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  • Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 5 maths textbook solution

Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 5 maths textbook solution

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Answer: P=R s \; 1822, t=12 \text { years }

Given:  Compound interest=6% per annum,

To find: The amount after 10 years for Rs 1000 and the time in which the amount doubles.

Hint: Use compound interest i.e. \frac{d P}{d t}=\frac{P r}{100}

Solution: Let P be the principal

        \begin{aligned} &=\frac{d P}{d t}=\frac{p r}{100} \\\\ &=\frac{d P}{P}=\frac{r}{100} d t \end{aligned}

Integrating on both sides we get,

        \begin{aligned} &=>\int \frac{d P}{P}=\int \frac{r}{100} d t \\\\ &=>\log P=\frac{r}{100} \times t+C \ldots(i) \end{aligned}

At t=0, we have initial principal P=P0

        \begin{aligned} &=>\log \left(P_{0}\right)=0+C \\\\ &=>C=\log P_{0} \end{aligned}

Now substituting value of C in equation (i)

        \begin{aligned} &=>\log P=\frac{r}{100} t+\log P_{0} \\\\ &=>\log P-\log P_{0}=\frac{r t}{100} \\\\ &=>\log \left(\frac{p}{p_{0}}\right)=\frac{r t}{100} \end{aligned}

Now, P0=1000, t=10 years and r=6%

        \begin{aligned} &=>\log \left(\frac{P}{1000}\right)=\frac{6 \times 10}{100} \\\\ &=>\log P-\log 1000=0.6 \\\\ &=>\log P=0.6+\log 1000 \end{aligned}

        =>\log P=\log e^{0.6}+\log 1000 \quad\quad\quad\left[\log \log e^{x}=x\right]

        \begin{aligned} &=>\log P=\log \left(e^{0.6} \times 1000\right) \\\\ &=>\log P=\log (1.822 \times 1000) \quad\quad\quad\left[\log \log e^{0.6}=1.822\right] \end{aligned}

        \begin{aligned} &=>\log P=\log (1822)\\\\ &=>P=1822 \end{aligned}[Using anti-logarithm on both sides]

Therefore P=Rs 1822

Thus Rs 1000 will be Rs 1822 in 10 years.

Now let t1 be the time taken to double Rs 1000 then, P=2000, P0=1000, r=6%

From equation (i)

        \begin{aligned} &\log \left(\frac{p}{p_{0}}\right)=\frac{r t}{100} \\\\ &=>\log \left(\frac{2000}{1000}\right)=\frac{6 t_{1}}{100} \\\\ &=>\log 2=\frac{6 t_{1}}{100} \end{aligned}

        =6 t_{1}=100 \log 2                [By cross multiplication]

 

        \begin{aligned} &=6 t_{1}=100 \times 0.6931\quad\quad\quad &[\log 2=0.6931]\\\\ &=t_{1}=\frac{69.31}{6}\\\\ &=t_{1}=11.55 \text { Years } \end{aligned}

        =12 years (approximately)

Hence, it will take 12 years to double the amount.

 

Posted by

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