Question

The mass of $1$ chocolate bar is $0.5kg$. What is the mass of $50$ chocolate in milligrams?
$\left( A \right)\,\,500$
$\left( B \right)\,\,5000$
$\left( C \right)\,\,500000$
$\left( D \right)\,\,25000000$

Answer 1

Hint: One can think of the unitary method to solve this question. The only thing we are using here is the concept of proportionality. The only thing we have to do is to store the given values in some variables and then we have to find the unknown variable for the required quantity.

Formula used: $\dfrac{{{x_1}}}{{{y_1}}} = \dfrac{{{x_2}}}{{{y_2}}}$

Complete step by step answer:
Here in the question, it is given that the mass of $1$ chocolate bar is $0.5\,kg.$
Let,
${x_1} = \,1\,chocolate\,\,\,\,\,\,\,\,\,\,\,\,\,{y_1} = 0.5kg$
${x_2} = 50\,chocolate\,\,\,\,\,\,\,\,\,\,\,\,\,{y_2} = ?$
We have to find the value of ${y_2}$.
Using unitary method,
$\dfrac{{{x_1}}}{{{y_1}}} = \dfrac{{{x_2}}}{{{y_2}}}$
On putting the values, we get
$\dfrac{1}{{0.5}} = \dfrac{{50}}{{{y_2}}}$
Using cross-multiplication method,
${y_2} = 50 \times 0.5$
${y_2} = 25kg$
But according to the question we have to find the mass of chocolate in milligrams.
We know that,
$1\,kg = \,1000\,grams - - - - \left( 1 \right)$
$1gm = 1000milligrams - - - - - \left( 2 \right)$
From $\left( 1 \right)\,\,and\,\,\left( 2 \right)$
$1\,kg\, = \,1000000\,milligrams$
Therefore,
${y_2} = 25kg$
${y_2} = 25 \times 1000000\,\,milligrams$
${y_2} = 25000000\,\,milligrams$
Hence, the mass of $50$ chocolate is $25000000$ milligrams.

Note:
> In the unitary method there are two types of variation. The first one is direct variation and the second one is indirect variation. In direct variation, increase or decrease in one quantity will cause an increase or decrease in another quantity. But indirect variation is the inverse of direct variation. If we increase a quantity, then the value of another quantity decreases.
> The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value. In essence, this method is used to find the value of a unit from the value of a multiple, and hence the value of a multiple. This method is majorly used for ratio and proportion concepts.