8) Arif took a loan of 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in
amounts he would be paying after 11/2 years if the interest is compounded annually.
Answers
Answer
1) Compounded Annually :
P=Rs.80000
R=10% p.a.
T=1 1/2 years ⟹n=1+ 2 1
Amount for 1st year.
A=P[1+ R/100 ] n
=Rs.80000[1+ 10/100 ]=Rs.88000
SI on Rs. 88000 for next 1/2 year
=Rs.88000× 10/100 × 1/2 =Rs.4400
Therefore, Amount = Rs.88000+Rs.4400 = 92400Rs.
2) Compounded half yearly :
P=Rs.80000
R=10% p.a.=5% per half year
T=1 1 /2 years ⟹n=3
A=Rs.80000[1+ 5/100 ] 3
A=Rs.92610
Thus, the difference between the two amounts = Rs.92610−Rs.92400 =Rs.210
Answer:
42984
Step-by-step explanation:
Principal = 80000
Rate = 10
Time period = 5 complete yrs. and 1 half year
Amount = P (100+r/100)^T
= 80000 (100+10/100)^5 × (100+5/100)^1
= 80000 (110/100)^5 × (105/100)
= 80000 (11/10)(11/10)(11/10)(11/10)(11/10)(105/100)
= 8 (121)(121)(105/100)
= 2 (121)(121)(105/25)
= 242(121)(21/5)
= 29282(21/5)
= 122984.4
= 122984
Difference = 122984 - 80000
= 42984