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The assets of a person are increasing at a rate proportional to the square root of the assets at a given…
The assets of a person are increasing at a rate proportional to the square root of the assets at a given time. His assets increase from Rupees 9 crores to Rupees 16 crores in 2 years, then his assets at the end of 5 more years will be
46.25 cores 42.25 crores 30.25 crores 56.25 crores Solution
$\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{dt}} \propto \sqrt{\mathrm{x}} \\ & \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k} \sqrt{\mathrm{x}} \\ & \Rightarrow \frac{\mathrm{dx}}{\sqrt{\mathrm{x}}}=\mathrm{kdt} \\ & \Rightarrow 2 \sqrt{\mathrm{x}}=\mathrm{kt}+\mathrm{C}\end{aligned}$
When $\mathrm{t}=0, \mathrm{x}=9$ i.e. $2 \sqrt{9}=\mathrm{k} \times 0+\mathrm{C}$
$\begin{aligned}
& \Rightarrow \mathrm{c}=6 \\
& \Rightarrow 2 \sqrt{\mathrm{x}}=\mathrm{kt}+6[\text { Putting } \mathrm{c}=6] \\
& \text { When } \mathrm{t}=2, \mathrm{x}=16 \text { i.e. } 2 \sqrt{16}=\mathrm{k} \times 2+6 \Rightarrow \mathrm{k}=1 \\
& \Rightarrow 2 \sqrt{\mathrm{x}}=\mathrm{t}+6 \text { [Putting } \mathrm{k}=1]
\end{aligned}$
After 5 years at $t=7$
$\begin{aligned} & 2 \sqrt{x}=7+6 \\ & \Rightarrow \sqrt{x}=6.5 \\ & \Rightarrow x=42.25\end{aligned}$
Asked in: MHT CET 2022 (06 Aug Shift 1)
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