Question

If $a$ , $b$ and $c$ are real numbers such that $\dfrac{a}{b} > 1$ and $\dfrac{a}{c} < 0$. Then which one of these is correct ?
A) $a + b - c > 0$
B) $a > b$
C) $(a - c)(b - c) > 0$
D) $a + b + c > 0$
E) $abc > 0$

Answer 1

Hint: It is given that $a$ , $b$ and $c$ are real numbers so consider that to the end. Then they also give us some conditions. So just look at that condition and imagine some values for given variables and see how those conditions can be satisfied. Then see the options and you will find the right answer.

Complete step by step answer:
Given that $a$ , $b$ and $c$ are real numbers.
Now, look at given conditions,
First condition,
$ \Rightarrow \dfrac{a}{b} > 1$
From above condition we can say that,
$a > 0$ if and only if $c < 0$ and also $b > 0$
We find the above terms by putting some random values of $a$ , $b$ and $c$ .
Now look at second condition,
Second condition,
$ \Rightarrow \dfrac{a}{c} < 0$
From above condition we can say that,
$a < 0$ if and only if $c > 0$ and also $b < 0$
Again we find the above terms by putting some random values of $a$ , $b$ and $c$ .
Now, let’s come to conclusion,
So, after studying both condition we can say that,
$ \Rightarrow $$a - c > 0$ and $b - c > 0$
Now put above terms together and we can see our answer,
\[ \Rightarrow (a - c)(b - c) > 0\]
Hence, option (C) satisfies the inequality.
Therefore, the correct answer is option (C) $(a - c)(b - c) > 0$.

Note:
In this question we see real numbers. So let’s see something more about real numbers. They are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they can be represented in the number line, also.