ML Aggarwal Compound Interest Exe-2.2 Class 9 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-Exe-2.2 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-2.2 Class 9 ICSE Maths Solutions
| Board | ICSE |
| Subject | Maths |
| Class | 9th |
| Chapter-2 | Compound Interest |
| Topics | Solution of Exe-2.2 Questions |
| Edition | 2024-2025 |
Compound Interest Exe-2.2
ML Aggarwal Class 9 ICSE Maths Solutions
Question 1. Find the amount and the compound interest on ₹ 5000 for 2 years at 6% per annum, interest payable yearly.
Answer :
It is given that
Principal (P) = ₹ 5000
Rate of interest (r) = 6% p.a.
Period (n) = 2 years
We know that
Amount = P (1 + r/100)n
Substituting the values
= 5000 (1 + 6/100)2
By further calculation
= 5000 × 53/50 × 53/50
= ₹ 5618
Here
CI = A – P
Substituting the values
= 5618 – 5000
= ₹ 618
Question 2. Find the amount and the compound interest on ₹ 8000 for 4 years at 10% per annum interest reckoned yearly.
Answer :
It is given that
Principal (P) = ₹ 8000
Rate of interest (r) = 10% p.a.
Period (n) = 4 years
We know that
Amount = P (1 + r/100)n
Substituting the values
= 8000 (1 + 10/100)4
By further calculation
= 8000 × 11/10 × 11/10 × 11/10 × 11/10
= ₹ 11712.80
Here
CI = A – P
Substituting the values
= 11712.80 – 8000
= ₹ 3712.80
Question 3. If the interest is compounded half yearly, calculate the amount when the principal is ₹ 7400, the rate of interest is 5% and the duration is one year.
Answer :
It is given that
Principal (P) = ₹ 7400
Rate of interest (r) = 5%
Period (n) = 1 year
We know that
A = P (1 + r/(2 × 100))2×n
Substituting the values
= 7400 (1 + 5/200)2
By further calculation
= 7400 × 205/200 × 205/200
= ₹ 7774.63
Question 4. Find the amount and the compound interest on ₹ 5000 at 10% p.a. for 1 ½ years, compound interest reckoned semi-annually.
Answer :
It is given that
Principal (P) = ₹ 5000
Rate of interest = 10% p.a. or 5% half-yearly
Period (n) = 1 ½ years or 3 half-years
We know that
A = P (1 + r/100)n
Substituting the values
= 5000 (1 + 5/100)3
By further calculation
= 5000 × 21/20 × 21/20 × 21/20
= ₹ 5788.12
Here
CI = A – P
Substituting the values
= 5788.12 – 5000
= ₹ 788.12
Question 5. Find the amount and the compound interest on ₹ 100000 compounded quarterly for 9 months at the rate of 4% p.a.
Answer :
It is given that
Principal (P) = ₹ 100000
Rate of interest = 4% p.a. or 1% quarterly
Period (n) = 9 months or 3 quarters
We know that
A = P (1 + r/100)n
Substituting the values
= 100000 (1 + 1/100)3
By further calculation
= 100000 × 101/100 × 101/100 × 101/100
= ₹ 103030.10
Here
CI = A – P
Substituting the values
= 103030.10 – 100000
= ₹ 3030.10
Question 6. Find the difference between CI and SI on sum of ₹ 4800 for 2 years at 5% per annum payable yearly.
Answer :
It is given that
Principal (P) = ₹ 4800
Rate of interest (r) = 5% p.a.
Period (n) = 2 years
We know that
SI = Prt/100
Substituting the values
= (4800 × 5 × 2)/ 100
= ₹ 480
If compounded yearly
A = P (1 + r/100)n
Substituting the values
= 4800 (1 + 5/100)2
By further calculation
= 4800 × 21/20 × 21/20
= ₹ 5292
Here
CI = A – P
Substituting the values
= 5292 – 4800
= ₹ 492
So the difference between CI and SI = 492 – 480 = ₹ 12
Page 59
Compound Interest Exe-2.2
ML Aggarwal Class 9 ICSE Maths Solutions
Question 7. Find the difference between the simple interest and compound interest on ₹ 2500 for 2 years at 4% per annum, compound interest being reckoned semi-annually.
Answer :
It is given that
Principal (P) = ₹ 2500
Rate of interest (r) = 4% p.a. or 2% half-yearly
Period (n) = 2 years or 4 half-years
We know that
SI = Prt/100
Substituting the values
= (2500 × 4 × 2)/100
= ₹ 200
If compounded semi-annually
A = P (1 + r/100)n
Substituting the values
= 2500 (1 + 2/100)4
By further calculation
= 2500 × 51/50 × 51/50 × 51/50 × 51/50
= ₹ 2706.08
We know that
CI = A – P
Substituting the values
= 2706.08 – 2500
= ₹ 206.08
So the difference between CI and SI = 206.08 – 200 = ₹ 6.08
Question 8. Find the amount and the compound interest on ₹ 2000 in 2 years if the rate is 4% for the first year and 3% for the second year.
Answer :
It is given that
Principal (P) = ₹ 2000
Rate of interest = 4% on the first year and 3% for the second year
Period (n) = 2 years
We know that
Amount = P (1 + r/100)n
Substituting the values
= 2000 (1 + 4/100) (1 + 3/100)
By further calculation
= 2000 × 26/25 × 103/100
= ₹ 2142.40
Here
CI = A – P
Substituting the values
= 2142.40 – 2000
= ₹ 142.40
Question 9. Find the compound interest on ₹ 3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum.
Answer :
It is given that
Principal (P) = ₹ 3125
Rate of interest for continuous = 4%, 5% and 6%
Period (n) = 3 years
We know that
Amount = P (1 + r/100)n
Substituting the values
= 3125 (1 + 4/100) (1 + 5/100) (1 + 6/100)
By further calculation
= 3125 × 26/25 × 21/50 × 53/50
= ₹ 3617.25
Here
CI = A – P
Substituting the values
= 3617.25 – 3125
= ₹ 492.25
Question 10. What sum of money will amount to ₹ 9261 in 3 years at 5% per annum compound interest?
Answer :
It is given that
Amount (A) = ₹ 9261
Rate of interest (r) = 5% per annum
Period (n) = 3 years
We know that
A = P (1 + r/100)n
Substituting the values
9261 = P (1 + 5/100)3
By further calculation
9261 = P (21/20)3
So we get
P = (9261 × 20 × 20 × 20)/ (21 × 21 × 21)
P = ₹ 8000
Therefore, the sum of money is ₹ 8000.
Question 11. What sum invested at 4% per annum compounded semi-annually amounts to ₹ 7803 at the end of one year?
Answer :
It is given that
Amount (A) = ₹ 7803
Rate of interest (r) = 4% p.a. or 2% semi-annually
Period (n) = 1 year or 2 half years
We know that
A = P (1 + r/100)n
Substituting the values
= 7803 + (1 + 2/100)2
By further calculation
= 7803 + (51/20)2
= 7803 × 50/51 × 50/51
= ₹ 7500
Hence, the principal is ₹ 7500.
Question 12. What sum invested for 1 ½ years compounded half yearly at the rate of 4% p.a. will amount to ₹132651?
Answer:
It is given that
Amount (A) = ₹ 132651
Rate of interest (r) = 4% p.a. or 2% half yearly
Period (n) = 1 ½ years or 3 half years
We know that
A = P (1 + r/100)n
It can be written as
P = A ÷ (1 + r/100)n
Substituting the values
= 132651 ÷ (1 + 2/100)3
By further calculation
= 132651 ÷ (51/50)3
So we get
= 132651 × (50/51)3
= 132651 × 50/51 × 50/51 × 50/51
= ₹ 125000
Hence, the principal amount is ₹ 125000.
Question 13. On what sum will the compound interest for 2 years at 4% per annum be ₹ 5712?
Answer:
It is given that
CI = ₹ 5712
Rate of interest (r) = 4% p.a.
Period (n) = 2 years
We know that
A = P (1 + r/100)n
It can be written as
CI = A – P = P (1 + r/100)n – P
= P [(1 + r/100)n – 1]
Substituting the values
5712 = P [(1 + 4/100)2 – 1]
= P [(26/25)2 – 1]
By further calculation
= P [676/625 – 1]
Taking LCM
= P [(676 – 625)/ 625]
= P × 51/625
Here
P = 5712 × 625/51
= 112 × 625
= ₹ 70000
Hence, the principal amount is ₹ 70000.
Question 14. A man invests ₹ 1200 for two years at compound interest. After one year the money amounts to ₹ 1275. Find the interest for the second year correct to the nearest rupee.
Answer :
It is given that
Principal = ₹ 1200
After one year, the amount = ₹ 1275
So the interest for one year = 1275 – 1200 = ₹ 75
We know that
Rate of interest = (SI × 100)/ (P × t)
Substituting the values
= (75 × 100)/ (1200 × 1)
By further calculation
= 75/12
= 25/4
= 6 ¼ % p.a.
