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Multiple Choice Questions

If alpha and beta are two real numbers such that where 1 p q r then which one of the following ...........

Quantitative Aptitude (CDS) Number System

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If α and β are two real numbers such that

$\alpha + \beta=-\frac{p}{q}$

$\alpha \beta=\frac{r}{p}$

where

1 p q r, then which one of the following is the greatest?

$\frac{1}{\alpha +\beta }$

$\frac{1}{\alpha }+\frac{1}{\beta }$

$-\frac{1}{\alpha \beta }$

$\frac{\alpha \beta }{\alpha +\beta }$

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Explanation:

As 1< p

$\alpha +\beta = -\frac{p}{q}\: is \: negative$

$\alpha \beta=\frac{r}{p}\: is\: both\: positive\: and\: its\: magnitude\: is\: greater\: than\: one.$

Option 1

$\frac{1}{\alpha +\beta } = -\frac{q}{p} \: is\: negative$ .

Option 2

$\frac{1}{\alpha } + \frac{1}{\beta }= \frac{\alpha +\beta }{\alpha \beta } \: = -\frac{q}{p}(\frac{p}{r})$

$-\frac{q}{r}\: is \: negative$

Option 3

$-\frac{1}{\alpha +\beta }= -\frac{p}{r}\:$

Option 4

$\frac{\alpha \beta }{\alpha +\beta }= \frac{r}{p}\times(-\frac{p}{q})=-\frac{r}{p}$

As r > p , Hence option 1 is greater than option 4.

As q > p , Hence option 3 is greater than option 2.

$As\: \: r > p, so \: \frac{1}{r} < \frac{1}{q}$

$Hence\: \: - \frac{p}{r} >- \frac{p}{q} , Thus \:\: option\: 3 \: is\: greatest$

By: Munesh Kumari

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