Find $3$ rational numbers between $0$ and $1$ .

\frac{3}{2}, \frac{1}{4}

and

\frac{3}{4}

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\frac{1}{2}, \frac{1}{4}

and

\frac{3}{4}

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\frac{1}{2}, \frac{5}{4}

and

\frac{3}{4}

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\frac{1}{2}, \frac{1}{4}

and

\frac{7}{4}

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Solution

The correct option is B $\frac{1}{2}, \frac{1}{4}$ and $\frac{3}{4}$
The mean of $0$ and $1$ is $\frac{0 + 1}{2} = \frac{1}{2}$ .

The mean of $0$ and $\frac{1}{2}$ is $(\begin{matrix} 0 + \frac{1}{2} \end{matrix}) \div 2 = \frac{1}{2} \div 2 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

The mean of $\frac{1}{2}$ and $1$ is $(\begin{matrix} \frac{1}{2} + 1 \end{matrix}) \div 2$

$= (\begin{matrix} \frac{1 + 2}{2} \end{matrix}) \div 2 = \frac{3}{2} \div 2 = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$ .

So, the $3$ rational numbers between $0$ and $1$ are $\frac{1}{2}, \frac{1}{4}$ and $\frac{3}{4}$ .

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Similar questions

< \frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4} > a n d < - \frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4} > .

f = {(1, 3), (2, 1), (3, 4), (4, 2)}

g = {(1, 2), (2, 3), (3, 4), (4, 1)}

n (f o g)

The set of solutions satisfying both $x^{2} + 5 x + 6 \geq 0 a n d x^{2} + 3 x - 4 < 0 i s$

\frac{1}{4}

\frac{3}{4}

\frac{1}{4}

\frac{1}{3}

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