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# ps value

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SVP
Joined: 07 Nov 2007
Posts: 1760
Location: New York

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13 Sep 2008, 14:10
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If P^2-qr=10,q^2+pr=10 , r^2+pq=10 , and rq , what is the value of p^2+q^2+r^2?

10
15
20
25
30

Would like to see how others will approach this problem.

I was able to solve this by plug in..

--== Message from GMAT Club Team ==--

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Manager
Joined: 09 Jul 2007
Posts: 240

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13 Sep 2008, 18:55
I can not solve this question .. I give up..

I guess plug in is the only option..
Senior Manager
Joined: 29 Mar 2008
Posts: 346

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13 Sep 2008, 19:01
x2suresh wrote:
If P^2-qr=0,q^2+pr=0 , r^2+pq=0 , and r<>q , what is the value of p^2+q^2+r^2?

10
15
20
25
30

Would like to see how others will approach this problem.

I was able to solve this by plug in..

I think I have seen this question before.... X2Suresh, are you sure the equations equate to 0 and not 10?
P^2-qr=10,q^2+pr=10 , r^2+pq=10

I remember guessing the answer to be 25 instead of 20 (strategic guess) which turned out to be wrong.
I think plugging in should be the best approach.
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"Leave no stone unturned."
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Senior Manager
Joined: 09 Oct 2007
Posts: 459

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13 Sep 2008, 20:10
x2suresh wrote:
If P^2-qr=0,q^2+pr=0 , r^2+pq=0 , and r<>q , what is the value of p^2+q^2+r^2?

10
15
20
25
30

Would like to see how others will approach this problem.

I was able to solve this by plug in..

I'm having a hard time understanding this problem. Is the problem written correctly? How do you solve by plugging in?

What I'm getting so far is that lpl = lql = lrl, and that r is not equal to q, but if this is the case, the p^2-qr cannot be possible. I'm confused :S
SVP
Joined: 07 Nov 2007
Posts: 1760
Location: New York

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13 Sep 2008, 21:54
leonidas wrote:
x2suresh wrote:
If P^2-qr=0,q^2+pr=0 , r^2+pq=0 , and r<>q , what is the value of p^2+q^2+r^2?

10
15
20
25
30

Would like to see how others will approach this problem.

I was able to solve this by plug in..

I think I have seen this question before.... X2Suresh, are you sure the equations equate to 0 and not 10? P^2-qr=10,q^2+pr=10 , r^2+pq=10

I remember guessing the answer to be 25 instead of 20 (strategic guess) which turned out to be wrong.
I think plugging in should be the best approach.

Sorry guys..I made typo mistake.. it should be 10.. Corrected in the original post .

leonidas, thanks for pointing out.
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Smiling wins more friends than frowning

VP
Joined: 17 Jun 2008
Posts: 1322

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13 Sep 2008, 23:16
x2suresh wrote:
If P^2-qr=10,q^2+pr=10 , r^2+pq=10 , and r<>q , what is the value of p^2+q^2+r^2?

10
15
20
25
30

Would like to see how others will approach this problem.

I was able to solve this by plug in..

I used this approach
P^2-qr=10 ---1
q^2+pr=10 ---2
r^2+pq=10---3
1-2 =>p-q=r => p=r+q -----3
substitute 3 in 2 => q^2+qr+r^2=10 => from 1 =>
q^2+p^2+r^2=10+10=20

IMO C
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cheers
Its Now Or Never

SVP
Joined: 30 Apr 2008
Posts: 1850
Location: Oklahoma City
Schools: Hard Knocks

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13 Sep 2008, 23:27
Can you go into more detail? I was not able to follow your explanation at all.

spriya wrote:
x2suresh wrote:
If P^2-qr=10,q^2+pr=10 , r^2+pq=10 , and r<>q , what is the value of p^2+q^2+r^2?

10
15
20
25
30

Would like to see how others will approach this problem.

I was able to solve this by plug in..

