Compound Interest Class 8 RS Aggarwal Exe-8B Goyal Brothers ICSE Maths Solutions Ch-8. We provide step by step Solutions of council prescribe textbook / publication to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.
Compound Interest Class 8 RS Aggarwal Exe-8B Goyal Brothers ICSE Maths Solutions Ch-8
| Board | ICSE |
| Publications | Goyal Brothers Prakashan |
| Subject | Maths |
| Class | 8th |
| writer | RS Aggarwal |
| Book Name | Foundation |
| Ch-8 | Simple Interest and Compound Interest |
| Exe-8A | Problems on Compound Interest |
| Academic Session | 2024-2025 |
How to Solve Problems on Compound Interest
The formula we use to find compound interest is A = P(1 + r/100)^t. In this formula, A stands for the total amount that accumulates. P is the original principal; that’s the money we start with. The r is the interest rate while t is for time.
Exercise- 8B
Que-1: Find the amount and the compound interest on Rs5000 for 2 years at 8% per annum, compounded annually.
Sol: Principal (P) = Rs. 5000
Rate (R) = 8% p.a.
Period (n) = 2 years
Amount (A) = P (1+(R/100))^n
= Rs. 5000(1+(8/100))²
= Rs.5000×(27/25)×(27/25)
= Rs.5832∴
C.I. = A – P
= Rs. 5832 – 5000
= Rs. 832
Que-2: Find the amount and the compound interest on Rs8000 for 2 years at 6% per annum, compounded annually.
Sol: Principal (P) = Rs. 8000
Rate (R) = 6% p.a.
Period (n) = 2 years
Amount (A) = P (1+(R/100))^n
= Rs. 8000(1+(6/100))²
= Rs.8000×(106/100)×(106/100)
= Rs.8988.80
∴ C.I. = A – P
= Rs. 8988.80 – 8000
= Rs. 988.80
Que-3: Find the amount and the compound interest on Rs2500 for 2 years, compounded annually, the rate of interest being 6% during the first year and 8% during the second year.
Sol: P for first year = 2500
Rate of first year = 6%
time = 1 year
I = PRT/100
I = (2500*6*1)/100
I = 150
Amount at the end of first year = 2500 + 150 = 2650
P for second year = 2650
Time = 1 year
Rate for second year = 8%
I = PRT/100
I = (2650*8*1)/100
I = 212
Amount = 2650 + 212 = 2862
C.I. = A – P
= 2862 – 2500
= 362.
Que-4: Find the amount and the compound interest on Rs25000 for 3 years at 6% per annum, compounded annually.
Sol: Given, Principal =Rs. 25,000
Time = 3 years
Rate = 6%
Amount compounded annually = 25000(1+(6/100))³
= 25,000×(106/100)×(106/100)×(106/100)
=29,775.4
Compound interest = Rs. 29,775.4 − Rs. 25,000
= Rs. 4775.4.
Que-5: Find the amount and the compound interest on Rs10000 for 3 years at 10% per annum, compounded annually.
Sol: Principal (P) = Rs. 10000
Rate (R) = 10% p.a.
Period (n) = 3 years
∴ Amount = P (1+(R/100))^n
= Rs. 10000 (1+(10/100))³
= Rs.10000×(11/10)×(11/10)×(11/10)
= Rs.13310
C.I. = A – P
= Rs. 13310 – 10000
= Rs. 3310
Que-6: Karim took a loan of Rs25000 from Corporation Bank at 12% per annum, compounded annually. How much amount, he will have to pay at the end of 3 years.
Sol: Principal = ₹25000
rate= 12%
time= 3years
Amount = P× ( 1+ r/100)^n
Amount = 25000 × ( 1+ 12/100)^3
Amount = 25000 × 28/25 × 28/25× 28/25
Amount = 175616/5
Amount = 35123.20
Que-7: Manoj deposited Rs15625 in a bank at 8% per annum, compounded annually. How much amount will he get after 3 years ?
Sol: P = 15625 rs
R = 8%
T = 3yrs
A = p ( 1+r/100)^t
= 15625( 1+8/100)³
15625( 1+2/25)³
15625( 27/25)³
15625*27/25*27/25*27/25
These all 25 are cancelled with 15625…
Remaining will be multiplied
27*27*27 = 19683
Que-8: A person lent out Rs16000 on simple interest and the same on the compound interest for 2 years at 12*(1/2)% per annum. Find the ratio of the amounts received by him as interest after 2 years.
Sol: From above solution,
simple interest = Rs 4000
compound interest
= Amount – principle
= 20250-16000
= 4250 Rs
Ratio of interest = 4000/4250
=16/17
–: Compound Interest Class 8 RS Aggarwal Exe-8B Goyal Brothers ICSE Maths Ch-8 :–
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