UG Applied Mathematics Booster Test 2 - Applications of Differential Equations -

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UG Applied Mathematics Booster Test 2 - Applications of Differential Equations

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

UG Applied Mathematics Booster Test 2 - Applications of Differential Equations

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QUESTION 1 OF 20

A cake cools from an initial temperature of 185°F in a room with a temperature of 75°F. What is the value of the constant λ evaluated at t = 0?

Answer Review

1 A cake cools from an initial temperature of 185°F in a room with a temperature of 75°F. What is the value of the constant λ evaluated at t = 0?

A 75

B 110

C 185

D 260

💡 Short Explanation

Newton's Law of Cooling uses \(T-A=\lambda e^{-kt}\) At \(t=0\), exponential term becomes 1 So λ = initial temperature difference

✅ Correct Answer Explanation

At \(t=0\), the model is: \(T-A=\lambda e^{0}=\lambda\) Given: Initial temperature \(T={185}^{\circ }F\) Surrounding temperature \(A={75}^{\circ }F\) So, \(\lambda =185-75=110\) Thus, λ represents the initial excess temperature above the surroundings.

❓ Why Other Options Are Incorrect

Option A → 75 is the ambient temperature, not the temperature difference
Option C → 185 is the initial body temperature, not λ
Option D → 260 is not relevant to the cooling equation

🎯 Strategy

Used: Substitution

Application: Substitute t = 0 into Newton's cooling equation to isolate λ

Final Logic: λ equals initial temperature difference (185 − 75)

🧠 Memory Trick

"λ = Start difference (Body − Surrounding)"

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Newton's Law of Cooling

2 Match the specific differential equation models to their respective application areas:

List I	List II
1. \(\frac{dT}{dt}=-k(T-A)\)	a. — Financial Application
2. \(\frac{dA}{dt}=rA\)	b. — Population Dynamics
3. \(\frac{dP}{dt}=(\alpha -\beta )P\)	c. — Physical Application (Heating/Cooling)
4. \(\frac{dx}{dt}=-k_{1}x\)	d. — Radioactive Decay

A 1-c, 2-a, 3-b, 4-d

B 1-b, 2-d, 3-c, 4-a

C 1-d, 2-a, 3-b, 4-c

D 1-c, 2-b, 3-a, 4-d

💡 Short Explanation

Each differential equation models a specific real-world process: Newton's Law of Cooling → Heating/Cooling Continuous Compounding → Finance Population Growth → Population Dynamics Carbon-14 Decay → Radioactive Decay

✅ Correct Answer Explanation

The applications of the given differential equations are: \(\frac{dT}{dt}=-k(T-A)\) describes Newton's Law of Cooling, so it corresponds to Physical Application (Heating/Cooling). \(\frac{dA}{dt}=rA\) models continuous compound interest, so it corresponds to Financial Application. \(\frac{dP}{dt}=(\alpha -\beta )P\) models the change in population, so it corresponds to Population Dynamics. \(\frac{dx}{dt}=-k_{1}x\) describes radioactive decay, such as Carbon-14 decay. Thus, the correct matching is: 1 → c 2 → a 3 → b 4 → d Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
Correct.
Every differential equation is matched with its correct application.
Option B)
Incorrect because it mismatches cooling, finance, and radioactive decay applications.
Option C)
Incorrect because Newton's Law of Cooling is not a radioactive decay model, and Carbon-14 decay is not a heating/cooling model.
Option D)
Incorrect because the applications of finance and population dynamics are interchanged.

🎯 Strategy

Used

Identify the Real-World Application of Each Differential Equation

Application:

1. Recognize the mathematical model.
2. Recall its physical or practical interpretation.
3. Match the equation to its corresponding application area.

🧠 Memory Trick

Mnemonic: "TAPX = Temperature, Amount, Population, X (Radioactivity)."

📚 Chapter: Differential Equations

3 Ms. Rajni deposited Rs. 10,000 in a bank paying 4% continuous interest. Which mathematical steps are correct in finding the amount after 10 years?
1. The equation formed is dA/dt = 0.04 A
2. The general solution integrates to A = A0 e^(0.04t)
3. At t = 10, the formula is 10000 e^(0.4)
4. The amount decreases exponentially over time.

