A rational number between 1-4 and 1-3 is

Explanation for correct options:We have: 1/4 and 1/3The rational numbers equivalent to 1/4 is 1× 6/4× 6=6/24 , 1× 12/4× 12=12/48The rational numbers equivalent ...

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Standard VIII

Mathematics

Rational Numbers between Two Rational Numbers

A rational nu...

Question

A rational number between $\frac{1}{4}$ and $\frac{1}{3}$ is

$\frac{16}{48}$

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$\frac{13}{48}$

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$\frac{7}{24}$

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All of the above

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Solution

The correct option is C
$\frac{7}{24}$

Explanation for correct options:
We have: $\frac{1}{4}$ and $\frac{1}{3}$
The rational numbers equivalent to $\frac{1}{4}$ is $\frac{1 \times 6}{4 \times 6} = \frac{6}{24}, \frac{1 \times 12}{4 \times 12} = \frac{12}{48}$
The rational numbers equivalent to $\frac{1}{3}$ is $\frac{1 \times 8}{3 \times 8} = \frac{8}{24}, \frac{1 \times 16}{3 \times 16} = \frac{16}{48}$
So,
The rational numbers between $\frac{6}{24}$ and $\frac{8}{24}$ is $\frac{7}{24}$
The rational numbers between $\frac{12}{48}$ and $\frac{16}{48}$ are $\frac{13}{48}, \frac{14}{48}, \frac{15}{48}$
Thus, the rational numbers between $\frac{1}{4}$ and $\frac{1}{3}$ are $\frac{7}{24},$ $\frac{13}{48}, \frac{14}{48}, \frac{15}{48}$
Hence, options (b) and (c) are correct.

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Determine whether the following numbers are in proportion or not $:$

$\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{7}$

Q. If

ω

is complex cube root of unity and a, b, c are three real numbers such that

\frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} = 2 ω^{2}

and

\frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} = 2 ω

then

\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}

Q. Among

\frac{1}{12}

\frac{1}{15}

and

\frac{1}{17}

, is the greatest fraction

Q. If

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

, where

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

equals

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

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