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A sum becomes Rs. 4500 after two years and Rs. 6750 after four years at the same rate of compound interest. The sum is:
|
A) Rs. 4000 |
|
B) Rs. 2500 |
|
C) Rs. 3000 |
|
D) Rs. 3050 |
Correct Answer:
| C) Rs. 3000 |
Description for Correct answer:
Amount(A1) = Rs.4500,
t1= 2 years
Amount(A2) = Rs.6750,
t1= 4 years
Let the Rate % = R %
Principal = Rs. P
According to the question,
\( \Large \textbf{Case1:} \) 4500
=\( \Large p \left(1+\frac{R}{100}\right)^{2}\)...........1
\( \Large \textbf{Case2:} \) 6750
=\( \Large p \left(1+\frac{R}{100}\right)^{4}\)...........1
By dividing eq. 2, by eq. 1
\( \Large \frac{6750}{4500}= \left(1+\frac{R}{100}\right)^{2}\)
\( \Large \frac{3}{2}= \left(1+\frac{R}{100}\right)^{2}\)..........3
From eq. 1
\( \Large 4500=P \times \frac{3}{2} \)
P= Rs.3000
Hence,Required principal
=Rs. 3000
Amount(A1) = Rs.4500,
t1= 2 years
Amount(A2) = Rs.6750,
t1= 4 years
Let the Rate % = R %
Principal = Rs. P
According to the question,
\( \Large \textbf{Case1:} \) 4500
=\( \Large p \left(1+\frac{R}{100}\right)^{2}\)...........1
\( \Large \textbf{Case2:} \) 6750
=\( \Large p \left(1+\frac{R}{100}\right)^{4}\)...........1
By dividing eq. 2, by eq. 1
\( \Large \frac{6750}{4500}= \left(1+\frac{R}{100}\right)^{2}\)
\( \Large \frac{3}{2}= \left(1+\frac{R}{100}\right)^{2}\)..........3
From eq. 1
\( \Large 4500=P \times \frac{3}{2} \)
P= Rs.3000
Hence,Required principal
=Rs. 3000
Part of solved Simple and compound interest questions and answers : >> Aptitude >> Simple and compound interest
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