A 1

B 1,2

C 1,2,3

D 1,2,3,4

💡 Short Explanation

Continuous compounding follows exponential growth Differential equation model applies Step C correctly evaluates at t = 10

✅ Correct Answer Explanation

1: \(\frac{dA}{dt}=0.04A\) is correct 2: Solution \(A=A_{0}e^{0.04t}\)is correct 3: Substituting \(t=10\Rightarrow 10000e^{0.4}\)is correct 4: Incorrect because amount does not decrease; it grows exponentially

❓ Why Other Options Are Incorrect

Option D → Incorrect since continuous interest leads to growth, not decay

🎯 Strategy

Used: Elimination

Application: Remove incorrect statement about decay

Final Logic: Continuous compounding always produces exponential growth

🧠 Memory Trick

"Bank growth = A e^(rt), always upward"

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Continuous Compound Interest

4 Identify the incorrect mathematical step when finding the doubling time for a continuously compounded bank deposit
\(A=A_{0}e^{rt}.\)
\(2A_{0}=A_{0}e^{rt}.\)
\(e^{rt}=2.\)
\(t=\frac{r}{\ln\,2}.\)
\(rt=ln⁡2.\)

A The equation is set to

B Simplifying yields

C The final derived formula for time is

D Applying logarithm gives

💡 Short Explanation

The correct doubling time formula is \(t=\frac{\ln\,2}{r},\) not \(\frac{r}{\ln\,2}.\) Hence, Option C is the incorrect step

✅ Correct Answer Explanation

For continuous compounding, \(A=A_{0}e^{rt}.\) To find the doubling time, \(2A_{0}=A_{0}e^{rt}.\) Dividing both sides by \(A_{0}\), \(e^{rt}=2.\) Taking the natural logarithm, \(rt=ln⁡2.\) Finally, \(t=\frac{\ln\,2}{r}.\) Therefore, the statement \(t=\frac{r}{\ln\,2}\) is incorrect. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A)
Incorrect as a choice because this is a correct mathematical step.
Setting the final amount equal to twice the principal correctly represents doubling.
Option B)
Incorrect as a choice because dividing both sides by \(A_{0}\)correctly gives
\(e^{rt}=2.\)
Option C)
Correct.
The numerator and denominator are interchanged.
The correct formula is
\(t=\frac{\ln\,2}{r}.\)
Option D)
Incorrect as a choice because taking the natural logarithm correctly gives
\(rt=ln⁡2.\)

🎯 Strategy

Used

Apply Logarithms to the Exponential Equation

Application:

1. Start with
\(A=A_{0}e^{rt}.\)
1. Set the amount equal to \(2A_{0}\).
2. Cancel \(A_{0}\).
3. Apply the natural logarithm.
4. Solve for \(t\).

🧠 Memory Trick

Mnemonic: "Log First, Rate Second."

📚 Chapter: Differential Equations

5 If the rate of change of a population \(P\) is given by
\(\frac{dP}{dt}=kP,\)
where
\(k=(\alpha -\beta ),\)
integrating this equation using separation of variables
\(\int \frac{dP}{P}=\int k dt\)
leads directly to:
\(P=kt+c\)
\(log⁡P=k+t+c\)
\(log⁡P=kt+c\)
\(P=log⁡(kt)\)

💡 Short Explanation

Separating the variables and integrating gives \(log⁡P=kt+c.\) This is the integrated form before converting to the exponential solution.

✅ Correct Answer Explanation

Given, \(\frac{dP}{dt}=kP,\) separate the variables: \(\frac{dP}{P}=k dt.\) Integrate both sides: \(\int \frac{dP}{P}=\int k dt.\) Using the standard integration formulas, \(\int \frac{dP}{P}=log⁡P,\) and \(\int k dt=kt+c.\) Hence, \(log⁡P=kt+c.\) Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A)
\(P=kt+c\)
Incorrect because integrating
\(\frac{1}{P}\)
produces a logarithmic function, not a linear function.
Option B)
\(log⁡P=k+t+c\)
Incorrect because
\(\int k dt=kt+c,\)
not \(k+t+c\).
Option C)
\(log⁡P=kt+c\)
Correct.
This is the correct integrated equation obtained after separation of variables.
Option D)
\(P=log⁡(kt)\)
Incorrect because the logarithm applies to \(P\), not to \(kt\).

🎯 Strategy

Used

Separation of Variables

Application:

1. Separate the variables:
\(\frac{dP}{P}=k dt.\)
1. Integrate both sides.
2. Apply the standard integral
\(\int \frac{1}{P} dP=log⁡P.\)
1. Write the integrated equation.

