Ex.8.1 Q1 Compairing Quantities - NCERT Maths Class 7

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Question

Find the ratio of:

(a) \(\rm Rs\, 5\) to \(50\) paise

(b) \(15\,\rm kg\) to \(210\,\rm g\)

(c) \(9 \,\rm m\) to \(27 \,\rm cm\)

(d) \(30\) days to \(36\) hours

 Video Solution
Comparing Quantities
Ex 8.1 | Question 1

Text Solution

What is Known?

Quantities in different units

What is Unknown?

Our task is to find the ratio of given quantities.

Reasoning:

These questions are based on the basic concept of comparing two quantities i.e. to find ratio. For finding ratio, first we should observe that both the quantities should be in same unit for e.g. we cannot compare metres with kilometres or grams with kilograms. So, we will have to convert both quantities into same units.

Steps:

(a) \(\rm{Rs} \, 5\) to \(50\) paise

Since, \(1 {\rm rupee} = 100 \,\rm paise\)

\[\begin{align}{\text{So, 5 rupees}}\,\,&= \frac{{{\rm{100}}}}{{\rm{1}}} \times {\rm{5}}\,\\&= \,{\rm{500}}\,{\rm{paisa}}\end{align}\]

 Therefore the ratio 

\[\begin{align}&=\frac{500}{50} \\  & =\frac{10}{1}\text{or }10\text{:}\,1 \end{align}\]

(b)\(15\,\rm kg\) to \(210\,\rm g\)

Since\(, 1 \,\rm kg = 1000\, grams\)

\[\begin{align}{\text{So,}}\,\,15\,{\rm{kg}} &= \frac{{1000}}{1} \times 15\\ &= 15000\,{\rm{grams}}\end{align}\]

 There for the ratio 

\[\begin{align}&= \frac{{15000}}{{210}}\\ &= \frac{{500}}{7} {\rm{or}}\,\,500:7\end{align}\]

(c) \(9 \,\rm m\) to \(27 \,\rm cm\)

\(1 \,\rm m = 100\, cm\)

\[\begin{align}{\rm{So,}}\,\,{\rm{9}}\,{\rm{m}} &= \frac{{100}}{1} \times 9\\ &= 900\,{\rm{m}}\end{align}\]

\[\begin{align}{\rm{Thus,}}\,{\rm{ratio}} &= \frac{{900}}{{27}}\\& = \frac{{100}}{3}{\rm{or}}\,\,100\,{\rm{:}}\,3\end{align}\]

(d) \(30\) days to \(36\) hours

Since, \(1\) day = \(24\) hrs

\[\begin{align}{\rm{So,}}\,{\rm{30}}\,{\rm{days}} = \frac{{24}}{1} \times 30\\{\rm{ = 720}}\,{\rm{hrs}}\end{align}\]

\[\begin{align}{\rm{So,}}\,{\rm{ratio}} &= \frac{{720}}{{36}}\\&= \frac{{20}}{1}{\rm{or}}\,\,20\,{\rm{:}}\,1\end{align}\]

  
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