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M61-03

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M61-03  [#permalink]

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New post 18 Jun 2018, 00:43
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A
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Difficulty:

  35% (medium)

Question Stats:

77% (02:01) correct 23% (02:41) 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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New post 18 Jun 2018, 00:43
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, 05:23
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, 23:09
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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New post 21 Dec 2018, 02:51
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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M61-03  [#permalink]

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New post 15 May 2019, 03:53
You can solve it by using the quadratic equation method which is somewhat longer but might help you understand.

Using the information above.

f(11)=f(20)

Thus:
(11-p)(11-q) = (20-p)(20-q)
Solving both sides of the question:

11^2 -11p - 11q + pq = 20^2 - 20p - 20q + pq
9p + 9q = 20^2 - 11^2
9 (p+q) = (20+11) (20-11)

p + q = 20 + 11

So in effect you get the values for both P and Q, which for the purpose of this question are reversible.
Therefore:

f(10) = (10 - 20)(10 - 11)
f(10) = 10
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Re: M61-03  [#permalink]

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New post 15 May 2019, 04:44
MathRevolution wrote:
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


The product of two numbers is zero when either of/both the numbers is zero.
So we can assume: (p,q) : (11,20) in any order.

So f(10)= (10-11)(10-20)
answer 10.
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Re: M61-03   [#permalink] 15 May 2019, 04:44
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