MHT CET · Maths · Basic of Mathematics
The money invested in a company is compounded continuously. If ₹ \(200\) invested today becomes ₹ \(400\) in 6 years, then at the end of 33 years it will become ₹
- A \(1600 \sqrt{2}\)
- B \(3200 \sqrt{2}\)
- C \(12800 \sqrt{2}\)
- D \(6400 \sqrt{2}\)
Answer & Solution
Correct Answer
(D) \(6400 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
Here, Amount \((A)=\) ₹ \(400\)
Principal \((P)=\) ₹ \(200, N=6\) years
\(A-P\left(1+\frac{R}{100}\right)^N \)
\( \Rightarrow 400=200\left(1+\frac{R}{100}\right)^6 \)
\( \Rightarrow\left(1+\frac{R}{100}\right)^6=2 \)
\( \Rightarrow 1+\frac{R}{100}=2^{\frac{1}{6}} \)
\( A =P\left(1+\frac{R}{100}\right)^N \)
\( =200\left(1+\frac{R}{100}\right)^{33} \)
\( =200\left(2^{\frac{1}{6}}\right)^{33} \)
\( =200\left(2^5 \cdot 2^{\frac{1}{2}}\right) \)
\( =200(32 \sqrt{2}) \)
\( =6400 \sqrt{2}\)
Principal \((P)=\) ₹ \(200, N=6\) years
\(A-P\left(1+\frac{R}{100}\right)^N \)
\( \Rightarrow 400=200\left(1+\frac{R}{100}\right)^6 \)
\( \Rightarrow\left(1+\frac{R}{100}\right)^6=2 \)
\( \Rightarrow 1+\frac{R}{100}=2^{\frac{1}{6}} \)
\( A =P\left(1+\frac{R}{100}\right)^N \)
\( =200\left(1+\frac{R}{100}\right)^{33} \)
\( =200\left(2^{\frac{1}{6}}\right)^{33} \)
\( =200\left(2^5 \cdot 2^{\frac{1}{2}}\right) \)
\( =200(32 \sqrt{2}) \)
\( =6400 \sqrt{2}\)
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