Amount and compound interest earned on Rs. 10000 for 1 year at 9% per annum compounded half
yearly is
Answers
Answer:
your answer
Step-by-step explanation:
Principal, P =Rs.10000, Rate, R = 8% per annum compounded half yearly for 1 year.
Now, There are two half years in a year. Therefore compounding has to be 2 times.
And rate = half of 10% = 5% half yearly.
Therefore, the required amount = 10000(1+\frac{4}{100})^{2} = Rs. 10816
And Compound Interest, C.I. = Amount - Principal = Rs. (10816 - 10000) = Rs. 816 give me a like and mark as me a brainlist
Answer:
Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly)
Time(n) = 1\ \frac{1}{2}1
2
1
years = 3 years (compounded half yearly)
Amount (A) = P\left(1+\frac{R}{100}\right)^nP(1+
100
R
)
n
= 10000\left(1+\frac{5}{100}\right)^310000(1+
100
5
)
3
= 10000\left(1+\frac{1}{20}\right)^310000(1+
20
1
)
3
= 10000\left(\frac{21}{20}\right)^310000(
20
21
)
3
= 10000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}10000×
20
21
×
20
21
×
20
21
= Rs. 11,576.25
Compound Interest (C.I.) = A – P
= Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25
If it is compounded annually, then
Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10%, Time (n) = 1\ \frac{1}{2}1
2
1
years.
Amount (A) for 1 year = P\left(1+\frac{R}{100}\right)^nP(1+
100
R
)
n
= 10000\left(1+\frac{10}{100}\right)^110000(1+
100
10
)
1
= 10000\left(1+\frac{1}{10}\right)^110000(1+
10
1
)
1
= 10000\left(\frac{11}{10}\right)^110000(
10
11
)
1
= 10000\times\frac{11}{10}10000×
10
11
= Rs. 11,000
Interest for \frac{1}{2}
2
1
year = \frac{11000\times1\times10}{2\times100}=RS.\ 550
2×100
11000×1×10
=RS. 550
\therefore∴ Total amount = Rs. 11,000 + Rs. 550 = Rs. 11,550
Now, C.I. = A – P = Rs. 11,550 – Rs. 10,000
= Rs. 1,550
Yes, interest Rs. 1,576.25 is more than Rs. 1,550.