Question 23 Marks
The population of a town in the year 2014 was 150,500 . If the annual increasing during three successive years he at the rate of $7 \%, 8 \%$ and $6 \%$ respectively, find the population at the end of 2017.
AnswerLet $P$ be the population at the end of 2017. Here,
$\mathrm{P}_{0}=150,500, \mathrm{r}_{1}=7, \mathrm{r}_{2}=8$ and $\mathrm{r}_{3}=6$
$\therefore \mathrm{P}=\mathrm{P}_{0}\left(1+\frac{r_{1}}{100}\right)\left(1+\frac{r_{2}}{100}\right)\left(1+\frac{r_{3}}{100}\right)$
$\Rightarrow \mathrm{P}=150,500\left(1+\frac{7}{100}\right)\left(1+\frac{8}{100}\right)\left(1+\frac{6}{100}\right)=150500\left(\frac{107}{100} \times \frac{108}{100} \times \frac{106}{100}\right)$
$\Rightarrow \mathrm{P}=\frac{1505 \times 107 \times 108 \times 106}{10000}$
$\Rightarrow \log \mathrm{P}=\log \left(\frac{1505 \times 107 \times 108 \times 106}{10^{4}}\right)$
$\Rightarrow \log \mathrm{P}=\log 1505+\log 107+\log 108+\log 106-\log 10^{4}$
$\Rightarrow \log \mathrm{P}=\log 1505+\log 107+\log 108+\log 106-4 \log 10$
$\Rightarrow \log \mathrm{P}=3.4775+2.0294+2.0334+2.0253-4=5.2656$
$\Rightarrow \mathrm{P}=\operatorname{antilog}(5.2656)=184,400$
View full question & answer→Question 33 Marks
Divide ₹ 21866 into two parts such that the amount of one in 3 years is same as the amount of the second in 5 years, the rate of compound interest being $5 \%$ per annum.
AnswerDivide ₹ 21866 into two parts such that the amount of one in 3 years is same as the amount of the second in 5 years, the rate of compound interest being $5 \%$ per annum.
Divide Rs. 21866 into two parts such that the amount of one part in 3 years is the same as the amount of the second part in 5 years.
Let, First part ( That put for 3 years ) $=x$
So, Second part ( That put for 5 years ) $=21866-x$
We know the formula for compound interest:
$A=P\left(1+\frac{R}{100}\right)^{n}$
So from the given condition, we get,
$x\left(1+\frac{5}{100}\right)^{3}=(21866-x)\left(1+\frac{5}{100}\right)^{5}$
$x=(21866-x)\left(1+\frac{5}{100}\right)^{5-3}$
$x=(21866-x)\left(\frac{105}{100}\right)^{2}$
$x=(21866-x)\left(\frac{21}{20}\right)\left(\frac{21}{20}\right)$
$\frac{400 x}{441}+x=21866$
$\frac{400 x+441 x}{441}=21866$
$\frac{841 x}{441}=21866$
$x=\frac{21866 \times 441}{841}$
$\mathrm{x}=11466$
Therefore,
The first part (That put for 3 years) $=₹ 11466$
So, the second part (That put for 5 years) $=21866-11466=₹ 10400$
View full question & answer→Question 43 Marks
If $\mathrm{A}=[1,2,3]$ and $\mathrm{f} g, \mathrm{~h}$ and s are relations corresponding to the subsets of $\mathrm{A} \times \mathrm{A}$ indicated against them,
which of $f, g, h$ and $s$ are functions? In case of a function, find its domain and range.
i. $\mathrm{f}=\{(2,1),(3,3)\}$
ii. $g=\{(1,2),(1,3),(2,3),(3,1)\}$
iii. $\mathrm{h}=\{(1,3),(2,1),(3,2)\}$
iv. $s=\{(1,2),(2,2),(3,1)\}$
Answeri. $f$ is not a function because the element 1 of A does not appear as the first component of ordered pairs of $f$, so 1 has no image in A.
ii. $g$ is not a function because the different pairs $(1,2)$ and $(1,3)$ of $g$ have the same first component i.e. the element 1 of $A$ has two different images in A.
iii. h is a function because each element of A has a unique image in A .
Domain of $h=\{1,2,3\}=A$ and range of $h=\{3,1,2\}=A$.
iv. $s$ is a function because each element of $A$ has a unique image in $A$.
Domain of $s=\{1,2,3\}=A$ and range of $s=\{2,1\}$.
View full question & answer→Question 53 Marks
Find the foot of the perpendicular from the point $(3,8)$ to the line $x+3 y=7$.
