If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other

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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

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GMATWhizTeam · Accepted Answer

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

CareerGeek · Answer

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

exc4libur · Answer

c=r(a+b), b=r(a+c), a=r(c+b)
a+b+c=r( + + )
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)

freedom128 · Answer

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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nick1816 · Answer

\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)

(2r-1)(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

unraveled · Answer

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}

Bunuel 
Can 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.

firas92 · Answer

If a=b=c, r=1/2
If a=-2, b=1, c=1, r=-1

Answer is (C)

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razorhell · Answer

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

lacktutor · Answer

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 (b-c) 
-->  r =-1

Well, we got two values of r:  \frac{1}{2}  and  -1 
 Answer (C).

meilimeagan · Answer

How do you know to check the case that a+b+c=0?

If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other

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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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