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by gmatmachoman » Tue Aug 17, 2010 9:37 am
In sequence P, P4 and P5 are 11 and 9 respectively. Each term after the first two terms in sequence P is either the sum of the previous two terms if that sum is odd, or half the sum of the previous two terms if the sum is even. What
is the largest possible product of P1 and P2?

(A) 40
(B) 14
(C) 12
(D) 10
(E) 7

Question is very much correct and needs bit of guess work too
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by uwhusky » Tue Aug 17, 2010 9:46 am
Is it A?

If P4 = 11, that means P3 + P2 has to be either 11 or 22.

P5 = 9, that means that P3 can only be 7, because 11 > 9 already.

So P2 has to be either 4 or 15, and since 15 > 2 x 7, P2 has to be 4.

If P2 is 4 and P3 is 7, then P1 could be either 10 or 3. 10 is larger, so together = 10 x 4 = 40?

I must admit, I chose E/D at one point before reaching A.
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by gmatmachoman » Tue Aug 17, 2010 9:56 am
how did u feel abtthe degree of difficulty of this one..is this solvable in less than 2 mins??

A is the answer.. as u rightly said the sequence can be : 10, 4,7,11,9 or 3,4,7,11,9
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by uwhusky » Tue Aug 17, 2010 10:02 am
I feel that my understanding of how to approach this question is there, and I can definitely find the numbers for P1, P2, and P3. But I can see that a lot of people will get this question wrong because they did not read the question carefully.

For example, a lot of people may choose E by adding 3 (P1) to 4 (P2), a lot might choose B by adding 10 to 4, and few may choose D by answering what the max value of P1 could be.

I think this is a very tricky question, but a GOOD tricky question. 2 minutes shouldn't be a problem, but the real test is on whether you read the question correctly.
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by beatthegmatinsept » Tue Aug 17, 2010 10:06 am
I for sure couldn't solve this question until I read uwhusky's response. But that just tells me I need a lot of work in Quant :)
Gmatmachoman - Whats the source of this question?
Being defeated is often only a temporary condition. Giving up is what makes it permanent.
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by kvcpk » Tue Aug 17, 2010 10:12 am
gmatmachoman wrote:In sequence P, P4 and P5 are 11 and 9 respectively. Each term after the first two terms in sequence P is either the sum of the previous two terms if that sum is odd, or half the sum of the previous two terms if the sum is even. What
is the largest possible product of P1 and P2?

(A) 40
(B) 14
(C) 12
(D) 10
(E) 7

Question is very much correct and needs bit of guess work too
P1,P2,P3,11,9,P5...
P3+11 =9
P3 = -2 [assuming P3 is even]
If P3 is negative, before terms will also be negative. We dont get the maximum product.
hence P3 cannotbe -ve.

P3 +11 = 18
P3 = 7 [assuming P3 is odd]

P2+p3 = 11
P2+7=11
P2=4 [assuming P2 is even]

P2 + P3 =22
P2 = 22-7 = 15 [Assuming P2 is odd]

P1+15 = 7
P1=-12 [assuming P1 is even and P2=15]
P1 = 14-15 =-1 [assuming P1 is odd and P2 = 15]
P1P2 = -ve

P1+4 = 7
P1=3[assuming P1 is odd and P2=4]
P1+4=14
P1=10 [assuming P1 is even and P2=4]
P1*P2 = 4*10 = 40

pick A.

I am still thinking of an easy way. Will post If I find some.
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by uwhusky » Tue Aug 17, 2010 10:18 am
kvcpk, you brought up a good point about negative value. This question could become more difficult if P1 and P2 have the option of being negative values with product that is larger than product of two positive values.

I think your approach and my approach are pretty much the same, the difference is that I didn't bother checking possible negative values. You covered all the bases!
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by selango » Tue Aug 17, 2010 10:20 am
P5=9

P4=11

P3+11=9

-->P3=-2 or 7[Ignore -2 as we need maximum product]

P3=7

P2+7=11

P2=4 or 15

P3=7

P2=4 or 15

P1=3 or 10 or -8 or -1

Maximum product =10*4=40[Only 40 is maximum in answer options]
--Anand--
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by gmatmachoman » Tue Aug 17, 2010 10:37 am
guys,,,, source is princeton....

praveen and anand did a good job...

praveen ..negative figures are not necessary as we are looking for maximum ..hope u agree...wat say?

approach is very muck like mine... and anand
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by kvcpk » Tue Aug 17, 2010 10:54 am
uwhusky wrote:kvcpk, you brought up a good point about negative value. This question could become more difficult if P1 and P2 have the option of being negative values with product that is larger than product of two positive values.

I think your approach and my approach are pretty much the same, the difference is that I didn't bother checking possible negative values. You covered all the bases!
I agree with you Uwhusky. It would be more complex if P1 and P2 are both negative and we get a product that is higher than 40.

I did some more analysis on this one. We need both P1 and P2 to be -ve as per our thought here.
We got P3=-2
Means Sum of 2 terms before it(P1,P2) can be -2 or -4

If sum of 2 -ve number is -4, there product cannot exceed 4. This is not in the options.
So we can ignore it.

What you say??
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don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
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by uwhusky » Tue Aug 17, 2010 11:01 am
I don't think the writer of this question had intention of including negative values for any of the sequence, because if he/she did, then this question would become progressively complicated to solve in 2 minutes.

Assuming that P3 could be 7 or -2, then P2 could be 4, 15, 13 or 24. Then P1 could have like...gosh...I'll just say a decent number of possibilities. Then you have to multiply them together and see which one is larger, assuming 40 is no longer the largest.

Yes, this question could become a "very difficult" question.
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by kvcpk » Tue Aug 17, 2010 11:45 am
gmatmachoman wrote:guys,,,, source is princeton....

praveen and anand did a good job...

praveen ..negative figures are not necessary as we are looking for maximum ..hope u agree...wat say?

approach is very muck like mine... and anand
Buddy,

If this is PR question, we can PM GmatguruNy, he should be able to help us if there is any better approach.

What say?
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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