Here
Interest for the second year on ₹ 1275 at the rate of 25/4% = Prt/100
Substituting the values
= (1275 × 25 × 1)/ (100 × 4)
By further calculation
= 1275/16
= ₹ 79.70
= ₹ 80
Question 15. At what rate percent per annum compound interest will ₹ 2304 amount to ₹ 2500 in 2 years?
Answer:
It is given that
Amount = ₹ 2500
Principal = ₹ 2304
Period (n) = 2 years
Consider r% p.a. as the rate of interest
We know that
A = P (1 + r/100)n
It can be written as
(1 + r/100)n = A/P
Substituting the values
(1 + r/100)2 = 2500/2304
By further calculation
(1 + r/100)2 = 625/576 = (25/24)2
So we get
1 + r/100 = 25/24
r/100 = 25/24 – 1
Taking LCM
r = 100/24 = 25/6 = 4 1/6
Hence, the rate of interest is 4 1/6% p.a.
Question 16. A sum compounded annually becomes 25/16 time of itself in two years. Determine the rate of interest per annum.
Answer:
Consider sum (P) = x
Amount (A) = 25/16x
Period (n) = 2 years
We know that
A/P = (1 + r/100)n
Substituting the values
25x/16x = (1 + r/100)2
By further calculation
(1 + r/100)2 = (5/4)2
So we get
1 + r/100 = 5/4
r/100 = 5/4 – 1/1 = 1/4
By cross multiplication
r = 100 × ¼ = 25
Hence, the rate of interest is 25% p.a.
Question 17. At what rate percent will ₹ 2000 amount to ₹ 2315.25 in 3 years at compound interest?
Answer:
It is given that
Principal (P) = ₹ 2000
Amount (A) = ₹ 2315.25
Period (n) = 3 years
Consider r% p.a. as the rate of interest
We know that
A/P = (1 + r/100)n
Substituting the values
2315.25/2000 = (1 + r/100)3
By further calculation
(1 + r/100)3 = 231525/(100 × 2000) = 9261/8000 = (21/20)3
So we get
1 + r/100 = 21/20
It can be written as
r/100 = 21/20 – 1 = 1/20
r = 100/20 = 5
Hence, the rate of interest is 5% p.a.
Question 18. If ₹ 40000 amounts to ₹ 48620.25 in 2 years, compound interest payable half-yearly, find the rate of interest per annum.
Answer:
It is given that
Principal (P) = ₹ 40000
Amount (A) = ₹ 48620.25
Period (n) = 2 years = 4 half years
Consider rate of interest = r% p.a. = r/2% half yearly
We know that
A/P = (1 + r/100)n
Substituting the values
48620.25/40000 = (1 + r/200)4
By further calculation
(1 + r/200)4 = 4862025/ (100 × 40000) = 194481/160000
So we get
(1 + r/200)4 = (21/20)4
It can be written as
1 + r/200 = 21/20
r/200 = 21/20 – 1 = 1/20
By cross multiplication
r = 200 × 1/20 = 10
Hence the rate of interest per annum is 10%.
Question 19. Determine the rate of interest for a sum that becomes 216/125 times of itself in 1 ½ years, compounded semi-annually.
Answer:
Consider principal (P) = x
Amount (A) = 216/125 x
Period (n) = 1 ½ years = 3 half years
Take rate percent per year = 2r% and r% half yearly
We know that
A/P = (1 + r/100)n
Substituting the values
216x/125x = (1 + r/100)3
By further calculation
(1 + r/100)3 = 216/125 = (6/5)3
So we get
1 + r/100 = 6/5
r/100 = 6/5 – 1 = 1/5
By cross multiplication
r = 100 × 1/5 = 20%
So the rate percent per year = 2 × 20 = 40%
Question 20. At what rate percent p.a. compound interest would ₹ 80000 amounts to ₹ 88200 in two years, interest being compounded yearly. Also find the amount after 3 years at the above rate of compound interest.
Answer:
It is given that
Principal (P) = ₹ 80000
Amount (A) = ₹ 88200
Period (n) = 2 years
Consider r% per annum as the rate of interest percent
We know that
A/P = (1 + r/100)n
Substituting the values
88200/80000 = (1 + r/100)2
By further calculation
(1 + r/100)2 = 441/400 = (21/20)2
So we get
1 + r/100 = 21/20
r/100 = 21/20 – 1 = 1/20
By cross multiplication
r = 1/20 × 100 = 5
Hence, the rate of interest is 5% per annum.
Question 21. A certain sum amounts to ₹ 5292 in 2 years and to ₹ 5556.60 in 3 years at compound interest. Find the rate and the sum.
