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Growth or decay dP/dt = kP + c: population, bacteria, cooling

This type of question, asked 4 times: 2021 Paper 1, 2021 Paper 25, 27 Jan 2024, 31 Jan 2024. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 1

    75
    The population P=P(t)P=P(t) at time ' tt ' of a certain species follows the differential equation dPdt=\frac{dP}{dt}= 0.5P-450. If P(0)=850P(0) = 850, then the time at which population becomes zero is :

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  2. Question 2

    2021 Paper 25

    64
    The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t=0t= 0. The number of bacteria is increased by 20%20\% in 2 hours. If the population of bacteria is 2000 after klog⁡e(65)\frac{k}{\log_{e}\left( \frac{6}{5} \right)} hours, then (klog⁡e2)2\left( \frac{k}{\log_{e}2} \right)^{2} is equal to

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  3. Question 3

    27 Jan 2024

    63
    Let x=x(t)x = x(t) and y=y(t)y = y(t) be solutions of the differential equations  dxdt+ax=0\ \frac{dx}{dt}+ ax = 0 and dydt+by=0\frac{dy}{dt}+ by = 0 respectively, a,b∈Ra,b \in R. Given that x(0)=2;y(0)=1x(0) = 2;y(0) = 1 and 3y(1)=2x(1)3y(1) = 2x(1), the value of tt, for which x(t)=y(t)x(t) = y(t), is :

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  4. Question 4

    31 Jan 2024

    158
    The temperature T(t)T(t) of a body at time t=0t = 0 is 160∘160^{\circ} FF and it decreases continuously as per the differential equation dTdt=−K(T−80)\frac{dT}{dt}= - K(T - 80), where KK is positive constant. If T(15)=120∘FT(15) =120^{\circ}F, then T(45)T(45) is equal to

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