ML Aggarwal Compound Interest Exe-2.1 Class 9 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-2.1 Questions for Compound Interest council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-9.
ML Aggarwal Compound Interest Exe-2.1 Class 9 ICSE Maths Solutions
| Board | ICSE |
| Subject | Maths |
| Class | 9th |
| Chapter-2 | Compound Interest |
| Topics | Solution of Exe-2.1 Questions |
| Edition | 2024-2025 |
Compound Interest Exe-2.1
Question 1. Find the amount and the compound interest on ₹ 8000 at 5% per annum for 2 years.
Answer :
It is given that
Principal = ₹ 8000
Rate of interest = 5% p.a.
We know that
Interest for the first year = Prt/100
Substituting the values
= (8000 × 5 × 1)/ 100
= ₹ 400
So the amount for the first year or principal for the second year = 8000 + 400 = ₹ 8400
Here
Interest for the second year = (8400 × 5 × 1)/ 100
So we get
= ₹ 420
We know that
Amount after the second year = 8400 + 420
= ₹ 8820
Total compound interest = 8820 + 8000
= ₹ 820
Question 2. A man invests ₹ 46875 at 4% per annum compound interest for 3 years. Calculate:
(i) the amount standing to his credit at the end of the second year.
(ii) the interest for the third year.
(iii) the interest for the first year.
Answer :
It is given that
Principal = ₹ 46875
Rate of interest = 4% p.a.
(i) Interest for the first year = Prt/100
Substituting the values
= (46875 × 4 × 1)/ 100
= ₹ 1875
So the amount after first year or principal for the second year = 46875 + 1875 = ₹ 48750
Here
Interest for the second year = (48750 × 4 × 1)/ 100
So we get
= ₹ 1950
(ii) We know that
Amount at the end of second year = 48750 + 1950
= ₹ 50700
(iii) Interest for the third year = (50700 × 4 × 1)/ 100 = ₹ 2028
Question 3. Calculate the compound interest for the second year on ₹ 8000 for three years at 10% p.a.
Also find the sum due at the end of third year.
Answer :
It is given that
Principal = ₹ 8000
Rate of interest = 10% p.a.
We know that
Interest for the first year = Prt/100
Substituting the values
= (8000 × 10 × 1)/ 100
= ₹ 800
So the amount after the first year or principal for the second year = 8000 + 800 = ₹ 8800
(i) Interest for the second year = (8800 × 10 × 1)/ 100
= ₹ 880
So the amount after second year or principal for the third year = 8800 + 880 = ₹ 9680
Interest for the third year = (9680 × 10 × 1)/ 100
= ₹ 968
(ii) Amount due at the end of the third year = 9680 + 968
= ₹ 10648
Question 4. Ramesh invests ₹ 12800 for three years at the rate of 10% per annum compound interest.
Find:
(i) the sum due to Ramesh at the end of the first year.
(ii) the interest he earns for the second year.
(iii) the total amount due to him at the end of three years.
Answer :
It is given that
Principal = ₹ 12800
Rate of interest = 10% p.a.
(i) We know that
Interest for the first year = (12800 × 10 × 1)/ 100
= ₹ 1280
So the sum due at the end of first year = 12800 + 1280
= ₹ 14080
(ii) Principal for second year = ₹ 14080
So the interest for the second year = (14080 × 10 × 1)/ 100
= ₹ 1408
(iii) We know that
Sum due at the end of second year = 14080 + 1408
= ₹ 15488
Here
Principal for third year = ₹ 15488
Interest for the third year = (15488 × 10 × 1)/ 100
= ₹ 1548.80
So the total amount due to him at the end of third year = 15488 + 1548.80
= ₹ 17036.80
Question 5. The simple interest on a sum of money for 2 years at 12% per annum is ₹ 1380. Find:
(i) the sum of money.
(ii) the compound interest on this sum for one year payable half-yearly at the same rate.
Answer :
It is given that
Simple Interest (SI) = ₹ 1380
Rate of interest (R) = 12% p.a.
