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The question is : if $x=y-\frac{50}{y}$ , where x and y are both > $0$ . If the value of y is doubled in the equation above the value of x will a)Decrease b)remain same c)increase four fold d)double e)Increase to more than double - Ans)e

Here is how I am solving it $x = \frac{y^2-50}{y}$ by putting y=$1$ . I get $-49$ and by putting y=2(Doubling) I get $y=-24$. Now how is $-24$ more than double of $-49$. Correct me if I am wrong double increment of $-49$ will be = $-49+49 = 0$ which would be greater than -24.

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2 Answers 2

up vote 4 down vote accepted

Well, the problem statement explicitly restricts itself away from cases where $x<0$, so I think your concern is eliminated by that restriction.

In any rate, consider $x_1 = y-\frac{50}{y}$ and $x_2 = 2y-\frac{50}{2y}$. Then, compute $\frac{x_2}{x_1}$ symbolically, and verify what it can be.

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Let $x_1=y-\frac{50}{y}$ and $x_2=2y-\frac{50}{2y}$. Then $x_2-x_1=(2y-y)-(\frac{50}{2y}-\frac{50}{y})=y+\frac{50}{2y}=y-\frac{50}{y}+(\frac{150}{2y})=x_1+(\frac{150}{2y})$

$\implies x_2-2x_1=(\frac{150}{2y})\gt 0$ (as $y\gt 0$)$\implies x_2\gt 2x_1\implies x_2/x_1\gt 2$ which is option (e).

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