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
Question 9. Find the amount and the compound interest on ₹ 2000 at 10% p.a. for 2 years, compounded annually.
Answer :
It is given that
Principal (P) = ₹ 2000
Rate of interest (r) = 10% p.a.
Period (n) = 2 ½ years
We know that
Amount = P (1 + r/100)n
Substituting the values
= 2000 (1 + 10/100)2 (1 + 10/(2 × 100))
By further calculation
= 2000 × 11/10 × 11/10 × 21/20
So we get
= ₹ 2541
Here
Interest = A – P
Substituting the values
= 2541 – 2000
= ₹ 541
Question 10. Find the amount and the compound interest on ₹ 50000 for 1 ½ years at 8% per annum, the interest being compounded semi-annually.
Answer :
It is given that
Principal (P) = ₹ 50000
Rate of interest (r) = 8% p.a. = 4% semi-annually
Period (n) = 1 ½ years = 3 semi-annually
We know that
Amount = P (1 + r/100)n
Substituting the values
= 50000 (1 + 4/100)3
By further calculation
= 50000 (26/25)3
= 50000 × 26/25 × 26/25 × 26/25
= ₹ 56243.20
Here
Compound Interest = A – P
Substituting the values
= 56243.20 – 50000
= ₹ 6243.20
Question 11. Calculate the amount and the compound interest on ₹ 5000 in 2 years when the rate of interest for successive years is 6% and 8%, respectively.
Answer :
It is given that
Principal = ₹ 5000
Period = 2 years
Rate of interest for the first year = 6%
Rate of interest for the second year = 8%
We know that
Amount for two years = P (1 + r/100)n
Substituting the values
= 5000 (1 + 6/100) (1 + 8/100)
By further calculation
= 5000 × 53/50 × 27/25
= ₹ 5724
Here
Interest = A – P
Substituting the values
= 5724 – 5000
= ₹ 724
Question 12. Calculate the amount and the compound interest on ₹ 17000 in 3 years when the rate of interest for successive years is 10%, 10% and 14%, respectively.
Answer :
It is given that
Principal = ₹ 17000
Period = 3 years
Rate of interest for 3 successive years = 10%, 10% and 14%
We know that
Amount after 3 years = P (1 + r/100)n
Substituting the values
= 17000 (1 + 10/100) (1 + 10/100) (1 + 14/100)
By further calculation
= 17000 × 11/10 × 11/10 × 57/50
= ₹ 23449.80
Here
Amount of compound interest = A – P
Substituting the values
= 23449.80 – 17000
= ₹ 6449.80
Question 13. A sum of ₹ 9600 is invested for 3 years at 10% per annum at compound interest.
(i) What is the sum due at the end of the first year?
(ii) What is the sum due at the end of the second year?
(iii) Find the compound interest earned in 2 years.
(iv) Find the difference between the answers in (ii) and (i) and find the interest on this sum for one year.
(v) Hence, write down the compound interest for the third year.
Answer :
It is given that
Principal = ₹ 9600
Rate of interest = 10% p.a.
Period = 3 years
We know that
Interest for the first year = Prt/100
Substituting the values
= (9600 × 10 × 1)/ 100
= ₹ 960
(i) Amount after one year = 9600 – 960 = ₹ 10560So the principal for the second year = ₹ 10560
Here the interest for the second year = (10560 × 10 × 1)/ 100
= ₹ 1056
(ii) Amount after two years = 10560 + 1056 = ₹ 11616
(iii) Compound interest earned in 2 years = 960 + 10560 = ₹ 2016
(iv) Difference between the answers in (ii) and (i) = 11616 – 10560 = ₹ 1056
We know that
Interest on ₹ 1056 for 1 year at the rate of 10% p.a. = (1056 × 10 × 1)/ 100
= ₹ 105.60
(v) Here
Principal for the third year = ₹ 11616
So the interest for the third year = (11616 × 10 × 1)/ 100
= ₹ 1161.60
Question 14. The simple interest on a certain sum of money for 2 years at 10% p.a. is ₹ 1600. Find the amount due and the compound interest on this sum of money at the same rate after 3 years, interest being reckoned annually.
