Question

Ria decided to bake a cake. For this, she used $\frac{3}{5}\text{kg}$ flour, $\frac{1}{6}\text{kg}$ of butter, $\frac{1}{3}\text{kg}$ of sugar, and $\frac{2}{9}\text{kg}$ of milk. What will be the total weight of cake?

• A. $\frac{7}{23}\text{kg}$
• B. $1\frac{29}{90}\text{kg}$
• C. $3\frac{2}{7}\text{kg}$
• D. $\frac{23}{7}\text{kg}$

Answer 1

The correct option is B $1\frac{29}{90}\text{kg}$

The fraction of different ingredients used by Ria is as follows,

$\text{Fraction of flour used}=\frac{3}{5}\text{kg}$

This fraction can be represented as,


$\text{Fraction of butter used}=\frac{1}{6}\text{kg}$, and this fraction can be represented as,


$\text{Fraction of sugar used}=\frac{1}{3}\text{kg}$

This fraction can be represented as,


And, $\text{Fraction of milk used}=\frac{2}{9}\text{kg}$

This fraction can be represented as,


Thus, the weight of cake, i.e., the combined fraction of all the given fractions will be calculated as,

$\text{Combined fraction}=\frac{3}{5}+\frac{1}{6}+\frac{1}{3}+\frac{2}{9}$

Since the fractions here are unlike fractions, therefore to compute the combined fraction, these fractions have to be converted to like fractions with same denominator.

The LCM of $5,6,3$, and $9$ is $90$, therefore the fractions can be represented as,

$\begin{array}{cc}\frac{3}{5}& =\frac{3}{5}\times \frac{18}{18}\\ & =\frac{54}{90}\\ & \\ \frac{1}{6}& =\frac{1}{6}\times \frac{15}{15}\\ & =\frac{15}{90}\\ & \\ \frac{1}{3}& =\frac{1}{3}\times \frac{30}{30}\\ & =\frac{30}{90}\\ & \\ \frac{2}{9}& =\frac{2}{9}\times \frac{10}{10}\\ & =\frac{20}{90}\end{array}$

Thus, the weight of cake will be calculated as,

$\begin{array}{cc}\text{Combined fraction}& =\frac{54}{90}+\frac{15}{90}+\frac{30}{90}+\frac{20}{90}\\ & =\frac{54+15+30+20}{90}\\ & =\frac{119}{90}\\ & =1\frac{29}{90}\end{array}$

Hence, $1\frac{29}{90}\text{kg}$ is the total weight of the cake.