Calculate amount and compound interest on rupees 1500 for 2 and a half years at 10% p.a compounded annually. Please ans thi fast!!!!!
Answers
Answer:
AMOUNT = 1905.75 , COMPOUND INTEREST = 405.75
Step-by-step explanation:
Given,
Principal,P =Rs.1500, Rate,R = 10% and time period,n = 2.5 years.
We know, Amount when interest is compounded annually = P(1+ R/100)^n
Amount after 2 years at 10% , A = 1500(1+ 10/100})^2 = Rs.1815
This acts as the principal amount for the next half year.
SI on next 1/2 year at = P*R*T/100 = 1815*10*1/2÷100= Rs. 90.75
Therefore, Total amount to be paid after 2.5 years = Rs. (1815+90.75) = Rs.1905.75
Now, Compound Interest after 2 years = A - P = Rs.(1815-1500) = Rs. 315
Therefore, Compound Interest after 2.5 years, CI = Rs. 315 + SI = Rs.405.75.
Answer:
Solution!!
The concept of compound interest has to be used here. The principal, rate of interest and time is given. We have to find the compound interest.
Principle (P)= Rs 1500
Rate of interest (R) = 10%
Time (T) = 2.5 years = 5/2 years
\sf \bold{\to A=P\left(1+\dfrac{R}{100}\right)^{T}}→A=P(1+
100
R
)
T
\sf \to A=1500\left(1+\dfrac{10}{100}\right)^{\frac{5}{2}}→A=1500(1+
100
10
)
2
5
\sf \to A=1500\left(\dfrac{110}{100}\right)^{\frac{5}{2}}→A=1500(
100
110
)
2
5
\sf \to A=1500\left(\dfrac{11}{10}\right)^{\frac{5}{2}}→A=1500(
10
11
)
2
5
\sf \to A=1500\times \dfrac{11^{\frac{5}{2}}}{10^{\frac{5}{2}}}→A=1500×
10
2
5
11
25
\sf \to A=\dfrac{1500\sqrt{11^{5}}}{\sqrt{10^{5}}}\quad (a^{\frac{m}{n}}=\sqrt[n]{a^{m}})→A=
10
5
1500
11
5
(a
n
m
=
n
a
m
)
\sf \to A=\dfrac{1500\times 11^{2}\sqrt{11}}{10^{2}\sqrt{10}}→A=
10
2
10
1500×11
2
11
A=
10
2
10
1500×11
2
11
\sf \to A=\dfrac{1500\times 121\sqrt{11}}{100\sqrt{10}}→A=
100
10
1500×121
11
\sf \to A=\dfrac{15\times 121\sqrt{11}}{\sqrt{10}}→A=
10
15×121
11
\sf \to A=\dfrac{1815\sqrt{11}}{\sqrt{10}}→A=
10
1815
11
\sf \to A=\dfrac{363\sqrt{110}}{2}→A=
2
363
110
\to→ A ≈ Rs 1903.58
CI = Amount - Principal
CI = Rs 1903.58 - Rs 1500
CI = Rs 403.58