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Re: If the ratio of x to y is 2 to 5, then (y-x)/(x+y)? [#permalink]
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I believe this problem can be solved using algebra.

The ratio of x to y is 2 to 5 means
(x/y)=(2/5)

Cross-multiply

2x = 5y
x = (2/5)*y

Then, substitute x in the original equation in the problem with the value we just found.

(y - (2/5)*y) / ((2/5)*y + y)
= ((3/5)*y) / ((7/5)*y)

Simplify by canceling the y in the numerator with the y in the denominator.
(3/5) / (7/5)
= (3/5) * (5/7)

The 5s cancel each other out, and you are left with 3/7.

I would love an expert to double check me on this, but this should be a sufficient way to complete the problem using algebra. However, plugging in numbers, in this case, is a more efficient way to complete the problem and is also less prone to error because you are eliminating variables.
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Re: If the ratio of x to y is 2 to 5, then (y-x)/(x+y)? [#permalink]
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MrWallSt wrote:
If the ratio of x to y is 2 to 5, then (y-x)/(x+y)?

A. -7/3
B. -3/7
C. 3/7
D. 1
E. 7/3

Can any one solve this using algebra and not plugging in?


In my point of view this is very easy question. You can use Componendo and dividendo method to reach the answer.

x/y= 2/5
x+y/x-y = 2+5/2-5
x+y/x-y = 7/-3
Can be written as -3/7 = x-y/x+y
also 3/7 = y-x/x+y


Thanks,
AB

+1 Kudos if you like and understand

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Re: If the ratio of x to y is 2 to 5, then (y-x)/(x+y)? [#permalink]
@sjmarinov and @bhatiamanu05, both of those methods make sense. Appreciate the responses.
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Re: If the ratio of x to y is 2 to 5, then (y-x)/(x+y)? [#permalink]
\(\frac{x}{y} = \frac{2}{5}\)

We require to find \(\frac{y-x}{x+y} = \frac{y}{x+y} - \frac{x}{x+y}\)

\(\frac{y}{x} = \frac{5}{2}\)

\(\frac{y}{x+y} = \frac{5}{5+2} = \frac{5}{7}\) ......... (a)

\(\frac{x}{x+y} = \frac{2}{5+2} = \frac{2}{7}\) .......... (b)

(a) - (b)

\(= \frac{5}{7} - \frac{2}{7}\)

\(= \frac{3}{7}\)

Answer = C
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Re: If the ratio of x to y is 2 to 5, then (y-x)/(x+y)? [#permalink]
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Re: If the ratio of x to y is 2 to 5, then (y-x)/(x+y)? [#permalink]
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