# CAT 2020 Set-2 | Question: 75

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For the same principal amount, the compound interest for two years at $5\%$ per annum exceeds the simple interest for three years at $3\%$ per annum by $\text{Rs } 1125.$ Then the principal amount in rupees is

Let the principal be $\text{P}.$

We know that, $\boxed{\text{A = P} \left( 1 + \dfrac{r}{100} \right)^{t}}$, where, $\text{A } =$ amount, $\text{P}=$ principal, $r =$ rate, $t =$ time.

And $\boxed{\text{Compound Interest (CI) = A – P}}$

Now, $\text{CI = P} \left( 1 + \dfrac{5}{100} \right)^{2} – \text{P}$

$\Rightarrow \text{CI = P} \left( 1+\dfrac{1}{20} \right)^{2} – \text{P}$

$\Rightarrow \text{CI = P} \left(\dfrac{21}{20} \right)^{2} – \text{P}$

$\Rightarrow \text{CI} = \dfrac{441}{400} – \text{P}$

$\Rightarrow \text{CI} = \dfrac{441\text{P} – 400\text{P}}{400}$

$\Rightarrow \boxed{\text{CI} = \dfrac{41\text{P}} {400}}$

We know that, Simple Interest $\boxed{\text{(SI)} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100}}$, where, $\text{P} =$ principal, $\text{R} =$ rate, $\text{T} =$ time.

$\Rightarrow \text{SI} = \dfrac{\text{P} \times 3 \times 3}{100}$

$\Rightarrow \text{SI} = \dfrac{9\text{P}}{100}$

Now, $\text{CI – SI} = 1125$

$\Rightarrow \dfrac{41\text{P}}{400} – \dfrac{9\text{P}}{100} = 1125$

$\Rightarrow \dfrac{41\text{P} – 36\text{P}}{400} = 1125$

$\Rightarrow \dfrac{5\text{P}}{400} = 1125$

$\Rightarrow \text{P} = 1125 \times 80$

$\Rightarrow \text{P} =$ ₹ $\; 90000$

$\therefore$ The principle amount is ₹ $\; 90000.$

Correct Answer $: 90000$
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