Question

If $a,b$ and $c$ are real numbers such that $\frac{a}{b}>1$ and $\frac{a}{c}<0$. Then, which one of the following 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

The correct option is C

$(a–c)(b–c)>0$

Explanation for the correct option:

Finding the correct value.

Given that, $\frac{a}{b}>1$ and $\frac{a}{c}<0$

We can write to conditions from these two conditions as:

$a>0$ if $c<0$ and also $b>0$

$$\begin{gathered} \Rightarrow a>c \\ \Rightarrow \left(a-c\right)>0 \end{gathered}$$

$a<0$ if $c>0$ and also $b<0$

$$\begin{gathered} \Rightarrow c>b \\ \Rightarrow \left(c-b\right)>0 \\ \Rightarrow \left(b-c\right)>0 \end{gathered}$$

$\mathbf{\therefore }\left(a-c\right)\left(b-c\right)\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$

Hence, option ‘C’ is correct.