the amount of certain principal is 6655rs in 3 years compounded annually at rate of 10 p.c.p.a. find the principal.
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Answer:
Amount (A) = 6655 rupees
Rate (R) = 10 p.c.p.a.
Time (N) = 3 years.
Solution:
\large{ \: A = P × (1 + \frac{r}{100})^{3}}A=P×(1+
100
r
)
3
\large{ \: 6655 = P × (1 + \frac{10}{100})^{3}}6655=P×(1+
100
10
)
3
\large{ \: = P × ( \frac{110}{100})^{3}}=P×(
100
110
)
3
so,
\huge{ \: P = \frac{6655 \times {10}^{3} }{11 \times 11 \times 11}}P=
11×11×11
6655×10
3
\huge{ \: P = 5 \times {10}^{3}}P=5×10
3
\huge{ \: = 5000}=5000
Hence,
The principal was 5000 rupees.
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