I used this approach
P^2-qr=10 ---1
q^2+pr=10 ---2
r^2+pq=10---3
1-2 =>p-q=r => p=r+q -----3
substitute 3 in 2 => q^2+qr+r^2=10 => from 1 =>
q^2+p^2+r^2=10+10=20

IMO C

_________________

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Intern Joined: 26 Aug 2008 Posts: 15 Re: ps value [#permalink] ### Show Tags 14 Sep 2008, 01:07 First of all, questions such as this one will not be in your screen on the test day... P^2-qr=10 (1) q^2+pr=10 (2) r^2+pq=10 (3) Okay first of all you need something to factor out and a^2-b^2 = (a+b) (a-b) is your friend here. How will you get something like this ? well if you look at the first 2 equations you will see that qr and pr has r in common..hmm.. So if you substrac 1 from 2: P^2-qr - (q^2+pr) = 10-10 = 0 P^2-qr-q^2-pr=0 So lets put the p^ 2 and q^2 to the left and take the rest to the right P^2 - q^2 = pr+qr Now remember the a^2-b^2 = (a+b) (a-b) formula (p-q)(p+q) = r( p+q) you can simply this by dividing both sides with (p+q) p-q=r ----> p=q+r we are not there yet so you need to use this..spider senses!! there is a P in the second equation lets use it q^2+pr=10 ===> q^2+(q+r)r=10 q^2+qr+r^2=10 hmm...so what is QR ? P^2-qr=10 (1) aha ...qr=p^2-10 lets insert this to what we just found q^2+qr+r^2=10 ---> q^2+(p^2-10)+r^2 =10 there you have it : q^2+p^2+r^2 = 20 well there was P in the third equation as well...since we spent so many time on this question and ruined our GMAT score and timing lets have fun to see if we can get the same result with the third equation: r^2+pq=10 (3) r^2+(q+r)q=10 r^2+q^2+qr=10 so we need that qr again!! P^2-qr=10 (1) ...qr=p^2-10 and one more time we found: q^2+p^2+r^2 = 20 YAY!!! Now do you really think you want to do this during your test ? I don't think so!! SVP Joined: 30 Apr 2008 Posts: 1850 Location: Oklahoma City Schools: Hard Knocks Re: ps value [#permalink] ### Show Tags 14 Sep 2008, 06:42 gmatdelight wrote: Now do you really think you want to do this during your test ? I don't think so!! It depends on how well you're doing. If you're doing really well 750+, I think this is a problem that could be for the very upper level test takers. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 25 Mar 2006
Posts: 9
Location: Kolkata, India

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14 Sep 2008, 08:43
Thanks for the question. I solved this question as below:

p^2 - qr = 10 ------(1) => qr = p^2 - 10 ------(4)
q^2 + pr = 10 -----(2)
r^2 + pq = 10 -----(3)

Adding (1), (2) & (3), we get,
(p^2 + q^2 + r^2) + (pq + pr - qr) = 30

Using (4) to substitute qr, we get,
(p^2 + q^2 + r^2) + (pq + pr - p^2 + 10) = 30

or, (p^2 + q^2 + r^2) + p (q + r - p) = 30 - 10 = 20 ------(5)

Now, from (1) & (2), we have,
p^2 - qr = q^2 + pr = 10
or, p^2 - q^2 = pr + qr
or, (p+q)(p-q)= r(p+q) => p-q=r => p=q+r

Using p=q+r in (5) above, we get,
(p^2 + q^2 + r^2) + p (p - p) = 20
(p^2 + q^2 + r^2) = 20.
SVP
Joined: 07 Nov 2007
Posts: 1760
Location: New York

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14 Sep 2008, 08:53
Thanks for discussion.

OA is C ( 20)

I solved this by pluggin the following values.

$$p=sqrt(10)$$
$$q=0$$
$$r=sqrt(10)$$
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VP
Joined: 05 Jul 2008
Posts: 1369

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14 Sep 2008, 11:34
x2suresh wrote:
Thanks for discussion.

OA is C ( 20)

I solved this by pluggin the following values.

$$p=sqrt(10)$$
$$q=0$$
$$r=sqrt(10)$$

These values are a little insane for me. Good for you.
SVP
Joined: 17 Jun 2008
Posts: 1504

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14 Sep 2008, 22:42
I solved it as below....

Since, the question is asking to get the value of p^2 + q^2 + r^2, first step for me was to add all the three equations.

Thus, (p^2 + q^2 + r^2) + (pr + pq -qr) = 30..............(A)

Now, I need to transform the second expression on the left into a numerical value. Also, from the question, I get a hint that q does not equal r.....that means, I should get an equation with one of the factors as (q-r) and right side as 0. This was possible by subtracting the third equation from the second in the question.

Thus ,q^2 - r^2 + pr - pq = 0
or, (q-r)(q+r-p) = 0.

But, since q does not equal r, hence q + r = p.

Now, the second expression in the earlier equation (A) will look like,
p(r + q) - qr = p^2 - qr = 10 (from the first equation in the question).

Hence, p^2 + q^2 + r^2 = 20.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: ps value   [#permalink] 14 Sep 2008, 22:42
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