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JEE Math Practice Question

The temperature $$T(t)$$ of a body at time $$t=0$$ is $$160^{\circ} \mathrm{F}$$ and it decreases continuously as per the differential equation $$\frac{d T}{d t}=-K(T-80)$$, where $$K$$ is a positive constant. If $$T(15)=120^{\circ} \mathrm{F}$$, then $$T(45)$$ is equal to

  1. A.90$$^\circ$$ F
  2. B.85$$^\circ$$ F
  3. C.80$$^\circ$$ F
  4. D.95$$^\circ$$ F

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