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Discuss the toughest problem solving questions.

by dkiran01 » Sun Apr 09, 2006 9:29 pm
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 Pearson-vue the standard of quant questions has increased exponentially :shock: , 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..
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question 1

by dkiran01 » Sun Apr 09, 2006 9:53 pm
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.
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by oxygen008 » Wed Apr 19, 2006 9:10 am
Whats the answer, don't you need at least one angle? or the height? Does involve Trig? Trig isn't tested, is it?
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Solution taken from a diff site

by dkiran01 » Wed Apr 19, 2006 12:41 pm
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)(x-y) = (27*7)(1) = 63(3) = 27(7) = 21(9)

We are required to find the sum of all 2x + 30.

From, (x+y)(x-y) = (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
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Out of the GMAT scope

by oxygen008 » Tue Apr 25, 2006 8:28 am
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...
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by Nat » Thu Aug 10, 2006 12:34 am
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 right-angled triangles of POX and QOX. Obviously, QX is PR/2-PQ=30/2-9=6. So:

QO^2=OX^2+36 and
PO^2=OX^2+225, therefore

PO^2=QO^2+189, therefore

PO^2-QO^2=189, therefore

(PO-QO)(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 PO-PQ is less than PO+PQ, we can check up only those divisors of 189 that suit PO-PQ 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 PO-QO 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.
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help

by anuroopa » Fri Mar 09, 2007 3:59 am
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
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by Nat » Fri Mar 09, 2007 4:59 am
a. why are we finding the distance QX?
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.
b. how did you figure the other values of 98 and 33
Please specify. Values other than 98 and 33? Or other values in triangles where PO is 98 or 33?
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by Stacey Koprince » Sun Mar 11, 2007 10:41 pm
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by Badri » Mon Mar 12, 2007 5:41 am
Assuming: PO = x and QO = y

x.x - 15.15 = y.y - 6.6 (equating the Perpendicular from O on PR)

=> (x+y).(x-y) = (15+6).(15-6)
= 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 (x-y) = 9 is not possible because it gives x=15 and y=6

*Also, Any solution (x+Y) < (x-Y) is not possible (x > y > 0)

Sol2:
(x+y) = 7.3.3 = 63
(x-y) = 3

=> x = 33 => Perimeter1 = 2.33+30 = 96

Sol3:
(x+y) = 21.3.3 = 189
(x-y) = 1

=> x = 95 => Perimeter2 = 2.95+30 = 220

Sol4:
(x+y) = 3.3.3 = 27
(x-y) = 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 !!!
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