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M61-03

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Math Revolution GMAT Instructor
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M61-03  [#permalink]

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New post 17 Jun 2018, 23:43
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

62% (01:22) correct 38% (01:37) wrong based on 13 sessions

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Let \(f(x)= (x-p)(x-q)\). If \(f(11)=f(20)=0\), then \(f(10)=\)?


A. -5
B. -1
C. 1
D. 5
E. 10

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Re M61-03  [#permalink]

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New post 17 Jun 2018, 23:43
Official Solution:


Let \(f(x)= (x-p)(x-q)\). If \(f(11)=f(20)=0\), then \(f(10)=\)?


A. -5
B. -1
C. 1
D. 5
E. 10


Since 11 and 20 are roots of \(f(x) = 0\), we must have \(f(x) = (x-11)(x-20)\). Thus, \(f(10) = (10-11)(10-20) = (-1)(-10) = 10\).


Answer: E
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M61-03  [#permalink]

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New post 20 Jun 2018, 04:23
Hi Bunuel ,

f(x)=(x−p)(x−q)

f(11)=f(20)

If I take p and q as 15.5 then also f(11) = F(20) , I was not given that p & Q are root .(x−p)(x−q)

F(11)=(11-15.5)(11.15.5) = F(20) = (20-11.5)(20-11.5)

Please correct me , I am confused .
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Re: M61-03  [#permalink]

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New post 05 Aug 2018, 22:09
baru wrote:
Hi Bunuel ,

f(x)=(x−p)(x−q)

f(11)=f(20)

If I take p and q as 15.5 then also f(11) = F(20) , I was not given that p & Q are root .(x−p)(x−q)

F(11)=(11-15.5)(11.15.5) = F(20) = (20-11.5)(20-11.5)

Please correct me , I am confused .


You didn't consider f(11) = 0 and f(20) = 0.
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Re M61-03  [#permalink]

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New post 21 Dec 2018, 01:51
I think this the explanation isn't clear enough, please elaborate. I thought 11 and 20 will replace x in the F(x) and I got answer 1. I equated f(11) to f(20) to get values in terms of p+q and pq and then substituted in answer
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Re M61-03 &nbs [#permalink] 21 Dec 2018, 01:51
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M61-03

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