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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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27 Mar 2020, 01:08
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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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Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other  [#permalink]

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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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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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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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27 Mar 2020, 06:35
1
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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27 Mar 2020, 09:59
2
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)

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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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)

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).

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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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27 Mar 2020, 11:14
If a=b=c, r=1/2
If a=-2, b=1, c=1, r=-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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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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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.
--> $$\frac{a}{a+a}= \frac{a}{a+a}=\frac{a}{a+a}=r$$
-->$$r =\frac{1}{2}$$

$$\frac{b}{c+a}=r$$ --> $$b =rc +ra$$
$$\frac{c}{a+b}=r$$ --> $$c = ra+ rb$$
--------------------
$$b -c = rc+ra -ra -rb = -r (b-c)$$
--> $$r =-1$$

Well, we got two values of r: $$\frac{1}{2}$$ and $$-1$$
Re: If a/(b + c) = b/(c + a) = c/(a + b) = r then r cannot take any other   [#permalink] 29 Mar 2020, 16:48