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if $\left(\dfrac{x+y}{x-y}=\dfrac{4}{3}\right)$ and $x ≠ 0$, then what percentage (to the nearest integer) of $x + 3y$ is $x – 3y$ ?









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Option(D) is correct

Dividing  both  the  numerator  and  the  denominator  of  the  given  equation :



Cross-multiplying this equation yields 


Solving for $x/y$ yields \(x/y=7\)

Now, the percentage of $x+3y$ the expression $x-3y$ makes is

\(\dfrac{x-3y}{x+3y}\times 100\)

Dividing both the numerator and the denominator of the expression by $y$ yields

\(=\dfrac{x/y-3}{x/y+3}\times 100\)

\(=\dfrac{7-3}{7+3}\times 100\)


Edit: For an alternative solution using Componendo & Dividendo method, check comment by Chirag Goyal.

Edit 2: For yet another alternative solution, check comment by Ameya.

(3) Comment(s)


x+y = 4k

x - y = 3k

by solving

x=7k/2 y = k/2

substitute for x and y in the required equation to get 40% as your answer


One solution would be,

The question is asking what percent of $x+3y$ is $x–3y$

So let that percent be $z$

therefore $\dfrac{z}{100}*(x + 3y) = x- 3y$ -------- (1)

From the equation in the question we get,

$3x + 3y = 4x - 4y$

$\Rightarrow x = 7y$

Substitute this in equation 1,

$z = \textbf{40%}$

Chirag Goyal

$\dfrac{x+y}{x-y}=\dfrac{4}{3}$ $or$ $\dfrac{x+y}{x-y}=\dfrac{8}{6}=\dfrac{7+1}{7-1}$

Using $\text{Componendo & Dividendo}$


$\dfrac{x-3y}{x+3y}=\dfrac{7-3\times1}{7+3\times1}=\dfrac{4}{10}\ or\ 40\%$