Compound Interest
Class 8th Mathematics (new) MHB Solution
Class 8th Mathematics (new) MHB Solution
- Find the amount and the compound interest.No.Principal (₹ )Rate (p.c.p.a.)Duration…
- Sameerrao has taken a loan of ₹ 12500 at a rate of 12 p.c.p.a. for 3 years. If the…
- To start a business Shalaka has taken a loan of ₹ 8000 at a rate of 10 {1}/{2}…
- On the construction work of a flyover bridge there were 320 workers initially. The…
- A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if…
- In a forest there are 40,000 trees. Find the expected number of trees after 3 years if…
- The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year…
- Find the compound interest if the amount of a certain principal after two years is₹…
- A loan of ₹ 15000 was taken on compound interest. If the rate of compound interest is…
- A principal amounts to ₹ 13924 in 2 years by compound interest at 18 p.c.p.a. Find the…
- The population of a suburb is 16000. Find the rate of increase in the population if the…
- In how many years ₹ 700 will amount to ₹ 847 at a compound interest rate of 10 p.c.p.a.…
- Find the difference between simple interest and compound interest on ₹ 20000 at 8…
Practice Set 14.1
Question 1.Find the amount and the compound interest.
Answer:
(a) Principal = 2000/-, Rate = 5% (p.c.p.a), Duration (n) = 2 years
2
A = 2000 (1+0.05)2
A = 2000 (1.05)2
A = 2000 (1.1025)
∴A= 2205/-
∴ C.I = A – P
∴ C.I = 2205 – 2000
C.I. = 205/-
Amount is 2205/- and Compound interest is 205/- .
b. Principal = 5000/-, Rate = 8% (p.c.p.a), Duration (n) = 3 years
3
A = 5000 (1 + 0.08)3
A = 5000 (1.08)3
A = 5000 (1.259712)
∴A= 6298.56/-
∵ C.I. = A - P
∴ C.I. = 6298.56 - 5000
C.I. = 1298.56/-
Amount is 6298.56/- and Compound interest is 1298.56/- .
c. Principal = 4000/-, Rate = 7.5% (p.c.p.a), Duration (n) = 2 years
A = 4000 (1 + 0.075)2
A = 4000 (1.075)2
A = 4000 (1.155625)
∴A= 4622.5/-
∵ C.I. = A - P
∴ C.I. = 4622.5 - 4000
C.I. = 622.5/-
Amount is 4622.5/- and Compound interest is 622.5/- .
Question 2.
Sameerrao has taken a loan of ₹ 12500 at a rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?
Answer:
Principal = 12500/-, Rate = 12% (p.c.p.a), Duration (n) = 3 years
3
A = 12500 (1+ 0.12)3
A = 12500 (1.12)3
A = 12500 (1.404928)
A = 17561.60/-
Sameerao has to pay an amount of 17561.60/- .
Question 3.
To start a business Shalaka has taken a loan of ₹ 8000 at a rate of 
p.c.p.a. After two years how much compound interest will she have to pay?
Answer:
Principal = 8000/-, Rate = 10.5% (p.c.p.a), Duration (n) = 2 years
2
A = 8000 (1+0.105)2
A = 8000 (1.105)2
A = 8000 (1.221025)
∴ A = 9768.2/-
∵ C.I. = A - P
∴ C.I. = 9768.2 - 8000
C.I. = 1768.2/-
∴ Shalaka has to pay a compound interest of 1768.2/- .
Practice Set 14.2
Question 1.On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.
Answer:
Present number of workers = 320 workers, Rate (increase) = 25% (p.c.p.a), Duration (n) = 2 years
2
A = 320 (1+0.25)2
A = 320 (1.25)2
A = 320 (1.5625)
∴A= 500/-
∴ The number of workers after 2 years will be 500.
Question 2.
A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.