Period (T) = 2 years
(i) We know that
Sum (P) = (SI × 100)/ (R × T)
Substituting the values
= (1380 × 100)/ (12 × 2)
= ₹ 5750
(ii) Here
Principal (P) = ₹ 5750
Rate of interest (R) = 12% p.a. or 6% half-yearly
Period (n) = 1 year – 2 half years
So we get
Amount (A) = P (1 + R/100)n
Substituting the values
= 5750 (1 + 6/100)2
By further calculation
= 5750 × (53/50)2
So we get
= 5750 × 53/50 × 53/50
= ₹ 6460.70
Here
Compound Interest = A – P
Substituting the values
= 6460.70 – 5750
= ₹ 710.70
Question 6. A person invests ₹ 10000 for two years at a certain rate of interest, compounded annually. At the end of one year this sum amounts to ₹ 11200. Calculate:
(i) the rate of interest per annum.
(ii) the amount at the end of second year.
Answer :
It is given that
Principal (P) = ₹ 10,000
Period (T) = 1 year
Sum amount (A) = ₹ 11,200
Rate of interest = ?
(i) We know that
Interest (I) = 11200 – 10000 = ₹ 1200
So the rate of interest
R = (I × 100)/ (P × T)
Substituting the values
R = (1200 × 100)/ (10000 × 1)
So we get
R = 12% p.a.
Therefore, the rate of interest per annum is 12% p.a.
(ii) We know that
Period (T) = 2 years
Rate of interest (R) = 12% p.a.
Here
A = P (1 + R/100)t
Substituting the values
A = 10000 (1 + 12/100)2
By further calculation
A = 10000 (28/25)2
We can write it as
A = 10000 × 28/25 × 28/25
So we get
A = 16 × 28 × 28
A = ₹ 12544
Therefore, the amount at the end of second year is ₹ 12544.
Question 7. Mr. Lalit invested ₹ 75000 at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to ₹ 5325. Calculate
(i) the rate of interest.
(ii) the amount at the end of second year, to the nearest rupee.
Answer :
It is given that
Investment of Mr. Lalit = ₹ 5000
Period (n) = 2 years
(i) We know that
Amount after one year = ₹ 5325
So the interest for the first year = A – P
Substituting the values
= 5325 – 5000
= ₹ 325
Here
Rate = (SI × 100)/ (P × T)
Substituting the values
= (325 × 100)/ (5000 × 1)
So we get
= 13/2
= 6.5 % p.a.
(ii) We know that
Interest for the second year = (5325 × 13 × 1)/ (100 × 2)
By further calculation
= (213 × 13)/ (4 × 2)
So we get
= 2769/8
= ₹ 346.12
So the amount after second year = 5325 + 346.12
We get
= ₹ 5671.12
= ₹ 5671 (to the nearest rupee)
Question 8. A man invests ₹ 5000 for three years at a certain rate of interest, compounded annually. At the end of one year it amounts to ₹ 5600. Calculate:
(i) the rate of interest per annum
(ii) the interest accrued in the second year.
(iii) the amount at the end of the third year.
Answer :
It is given that
Principal = ₹ 5000
Consider r% p.a. as the rate of interest
(i) We know that
At the end of one year
Interest = Prt/100
Substituting the values
= (5000 × r × 1)/ 100
= 50r
Here
Amount = 5000 + 50r
We can write it as
5000 + 50r = 5600
By further calculation
50r = 5600 – 5000 = 600
So we get
r = 600/50 = 12
Hence, the rate of interest is 12% p.a.
(ii) We know that
Interest for the second year = (5600 × 12 × 1)/ 100
= ₹ 672
So the amount at the end of second year = 5600 + 672
= ₹ 6272
(iii) We know that
Interest for the third year = (6272 × 12 × 1)/ 100
= ₹ 752.64
So the amount after third year = 6272 + 752.64
= ₹ 7024.64
6 thoughts on “ML Aggarwal Compound Interest Exe-2.1 Class 9 ICSE Maths Solutions”
In question number 17
Line no. 18th and 19th are bit difficult to understand because there is a mistake it should be written like this :–
By further calculation,
= 70000 + 7700
= 77700
Some sums are solved but incorrect. Please carfully solve the sums. Or please stop to provide the solutions
update soon if any
In, Ex 2.1 here it is given Me Lalit invested ₹75000 but in the solutions there is written ₹5000……… I think the question is wrong as I have the same ML Aggarwal book of same edition
is this book applicable for 2023-24 session if yes send screen shot on 8948221203 whatsapp number
Yes. This is current year book