AnswerThe given equation of the line is
$\mathrm{x}+3 \mathrm{y}-7$
$\Rightarrow y=-\frac{1}{3} x+\frac{7}{3}$
$\therefore$ Slope of the line, $\mathrm{m}_{1}=-\frac{1}{3}$
Let $m_{2}$ be the slope of the perpendicular line.
$\therefore \mathrm{m}_{1} \mathrm{~m}_{2}=-1$
$\Rightarrow-\frac{1}{3} \times m_{2}=-1$
$\Rightarrow \mathrm{m}_{2}=3$
$\therefore$ Equation of the perpendicular line with slope 3 and passing through $(3,8)$ is
$y-8=3(x-3)$
$\Rightarrow 3 \mathrm{x}-\mathrm{y}-1=0$
$\therefore$ The foot of perpendicular is the point of intersection of the lines $x+3 y-7=0$ and $3 x-y-1=0$.
Solving these equations, we get
$x=1, y=2$, So $(1,2)$ is the foot of the perpendicular.
View full question & answer→Question 63 Marks
Given a G.P. with $\mathrm{a}=729$ and 7 th term $=64$, determine $\mathrm{S}_{7}$.
AnswerLet $r$ be the common ratio of G.P. Then
$\mathrm{a}_{7}=64 \Rightarrow \mathrm{ar}^{7-1}=64 \Rightarrow 729 \mathrm{r}^{6}=64$
$\Rightarrow \quad r^{6}=\frac{64}{729} \Rightarrow r^{6}=\left(\frac{2}{3}\right)^{6}$ or $\left(-\frac{2}{3}\right)^{6}$
$\Rightarrow \quad r=\frac{2}{3}$ or $-\frac{2}{3}$
When $\mathrm{r}=\frac{2}{3}, S_{7}=\frac{a\left(1-r^{7}\right)}{1-r}=\frac{729\left(1-\left(\frac{2}{3}\right)^{7}\right)}{1-\frac{2}{3}} 3 \times 729\left(1-\left(\frac{2}{3}\right)^{7}\right)$
$=2187-128=2059$
When $\mathrm{r}=\frac{2}{3}, S_{7}=\frac{729\left(1-\left(-\frac{2}{3}\right)^{7}\right)}{1-\left(-\frac{2}{3}\right)}=\frac{3}{5} \times 729\left(1+\frac{2^{7}}{3^{7}}\right)$
$=\frac{1}{5}(2187+128)=\frac{2315}{5}=463$
View full question & answer→Question 73 Marks
Find $n$ so that $\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ maybe the A.M. between a and b .
AnswerThe arithmetic mean between a and b is $\frac{a+b}{2}$
According to given, $\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}=\frac{a+b}{2}$
$\Rightarrow 2 \mathrm{a}^{\mathrm{n}}+2 \mathrm{~b}^{\mathrm{n}}=\mathrm{a}^{\mathrm{n}}+\mathrm{ab}^{\mathrm{n}-1}+\mathrm{a}^{\mathrm{n}-1} \mathrm{~b}+\mathrm{b}^{\mathrm{n}}$
$\Rightarrow \mathrm{a}^{\mathrm{n}}+\mathrm{b}^{\mathrm{n}}-\mathrm{ab}^{\mathrm{n}-1}-\mathrm{a}^{\mathrm{n}-1} \mathrm{~b}=0$
$\Rightarrow\left(\mathrm{a}^{\mathrm{n}}-\mathrm{a}^{\mathrm{n}-1} \mathrm{~b}\right)+\left(\mathrm{b}^{\mathrm{n}}-\mathrm{ab}^{\mathrm{n}-1}\right)=0$
$\left.\Rightarrow \mathrm{a}^{\mathrm{n}-1}(\mathrm{a}-\mathrm{b})+\mathrm{b}^{\mathrm{n}-1}\right)(\mathrm{b}-\mathrm{a})=0$
$\Rightarrow(\mathrm{a}-\mathrm{b})\left(\mathrm{a}^{\mathrm{n}-1}-\mathrm{b}^{\mathrm{n}-1}\right)=0$
$\Rightarrow \mathrm{a}^{\mathrm{n}-1}-\mathrm{b}^{\mathrm{n}-1}=0(\because \mathrm{a} \neq \mathrm{b})$
$\Rightarrow \mathrm{a}^{\mathrm{n}-1}=\mathrm{b}^{\mathrm{n}-1} \Rightarrow \frac{a^{n-1}}{b^{n-1}}=1$
$\Rightarrow \quad\left(\frac{a}{b}\right)^{n-1}=1=\left(\frac{a}{b}\right)^{0}$
$\Rightarrow \mathrm{n}-1=0 \Rightarrow \mathrm{n}=1$
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