Answer:
It is given that
Amount after 2 years = ₹ 5292
Amount after 3 years = ₹ 5556.60
So the difference = 5556.60 – 5292 = ₹ 264.60
Here ₹ 264.60 is the interest on ₹ 5292 for one year
We know that
Rate % = (SI × 100)/ (P × t)
Substituting the values
= (264.60 × 100)/ (5292 × 1)
Multiply and divide by 100
= (26460 × 100)/ (100 × 5292)
= 5%
Here
A = P (1 + r/100)n
Substituting the values
5292 = P (1 + 5/100)2
By further calculation
P = 5292 ÷ (1 + 5/100)2
So we get
P = 5292 ÷ (21/20)2
P = 5292 × 21/20 × 21/20
P = ₹ 4800
Hence, the rate is 5% and the sum is ₹ 4800.
Question 22. A certain sum amounts to ₹ 798.60 after 3 years and ₹ 878.46 after 4 years. Find the interest rate and the sum.
Answer:
It is given that
Amount after 3 years = ₹ 798.60
Amount after 4 years = ₹ 878.46
So the difference = 878.46 – 798.60 = ₹ 79.86
Here ₹ 79.86 is the interest on ₹ 798.60 for 1 year.
We know that
Rate = (SI × 100)/ (P × t)
Substituting the values
= (79.86 × 100)/ (798.60 × 1)
Multiply and divide by 100
= (7986 × 100 × 100)/ (79860 × 100 × 1)
= 10%
Here
A = P (1 + r/100)n
It can be written as
P = A ÷ (1 + r/100)n
Substituting the values
P = 798.60 ÷ (1 + 10/100)3
By further calculation
P = 79860/100 × 10/11 × 10/11 × 10/11
P = ₹ 600
Question 23. In what time will ₹ 15625 amount to ₹ 17576 at 4% per annum compound interest?
Answer:
It is given that
Amount (A) = ₹ 17576
Principal (P) = ₹ 15625
Rate = 4% p.a.
Consider n years as the period
We know that
A/P = (1 + r/100)n
Substituting the values
17576/15625 = (1 + 4/100)n
By further calculation
(26/25)3 = (26/25)n
So we get
n = 3
Question 24. (i) In what time will ₹ 1500 yield ₹ 496.50 as compound interest at 10% per annum compounded annually?
(ii) Find the time (in years) in which ₹ 12500 will produce ₹ 3246.40 as compound interest at 8% per annum, interest compounded annually.
Answer :
(i) It is given that
Principal (P) = ₹ 1500
CI = ₹ 496.50
So the amount (A) = P + SI
Substituting the values
= 1500 + 496.50
= ₹ 1996.50
Rate (r) = 10% p.a.
We know that
A = P (1 + r/100)n
It can be written as
A/P = (1 + r/100)n
Substituting the values
1996.50/1500 = (1 + 10/100)n
By further calculation
199650/(1500 × 100) = (11 /10)n
So we get
1331/1000 = (11/10)n
(11/10)3 = (11/10)n
Here Time n = 3 years
(ii) It is given that
Principal (P) = ₹ 12500
CI = ₹ 3246.40
So the amount (A) = P + CI
Substituting the values
= 12500 + 3246.40
= ₹ 15746.40
Rate (r) = 8% p.a.
We know that
A = P (1 + r/100)n
It can be written as
A/P = (1 + r/100)n
Substituting the values
15746.40/12500 = (1 + 8/100)n
Multiply and divide by 100
1574640/ (12500 × 100) = (27/25)n
By further calculation
78732/ (12500 × 5) = (27/ 25)n
19683/ (3125 × 5) = (27/25)n
So we get
19683/15625 = (27/25)n
(27/25)3 = (27/25)n
Here Period = 3 years
Question 25. ₹ 16000 invested at 10% p.a., compounded semi-annually, amounts to ₹ 18522, find the time period of investment.
Answer :
It is given that
Principal (P) = ₹ 16000
Amount (A) = ₹ 18522
Rate = 10% p.a. or 5% semi-annually
Consider period = n half years
We know that
A/P = (1 + r/100)n
Substituting the values
18522/16000 = (1 + 5/100)n
By further calculation
9261/8000 = (21/20)n
So we get
(21/20)3 = (21/20)n
n = 3 half years
Here
Time = 3/2 = 1 ½ years
Question 26. What sum will amount to ₹ 2782.50 in 2 years at compound interest, if the rates are 5% and 6% for the successive years?
Answer:
It is given that
Amount (A) = ₹ 2782.50
Rate of interest for two successive years = 5% and 6%
We know that
A = P (1 + r/100)n
Substituting the values
2782.50 = P (1 + 5/100) (1 + 6/100)
By further calculation
2782.50 = P × 21/20 × 53/50
So we get
P = 2782.50 × 20/21 × 50/53
Multiply and divide by 100
P = 278250/100 × 20/21 × 50/53
P = ₹ 2500
Hence, the principal is ₹ 2500.
