posted tomm.

 Junior  Next Rank: 30 Posts
 Posts: 27
 Joined: 09 Apr 2006
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Let PO = x = RO, and QO = y,
then by stewart's theorem
(https://planetmath.org/encyclopedia/Proo ... eorem.html)
21*x^2 + 9*x^2 = 30*y^2 + 30*9*21 => x^2  y^2 = (3^3)*(7)
(x+y)(xy) = (27*7)(1) = 63(3) = 27(7) = 21(9)
We are required to find the sum of all 2x + 30.
From, (x+y)(xy) = (27*7)(1) = 63(3) = 27(7) = 21(9)
(x,y) can take values as (95,94) (33,30), (17,10) (15,6)
but (x,y) as (15,6) is not possible as QO + RO > QR.
Hence, the required sum is 2(95 + 33 + 17) + 3*30 = 380
Hence, choice (c) is the correct answer.
Note that remembering the result from stewart's theorem can make
you derive the length of median, angle bisector easily
then by stewart's theorem
(https://planetmath.org/encyclopedia/Proo ... eorem.html)
21*x^2 + 9*x^2 = 30*y^2 + 30*9*21 => x^2  y^2 = (3^3)*(7)
(x+y)(xy) = (27*7)(1) = 63(3) = 27(7) = 21(9)
We are required to find the sum of all 2x + 30.
From, (x+y)(xy) = (27*7)(1) = 63(3) = 27(7) = 21(9)
(x,y) can take values as (95,94) (33,30), (17,10) (15,6)
but (x,y) as (15,6) is not possible as QO + RO > QR.
Hence, the required sum is 2(95 + 33 + 17) + 3*30 = 380
Hence, choice (c) is the correct answer.
Note that remembering the result from stewart's theorem can make
you derive the length of median, angle bisector easily
i've been prepping for about 6 months now, i think i have a pretty good feel for the type of material that is "fair game" for the test.
That question is well beyond the scope of the GMAT...
Lets try to keep material thats posted, to material thats relevant!
The site is just getting started, and posting questions that aren't relevant doesn't add any value...
That question is well beyond the scope of the GMAT...
Lets try to keep material thats posted, to material thats relevant!
The site is just getting started, and posting questions that aren't relevant doesn't add any value...
I disagree with the previous message. The problem is actually not that complex as it looks, and it does not require any specific knowledge, such as the Stewart's theorem. Therefore, I can easily imagine hitting at this problem during GMAT examination, and thanks dkiran01 for giving us this sample.
One can solve this problem using the banal Pythagorean theorem: the first step will be to draw down the height of the isosceles POR triangle, say, to point X on PR, and get two rightangled triangles of POX and QOX. Obviously, QX is PR/2PQ=30/29=6. So:
QO^2=OX^2+36 and
PO^2=OX^2+225, therefore
PO^2=QO^2+189, therefore
PO^2QO^2=189, therefore
(POQO)(PO+QO)=189.
189 is 1*3*3*3*7, and we know from the formulation of the given problem that both brackets result in integer values. It logically means that values are integrally divisible by divisors of 189.
As POPQ is less than PO+PQ, we can check up only those divisors of 189 that suit POPQ so that it is less than PO+PQ. Easily, it's 1, 3, 7 and 9. As dkiran01 showed, the last option is invalid because the sum of any two legs of any triangle is bigger than the thrid leg. So, we get suiting values of POQO and, therefore, PO: 98, 33 and 17. That's what we needed to calculate the sum of all possible perimeters of the triangle POR. Easy? I think, as soon as you grip the idea of how to solve this problem, it will take up to 1 minute to make a calculation. Suitable for GMAT.
One can solve this problem using the banal Pythagorean theorem: the first step will be to draw down the height of the isosceles POR triangle, say, to point X on PR, and get two rightangled triangles of POX and QOX. Obviously, QX is PR/2PQ=30/29=6. So:
QO^2=OX^2+36 and
PO^2=OX^2+225, therefore
PO^2=QO^2+189, therefore
PO^2QO^2=189, therefore
(POQO)(PO+QO)=189.
