Solve the question with steps.-Pls frnds it's urgent.
1) Find using the compound interest formula p[1+(r/100)]^n:
a)Find the principal f the Compound interest is compounded annually at the rate of 10% per annum for 3 yrs is Rs.331.
b) What sum of money will amount to Rs.2704 in 2 years at 4% per annum.
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a)
![A=P[1+ \frac{r}{100} ]^{n}\\ \\P+331=P[1+ \frac{10}{100} ]^{3}\\ \\P+331=P[1.1]^{3}\\ \\P+331=1.331P\\ \\1.331P-P=331\\ \\0.331P=331\\ \\P= \frac{331}{0.331} =\boxed{Rs.\ 1000}](https://tex.z-dn.net/?f=A%3DP%5B1%2B+%5Cfrac%7Br%7D%7B100%7D+%5D%5E%7Bn%7D%5C%5C+%5C%5CP%2B331%3DP%5B1%2B+%5Cfrac%7B10%7D%7B100%7D+%5D%5E%7B3%7D%5C%5C+%5C%5CP%2B331%3DP%5B1.1%5D%5E%7B3%7D%5C%5C+%5C%5CP%2B331%3D1.331P%5C%5C+%5C%5C1.331P-P%3D331%5C%5C+%5C%5C0.331P%3D331%5C%5C+%5C%5CP%3D+%5Cfrac%7B331%7D%7B0.331%7D+%3D%5Cboxed%7BRs.%5C%C2%A01000%7D "A=P[1+ \frac{r}{100} ]^{n}\\ \\P+331=P[1+ \frac{10}{100} ]^{3}\\ \\P+331=P[1.1]^{3}\\ \\P+331=1.331P\\ \\1.331P-P=331\\ \\0.331P=331\\ \\P= \frac{331}{0.331} =\boxed{Rs.\ 1000}")
Principal = Rs. 1000
b)
![A=P[1+ \frac{r}{100} ]^{n}\\ \\2704=P[1+ \frac{4}{100} ]^{2}\\ \\2704=P[1.04]^{2}\\ \\2704=1.0816P\\ \\P= \frac{2704}{1.0816} =\boxed{Rs.\ 2500}](https://tex.z-dn.net/?f=A%3DP%5B1%2B+%5Cfrac%7Br%7D%7B100%7D+%5D%5E%7Bn%7D%5C%5C+%5C%5C2704%3DP%5B1%2B+%5Cfrac%7B4%7D%7B100%7D+%5D%5E%7B2%7D%5C%5C+%5C%5C2704%3DP%5B1.04%5D%5E%7B2%7D%5C%5C+%5C%5C2704%3D1.0816P%5C%5C+%5C%5CP%3D+%5Cfrac%7B2704%7D%7B1.0816%7D+%3D%5Cboxed%7BRs.%5C+2500%7D "A=P[1+ \frac{r}{100} ]^{n}\\ \\2704=P[1+ \frac{4}{100} ]^{2}\\ \\2704=P[1.04]^{2}\\ \\2704=1.0816P\\ \\P= \frac{2704}{1.0816} =\boxed{Rs.\ 2500}")
The sum of money is Rs. 2500
Principal = Rs. 1000
b)
The sum of money is Rs. 2500
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