Question

If the cost of 1 litre of milk is RS. \[19\dfrac{1}{5}\] then, what is the cost of \[6\dfrac{1}{4}\] litres of milk?
(a) RS.. 820
(b) RS. 320
(c) RS. 120
(d) RS. 220

Answer 1

Hint: We solve this problem first by converting the given mixed fractions into proper fractions.
The conversion of mixed fraction to proper fraction can be done as
\[a\dfrac{b}{c}=\dfrac{\left( a\times c \right)+b}{c}\]
After converting the mixed fractions to proper fractions we can find the required value easily.
We use the condition that if cost of one litre is \['k'\] then the cost of \['n'\] litres is \[n\times k\]

Complete step by step answer:
We are given that the cost of milk of 1 litre as RS. \[19\dfrac{1}{5}\]
Let us assume that the cost of 1 litre of milk as
\[\Rightarrow C=19\dfrac{1}{5}\]
We know that the conversion of mixed fraction to proper fraction can be done as
\[a\dfrac{b}{c}=\dfrac{\left( a\times c \right)+b}{c}\]
Now, let us convert the above mixed fraction into proper fraction then we get
\[\begin{align}
  & \Rightarrow C=\dfrac{\left( 19\times 5 \right)+1}{5} \\
 & \Rightarrow C=\dfrac{95+1}{5}=\dfrac{96}{5} \\
\end{align}\]
We are asked to find the cost of \[6\dfrac{1}{4}\] litres of milk.
Let us assume that the value of \[6\dfrac{1}{4}\] litres of milk as
\[\Rightarrow x=6\dfrac{1}{4}\]
Now, let us convert the above mixed fraction into proper fraction then we get
\[\begin{align}
  & \Rightarrow x=\dfrac{\left( 6\times 4 \right)+1}{4} \\
 & \Rightarrow x=\dfrac{24+1}{4}=\dfrac{25}{4} \\
\end{align}\]
Now let us assume that the cost of \[6\dfrac{1}{4}\] litres of milk as \[S\]
We know that if cost of one litre is \['k'\] then the cost of \['n'\] litres is \[n\times k\]
By using the above condition we get
\[\Rightarrow S=x\times C\]
Now, by substituting the required values in above equation we get
\[\begin{align}
  & \Rightarrow S=\dfrac{25}{4}\times \dfrac{96}{5} \\
 & \Rightarrow S=5\times 24 \\
 & \Rightarrow S=120 \\
\end{align}\]
Therefore we can say that the cost of \[6\dfrac{1}{4}\] litres of milk is RS. 120

So, the correct answer is “Option c”.

Note: Students may make mistakes in multiplying the mixed fractions.
We can directly add or subtract the mixed fractions as
\[\Rightarrow a\dfrac{b}{c}\pm x\dfrac{y}{z}=\left( a\pm x \right)\left( \dfrac{b}{c}\pm \dfrac{y}{z} \right)\]
Here \[a\pm x\] will be in the form of mixed fraction.
But we cannot multiply or divide the mixed fractions directly.
But some students may do the multiplication as
\[\Rightarrow a\dfrac{b}{c}\times x\dfrac{y}{z}=\left( a\times x \right)\left( \dfrac{b}{c}\times \dfrac{y}{z} \right)\]
This gives the wrong answer because we are not defined the mixed fractions for multiplication or division directly.