189 is 1*3*3*3*7, and we know from the formulation of the given problem that both brackets result in integer values. It logically means that values are integrally divisible by divisors of 189.
As POPQ is less than PO+PQ, we can check up only those divisors of 189 that suit POPQ so that it is less than PO+PQ. Easily, it's 1, 3, 7 and 9. As dkiran01 showed, the last option is invalid because the sum of any two legs of any triangle is bigger than the thrid leg. So, we get suiting values of POQO and, therefore, PO: 98, 33 and 17. That's what we needed to calculate the sum of all possible perimeters of the triangle POR. Easy? I think, as soon as you grip the idea of how to solve this problem, it will take up to 1 minute to make a calculation. Suitable for GMAT.
please help
can somebody explain how this question is solved  As i understand it  the angle bisector of POR is perpendicular to PR and thereby divides PR into 2 equal halves  hence if PX = 15, OP has to be 17 (pythagorean triplets)
now, here are my questions?
a. why are we finding the distance QX?
b. how did you figure the other values of 98 and 33
can somebody explain how this question is solved  As i understand it  the angle bisector of POR is perpendicular to PR and thereby divides PR into 2 equal halves  hence if PX = 15, OP has to be 17 (pythagorean triplets)
now, here are my questions?
a. why are we finding the distance QX?
b. how did you figure the other values of 98 and 33
There is no problem with finding it, indeed, it's just a step of calculations. Obviously, QX is 6. Why we want to know it  because there is a crucial condition that QO is an integral, and we want to find such PRO triangles that satisfy this condition. Knowing that QX in triangle QXO is 6, we can find possible XOs that make QO integral.a. why are we finding the distance QX?
Please specify. Values other than 98 and 33? Or other values in triangles where PO is 98 or 33?b. how did you figure the other values of 98 and 33
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Assuming: PO = x and QO = y
x.x  15.15 = y.y  6.6 (equating the Perpendicular from O on PR)
=> (x+y).(xy) = (15+6).(156)
= 21.9
= 21.3.3 = 7.3.3.3
Since x and y are integers hence their 'addition' and 'difference' will also be integer Therefore:
Sol1:
*Since, O is not on PR hence (x+y) = 21 and (xy) = 9 is not possible because it gives x=15 and y=6
*Also, Any solution (x+Y) < (xY) is not possible (x > y > 0)
Sol2:
(x+y) = 7.3.3 = 63
(xy) = 3
=> x = 33 => Perimeter1 = 2.33+30 = 96
Sol3:
(x+y) = 21.3.3 = 189
(xy) = 1
=> x = 95 => Perimeter2 = 2.95+30 = 220
Sol4:
(x+y) = 3.3.3 = 27
(xy) = 7
=> x = 17 => Perimeter3 = 2.17+30 = 64
Sum of Total perimeters: 96+ 220 + 64 = 380
Isn't this easy?
Not necessary to remember the big theorem !!!
Enjoy !!!
x.x  15.15 = y.y  6.6 (equating the Perpendicular from O on PR)
=> (x+y).(xy) = (15+6).(156)
= 21.9
= 21.3.3 = 7.3.3.3
Since x and y are integers hence their 'addition' and 'difference' will also be integer Therefore:
Sol1:
*Since, O is not on PR hence (x+y) = 21 and (xy) = 9 is not possible because it gives x=15 and y=6
*Also, Any solution (x+Y) < (xY) is not possible (x > y > 0)
Sol2:
(x+y) = 7.3.3 = 63
(xy) = 3
=> x = 33 => Perimeter1 = 2.33+30 = 96
Sol3:
(x+y) = 21.3.3 = 189
(xy) = 1
=> x = 95 => Perimeter2 = 2.95+30 = 220
Sol4:
(x+y) = 3.3.3 = 27
(xy) = 7
=> x = 17 => Perimeter3 = 2.17+30 = 64
Sum of Total perimeters: 96+ 220 + 64 = 380
Isn't this easy?
Not necessary to remember the big theorem !!!
Enjoy !!!