# Find two fraction, having 7 and 9 for their denominators, such that the sum is $1\dfrac{10}{63}.$

Answer

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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

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