Arif took a loan of rs.80000 at a interest of rate of 10% for 1½ years compounded annually.find the amount he would be paying after the interest
Answers
Answer:
me (n) = 1\ \frac{1}{2}1
2
1
years, Rate of interest (R) = 10%
Amount for 1 year (A) = P\left(1+\frac{R}{100}\right)^nP(1+
100
R
)
n
= 80000\left(1+\frac{10}{100}\right)^180000(1+
100
10
)
1
= 80000\left(1+\frac{1}{10}\right)^180000(1+
10
1
)
1
= 80000\left(\frac{11}{10}\right)^180000(
10
11
)
1
= Rs. 88,000
Interest for \frac{1}{2}
2
1
year = \frac{88000\times10\times1}{100\times2}
100×2
88000×10×1
= Rs. 4,400
Total amount = Rs. 88,000 + Rs. 4,400 = Rs. 92,400
(ii) Here, Principal (P) = Rs. 80,000
Time(n) = 1\ \frac{1}{2}1
2
1
year = 3 years (compounded half yearly)
Rate of interest (R) = 10% = 5% (compounded half yearly)
Amount (A) = P\left(1+\frac{R}{100}\right)^nP(1+
100
R
)
n
= 80000\left(1+\frac{5}{100}\right)^380000(1+
100
5
)
3
= 80000\left(1+\frac{1}{20}\right)^380000(1+
20
1
)
3
= 80000\left(\frac{21}{20}\right)^380000(
20
21
)
3
= 80000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}80000×
20
21
×
20
21
×
20
21
= Rs. 92,610
Difference in amounts
= Rs. 92,610 – Rs. 92,400 = Rs. 210