🧠 Memory Trick

Mnemonic: "Divide by the Variable → Think Logarithm."

📚 Chapter: Differential Equations

6 A bacterial culture increases from 10,000 at 3 hours to 40,000 at 5 hours. To find the growth constant, dividing the two equations
\(\lambda e^{5k}=40,000\)
and
\(\lambda e^{3k}=10,000\)
isolates \(e^{2k}\). What is its value?

A 2

B 4

C 8

D 10

💡 Short Explanation

Divide the two equations to eliminate the constant \(\lambda\): \(\frac{\lambda e^{5k}}{\lambda e^{3k}}=\frac{40,000}{10,000}=4.\) Hence, \(e^{2k}=4.\)

✅ Correct Answer Explanation

The bacterial growth model is \(P(t)=\lambda e^{kt}.\) Given, \(\lambda e^{3k}=10,000\) and \(\lambda e^{5k}=40,000.\) Divide the second equation by the first: \(\frac{\lambda e^{5k}}{\lambda e^{3k}}=\frac{40,000}{10,000}.\) The constant \(\lambda\) cancels: \(e^{5k-3k}=4.\) Therefore, \(e^{2k}=4.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) 2
Incorrect because
\(\frac{40,000}{10,000}=4,\)
not 2.
Option B) 4
Correct.
Dividing the equations gives
\(e^{2k}=4.\)
Option C) 8
Incorrect because 8 would result from cubing 2, which is not required here.
Option D) 10
Incorrect because the ratio of the populations is 4, not 10.

🎯 Strategy

Used

Division of Exponential Equations

Application:

1. Write the two population equations.
2. Divide one equation by the other.
3. Cancel the common constant \(\lambda\).
4. Apply the exponent law
\(\frac{e^{a}}{e^{b}}=e^{a-b}.\)
1. Simplify to obtain \(e^{2k}\).

🧠 Memory Trick

Mnemonic: "Divide the Values, Subtract the Powers."

📚 Chapter: Differential Equations

7 The differential equation for the decay of Carbon-14 is
\(\frac{dx}{dt}=-k_{1}x.\)
Upon integration and applying the initial condition
\(x=x_{0}att=0,\)
the explicit solution becomes:
\(x=x_{0}e^{k_{1}t}\)
\(x=x_{0}ln⁡(k_{1}t)\)
\(x=x_{0}e^{-k_{1}t}\)
\(x=x_{0}-k_{1}t\)

💡 Short Explanation

Carbon-14 undergoes exponential decay. Solving \(\frac{dx}{dt}=-k_{1}x\) using separation of variables gives \(x=x_{0}e^{-k_{1}t}.\)

✅ Correct Answer Explanation

Given, \(\frac{dx}{dt}=-k_{1}x.\) Separate the variables: \(\frac{dx}{x}=-k_{1}dt.\) Integrate both sides: \(\int \frac{dx}{x}=\int -k_{1}dt.\) This gives \(ln⁡x=-k_{1}t+c.\) Exponentiating, \(x=e^{c}e^{-k_{1}t}.\) Let \(e^{c}=A,\) then \(x=Ae^{-k_{1}t}.\) Using the initial condition \(x=x_{0}whent=0,\) we obtain \(A=x_{0}.\) Hence, \(x=x_{0}e^{-k_{1}t}.\) Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A)
\(x=x_{0}e^{k_{1}t}\)
Incorrect because this represents exponential growth, whereas Carbon-14 undergoes exponential decay.
Option B)
\(x=x_{0}ln⁡(k_{1}t)\)
Incorrect because the solution of the differential equation is exponential, not logarithmic.
Option C)
\(x=x_{0}e^{-k_{1}t}\)
Correct.
This is the standard exponential decay model obtained after integration and applying the initial condition.
Option D)
\(x=x_{0}-k_{1}t\)
Incorrect because this is a linear equation and does not satisfy the differential equation
\(\frac{dx}{dt}=-k_{1}x.\)

🎯 Strategy

Used

Separation of Variables

Application:

1. Separate the variables.
2. Integrate both sides.
3. Exponentiate to remove the logarithm.
4. Apply the initial condition to determine the constant.

🧠 Memory Trick

Mnemonic: "Minus Sign Means Material Disappears."