Answer :
It is given that
Period = 2 years
Rate = 10% p.a.
We know that
Sum = (SI × 100)/ (r × n)
Substituting the values
= (1600 × 100)/ (10 × 2)
= ₹ 8000
Here
Amount after 3 years = P (1 + r/100)n
Substituting the values
= 8000 (1 + 10/100)3
By further calculation
= 8000 × 11/10 × 11/10 × 11/10
= ₹ 10648
So the compound interest = A – P
Substituting the values
= 10648 – 8000
= ₹ 2648
Question 15. Vikram borrowed ₹ 20000 from a bank at 10% per annum simple interest. He lent it to his friend Venkat at the same rate but compounded annually. Find his gain after 2 ½ years.
Answer :
First case-
Principal = ₹ 20000
Rate = 10% p.a.
Period = 2 ½ = 5/2 years
We know that
Simple interest = Prt/100
Substituting the values
= (20000 × 10 × 5)/ (100 × 2)
= ₹ 5000
Second case-
Principal = ₹ 20000
Rate = 10% p.a.
Period = 2 ½ years at compound interest
We know that
Amount = P (1 + r/100)n
Substituting the values
= 20000 (1 + 10/100)2 (1 + 10/ (2 × 100))2
By further calculation
= 20000 × 11/10 × 11/10 × 21/20
= ₹ 25410
Here
Compound Interest = A – P
Substituting the values
= 25410 – 20000
= ₹ 5410
So his gain after 2 years = CI – SI
We get
= 5410 – 5000
= ₹ 410
Question 16. A man borrows ₹ 6000 at 5% compound interest. If he repays ₹ 1200 at the end of each year, find the amount outstanding at the beginning of the third year.
Answer :
It is given that
Principal = ₹ 6000
Rate of interest = 5% p.a.
We know that
Interest for the first year = Prt/100
Substituting the values
= (6000 × 5 × 1)/ 100
= ₹ 300
So the amount after one year = 6000 + 300 = ₹ 6300
Principal for the second year = ₹ 6300
Amount paid = ₹ 1200
So the balance = 6300 – 1200 = ₹ 5100
Here
Interest for the second year = (5100 × 5 × 1)/ 100 = ₹ 255
Amount for the second year = 5100 + 255 = ₹ 5355
Amount paid = ₹ 1200
So the balance = 5355 – 1200 = ₹ 4155
Question 17. Mr. Dubey borrows ₹ 100000 from State Bank of India at 11% per annum compound interest. He repays ₹ 41000 at the end of first year and ₹ 47700 at the end of second year. Find the amount outstanding at the beginning of the third year.
Answer :
It is given that
Borrowed money (P) = ₹ 100000
Rate = 11% p.a.
Time = 1 year
We know that
Amount after first year = Prt/100
Substituting the values
= (100000 × 11 × 1)/ 100
By further calculation
= 100000 + 11000
= ₹ 111000
Amount paid at the end of first year = ₹ 41000
So the principal for second year = 111000 – 41000
= ₹ 70000
We know that
Amount after second year = P + (70000 × 11)/ 100
By further calculation
= 70000 + 700
= ₹ 77700
So the amount paid at the end of second year = ₹ 47700
Here the amount outstanding at the beginning year = 77700 – 47700
= ₹ 30000
Question 18. Jaya borrowed ₹ 50000 for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays ₹ 33000 at the end of first year. Find the amount she must pay at the end of second year to clear her debt.
Answer :
It is given that
Amount borrowed by Jaya = ₹ 50000
Period (n) = 2 years
Rate of interest for two successive years are 12% and 15% respectively
We know that
Interest for the first year = Prt/100
Substituting the values
= (50000 × 12 × 1)/ 100
= ₹ 6000
So the amount after first year = 50000 + 6000 = ₹ 56000
Amount repaid = ₹ 33000
Here
Balance amount for the second year = 56000 – 33000 = ₹ 23000
Rate = 15%
So the interest for the second year = (230000 × 15 × 1)/ 100
= ₹ 3450
Amount paid after second year = 23000 + 3450 = ₹ 26450
— : End of ML Aggarwal Compound Interest Exe-2.1 Class 9 ICSE Maths Solutions :–
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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