A person borrows Rs. 2048 at 61/4% compound interest. Immediately the sum borrowed is given to another person at the same rate on the condition that the interest is compounded for every half year. Find the amount gained by the first person in 1 1/2 years. Please solve it urgent please will mark as brainliest
Answers
Answer:
2.0625
Step-by-step explanation:
Therefore he gain after 1\frac121
2
1
years is Rs 2.0625.
Explanation:
Compound interest formula:
A=P(1+r)^tA=P(1+r)
t
A= Amount after n years
P=Initial amount
r= Rate of interest
t= time.
Given that, a person borrow a sum Rs 2048 at 6\frac{1}{4}6
4
1
% per annum, compounded annually for 1\frac121
2
1
years.
Here P= Rs 2048 , r= 6\frac{1}{4}6
4
1
% =0.0625
Since 1\frac121
2
1
is not an integer. So, first we have find out the interest for 1 year. Then we need to find out the interest for \frac12
2
1
.
Simple interest for the first year interest = Compound interest for the first year interest.
The interest for 1 year is =Prt=Prt
=2048×0.0625×1
=Rs 128.
Now Principal for \frac12
2
1
year is = Rs(2048+128)
=Rs 2176
The interest for \frac12
2
1
year is =Prt=Prt
=2176 \times 0.625\times \frac12=2176×0.625×
2
1
=Rs 68.
Therefore total amount that he will pay after 1\frac121
2
1
years is = Rs (2176+68) = Rs 2244.
Again given that on the same day he lent out his money to another person same rate of interest but compounded semiannually.
Formula for semi annually compound interest:
A=P(1+\frac rn)^{nt}A=P(1+
n
r
)
nt
A= Amount after n years
P=Initial amount
r= Rate of interest
t= time.
Here P= Rs 2048 , r= 6\frac{1}{4}6
4
1
% =0.0625, t=1\frac12=\frac32t=1
2
1
=
2
3
years, n=2[ semi annually]
A= 2048(1+\frac {0.0625}{2})^{\frac32 \times 2}A=2048(1+
2
0.0625
)
2
3
×2
=2048(1.03125)^3=2048(1.03125)
3
=Rs 2246.0625
Therefore he gain after 1\frac121
2
1
years is Rs(2246.0625-2244)=Rs 2.0625