📚 Chapter: Differential Equations

8 Radium decomposes at a rate proportional to the amount present. If half the original amount disappears in 1600 years, what equation properly defines the relationship to find \(k\)?
\(\frac{A_{0}}{2}=A_{0}e^{1600k}\)
\(\frac{A_{0}}{2}=A_{0}e^{-1600k}\)
\(2A_{0}=A_{0}e^{1600k}\)
\(\frac{A_{0}}{4}=A_{0}e^{-1600k}\)

💡 Short Explanation

Radioactive decay follows the exponential decay model \(A=A_{0}e^{-kt}.\) Since half of the original amount remains after 1600 years, \(A=\frac{A_{0}}{2},t=1600,\) giving \(\frac{A_{0}}{2}=A_{0}e^{-1600k}.\)

✅ Correct Answer Explanation

The radioactive decay model is \(A=A_{0}e^{-kt},\) where \(A_{0}\)= initial amount, \(A\)= amount remaining after time \(t\), \(k>0\)= decay constant. Given that half the original amount remains after 1600 years, \(A=\frac{A_{0}}{2}\) and \(t=1600.\) Substituting into the decay equation, \(\frac{A_{0}}{2}=A_{0}e^{-1600k}.\) This is the required equation used to determine the decay constant. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\frac{A_{0}}{2}=A_{0}e^{1600k}\)
Incorrect because radioactive decay requires a negative exponent.
Option B)
\(\frac{A_{0}}{2}=A_{0}e^{-1600k}\)
Correct.
This follows directly from the exponential decay model.
Option C)
\(2A_{0}=A_{0}e^{1600k}\)
Incorrect because the quantity decreases to half, not doubles.
Option D)
\(\frac{A_{0}}{4}=A_{0}e^{-1600k}\)
Incorrect because the problem states half the original amount remains, not one-fourth.

🎯 Strategy

Used

Apply the Exponential Decay Model

Application:

1. Write the decay equation:
\(A=A_{0}e^{-kt}.\)
1. Substitute the given remaining amount and time.
2. Obtain the equation for determining \(k\).

🧠 Memory Trick

Mnemonic: "Half Remaining → Negative Exponential."

📚 Chapter: Differential Equations

9 Charcoal from an ancient pit contained \(\frac{1}{4}\)of the Carbon-14 found in a living sample. Given
\(x(t)=\frac{x_{0}}{4},\)
the initial setup to find the age \(t\) is:
\(\frac{x_{0}}{4}=x_{0}e^{-k_{1}t}\)
\(\frac{x_{0}}{4}=x_{0}e^{k_{1}t}\)
\(4x_{0}=x_{0}e^{-k_{1}t}\)
\(\frac{x_{0}}{4}=e^{-k_{1}}\)

💡 Short Explanation

Carbon-14 follows the exponential decay model \(x=x_{0}e^{-k_{1}t}.\) Since only one-fourth of the original Carbon-14 remains, \(x=\frac{x_{0}}{4},\) which gives \(\frac{x_{0}}{4}=x_{0}e^{-k_{1}t}.\)

✅ Correct Answer Explanation

The exponential decay equation for Carbon-14 is \(x=x_{0}e^{-k_{1}t},\) where \(x_{0}\)= initial Carbon-14, \(x\)= remaining Carbon-14 after time \(t\), \(k_{1}\)= decay constant. The problem states that the sample contains \(\frac{1}{4}\) of the original Carbon-14. Therefore, \(x=\frac{x_{0}}{4}.\) Substituting into the decay equation, \(\frac{x_{0}}{4}=x_{0}e^{-k_{1}t}.\) This is the correct initial equation used to determine the age of the sample. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\frac{x_{0}}{4}=x_{0}e^{-k_{1}t}\)
Correct.
This is obtained directly from the exponential decay model.
Option B)
\(\frac{x_{0}}{4}=x_{0}e^{k_{1}t}\)
Incorrect because the exponent must be negative for radioactive decay.
Option C)
\(4x_{0}=x_{0}e^{-k_{1}t}\)
Incorrect because the sample contains one-fourth of the original amount, not four times the original amount.
Option D)
\(\frac{x_{0}}{4}=e^{-k_{1}}\)
Incorrect because the equation omits both the time variable \(t\) and the initial amount \(x_{0}\).

🎯 Strategy

Used

Substitute the Remaining Amount into the Decay Model

Application:

1. Write the exponential decay equation:
\(x=x_{0}e^{-k_{1}t}.\)
1. Substitute the remaining amount:
\(x=\frac{x_{0}}{4}.\)
1. Obtain the equation required to solve for \(t\).

