Fill in the blanks-1-2-1-3-1-4-

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Standard IX

Mathematics

Fundamental Law of Fractions

Fill in the b...

Question

Fill in the blanks:
$\frac{1}{2} + \frac{1}{3} + \frac{1}{4} =___$

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Solution

Step 1: The least common denominator (LCD) of $2$ , $3$ and $4$ is $12$ . Make the denominator as $12$ for all the fractions.
$\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12}$ ;
$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$ ;
$\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$
Step 2: Add the numerators and write the sum over the common denominator.
$\frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{6 + 4 + 3}{12} = \frac{13}{12}$
Hence, $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{12}$ .

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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

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. Among

\frac{1}{12}

\frac{1}{15}

and

\frac{1}{17}

, is the greatest fraction

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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Fill in the blanks:12+13+14=___

Step 1: The least common denominator (LCD) of 2, 3 and 4 is 12. Make the denominator as 12 for all the fractions.12=1×62×6=612;13=1×43×4=412;14=1×34×3=312Step 2: Add the numerators and write the sum over the common denominator.612+412+312=6+4+312=1312Hence, 12+13+14=1312.

Fill in the blanks:
$\frac{1}{2} + \frac{1}{3} + \frac{1}{4} =___$