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If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff

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If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff  [#permalink]

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New post 16 Jan 2015, 05:36
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If \(\frac{b}{(a-c)}\) \(= \frac{(a+b)}{c} = \frac{a}{b}\) for three positive number a,b, and c , all different, then \(\frac{a}{b}=\)

a) 1/2
b) 3/5
c) 2/3
d) 5/3
e) 2
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Re: If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff  [#permalink]

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New post 21 Jan 2015, 21:39
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\(\frac{b}{(a-c)} = \frac{a}{b}\)

\(b^2 = a^2 - ac\) ............. (1)

\(\frac{(a+b)}{c} = \frac{a}{b}\)

\(b^2 = ac - ab\) .................. (2)

As LHS is same, equating RHS of (1) & (2)

\(a^2 - ac = ac - ab\)

a - c = c - b

a+b = 2c .............. (3)

Replacing value of a+b in the original equation

\(\frac{a}{b} =\frac{2}{1}\)

Answer = E
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Re: If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff  [#permalink]

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New post 16 Jan 2015, 06:29
manpreetsingh86 wrote:
If \(\frac{b}{(a-c)}\) \(= \frac{(a+b)}{c} = \frac{a}{b}\) for three positive number a,b, and c , all different, then \(\frac{a}{b}=\)

a) 1/2
b) 3/5
c) 2/3
d) 5/3
e) 2


I got answer correct, but want to know other way to solve.
So Kindly provide OE
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Re: If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff  [#permalink]

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New post 16 Jan 2015, 09:07
anupamadw wrote:
manpreetsingh86 wrote:
If \(\frac{b}{(a-c)}\) \(= \frac{(a+b)}{c} = \frac{a}{b}\) for three positive number a,b, and c , all different, then \(\frac{a}{b}=\)

a) 1/2
b) 3/5
c) 2/3
d) 5/3
e) 2


I got answer correct, but want to know other way to solve.
So Kindly provide OE


hi, please post your approach.
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Re: If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff  [#permalink]

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New post 16 Jan 2015, 09:48
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manpreetsingh86 wrote:
If \(\frac{b}{(a-c)}\) \(= \frac{(a+b)}{c} = \frac{a}{b}\) for three positive number a,b, and c , all different, then \(\frac{a}{b}=\)

a) 1/2
b) 3/5
c) 2/3
d) 5/3
e) 2



b/(a-c) = (a+b)/c
bc=(a-c)(a+b)
bc=a^2+ab-ac-bc
2bc = a^2+ab-ac ---1

b/(a-c) = a/b
b^2=a^2-ac ----2

(a+b)/c=a/b
ab+b^2=ac
b^2=ac-ab -----3

From 2 and 3
a^2-ac = ac-ab
a^2+ab=2ac

Substitute back in (1)
2bc=a^2+ab-ac
2bc=2ac-ac
2bc=ac ( c is +ve number )
a/b=2

Ans E

But this is not nice way, I know :(
Could you please provide the quicker way to solve.
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Re: If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff  [#permalink]

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New post 16 Jan 2015, 11:28
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\(\frac{b}{a-c} = \frac{a+b}{c} = \frac{a}{b} =k\)

so, here we have to find the value of k.

b= ak-ck ------------1)

a+b =ck ---------------2)

adding 1) and 2) we have

a+2b = ak

a(k-1)= 2b

a/b = 2/k-1

also as per the question a/b =k

thus we have k = 2/k-1 ( after this step we can also plug-in the answer choices to get the final answer).

or k^2-k-2=0
(k+1)(k-2)=0
k=-1 or k=2.

out of these values only k=2 is given in the option. hence answer must be E.
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Re: If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff  [#permalink]

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New post 16 Jan 2015, 11:55
anupamadw wrote:
manpreetsingh86 wrote:
If \(\frac{b}{(a-c)}\) \(= \frac{(a+b)}{c} = \frac{a}{b}\) for three positive number a,b, and c , all different, then \(\frac{a}{b}=\)

a) 1/2
b) 3/5
c) 2/3
d) 5/3
e) 2


I got answer correct, but want to know other way to solve.
So Kindly provide OE

\(b/(a-c)=(a+b)/c=a/b\)
(I) (II) (III)

Using equations (I) & (III)
\(b^2 = a^2 - ac\) …………….. (IV)

Using equations (I) & (II)
\(bc= a^2-ac+ab-bc\)
\(2bc= [(a^2)-(ac)]+ab\)

Using equation (IV)
\(2bc= (b^2)+ab\)

Thus,
Either, b=0 (This is not possible due to given equality in question)
Or, \(2c = a + b\)
\(2=(a+b)/c\)
Using equality given in question
\(a/b=2\)

I hope it is a simple solution
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Re: If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff  [#permalink]

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New post 01 Feb 2017, 04:54
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manpreetsingh86 wrote:
If \(\frac{b}{(a-c)}\) \(= \frac{(a+b)}{c} = \frac{a}{b}\) for three positive number a,b, and c , all different, then \(\frac{a}{b}=\)

a) 1/2
b) 3/5
c) 2/3
d) 5/3
e) 2


\(\frac{b}{(a-c)}\) \(= \frac{(a+b)}{c} = \frac{a}{b}\)

The value of a/b is something from the 5 options.

Note that \(\frac{b}{(a-c)}\) \(= \frac{a}{b}\)
b * b = a * (a - c)
a, b and c are all positive numbers. (a - c) will be less than a so b must be less than a.
So only two options are possible - 5/3 or 2/1

Try 2/1 since it is easier.
If b = 2, a = 4. Then a - c = 4 - 3 = 1

Plug these values in here: \(\frac{(a+b)}{c}\), you get 6/3 = 2/1

It fits. This is the answer.

Answer (E)
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If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff  [#permalink]

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New post 01 Feb 2017, 21:07
manpreetsingh86 wrote:
If \(\frac{b}{(a-c)}\) \(= \frac{(a+b)}{c} = \frac{a}{b}\) for three positive number a,b, and c , all different, then \(\frac{a}{b}=\)

a) 1/2
b) 3/5
c) 2/3
d) 5/3
e) 2


equation1: b/(a-c)=a/b➡ac=a^2-b^2
equation2: (a+b)/c=a/b➡ac=ab+b^2
thus, a^2-b^2=ab+b^2➡
a^2-ab-2b^2=0➡
(a+b)(a-2b)=0
a+b=0 won't work as both numbers are positive
a-2b=0➡a=2b
only possible choice is a=2 and b=1
e
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Re: If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff  [#permalink]

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Re: If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff   [#permalink] 12 Jun 2019, 03:47
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