🧠 Memory Trick

Mnemonic: "Fraction Left, Exponential Right."

📚 Chapter: Differential Equations

10 Using the half-life of 5700 years for Carbon-14, substituting
\(x=\frac{x_{0}}{2}\)
and
\(t=5700\)
into
\(x=x_{0}e^{-k_{1}t}\)
yields the exact value for the decay constant \(k_{1}\)as:
\(\frac{5700}{\ln\,2}\)
\(\frac{\ln\,2}{5700}\)
\(-(ln⁡2)\times 5700\)
\(\ln\,\left(\frac{5700}{2}\right)\)

💡 Short Explanation

Using the half-life condition, \(\frac{x_{0}}{2}=x_{0}e^{-5700k_{1}},\) which simplifies to \(k_{1}=\frac{\ln\,2}{5700}.\)

✅ Correct Answer Explanation

The Carbon-14 decay model is \(x=x_{0}e^{-k_{1}t}.\) At the half-life, \(x=\frac{x_{0}}{2}\) \(t=5700\) Substitute these values: \(\frac{x_{0}}{2}=x_{0}e^{-5700k_{1}}.\) Divide both sides by \(x_{0}\): \(\frac{1}{2}=e^{-5700k_{1}}.\) Take the natural logarithm: \(ln⁡\left(\frac{1}{2}\right)=-5700k_{1}.\) Since \(ln⁡\left(\frac{1}{2}\right)=-ln⁡2,\) we get \(-ln⁡2=-5700k_{1}.\) Therefore, \(k_{1}=\frac{\ln\,2}{5700}.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\frac{5700}{\ln\,2}\)
Incorrect because the numerator and denominator are interchanged.
Option B)
\(\frac{\ln\,2}{5700}\)
Correct.
This is the decay constant obtained from the half-life equation.
Option C)
\(-(ln⁡2)\times 5700\)
Incorrect because \(k_{1}\)is obtained by dividing by 5700, not multiplying.
Option D)
\(\ln\,\left(\frac{5700}{2}\right)\)
Incorrect because the logarithm is applied to the ratio \(\frac{1}{2}\), not to \(\frac{5700}{2}\).

🎯 Strategy

Used

Apply the Half-Life Condition

Application:

1. Use the decay model:
\(x=x_{0}e^{-k_{1}t}.\)
1. Substitute
\(x=\frac{x_{0}}{2},t=5700.\)
1. Cancel \(x_{0}\).
2. Take the natural logarithm.
3. Solve for \(k_{1}\).

🧠 Memory Trick

Mnemonic: "Log 2 Above, Half-Life Below."

📚 Chapter: Differential Equations

11 When modeling the diffusion of a drug from the GI-tract, the rate of change is given by
\(\frac{dx}{dt}=-k_{1}x.\)
What mathematical logic justifies the negative sign?

A The drug is growing rapidly in the stomach.

B The drug is leaving the GI-tract, so the amount is decreasing.

C The GI-tract temperature is below surrounding temperature.

D The dosage was applied intravenously.

💡 Short Explanation

Drug leaves GI-tract over time. Amount x decreases continuously. Hence negative rate.

✅ Correct Answer Explanation

Since the drug diffuses out of the GI-tract into the bloodstream, the quantity x inside the GI-tract decreases with time. This is modeled as a decay process, hence dx/dt = -k₁x where the negative sign indicates reduction of the substance in the GI compartment.

❓ Why Other Options Are Incorrect

Option A → Growth contradicts diffusion out of GI-tract.
Option C → Temperature is irrelevant to drug diffusion sign.
Option D → Route of administration does not determine sign.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Identify physical meaning of negative rate.

Final Logic: Leaving compartment ⇒ decreasing quantity.

🧠 Memory Trick

"Leaving means minus."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Drug Diffusion Model

12 If \(y(t)\)represents the amount of drug in the bloodstream, it increases as drug diffuses from the GI-tract \(\left(k_{1}x\right)\)and decreases as tissues remove it \(\left(k_{2}y\right)\). The complete differential equation for the blood is:
\(\frac{dy}{dt}=k_{1}x+k_{2}y\)
\(\frac{dy}{dt}=-k_{1}x-k_{2}y\)
\(\frac{dy}{dt}=k_{1}x-k_{2}y\)
\(\frac{dy}{dt}=x-y\)

💡 Short Explanation

The amount of drug in the bloodstream increases due to absorption from the GI-tract and decreases due to elimination by body tissues. Therefore, \(\frac{dy}{dt}=k_{1}x-k_{2}y.\)

