Fill in the blanks-1-4-1-5-1-2-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Fundamental Law of Fractions

Fill in the b...

Question

Fill in the blanks:
$\frac{1}{4} + \frac{1}{5} + \frac{1}{2} =___$

Open in App

Solution

Step 1: The least common denominator (LCD) of $2$ , $4$ and $5$ is $20$ . Make the denominator as $20$ for all the fractions.
$\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$ ;
$\frac{1}{5} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}$ ;
$\frac{1}{2} = \frac{1 \times 10}{2 \times 10} = \frac{10}{20}$
Step 2: Add the numerators and write the sum over the common denominator.
$\frac{5}{20} + \frac{4}{20} + \frac{10}{20} = \frac{5 + 4 + 10}{20} = \frac{19}{20}$
Hence, $\frac{1}{4} + \frac{1}{5} + \frac{1}{2} = \frac{19}{20}$ .

Suggest Corrections

Similar questions

${\frac{1}{(s e c^{2} θ - c o s^{2} θ)} + \frac{1}{(c o s e c^{2} θ - s i n^{2} θ)}} (s i n^{2} θ c o s^{2} θ) = \frac{1 - s i n^{2} θ c o s^{2} θ}{2 + s i n^{2} θ c o s^{2} θ}$

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.

Join BYJU'S Learning Program

Fill in the blanks:14+15+12=___

Step 1: The least common denominator (LCD) of 2, 4 and 5 is 20. Make the denominator as 20 for all the fractions.14=1×54×5=520;15=1×45×4=420;12=1×102×10=1020Step 2: Add the numerators and write the sum over the common denominator.520+420+1020=5+4+1020=1920Hence, 14+15+12=1920.

Fill in the blanks:
$\frac{1}{4} + \frac{1}{5} + \frac{1}{2} =___$