Consider the following population and year graph (Fig.),find the slope of the line $AB$ and using it,find what will be the population in the year $2010$?
(N/A) Since line $AB$ passes through points $A(1985, 92)$ and $B(1995, 97)$,its slope is
$\frac{97-92}{1995-1985} = \frac{5}{10} = \frac{1}{2}$
Let $y$ be the population in the year $2010$. Then,according to the given graph,line $AB$ must pass through point $C(2010, y)$.
$\therefore$ Slope of $AB = \text{Slope of } BC$
$\Rightarrow \frac{1}{2} = \frac{y-97}{2010-1995}$
$\Rightarrow \frac{1}{2} = \frac{y-97}{15}$
$\Rightarrow \frac{15}{2} = y-97$
$\Rightarrow y-97 = 7.5$
$\Rightarrow y = 7.5 + 97 = 104.5$
Thus,the slope of line $AB$ is $\frac{1}{2}$,while in the year $2010$,the population will be $104.5$ crores.
$\frac{97-92}{1995-1985} = \frac{5}{10} = \frac{1}{2}$
Let $y$ be the population in the year $2010$. Then,according to the given graph,line $AB$ must pass through point $C(2010, y)$.
$\therefore$ Slope of $AB = \text{Slope of } BC$
$\Rightarrow \frac{1}{2} = \frac{y-97}{2010-1995}$
$\Rightarrow \frac{1}{2} = \frac{y-97}{15}$
$\Rightarrow \frac{15}{2} = y-97$
$\Rightarrow y-97 = 7.5$
$\Rightarrow y = 7.5 + 97 = 104.5$
Thus,the slope of line $AB$ is $\frac{1}{2}$,while in the year $2010$,the population will be $104.5$ crores.
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DifficultAP EAMCET 2021
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