✅ Correct Answer Explanation

In the two-compartment drug model: \(x(t)\)= amount of drug in the GI-tract. \(y(t)\)= amount of drug in the bloodstream. The bloodstream gains drug at the rate \(k_{1}x,\) and loses drug at the rate \(k_{2}y.\) Hence, the net rate of change is \(\frac{dy}{dt}=Rate of Entry-Rate of Removal=k_{1}x-k_{2}y.\) Therefore, \(\frac{dy}{dt}=k_{1}x-k_{2}y.\) Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\frac{dy}{dt}=k_{1}x+k_{2}y\)
Incorrect because \(k_{2}y\) represents drug removal from the bloodstream and must have a negative sign.
Option B)
\(\frac{dy}{dt}=-k_{1}x-k_{2}y\)
Incorrect because absorption from the GI-tract increases the drug in the bloodstream, so \(k_{1}x\) should be positive.
Option C)
\(\frac{dy}{dt}=k_{1}x-k_{2}y\)
Correct.
It correctly represents drug entering the bloodstream and drug leaving the bloodstream.
Option D)
\(\frac{dy}{dt}=x-y\)
Incorrect because it ignores the proportionality constants \(k_{1}\)and \(k_{2}\), which determine the rates of absorption and removal.

🎯 Strategy

Used

Net Rate = Inflow − Outflow

Application:

1. Identify the quantity entering the bloodstream.
2. Identify the quantity leaving the bloodstream.
3. Write:
\(Net Rate of Change=Inflow-Outflow.\)
1. Substitute the given rates.

🧠 Memory Trick

Mnemonic: "Entry Positive, Exit Negative."

📚 Chapter: Differential Equations

13 A 50 kg dog needs 45 mg of Nembutal per kg to be anesthetized. How many total milligrams of the drug must be present in the dog's bloodstream?

A 2585 mg

B 2250 mg

C 1250 mg

D 5000 mg

💡 Short Explanation

Multiply dose per kg by weight. 45 × 50 = 2250. Direct proportional calculation.

✅ Correct Answer Explanation

Required dose = 45 mg/kg × 50 kg = 2250 mg. This is the minimum bloodstream concentration needed for anesthesia.

❓ Why Other Options Are Incorrect

Option A → Incorrect multiplication.
Option C → Underestimates required dose.
Option D → Overestimates dosage.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: mg/kg × kg → mg.

Final Logic: Direct unit multiplication gives 2250 mg.

🧠 Memory Trick

"45 × 50 = 2250."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Drug Dosage Model

14 The half-life of Nembutal in a dog is 5 hours. To find the decay constant \(k\) using
\(x=x_{0}e^{-kt},\)
the correct substitution is:
\(\frac{x_{0}}{2}=x_{0}e^{-5k}\)
\(2x_{0}=x_{0}e^{-5k}\)
\(\frac{x_{0}}{2}=x_{0}e^{5k}\)
\(x_{0}=2e^{-5k}\)

💡 Short Explanation

At the half-life, the remaining amount is one-half of the initial amount. Substituting \(t=5,x=\frac{x_{0}}{2}\) into the decay model gives \(\frac{x_{0}}{2}=x_{0}e^{-5k}.\)

✅ Correct Answer Explanation

The exponential decay model is \(x=x_{0}e^{-kt},\) where \(x_{0}\)= initial amount, \(x\)= amount remaining after time \(t\), \(k\)= decay constant. Since the half-life is 5 hours, \(t=5\) and \(x=\frac{x_{0}}{2}.\) Substituting into the decay equation, \(\frac{x_{0}}{2}=x_{0}e^{-5k}.\) This equation is then used to determine the value of \(k\). Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\frac{x_{0}}{2}=x_{0}e^{-5k}\)
Correct.
It correctly applies the half-life condition to the exponential decay model.
Option B)
\(2x_{0}=x_{0}e^{-5k}\)
Incorrect because the quantity decreases to half, not doubles.
Option C)
\(\frac{x_{0}}{2}=x_{0}e^{5k}\)
Incorrect because radioactive or drug decay requires a negative exponent.
Option D)
\(x_{0}=2e^{-5k}\)
Incorrect because the initial amount \(x_{0}\)is incorrectly isolated and the equation does not follow from the decay model.