Answer:
Present number of sheeps (P) = 200 sheeps, Rate = 8% (p.c.p.a), Duration (n) = 3 years
3
A = 200 (1+0.08)3
A = 200 (1.08)3
A = 200 (1.259712)
∴A= 251.9424
A = 252 sheeps (Rounded off)
∴ The number of sheeps after 3 years is 252.
Question 3.
In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.
Answer:
Present Trees (P) = 40000 trees, Rate = 5% (p.c.p.a), Duration (n) = 3 years
3
A = 40000 (1+0.05)3
A = 40000 (1.05)3
A = 40000 (1.157625)
∴A= 46305/-
∴ The expected number of trees after 3 years will be 46305.
Question 4.
The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year find the depreciation in price of the machine after two years.
Answer:
Principal = 250000/-, Rate (decrement) = 10% (p.c.p.a), Duration (n) = 2 years
2
2
A = 250000 (1-0.1)2
A = 250000 (0.9)2
A = 250000 (0.81)
∴A= 202500/-
∵ C.I. = A - P
∴ Depreciation in Price (C.I.) = 202500 – 250000
Depreciation in Price (C.I.) = -47500/-
(-) sign denotes the depreciation in amount.
∴ Depreciation in Price of the machine after 2 years will be 47500/- .
Question 5.
Find the compound interest if the amount of a certain principal after two years is
₹ 4036.80 at the rate of 16 p.c.p.a.
Answer:
Amount= 4036.80/-, Rate = 16% (p.c.p.a), Duration (n) = 2 years
2
4036.80 = P (1+0.16)2
4036.80 = P (1.16)2
4036.80 = P (1.3456)
∴ P = 3000/-
∵ C.I. = A - P
∴ C.I. = 4036.80 - 3000
C.I. = 1036.80/-
Compound interest is 1036.80/- .
Question 6.
A loan of ₹ 15000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.
Answer:
Principal = 15000/-, Rate = 12% (p.c.p.a), Duration (n) = 3 years
3
A = 15000 (1+0.12)3
A = 15000 (1.12)3
A = 15000 (1.404928)
∴A= 21073.92/-
Amount to settle the loan after 3 years is 21073.92/- .
Question 7.
A principal amounts to ₹ 13924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.
Answer:
Amount= 13924/-, Rate = 18% (p.c.p.a), Duration (n) = 2 years
2
13924 = P (1+0.18)2
13924 = P (1.18)2
13924 = P (1.3924)
∴ A = 10000/-
∴ The principal is 10000/- .
Question 8.
The population of a suburb is 16000. Find the rate of increase in the population if the population after two years is 17640.
Answer:
Present Population (P) = 16000/-, Rate = R% (p.c.p.a), Duration (n) = 2 years
Population after 2 years (A) =17640/-
2
∴R= 5%
∴ The population of that suburb will increase at the rate of 5% .
Question 9.
In how many years ₹ 700 will amount to ₹ 847 at a compound interest rate of 10 p.c.p.a.
Answer:
Principal = 700/-, Rate = 10% (p.c.p.a), Duration (n) = n years Amount = 847/-
1.21 = 
1.21 = 
∴ n = 2 years
∴ The number of years required to gain an amount of 847/- from a principal of 700/- is 2 .
Question 10.
Find the difference between simple interest and compound interest on ₹ 20000 at 8 p.c.p.a.
Answer:
Principal = 20000/-, Rate = 8% (p.c.p.a), Duration (n) = n years
For the first year, compound interest and simple interest will be same, so it will vary from second year, therefore assuming the duration as 2 years in the same case.
A = 20000 (1+0.08)2
A = 20000 (1.08)2
A = 20000 (1.1664)
∴ A = 23328/-
∵ C.I. = A - P
C.I. = 23328 – 20000
C.I. = 3328/-
S.I. = 3200/-
∴ Difference = C.I. – S.I.
Difference = 3328 – 3200
Difference = 128 /-
∴ The difference between simple interest and compound interest is 128/- .
