Question

The least fraction that must be added to $1\dfrac{1}{3}\div 1\dfrac{1}{2}\div 1\dfrac{1}{9}$ to make the result as an integer is:
(a) $\dfrac{4}{5}$
(b) $\dfrac{3}{5}$
(c) $\dfrac{1}{81}$
(d) $\dfrac{1}{5}$

Answer 1

Hint: First of all convert all the mixed fractions into proper fractions. We are going to show the conversion of mixed fraction to proper fraction as follows $1\dfrac{1}{3}=1+\dfrac{1}{3}=\dfrac{4}{3}$. Similarly, do for other fractions as well. Then remove the division sign between the fractions and instead write a multiplication sign after that you can easily multiply the numerator and the denominator of a fraction. Then subtract the fraction you have got from 1. The result of subtraction is the answer.

Complete step-by-step answer:
The division of improper fraction is given in the question as:
$1\dfrac{1}{3}\div 1\dfrac{1}{2}\div 1\dfrac{1}{9}$……….eq. (1)
Now, we are going to convert all these improper fractions into proper fractions.
We can write $1\dfrac{1}{3}$ as:
$1\dfrac{1}{3}=1+\dfrac{1}{3}$
Taking L.C.M as 3 on the right hand side of the above equation we get,
$1\dfrac{1}{3}=\dfrac{3+1}{3}=\dfrac{4}{3}$
Hence, we have converted $1\dfrac{1}{3}$ to $\dfrac{4}{3}$.
Similarly, we can convert $1\dfrac{1}{2}$ and $1\dfrac{1}{9}$ to proper fractions as follows:
$\begin{align}
  & 1\dfrac{1}{2}=1+\dfrac{1}{2}=\dfrac{3}{2} \\
 & 1\dfrac{1}{9}=1+\dfrac{1}{9}=\dfrac{10}{9} \\
\end{align}$
Hence, we have converted $1\dfrac{1}{2}$ to $\dfrac{3}{2}$ and $1\dfrac{1}{9}$ to $\dfrac{10}{9}$.
Substituting these proper fractions in eq. (1) we get,
$\dfrac{4}{3}\div \dfrac{3}{2}\div \dfrac{10}{9}$
Now, eliminating division sign between the fractions and substituting the multiplication sign we get,
$\dfrac{4}{3}\times \dfrac{2}{3}\times \dfrac{10}{9}$
The multiplication of the numerator and the denominator in the above expression will give:
$\dfrac{80}{81}$
The least fraction which in addition will give integer is one in addition to which the above fraction will give 1. Let us assume that the fraction is x.
$x+\dfrac{80}{81}=1$
Subtracting $\dfrac{80}{81}$ on both the sides of the above equation we get,
$x=1-\dfrac{80}{81}$
$\Rightarrow x=\dfrac{81-80}{81}$
$\Rightarrow x=\dfrac{1}{81}$
From the above, the least fraction is $\dfrac{1}{81}$ which in addition with the given fraction will give the least integer.
Hence, the correct option is (c).

Note: The other way of solving the above problem is simplify in the same way as we did above till the proper fraction and its multiplication which is given below:
$\dfrac{80}{81}$
Then add the options given in the question to the above fraction and see in addition of which option will give you the least integer.
We are checking the option (a) so adding option (a) in $\dfrac{80}{81}$ we get,
$\dfrac{4}{5}+\dfrac{80}{81}$
Taking 240 as a L.C.M in the above expression we get,
$\dfrac{324+400}{405}=\dfrac{724}{405}$
This option is incorrect because we are not getting an integer.
Now, we are checking the correct option i.e. (c). Adding option (c) in $\dfrac{80}{81}$ we get,
$\begin{align}
  & \dfrac{1}{81}+\dfrac{80}{81} \\
 & =\dfrac{81}{81}=1 \\
\end{align}$
This option is correct because it has given the integer.