Quantitative comparision between x and 1/x

Question

Following is a GRE question called quantitative comparison.

Quantity A = $x$

Quantity B = $1/x$

$-1<x<0$

Options are as follows:

• a) Quantity A is bigger
• b) Quantity B is bigger
• c) Quantity A and B are equal
• d) Cannot determine

The answer when solved as it is A which I verified. But the way I did the sum was to simplify it first by removing fractions. So I multiplied $x$ on both sides.

After multiplication, I get

Quantity A = $x^2$ Quantity B = $1$

After simplification when I applied the values I am getting Quantity B as big.

1) What am I doing wrong?

2) Is there specific conditions under which I can only simplify?

Answer 1

If $-1<x<0$ then $B>A$, the correct answer is $b)$.

Indeed, suppose $A\geq B$, then $x^2\geq 1$ ans so $x\leq -1$ which is a contradiction.

Answer 2

When $X$ is between $-1$ and $0$, except when $X = 0$, then $A>B$.

Try it with, say, $X = -1/2$, as an example:

$A = \frac {-1/2}1 = -1/2,$ and $B = \frac 1{-1/2} = -2$. Hence, since $-1/2 > -2$, $A\gt B$.

If $x$ is between $-1$ and $0$, as the edited post claims, then the correct answer is $(a)$.

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If $x < -1$, Then $B>A$.

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If $X = 1$ or $X = -1$, then $\;A=B$.

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If $X \gt 0$, but $X<1$, Then $B>A$.

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When $X > 1$, then $A>B$.

Answer 3

The answer has to be a) Quantity A is greater than B Here's why:

Since |x| is between 0 and 1 that itself should convince you that 1/x > x

So magnitude of B is greater than A. But since x is also negative the greater quantity will in fact be smaller. Hence Quantity A is greater.

What the other guy did wrong was:

Assume A>B

=> x > 1/x

But at this point when you multiply by 'x' you must remember that x is a negative number

Hence the inequality also flips:

$x^2$ < 1

Which is consistent.

Answer 4

The correct answer is (a), that Quantity A is bigger. This is equivalent to saying that on the given range, that $x > \frac{1}{x}$. Let's look at why.

For $-1 < x < 0$, we really need to focus just on the left-hand inequality, namely that $-1 < x$. If we divide both sides by $x$ noting that $x$ is negative, then we find that $$ \frac{-1}{x} > \frac{x}{x} = 1 $$ where we have reversed the inequality since $x < 0$. Now, mulitply both sides by -1 to get (again, reversing the inequality that $$ -\frac{1}{x} < -1 < x $$ but this is exactly saying that quantity B is less than quantity A on the given range. Hence (a) is the answer.