Find two fraction, having 7 and 9 for their denominators, such that the sum is $1\dfrac{10}{63}.$
Answer
364.2k+ views
Hint: Convert the mixed fraction $1\dfrac{10}{63}$ to complete the fraction form. Assume the numerators in two fractions and use 7 and 9 as denominators, then use LCM to find out the two fractions.
According to the question, we have to add two fractions with denominators 7 and 9, such that the result is \[1\dfrac{10}{63}\].
First lets convert the mixed fraction into proper fraction form, i.e.,
$1\dfrac{10}{63}=\dfrac{63\times 1+10}{63}$
$1\dfrac{10}{63}=\dfrac{73}{63}..........\left( 1 \right)$
Now, let the first number be $\dfrac{n}{7}$ and the second number be $\dfrac{m}{9}.$
Now as per given condition, the sum of these two fraction is equal to \[1\dfrac{10}{63}\], i.e.,
$\dfrac{n}{7}+\dfrac{m}{9}=1\dfrac{10}{63}$
Using LCM rule, $\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+cb}{bd}$, the above equation can be written as,
\[\dfrac{9n+7m}{63}=1\dfrac{10}{63}\]
Substituting value from equation (1), we get
$\dfrac{9n+7m}{63}=\dfrac{73}{63}$
Cancelling the like terms, we get
$9n+7m=73...........(2)$
Now, n and m are both natural numbers greater than zero, i.e., they cannot be decimal numbers. Since, if n or m is zero then fraction won’t exist.
Now let n = 1, then substituting this value in equation (2), we get
$\begin{align}
& 9(1)+7m=73 \\
& \Rightarrow 7m=73-9 \\
& \Rightarrow m=\dfrac{64}{7} \\
\end{align}$
This is not a natural number.
So, let n = 2, then substituting this value in equation (2), we get
$\begin{align}
& 9(2)+7m=73 \\
& \Rightarrow 7m=73-18 \\
& \Rightarrow m=\dfrac{55}{7} \\
\end{align}$
This is also not a natural number.
So, let n = 3, then substituting this value in equation (2), we get
$\begin{align}
& 9(3)+7m=73 \\
& \Rightarrow 7m=73-27 \\
& \Rightarrow m=\dfrac{46}{7} \\
\end{align}$
This is also not a natural number.
So, let n = 4, then substituting this value in equation (2), we get
$\begin{align}
& 9(4)+7m=73 \\
& \Rightarrow 7m=73-36 \\
& \Rightarrow m=\dfrac{37}{7} \\
\end{align}$
This is also not a natural number.
So, let n = 5, then substituting this value in equation (2), we get
$\begin{align}
& 9(5)+7m=73 \\
& \Rightarrow 7m=73-45 \\
& \Rightarrow m=\dfrac{28}{7}=4 \\
\end{align}$
This is a natural number.
Hence we get m = 4, n = 5 as the values of the numerator.
Therefore, the two required fractions are $\dfrac{5}{7}$ and $\dfrac{4}{9}.$
Note: We can verify our answer,
$\dfrac{5}{7}+\dfrac{4}{9}=\dfrac{5\times 9\times 4\times 7}{63}=\dfrac{45+28}{63}=\dfrac{73}{63}=1\dfrac{10}{63}$
Hence, we are right.
Also, an alternate method is to right $m=\dfrac{1}{7}\left[ 73-9n \right]$ using equation (2) and then use trial and error method.
According to the question, we have to add two fractions with denominators 7 and 9, such that the result is \[1\dfrac{10}{63}\].
First lets convert the mixed fraction into proper fraction form, i.e.,
$1\dfrac{10}{63}=\dfrac{63\times 1+10}{63}$
$1\dfrac{10}{63}=\dfrac{73}{63}..........\left( 1 \right)$
Now, let the first number be $\dfrac{n}{7}$ and the second number be $\dfrac{m}{9}.$
Now as per given condition, the sum of these two fraction is equal to \[1\dfrac{10}{63}\], i.e.,
$\dfrac{n}{7}+\dfrac{m}{9}=1\dfrac{10}{63}$
Using LCM rule, $\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+cb}{bd}$, the above equation can be written as,
\[\dfrac{9n+7m}{63}=1\dfrac{10}{63}\]
Substituting value from equation (1), we get
$\dfrac{9n+7m}{63}=\dfrac{73}{63}$
Cancelling the like terms, we get
$9n+7m=73...........(2)$
Now, n and m are both natural numbers greater than zero, i.e., they cannot be decimal numbers. Since, if n or m is zero then fraction won’t exist.
Now let n = 1, then substituting this value in equation (2), we get
$\begin{align}
& 9(1)+7m=73 \\
& \Rightarrow 7m=73-9 \\
& \Rightarrow m=\dfrac{64}{7} \\
\end{align}$
This is not a natural number.
So, let n = 2, then substituting this value in equation (2), we get
$\begin{align}
& 9(2)+7m=73 \\
& \Rightarrow 7m=73-18 \\
& \Rightarrow m=\dfrac{55}{7} \\
\end{align}$
This is also not a natural number.
So, let n = 3, then substituting this value in equation (2), we get
$\begin{align}
& 9(3)+7m=73 \\
& \Rightarrow 7m=73-27 \\
& \Rightarrow m=\dfrac{46}{7} \\
\end{align}$
This is also not a natural number.
So, let n = 4, then substituting this value in equation (2), we get
$\begin{align}
& 9(4)+7m=73 \\
& \Rightarrow 7m=73-36 \\
& \Rightarrow m=\dfrac{37}{7} \\
\end{align}$
This is also not a natural number.
So, let n = 5, then substituting this value in equation (2), we get
$\begin{align}
& 9(5)+7m=73 \\
& \Rightarrow 7m=73-45 \\
& \Rightarrow m=\dfrac{28}{7}=4 \\
\end{align}$
This is a natural number.
Hence we get m = 4, n = 5 as the values of the numerator.
Therefore, the two required fractions are $\dfrac{5}{7}$ and $\dfrac{4}{9}.$
Note: We can verify our answer,
$\dfrac{5}{7}+\dfrac{4}{9}=\dfrac{5\times 9\times 4\times 7}{63}=\dfrac{45+28}{63}=\dfrac{73}{63}=1\dfrac{10}{63}$
Hence, we are right.
Also, an alternate method is to right $m=\dfrac{1}{7}\left[ 73-9n \right]$ using equation (2) and then use trial and error method.
Last updated date: 26th Sep 2023
•
Total views: 364.2k
•
Views today: 7.64k
Recently Updated Pages
What do you mean by public facilities

Please Write an Essay on Disaster Management

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE
