potato is used to prepare an osmoscope- justify
define permeability.
Answers
Answer:
Given:-
Principal = 80000
Rate of interest = 10%
Times = 1.5 years
To Find:-
The amount after the interest is compounded annually.
The amount after the interest is compounded half yearly.
Solution:-
(i) The amount when the interest is compounded annually.
We know,
The formula of amount:-
\sf{A = P\bigg(1+\dfrac{r}{100}\bigg)^n}A=P(1+
100
r
)
n
Hence,
= \sf{A = 80000\bigg(1+\dfrac{10}{100}\bigg)^1 \bigg(1+\dfrac{10}{200}\bigg)}A=80000(1+
100
10
)
1
(1+
200
10
)
= \sf{A = 80000\bigg(1+\dfrac{1}{10}\bigg)^1 \bigg(1+\dfrac{1}{20}\bigg)}A=80000(1+
10
1
)
1
(1+
20
1
)
= \sf{A = 80000\bigg(\dfrac{10+1}{10}\bigg)\bigg(\dfrac{20+1}{20}\bigg)}A=80000(
10
10+1
)(
20
20+1
)
= \sf{A = 80000\bigg(\dfrac{11}{10}\bigg)\bigg(\dfrac{21}{20}\bigg)}A=80000(
10
11
)(
20
21
)
= \sf{A = 400 \times 11\times 21}A=400×11×21
= \sf{A = 92400}A=92400
Therefore, Amount after 1.5 years when the interest is compounded annually will be Rs.92400.
______________________________________
(ii) The amount when the interest is compounded half-yearly.
We know,
The formula of amount when the interest is compounded annually:-
\sf{A = P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}A=P(1+
200
r
)
2n
Now,
Since we are given the times as 1.5 years we can also write it like:-
\sf{1\dfrac{1}{2} = \dfrac{3}{2}}1
2
1
=
2
3
years
Hence,
\sf{A = 80000\bigg(1+\dfrac{10}{200}\bigg)^{2\times\dfrac{3}{2}}}A=80000(1+
200
10
)
2×
2
3
\sf{A = 80000\bigg(1+\dfrac{1}{20}\bigg)^3}A=80000(1+
20
1
)
3
\sf{A = 80000\bigg(\dfrac{20+1}{20}\bigg)^3}A=80000(
20
20+1
)
3
\sf{A = 80000\bigg(\dfrac{21}{20}\bigg)^3}A=80000(
20
21
)
3
\sf{A = 80000\times \dfrac{21}{20}\times \dfrac{21}{20}\times \dfrac{21}{20}}A=80000×
20
21
×
20
21
×
20
21
\sf{A = 10\times 21\times 21\times 21}A=10×21×21×21
\sf{A = 92610}A=92610
Therefore, Amount after 1.5 years when the interest is compounded half-yearly will be Rs.92610.
______________________________________