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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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If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other  [#permalink]

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New post 27 Mar 2020, 01:08
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A
B
C
D
E

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(N/A)

Question Stats:

20% (01:25) correct 80% (01:53) wrong based on 51 sessions

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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other  [#permalink]

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New post Updated on: 29 Mar 2020, 07:14
1
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  [#permalink]

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New post 27 Mar 2020, 02:23
2

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  [#permalink]

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New post 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  [#permalink]

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New post 27 Mar 2020, 06:35
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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  [#permalink]

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New post 27 Mar 2020, 09:59
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1
\(\frac{a}{b+c} = \frac{b}{c+a }=\frac{ c}{a+b} = r\)

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
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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other  [#permalink]

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New post 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 \(r = \frac{1}{2}\)

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.
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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 29 Mar 2020, 16:48
1
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).
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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other   [#permalink] 29 Mar 2020, 16:48

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

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