If a, b and c are positive and a/(2b - c)=2b/(3a + c) = a/b. What is

Question

If a, b and c are positive and  \frac{a}{2b-c}=\frac{2b}{3a+c}=\frac{a}{b} . What is the value of  \frac{a}{b} ?

A.  \frac{1}{3}   
 
B.  \frac{2}{3}  
 
C.  1 
 
D.  \frac{4}{3} 
 
E.  \frac{5}{3}

$!vakumar.m · Answer

a/(2b-c) = a/b
ab = 2ab - ac
=> ab = ac (since a is positive we can cancel it out)
=> b =c
Now, 2b/(3a+c) = a/b
=> 2b/(3a+b) = a/b
=> 2b^2 = 3a^2 +ab
divide both sides by ab
2b/a = 3a/b + 1
Now, if we substitute the value of a/b given in options - only Option B satisfies this equation.
Answer B

Aryaa03 · Answer

Sol :

\frac{a}{2b-c}=\frac{a}{b} .
As a is non-zero, cancel from each side and take reciprocal;
This gives, 2b-c = b or b = c;

\frac{2b}{3a+c}=\frac{a}{b} 
Put c = b and cross-multiply
2b*(b) = a*(3a+b)
2b^2 = 3a^2 + ab
3a^2 - 2b^2 +ab =0
3a^2 - 3b^2 +b^2 + ab =0
3(a^2-b^2) + b(a+b) = 0
3(a+b)(a-b) + b(a+b) = 0
(a+b)(3a-3b+b) = 0
(a+b)(3a-2b) = 0
As a, b and c are non-zero , a+b can't be 0.

Therefore, 3a-2b = 0 ==> 3a=2b or a/b = 2/3

Option-B

sahayrahul · Answer

I did b = c and then I evaluated two equation as below
a/2b-c = 2b/3a+c ==> 3a^2+ac = 4b^2-2bc ------1
now 2b/3a+c = a/b ==> 2b^2 =3a^2+ac ------2
Now I substitute eqn 2 in 1 for 2b^2 *2 = 4b^2
==> 3a^2+ac = 6a^2+2ac
==> -3a^2 = ac ==> a/c = -1/3

now we know b=c so we can say a/b == -1/3

what is my mistake in above approach

ZIX · Answer

I substituted the values of a/b from the options as I find solving through options much easier for such question types.

For example, option A:

\frac{a}{2b-c} = \frac{2}{3}; \frac{2b}{3a+c} = \frac{2}{3}

3a = 2b-c; 6b = 3a+c   (add these two together)
 6b=3b  (not possible)

Then using Option B, substituting  \frac{2}{3}  gave back  \frac{a}{b} = \frac{2}{3} . So this is our solution.

Krunaal · Answer

When you substituted (2) in (1), you did not include "-2bc" of (1) as highlighted above.

Praveena_10 · Answer

let the ratio is equal to k.

a/b=k from this a=bk

a/2b-c=k from this a=ck

2b/3a+c=k
2b=3ak+ck
2(a/k)=3ak+a
2=3k^2+k
3k^2+k-2=0

solving the eq we get k=2/3,-1. since both are positive numbers the ratio will be positive so k=2/3.

If a, b and c are positive and a/(2b - c)=2b/(3a + c) = a/b. What is

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If a, b and c are positive and a/(2b - c)=2b/(3a + c) = a/b. What is

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