Case study (4 Marks)

Questions

Question 14 Marks

Mr. Lal was 75 year old businessman. He had two sons and three daughters. On the silver jubilee of his company, he gifted an equal amount of money among his children. One of his daughters buys jewellery from the gifted money whereas the other daughter invested his money in bank.
It is known that if the interest is compounded continuously, the principal changes at the rate equal to the product of the rate of bank interest per annum and principal.

Based on the above given information answer the following questions:
Q.1. If P denotes the principal at time t and the rate of interest be r% per annum compounded continuously, then the differential equation that models this scenario is:
(A) $\frac{d P}{d t}=\frac{P r}{100}$
(B) $\frac{d P}{d t}=\frac{100}{P_r}$
(C) $\frac{d P}{d t}=P r$
(D) $\frac{d t}{d P}=\frac{P r}{100}$
Q.2. The particular solution of the associated differential equation is:
(A) $\log \left(\frac{P}{P_0}\right)=\frac{100}{r t}$
(B) $\log \left(\frac{P}{P_0}\right)=\frac{r t}{100}$
(C) $\log \left(\frac{P_0}{P}\right)=\frac{100}{r t}$
(D) $\log \left(\frac{P_0}{P}\right)=\frac{r t}{100}$

Q.3. If the interest is compounded continuously at 5% per annum, in how many years will ₹100 double itself? $\left(\log _e 2=0.6931\right)$
(A) 13.86 (B) 13 (C) 14 (D) 14.82

Q.4. At what rate interest rate will₹ 100 double itself in 10 years? $\left(\log _e 2=0.6931\right)$
(A) 6% (B) 7% (C) 6.93% (D) 7.36%

Answer

(1) (A) $\frac{d P}{d t}=\frac{P r}{100}$
Explanation:
If P denotes the principal at time f and the rate of interest be r% per annum compounded continuously, then according to the law given in the problem, we get
$\frac{d P}{d t}=\frac{P r}{100}$

(2) (B) $\log \left(\frac{P}{P_0}\right)=\frac{r t}{100}$
Explanation:
$\begin{array}{l}\text { Since, } \frac{d P}{d t}=\frac{P r}{100} \\ \Rightarrow \quad \frac{d P}{P}=\frac{r}{100} d t \\ \Rightarrow \quad \int \frac{d P}{P}=\frac{r}{100} \int d t \\ \Rightarrow \log P=\frac{r t}{100}+C\end{array}$
Let $P_0$ be the initial principal i.e., at $t=0, P=P_0$.
Putting $P=P_0$ in eq. (i), we get
$\log P_0=C$
Putting $C=\log P_0$ in eq. (i), we get
$\log P=\frac{r t}{100}+\log P_0$
$\Rightarrow \quad \log \left(\frac{P}{P_0}\right)=\frac{r t}{100}$

(3) (A) 13.86
We have, $\log \left(\frac{P}{P_0}\right)=\frac{r t}{100}$
Put r = 5, $P_0=$₹ 100 and P = ₹ 200 = $2 P_0$ in the above equation, we get
$\log 2=\frac{5 t}{100}$
$\Rightarrow \quad t=20 \log _e 2$
$=20 \times 0.6931$
= 13.86 years

(4) (C) 6.93%
Explanation:
We have,
$\log \left(\frac{P}{P_0}\right)=\frac{r t}{100}$
Put r = 10, $P_0=$₹ 100 and P = ₹ 200 = $2 P_0$ in the above equation, we get
$\log 2=\frac{10 r}{100}$
$\Rightarrow \quad r=10 \log _e 2$
$=10 \times 0.6931$
= 6.931% or 6.93% per annum

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Question 24 Marks

Read the following text and answer the following questions on the basis of the same:
Riya a biologist runs an experiment in the laboratory involving a culture of bacteria. She is doing this experiment first time. She notices that the mass of the bacteria in culture increases exponentially. The bacteria count in the culture is 1,00,000. The rate of growth of bacteria is proportional to the number of bacteria present in the culture. Also, the number of bacteria is increased by 10% in 2 hours.

Q.1. If y be the number of bacteria at time t, then the differential equation that models this scenario is
(A) $\frac{d y}{d x}=k y$
(B) $\frac{d y}{d t}=k y$
(C) $\frac{d y}{d t}=k t$
(D) $\frac{d y}{d x}=k t$

Q.2. The general solution of the associated differential equation is:
(A) $\log y=k t+c$
(B) $\log x=k t+c$
(C) $\log y=k x+c$
(D) $\log y=k t$

Q.3. The value of integration constant 'c' is:
(A) 1,00,000
(B) log 1000
(C) log 1 ,00,000
(D) 1,10,000

Q.4. Count of bacteria after 2 hours:
(A) 1,00,000
(B) log 1000
(C) log 1,00,000
(D) 1,10,000

Answer

(1) (B) $\frac{d y}{d t}=k y$
If y be the number of bacteria at time t, then
$\frac{d y}{d t} \propto y$
$\Rightarrow \quad \frac{d y}{d t}=k y$
(2) (A) log y = kt + c
Explanation :
$\frac{d y}{d t}=k y$
$\Rightarrow \quad \frac{d y}{y}=k d t$
Integrating both sides, we get
log y = kt + c
(3) (C) log 1,00,000
Explanation: Put t = 0 and y = 1,00,000 in equation log y = kt + c,
we get
log 1,00,000=k x 0+c
$c=\log 1,00,000$
(4) (D) 1,10,000
Explanation:
When, t = 2 hours
$y=1,00,000+10 \% \times 1,00,000$
= 1 ,10,000

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Applied Maths Case study (4 Marks)

Case study (4 Marks)

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