🎯 Strategy

Used

Apply the Half-Life Condition

Application:

1. Write the exponential decay equation:
\(x=x_{0}e^{-kt}.\)
1. Substitute the half-life values:
\(x=\frac{x_{0}}{2},t=5.\)
1. Obtain the equation used to calculate the decay constant.

🧠 Memory Trick

Mnemonic: "Half Time, Half Amount."

📚 Chapter: Differential Equations

15 For the satellite power supply
\(y=50e^{-0.004t},\)
what mathematical operation is required to find the time when the power is half of its original strength?
\(50=25e^{-0.004t}\)
\(25=50e^{-0.004t}\)
\(0=50e^{-0.004t}\)
\(100=50e^{-0.004t}\)

💡 Short Explanation

The initial power is \(50.\) Half of the original power is \(25.\) Therefore, substitute \(y=25\) into the given exponential decay equation.

✅ Correct Answer Explanation

The satellite power is modeled by \(y=50e^{-0.004t},\) where 50 = initial power, \(t\)= time, \(0.004\)= decay constant. To determine when the power is half of its original value, \(\frac{50}{2}=25.\) Substitute \(y=25\) into the model: \(25=50e^{-0.004t}.\) This equation is then solved (using logarithms) to determine the time \(t\). Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
\(50=25e^{-0.004t}\)
Incorrect because it reverses the original equation.
The dependent variable \(y\) should be replaced by 25, not 50.
Option B)
\(25=50e^{-0.004t}\)
✅ Correct.
Half of the initial power is substituted into the decay model.
Option C)
\(0=50e^{-0.004t}\)
Incorrect because exponential decay never reaches exactly zero.
Option D)
\(100=50e^{-0.004t}\)
Incorrect because 100 is twice the initial power and is not related to the half-life condition.

🎯 Strategy

Used

Substitute the Required Amount

Application:

1. Identify the initial amount.
2. Calculate half of the initial amount.
3. Replace \(y\) with the required value.
4. Form the equation to solve for time.

🧠 Memory Trick

Mnemonic: "Half Left, Model Right."

📚 Chapter: Differential Equations

16 Evaluating the power available at the end of 90 days for the satellite model
\(y=50e^{-0.004t}\)
requires calculating the exponent as:

A -0.004 × 90 = -0.36

B -0.004 × 90 = -3.6

C 0.004 × 90 = 0.36

D -0.04 × 90 = -3.6

💡 Short Explanation

Multiply decay constant by time. Keep correct decimal placement. Apply exponent rule.

✅ Correct Answer Explanation

Exponent is -0.004t. Substituting t = 90 gives -0.004 × 90 = -0.36, which is the correct exponent value used in evaluating the model.

❓ Why Other Options Are Incorrect

Option B → Incorrect multiplication.
Option C → Missing negative sign.
Option D → Wrong coefficient magnitude.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Check coefficient × time consistency.

Final Logic: Correct exponent calculation gives -0.36.

🧠 Memory Trick

"4×9 = 36 → shift decimals."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Exponential Decay Computation

17 If a bacterial culture doubled its initial population in 10 hours, the specific growth constant k can be written as:
\(\frac{\ln\,2}{10}\)
\(\frac{10}{\ln\,2}\)
\(\frac{\ln\,10}{2}\)
\(\frac{2}{10}\)

💡 Short Explanation

For exponential growth, \(P=P_{0}e^{kt}.\) If the population doubles in 10 hours, \(2P_{0}=P_{0}e^{10k},\) which gives \(k=\frac{\ln\,2}{10}.\)

✅ Correct Answer Explanation

The exponential growth model is \(P=P_{0}e^{kt},\) where \(P_{0}\)= initial population, \(k\)= growth constant. Since the population doubles in 10 hours, \(2P_{0}=P_{0}e^{10k}.\) Dividing both sides by \(P_{0}\), \(2=e^{10k}.\) Taking the natural logarithm, \(ln⁡2=10k.\) Therefore, \(k=\frac{\ln\,2}{10}.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\frac{\ln\,2}{10}\)
Correct.
It follows directly from
\(2=e^{10k}.\)
Option B)
\(\frac{10}{\ln\,2}\)
Incorrect because the numerator and denominator are interchanged.
Option C)
\(\frac{\ln\,10}{2}\)
Incorrect because the logarithm should be taken of 2 (doubling), not 10 (time).
Option D)
\(\frac{2}{10}\)
Incorrect because the exponential equation requires the natural logarithm, not simple division.

