1. Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at 12½ % per annum compounded annually.
Solution:
Given,
Principal (P) = ₹ 10,800
Rate (R) = 12½ % = 25/2 % (annual)
Number of years (n) = 3
Now,
Amount (A) = P(1 + R/100)n
= 10800(1 + 25/200)3
= 10800(225/200)3
= 15377.34375
= ₹ 15377.34 (approximately)
Compound Interest = A – P = ₹ (15377.34 – 10800) = ₹ 4,577.34
(b) ₹ 18000 for 2½ years at 10% per annum compounded annually.
Solution:
Given,
Principal (P) = ₹ 18,000
Rate (R) = 10% annual
Number of years (n) = 2½
[We can calculate the amount for 2 years and 6 months by calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 2 years]
Now,
First, the amount for 2 years has to be calculated
Amount, A = P(1 + R/100)n
= 18000(1 + 1/10)2
= 18000(11/10)2
= ₹ 21780
By taking ₹ 21780 as principal, the Simple Interest for the next ½ year will be calculated
Simple interest = (21780 x ½ x 10)/100
= ₹ 1089
Hence, the interest for the first 2 years = ₹ (21780 – 18000) = ₹ 3780
And, interest for the next ½ year = ₹ 1089
Total Compound Interest = ₹ 3780 + ₹ 1089
= ₹ 4,869
Therefore,
Amount, A = P + C.I.
= ₹ 18000 + ₹ 4869
= ₹ 22,869
(c) ₹ 62500 for 1½ years at 8% per annum compounded half yearly.
Solution:
Given,
Principal (P) = ₹ 62,500
Rate = 8% per annum or 4% per half year
Number of years = 1½
There will be 3 half years in 1½ years
Amount, A = P(1 + R/100)n
= 62500(1 + 4/100)3
= 62500(104/100)3
= 62500(26/25)3
= ₹ 70304
Compound Interest = A – P = ₹ 70304 – ₹ 62500 = ₹ 7,804
(d) ₹ 8000 for 1 year at 9% per annum compound half yearly.
(You could use the year by year calculation using SI formula to verify)
Solution:
Given,
Principal (P) = ₹ 8000
Rate of interest = 9% per annum or 9/2% per half year
Number of years = 1 year
There will be 2 half years in 1 year
Now,
Amount, A = P(1 + R/100)n
= 8000(1 + 9/200)2
= 8000(209/200)2
= 8736.20
Compound Interest = A – P = ₹ 8736.20 – ₹ 8000 = ₹ 736.20
(e) ₹ 10000 for 1 year at 8% per annum compounded half yearly.
Solution:
Given,
Principal (P) = ₹ 10,000
Rate = 8% per annum or 4% per half year
Number of years = 1 year
There are 2 half years in 1 year
Now,
Amount, A = P(1 + R/100)n
= 10000(1 + 4/100)2
= 10000(1 + 1/25)2
= 10000(26/25)2
= ₹ 10816
Compound Interest = A – P = ₹ 10816 – ₹ 10000 = ₹ 816
2. Kamala borrowed ₹ 26400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
(Hint: Find A for 2 years with interest is compounded yearly and then find S.I. on the 2nd year amount for 4/12 years.)
Solution:
Given,
Principal (P) = ₹ 26,400
Rate (R) = 15% per annum
Number of years (n) = 2 4/12
[We can calculate the amount for 2 years and 4 by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of 2 years]
First, the amount for 2 years has to be calculated
Amount, A = P(1 + R/100)n
= 26400(1 + 15/100)2
= 26400(1 + 3/20)2
= 26400(23/20)2
= ₹ 34914
Now,
Taking ₹ 34,914 as principal, the S.I. for the next 1/3 years will be-
S.I. = (34914 × 1/3 x 15)/100 = ₹ 1745.70
Interest for the first two years = ₹ (34914 – 26400) = ₹ 8,514
And interest for the next 1/3 year = ₹ 1,745.70
Total C.I. = ₹ (8514 + ₹ 1745.70) = ₹ 10,259.70
Amount = P + C.I. = ₹ 26400 + ₹ 10259.70 = ₹ 36,659.70
3. Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
Given,
For Fabina-
P = ₹ 12,500
R = 12%
T = 3 years
Interest paid by Fabina = (P x R x T)/100
= (12500 x 12 x 3)/100
= 4500
Amount paid by Radha at the end of 3 years
A = P(1 + R/100)n
A = 12500(1 + 10/100)3
= 12500(110/100)3
= ₹ 16637.50
C.I. = A – P = ₹ 16637.50 – ₹ 12500 = ₹ 4,137.50
The interest paid by Fabina is ₹ 4,500 and by Radha is ₹ 4,137.50
Thus, Fabina pays more interest
₹ 4500 − ₹ 4137.50 = ₹ 362.50
Hence, Fabina will have to pay ₹ 362.50 more.
4. I borrowed ₹ 12000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Solution:
P = ₹ 12000
R = 6% per annum
T = 2 years
S.I. = (P x R x T)/100
= (12000 x 6 x 2)/100
= ₹ 1440
To find the compound interest, first we calculate the amount (A).
Amount, A = P(1 + R/100)n
= 12000(1 + 6/100)2
= 12000(106/100)2
= 12000(53/50)2
= ₹ 13483.20
C.I. = A − P
= ₹ 13483.20 − ₹ 12000
= ₹ 1,483.20
C.I. − S.I. = ₹ 1,483.20 − ₹ 1,440
= ₹ 43.20
Hence, the extra amount to be paid is ₹ 43.20.
