# Ex.13.1 Q7 Direct and Inverse Proportions Solution - NCERT Maths Class 8

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## Question

Suppose $$2\,\rm{kg}$$ of sugar contains $$9 \times {10^6}$$ crystals. How many sugar crystals are there in

(1)  $$5\,\rm{kg}$$  of sugar?

(2) $$1.2\,\rm{kg}$$ of sugar?

Video Solution
Direct And Inverse Proportions
Ex 13.1 | Question 7

## Text Solution

(i) How many crystals are there in $$5\, \rm{kg}$$ of crystals?

What is Known?

$$2\,\rm{kg}$$ of sugar contains $$9 × 10^6$$ crystals.

What is Unknown?

(i) $$5\,\rm{kg}$$ of sugar contains how many crystals?

Reasoning:

Two numbers $$x$$ and $$y$$ are said in direct proportion if,

\begin{align}\frac{x}{y} = k,\quad\;\; x = yk\,\end{align}

Where $$k$$ is a constant.

Steps (i):

 Amount of sugar No. of crystals $$2$$ $${9 \times 10^6}$$ $$5$$ ?

More the amount of sugar more will be the number of crystals. Hence this is a direct proportion.

\begin{align}\frac{{{x_1}}}{{{y_1}}} &= \frac{{{x_2}}}{{{y_2}}}\\\frac{2}{{9 \times {{10}^6}}}&= \frac{5}{{{y_2}}}\\2 \times y_2 &= 9 \times {10^6} \times 5\\ y_2 &= \frac{{9 \times {{10}^6} \times 5}}{2}\\{y_2} &= 22.5 \times {10^6}\\{y_2} &= 2.25 \times {10^7}\end{align}

Hence there are $$2.25 \times {10^7}$$ crystals.

(ii) How many crystals are there in $$1.2\,\rm{kg}$$ of crystals?

What is Known?

$$2\,\rm{kg}$$ of sugar contains $$9 × 10^6$$ crystals.

What is UnKnown?

How many crystals are there in $$1.2\,\rm{kg}$$ of crystals?

Steps (ii):

\begin{align}\frac{{{x_1}}}{{{y_1}}} &= \frac{{{x_2}}}{{{y_2}}}\\\frac{2}{{9 \times {{10}^6}}} &= \frac{{1.2}}{{{y_2}}}\\2 \times y_2 &= 9 \times {10^6} \times 1.2\\ y_2 &=\frac{{9 \times {{10}^6} \times 5}}{2}\\{y_2} &= 5.4 \times {10^6}\end{align}

Hence there are $$5.4 \times {10^6}$$ crystals.

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