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# M61-03

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7999
GMAT 1: 760 Q51 V42
GPA: 3.82

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18 Jun 2018, 00:43
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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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7999 GMAT 1: 760 Q51 V42 GPA: 3.82 Re M61-03 [#permalink] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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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 .
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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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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Joined: 06 Feb 2018
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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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Joined: 23 Apr 2019
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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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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)
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Re: M61-03   [#permalink] 15 May 2019, 04:44
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# M61-03

Moderators: chetan2u, Bunuel