5. Vasudevan invested ₹ 60000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
Solution:
P = ₹ 60,000
Rate = 12% per annum = 6% per half year
n = 6 months = 1 half year
Now,
A = P(1 + R/100)n
= 60000(1 + 6/100)1
= 60000(106/100)
= 60000(53/50)
= ₹ 63600
(ii) after 1
Solution:
There are 2 half years in 1 year
So, n = 2
Now,
A = P(1 + R/100)n
= 60000(1 + 6/100)2
= 60000(106/100)2
= 60000(53/50)2
= ₹ 67416
6. 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 1½ years if the interest is
(i) Compounded annually
Solution:
P = ₹ 80,000
R = 10% per annum
n = 1½ years
We can calculate the amount for 1 year and 6 months by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.
Now, the amount for 1 year has to be calculated
Amount, A = P(1 + R/100)n
= 80000(1 + 10/100)1
= 80000 x 11/100
= ₹ 88000
Now, taking ₹ 88,000 as principal, the S.I. for the next ½ year will be calculated as
S.I. = (P x R x T)/100
= (88000 x 10 x ½)/100
= ₹ 4400
Interest for the first year = ₹ 88000 – ₹ 80000 = ₹ 8000
And interest for the next ½ year = ₹ 4,400
Total C.I. = ₹ 8,000 + ₹ 4,400 = ₹ 12,400
A = P + C.I.= ₹ (80000 + 12400)
= ₹ 92,400
(ii) Compounded half yearly
Solution:
The interest is compounded half yearly
Rate = 10% per annum = 5% per half year
There will be three half years in 1½ years
Now,
Amount, A = P(1 + R/100)n
= 80000(1 + 5/100)3
= 80000(105/100)3
= ₹ 92610
Thus, the difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210
7. Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year
Solution:
Given,
P = ₹ 8,000
R = 5% per annum
n = 2 years
Amount, A = P(1 + R/100)n
= 8000(1 + 5/100)2
= 8000(105/100)2
= ₹ 8820
(ii) The interest for the 3rd year
Solution:
The interest for the next one year, i.e. the third year, has to be calculated by taking ₹8,820 as principal, the S.I. for the next year will be calculated.
S.I. = (P x R x T)/100
= (8820 x 5 x 1)/100
= ₹ 441
8. Find the amount and the compound interest on ₹ 10,000 for 1½ years at 10% per annum, compounded half yearly. Would this interest be more that the interest he would get if it was compounded annually?
Solution:
Given,
P = ₹ 10,000
Rate = 10% per annum = 5% per half year
n = 1½ years
There will 3 half years in 1½ years
Now,
A = P(1 + R/100)n
= 10000(1 + 5/100)3
= 10000(105/100)3
= ₹ 11576.25
C.I. = A − P
= ₹ 11576.25 − ₹ 10000
= ₹ 1,576.25
We can calculate the amount for 1 year and 6 months by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.
Amount, A = P(1 + R/100)n
= 10000(1 + 10/100)1
= 10000(110/100)
= ₹ 11000
By taking ₹ 11,000 as the principal, the S.I. for the next ½ year will be calculated as
S.I. = (P x R x T)/100
= (11000 x 10 x ½)/100
= ₹ 550
So, the interest for the first year = ₹ 11000 − ₹ 10000 = ₹ 1,000
Hence, Total compound interest = ₹ 1000 + ₹ 550 = ₹ 1,550
So, the difference between two interests = 1576.25 – 1550 = 26.25
Hence, the interest would be 26.25 more when compounded half yearly than the interest when compounded annually.
9. Find the amount which Ram will get on ₹ 4,096, he gave it for 18 months at 12½ per annum, interest being compounded half yearly.
Solution:
Given,
P = ₹ 4,096
R = 12½ per annum = 25/2 per annum = 25/4 per half year
n = 18 months
There will be 3 half years in 18 months
So,
A = P(1 + R/100)n
= 4096(1 + 25/(4 x 100)) 3
= 4096 x (1 + 1/16) 3
= 4096 x (17/16) 3
= ₹ 4913
Hence, the required amount is ₹ 4,913.
10. The population of a place increased to 54000 in 2003 at a rate of 5% per annum
(i)find the population in 2001
Solution:
Given, population in the year 2003 = 54,000
Now,
54,000 = (Population in 2001) (1 + 5/100)2
54,000 = (Population in 2001) (105/100)2
Population in 2001 = 54000 x (100/105)2
= 48979.59
Hence, the population in the year 2001 was approximately 48,980
(ii) what would be its population in 2005?
Solution:
Population in 2005 = 54000(1 + 5/100)2
= 54000(105/100)2
= 54000(21/20)2
= 59535
Hence, the population in the year 2005 would be 59,535.
11. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
Given the initial count of bacteria = 5,06,000
Bacteria at the end of 2 hours = 506000(1 + 2.5/100)2
= 506000(1 + 1/40)2
= 506000(41/40)2
= 531616.25
Hence, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.).
12. A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
Principal = ₹ 42,000
Depreciation = 8% of ₹ 42,000 per year
= (P x R x T)/100
= (42000 x 8 x 1)/100
= ₹ 3360
Hence, the value after 1 year = ₹ 42000 − ₹ 3360 = ₹ 38,640.
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