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If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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27 Mar 2020, 01:08
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Competition Mode Question If \(\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} = r\), then r cannot take any other value except A. 1/2 B. –1 C. 1/2 or –1 D. –1/2 or –1 E. 1/2 or 1 Are You Up For the Challenge: 700 Level Questions
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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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Updated on: 29 Mar 2020, 07:14
Given, \(\frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} = r\) > \(\frac{a}{b+c} = r; \frac{b}{c+a} = r; \frac{c}{a+b} = r\) > \(a = (b + c)*r\) ....... (1) \(b = (a + c)*r\) ....... (2) \(c = (a + b)*r\) ........ (3)
(1) + (2) + (3), > \(a + b + c = (b + c + a + c + a + b+)r\) > \((a + b + c) = 2(a + b + c)*r\) > \(r = \frac{1}{2}\)
Also, If \(r = 1\), We get From (1), (2) & (3), \(a = (b + c)\); \(b = (c + a)\) & \(c = (a + b)\) or \(a + b + c = 0\) for all cases > \(r = 1\) is also possible
Option C
Originally posted by Dillesh4096 on 27 Mar 2020, 02:19.
Last edited by Dillesh4096 on 29 Mar 2020, 07:14, edited 1 time in total.



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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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27 Mar 2020, 02:23
Solution: Case 1: a + b + c ≠ 0• \(\frac{a}{(b+c)} =r\)
o \(a = r(b+c) = br + cr\)…(i) • \(\frac{b}{(c+a)}=r\)
o \(b =r(c+a) = cr + ar\) …(ii) • \(\frac{c}{(a+b)}=r\)
o \(c =r(a+b) = ar + br\)…(iii) By adding equation (i), (ii), and (iii), we get • \(a+b+c = br+cr+cr+ar+ar+br\)
o \(a+b+c=2r(a+b+c)\) o \(r = \frac{(a+b+c)}{2(a+b+c)}\) o \(r =\frac{1}{2}\) Case 2: a + b + c = 0• In this case \(b + c = a \)
o We know that \(r = \frac{a}{(b + c)} = \frac{a}{a} = 1\) o Thus, r can be 1/2 or 1 Hence, the correct answer is Option C
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If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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Updated on: 30 Mar 2020, 04:36
Quote: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other value except
A. 1/2 B. –1 C. 1/2 or –1 D. –1/2 or –1 E. 1/2 or 1 c=r(a+b), b=r(a+c), a=r(c+b) a+b+c=r([a+b]+[a+c]+[c+b]) a+b+c=r(2a+2b+2c) a+b+c=2r(a+b+c) r=1/2 a+b+c=0, b+c=a r=a/(b+c)=a/ r=1 Ans (D)
Originally posted by exc4libur on 27 Mar 2020, 06:28.
Last edited by exc4libur on 30 Mar 2020, 04:36, edited 1 time in total.



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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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27 Mar 2020, 06:35
If a=b=c=1, then a/(b + c) = b/(c + a) = c/(a + b) =1/2. Eliminate choice B and D
If a=b=1/2 and c=1, then a/(b + c) = b/(c + a) = c/(a + b) =1. Eliminate choice A and E
FINAL ANSWER IS (C) 1/2 or –1
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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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27 Mar 2020, 09:59
\(\frac{a}{b+c} = \frac{b}{c+a }=\frac{ c}{a+b} = r\)
a= br+cr.....(1) b= cr+ar.....(2) c= ar+br.....(3)
add (1), (2) and (3) a+b+c= 2(ar+br+cr)
a+b+c= 2r(a+b+c)
(2r1)(a+b+c) = 0
either r = 1/2, or a+b+c = 0
If a+b+c = 0 b+c= a
\(r= \frac{a}{b+c} = \frac{a}{a} =1\)
r can be 1/2 or 1



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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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27 Mar 2020, 10:53
If \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\) = r, then r cannot take any other value except A. 1/2 B. –1 C. 1/2 or –1 D. –1/2 or –1 E. 1/2 or 1 \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\) = r Hence a = r(b+c) b = r(c+a) c = r(a+b) Now adding all we have a + b + c = 2r(a + b + c) It can only be possible if 2r = 1 Hence \(r = \frac{1}{2}\) BunuelCan i eliminate C, D and E since they each have two values? As question is about r having no other values EXCEPT(except signifying that r can have only one value). Answer A.
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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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27 Mar 2020, 11:14
If a=b=c, r=1/2 If a=2, b=1, c=1, r=1
Answer is (C)
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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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28 Mar 2020, 06:01
Ans D
If ab+c=bc+a=ca+b=rab+c=bc+a=ca+b=r, then r cannot take any other value except
A. 1/2 B. –1 C. 1/2 or –1 D. –1/2 or –1 E. 1/2 or 1
If a=b=c we get r = 1/2
If (a,b,c) equals (3,1,2) we get r=1
Going through options ...... Only D satisfies ...... No need to check for other values



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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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29 Mar 2020, 16:48
If \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=r\), then \(r\) cannot take any other value except
Since nothing tells us about what a,b and c are, we can say that \(a=b =c\) in order to satisfy that expression. > \(\frac{a}{a+a}= \frac{a}{a+a}=\frac{a}{a+a}=r\) >\( r =\frac{1}{2}\)
In addition, \(\frac{b}{c+a}=r\) > \(b =rc +ra\) \(\frac{c}{a+b}=r\) > \(c = ra+ rb \)  \(b c = rc+ra ra rb = r (bc)\) > \(r =1\)
Well, we got two values of r: \(\frac{1}{2}\) and \(1\) Answer (C).




Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other
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29 Mar 2020, 16:48




