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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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