🎯 Strategy

Used

Apply the Doubling-Time Formula

Application:

1. Write the growth model:
\(P=P_{0}e^{kt}.\)
1. Substitute the doubling condition:
\(P=2P_{0},t=10.\)
1. Cancel \(P_{0}\).
2. Take the natural logarithm.
3. Solve for \(k\).

🧠 Memory Trick

Mnemonic: "Log 2 Above, Time Below.

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Growth Models

18 An old bone contains 75% of its original Carbon-14. In decimal format, what is the ratio of \(x(t)\)to \(x_{0}\)used to solve the equation
\(0.75x_{0}=x_{0}e^{-k_{1}t}?\)

A 0.25

B 1.75

C 0.75

D 7.50

💡 Short Explanation

If 75% of the original Carbon-14 remains, then \(75\%=\frac{75}{100}=0.75.\) Hence, \(\frac{x(t)}{x_{0}}=0.75.\)

✅ Correct Answer Explanation

The Carbon-14 decay model is \(x(t)=x_{0}e^{-k_{1}t}.\) The problem states that the old bone contains 75% of its original Carbon-14. Converting the percentage to decimal, \(75\%=\frac{75}{100}=0.75.\) Therefore, \(x(t)=0.75x_{0}.\) Substituting into the decay equation, \(0.75x_{0}=x_{0}e^{-k_{1}t}.\) Thus, the ratio \(\frac{x(t)}{x_{0}}=0.75.\) Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A) 0.25
Incorrect because 0.25 represents 25%, not 75%.
Option B) 1.75
Incorrect because the remaining amount cannot exceed the original amount in a radioactive decay process.
Option C) 0.75
Correct.
\(75\%\)expressed as a decimal is 0.75.
Option D) 7.50
Incorrect because this is 750%, not 75%.

🎯 Strategy

Used

Convert Percentage to Decimal

Application:

1. Express the percentage as a fraction over 100.
2. Convert to decimal.
3. Substitute the resulting ratio into the exponential decay equation.

🧠 Memory Trick

Mnemonic: "Move the Decimal Two Places Left."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Carbon Dating Model

📄 Passage

We readily take pills for diseases like common cold, headaches etc. without having a good understanding of how these medicines are absorbed into the blood or for how long these have effects on our body. When a single pill is taken, it first dissolves into the gastrointestinal tract (GI-tract) and each ingredient is diffused into the blood. These are carried to different body parts on which they act and are removed from the blood stream by kidneys and liver with different rates. Rate of change of drug in GI-tract = (Rate of drug intake) - (Rate at which drug leaves the GI Tract).

Question:
According to the passage, after a pill is dissolved in the GI-tract, what biological structures are responsible for removing the drug from the bloodstream?

A Heart and Lungs

B Kidneys and Liver

C Brain and Spine

D GI-tract and Stomach

💡 Short Explanation

Drug metabolism organs. Responsible for filtration and breakdown. Remove drug from blood.

✅ Correct Answer Explanation

The kidneys filter waste from blood, and the liver metabolizes drugs. Together they are responsible for removing drugs from the bloodstream after absorption.

❓ Why Other Options Are Incorrect

Option A → Not primary drug elimination organs.
Option C → Nervous system has no detox role.
Option D → GI tract is absorption site, not removal from blood.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match biological function to drug elimination.

Final Logic: Detoxification = kidney + liver.

🧠 Memory Trick

"Liver cleans, kidneys filter."

📚 Class XII – NCERT Biology (Applied Context) – Chapter: Human Physiology – Topic: Excretion and Drug Metabolism

📄 Passage

Question:
Based on the drug assimilation model passage, if the drug intake stops completely after swallowing one pill, the mathematical "Rate of drug intake" evaluates to:

A A constant positive value

B An exponentially decaying value

C Zero

D Infinity

💡 Short Explanation

No further drug input. Single-dose model. Intake rate stops completely.

✅ Correct Answer Explanation

After a single pill is swallowed, no additional drug enters the GI tract. Therefore, the rate of drug intake becomes zero, while only absorption and elimination processes continue.

❓ Why Other Options Are Incorrect

Option A → No continuous supply exists.
Option B → Decay applies to concentration, not intake.
Option D → Physically impossible intake rate.

🎯 Strategy

Used: Extreme Word Filter

Application: Identify "stops completely" condition.

Final Logic: No input ⇒ rate = 0.

🧠 Memory Trick

"No intake after pill → zero rate."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Drug Intake Models

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