Compound interest questions are provided here to help the students understand the applications of compound interest in our daily existence. As we know, compound interest is one of the important mathematical concepts that can be applied in many financial transactions.
Below are some situations where we can use the formula of CI to calculate the required results.
What is Compound Interest?
Interest is the additional money paid by organisations like banks or post offices on money deposited (kept) with them. Interest is also paid by people when they borrow money. When the interest is calculated on the previous year’s amount, the interest is called compounded or Compound Interest (C.I.).
The formula for finding the amount on compound interest is given by:
This is the amount when interest is compounded annually.
Compound interest (CI) = A – P
1. Find the compound interest (CI) on Rs. 12,600 for 2 years at 10% per annum compounded annually.
Solution:
Principal (P) = Rs. 12,600
Number of years (n) = 2
= 12600[1 + (10/100)] 2
= 12600 [(10 + 1)/10] 2
= 12600 × (11/10) × (11/10)
Total amount, A = Rs. 15,246
Compound interest (CI) = A – P
= Rs. 15,246 – Rs. 12,600
2. At what rate of compound interest per annum, a sum of Rs. 1200 becomes Rs. 1348.32 in 2 years?
Solution:
Let R% be the rate of interest per annum.
Principal (P) = Rs. 1200
Total amount after 2 years (A) = Rs. 1348.32
Rs. 1348.32 = Rs. 1200[1 + (R/100)] 2
1348.32/1200 = [1 + (R/100)] 2
[1 + (R/100)] 2 = 134832/120000
[1 + (R/100)] 2 = 2809/2500
[1 + (R/100)] 2 = (53/50) 2
Hence, the rate of interest is 6%.
R/200 = half-yearly rate
3. A TV was bought for Rs. 21,000. The value of the TV was depreciated by 5% per annum. Find the value of the TV after 3 years. (Depreciation means the reduction of value due to use and age of the item)
Solution:
Principal (P) = Rs. 21,000
Rate of depreciation (R) = 5%
Using the formula of CI for depreciation,
A = Rs. 21,000[1 (5/100)] 3
= Rs. 21,000[1 – (1/20)] 3
= Rs. 21,000[(20 – 1)/20] 3
= Rs. 21,000 × (19/20) × (19/20) × (19/20)
Therefore, the value of the TV after 3 years = Rs. 18,004.875.
4. Find the compound interest on Rs 48,000 for one year at 8% per annum when compounded half-yearly.
Solution:
Principal (P) = Rs 48,000
Also, the interest is compounded half-yearly.
So, A = P[1 + (R/200)] 2n
= Rs. 48000[1 + (8/200)] 2(1)
= Rs. 48000[1 + (1/25)] 2
= Rs. 48000[(25 + 1)/25] 2
= Rs. 48,000 × (26/25) × (26/25)
Therefore, the compound interest = A – P
= Rs (519,16.80 – 48,000)
5. Find the compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.
Solution:
Principal (P) = Rs. 8000
Rate of interest (R) = 15% p.a
Time (n) = 2 years 4 months
4 months = 4/12 years = 1/3 years
= Rs. 8000 [1 + (15/100)] 2 [1 + (1/3) × (15/100)]
= Rs. 8000 [1 + (3/20)] 2 [1 + (3/20 × 3)]
= Rs. 8000 [(20 + 3)/20] 2 [(20 + 1)/20]
= Rs. 8000 × (23/20) × (23/20) × (21/20)
Therefore, the compound interest = A – P = Rs. 11,109 – Rs. 8000 = Rs. 3109
6. If principal = Rs 1,00,000. rate of interest = 10% compounded half-yearly. Find
(i) Interest for 6 months.
(ii) Amount after 6 months.
(iii) Interest for the next 6 months.
(iv) Amount after one year.
Solution:
(i) A = P[1 + (R/200)] 2n
Here, 2n is the number of half years.
Let us find the interest compounded half-yearly for 6 months, i.e., one half year.
So, A = Rs. 1,00,000 [1 + (10/200)] 1
= Rs. 1,00,000 [(20 + 1)/20]
= Rs. 1,00,000 × 21/20
Compounded interest for 6 months = Rs. 1,05,000 – Rs. 1,00,000 = Rs. 5000
(ii) Amount after 6 months = Rs. 1,05,000
(iii) To find the interest for the next 6 months, we should consider the principal amount as Rs. 1,05,000.
Thus, A = Rs. 1,05,000 [1 + (10/200)] 1
= Rs. 1,05,000 × (21/20)
Compound interest for next 6 months = Rs. 1,10,250 – Rs. 1,05,000 = Rs. 5250
(iv) Amount after one year = Rs. 1,10,250
7. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.
(i) Find the population in 2001.
(ii) What would be its population in 2005?
Solution:
(i) Let P be the population in the year 2001.
Thus, population in the year 2003 = A = 54000 (given)
54000 = P[1 + (5/100)] 2
54000 = P[1 + (1/20)] 2
54000 = P × [(20 + 1)/20] 2
54000 = P × (21/20) × (21/20)
P = 54000 × (20/21) × (20/21)
The population in 2001 = 48980 (approx.)
(ii) Given that the population in the year 2003 = P = 54000
= 54000 × [(20 + 1)/20] 2
= 54000 × (21/20) × (21/20)
Therefore, the population in 2005 = 59535
R/400 = Quarterly rate
8. What is the difference between the compound interests on Rs. 5000 for 1 ½ year at 4% per annum compounded yearly and half-yearly?
Solution:
Time (n) = 1 ½ years
When the interest is compounded yearly,
= Rs. 5000 [1 + (4/100)] [1 + (1/2 × 4/100)]
= Rs. 5000 [1 + (1/25)] [1 + (1/50)]
= Rs. 5000 [(25 + 1)/25] [(50 + 1)/50]
= Rs. 5000 × (26/25) × (51/50)
CI = A – P = Rs. 5304 – Rs. 5000 = Rs. 304
When the interest is compounded half-yearly,
n = 1 ½ years = 3 half-years
A = Rs. 5000 [1 + (4/200)] 3
= Rs. 5000 [1 + (1/50)] 3
= Rs. 5000 [(50 + 1)/50] 3
= Rs. 5000 × (51/50) × (51/50) × (51/50)
CI = A – P = Rs. 5306.04 – Rs. 5000 = Rs. 306.04
Difference between compound interest = Rs. 306.04 – Rs. 304 = Rs. 2.04
9. The population of a town decreased every year due to migration, poverty and unemployment. The present population of the town is 6,31,680. Last year the migration was 4%, and the year before last, it was 6%. What was the population two years ago?
Solution:
The present population of the town (A) = 631680
Last year migration rate was 4%, and the year before, the previous migration rate was 6%.
Let P be the population of a town, two years ago.
According to the given situation, the total population is:
631680 = P [1 – (4/100)] [1 – (6/100)]
631680 = P [1 – (1/25)] [1 – (3/50)]
631680 = P[(25 – 1)/25] [(50 – 3)/50]
631680 = P × (24/25) × (47/50)
P = 631680 × (25/24) × (50/47)
Therefore, the population of the town, two years ago = 700000
10. Find the amount and the compound interest on Rs. 1,00,000 compounded quarterly for 9 months at the rate of 4% per annum.
Solution:
Here, R/400 is the quarterly interest rate.
4n = 9 months = 3 quarters
So, A = Rs. 1,00,000 [1 + (4/400)] 3
= Rs. 1,00,000 [1 + (1/100)] 3
= Rs. 1,00,000 [(100 + 1)/100] 3
= Rs. 1,00,000 × (101/100) × (101/100) × (101/100)
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