Quantitative Comparison – Absolute Values

A very interesting question testing concepts in absolute values, number line, and the meaning of arithmetic mean of a set of numbers. A good question to understand the meaning of the absolute values of the difference between two numbers.

Question

p, q, and r are three points on the real number line where q = $latex \frac{2pr}{p+r}$.

Quantity A: $latex \left | \frac{1}{p} – \frac{1}{r}\right |$

Quantity B: $latex \left | \frac{1}{q} – \frac{1}{p}\right |$

Correct Answer

Choice A – Quantity A is greater

Explanatory Answer

Given: q = $latex \frac{2pr}{p + r}$

Let’s make sense of this information.
Cross multiplying, we get q(p + r) = 2pr
i.e., qp + qr = 2pr

Divide the equation by pqr
$latex \frac{1}{r}$ + $latex \frac{1}{p}$ = $latex \frac{2}{q}$

Divide the equation by 2
$latex \frac{\frac{1}{r} + \frac{1}{p}}{2}$ = $latex \frac{1}{q}$

What can we infer from the above equation?

The same inference that we can draw if someone told us that z = $latex \frac{x + y}{2}$
The inference we will draw is that z is the arithmetic mean (average) of x and y.

Using the same logic, we can infer that $latex \frac{1}{q}$ is the arithmetic mean (average) of $latex \frac{1}{p}$ and $latex \frac{1}{r}$.

A few key insights from this inference

If z is the average of x and y, it is evident that z will lie between x and y on the number line.

By the same logic, $latex \frac{1}{q}$ will lie between $latex \frac{1}{p}$ and $latex \frac{1}{r}$ on the number line.

What are we trying to compare?

$latex \left | \frac{1}{p} – \frac{1}{r}\right |$ and $latex \left | \frac{1}{q} – \frac{1}{p}\right |$

If x and y are two points on the number line, what does |x – y| measure?

|x – y| measures the distance between the two points x and y.

Extending that logic, $latex \left | \frac{1}{p} – \frac{1}{r}\right |$ measures the distance between points $latex \frac{1}{p}$ and $latex \frac{1}{r}$

Similarly, quantity B measures the distance between points $latex \frac{1}{p}$ and $latex \frac{1}{q}$

We deduced that $latex \frac{1}{q}$ lies between $latex \frac{1}{p}$ and $latex \frac{1}{r}$.

Therefore, the distance between $latex \frac{1}{p}$ and $latex \frac{1}{r}$ (quantity A) will be greater than the distance between $latex \frac{1}{p}$ and $latex \frac{1}{q}$

Hence, Quantity A is greater than Quantity B