Question 27. A sum of money is invested at compound interest payable annually. The interest in two successive years is ₹ 225 and ₹ 240. Find:
(i) the rate of interest
(ii) the original sum
(iii) the interest earned in the third year.
Answer :
It is given that
Interest for the first year = ₹ 225
Interest for the second year = ₹ 240
So the difference = 240 – 225 = ₹ 15
Here ₹ 15 is the interest on ₹ 225 for 1 year
(i) Rate = (SI × 100)/ (P × t)
Substituting the values
= (15 × 100)/ (225 × 1)
So we get
= 20/3
= 6 2/3% p.a.
(ii) We know that
Sum = (SI × 100)/ (R × t)
Substituting the values
= (225 × 100)/ (20/3 × 1)
It can be written as
= (225 × 100 × 3)/ (20 × 1)
So we get
= 225 × 15
= ₹ 3375
(iii) Here
Amount after second year = 225 + 240 + 3375 = ₹ 3840
So the interest for the third year = Prt/100
Substituting the values
= (3840 × 20 × 1)/ (100 × 3)
= ₹ 256
Page 60
Compound Interest Exe-2.2
ML Aggarwal Class 9 ICSE Maths Solutions
Question 28. On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to ₹ 25 if the rate of interest charged for both is 5% p.a.?
Answer:
It is given that
Sum (P) = ₹ 100
Rate (R) = 5% p.a.
Period (n) = 2 years
We know that
SI = PRT/100
Substituting the values
= (100 × 5 × 2)/ 100
= ₹ 10
So the amount when interest is compounded annually = P (1 + R/100)n
Substituting the values
= 100 (1 + 5/100)2
By further calculation
= 100 × (21/20)2
= 100 × 21/20 × 21/20
So we get
= ₹ 441/4
Here
CI = A – P
Substituting the values
= 441/4 – 100
= ₹ 41/4
So the difference between CI and SI = 41/4 – 10 = ₹ ¼
If the difference is ₹ ¼ then sum = ₹ 100
If the difference is ₹ 25 then sum = (100 × 4)/ 1 × 25 = ₹ 10000
Question 29. The difference between the compound interest for a year payable half-yearly and the simple interest on a certain sum of money lent out at 10% for a year is ₹ 15. Find the sum of money lent out.
Answer :
It is given that
Sum = ₹ 100
Rate = 10% p.a. or 5% half yearly
Period = 1 years or 2 half years
We know that
A = P (1 + R/100)n
Substituting the values
= 100 (1 + 5/100)2
By further calculation
= 100 × 21/20 × 21/20
= ₹ 441/4
Here
CI = A – P
Substituting the values
= 441/4 – 100
= ₹ 41/4
SI = PRT/100
Substituting the values
= (100 × 10 × 1)/ 100
= ₹ 10
So the difference between CI and SI = 41/4 – 10 = ₹ ¼
Here if the difference is ₹ ¼ then sum = ₹ 100
If the difference is ₹ 15 then sum = (100 × 4 × 15)/ 1 = ₹ 6000
Question 30. The amount at compound interest which is calculated yearly on a certain sum of money is ₹ 1250 in one year and ₹ 1375 after two years. Calculate the rate of interest.
Answer:
It is given that
Amount after one year = ₹ 1250
Amount after two years = ₹ 1375
Here the difference = 1375 – 1250 = ₹ 125
So ₹ 125 is the interest on ₹ 1250 for 1 year
We know that
Rate of interest = (SI × 100)/ (P × t)
Substituting the values
= (125 × 100)/ (1250 × 1)
= 10%
Question 31. The simple interest on a certain sum for 3 years is ₹ 225 and the compound interest on the same sum at the same rate for 2 years is ₹ 153. Find the rate of interest and the principal.
Answer:
It is given that
SI for 3 years = ₹ 225
SI for 2 years = (225 × 2)/ 3 = ₹ 150
CI for 2 years = ₹ 153
So the difference = 153 – 150 = ₹ 3
Here ₹ 3 is interest on one year i.e. ₹ 75 for one year
We know that
Rate = (SI × 100)/ (P × t)
Substituting the values
= (3 × 100)/ (75 × 1)
= 4%
SI for 3 years = ₹ 225
Rate = 4% p.a.
So principal = (SI × 100)/ (R × t)
Substituting the values
= (225 × 100)/ (4 × 3)
= ₹ 1875
Question 32. Find the difference between compound interest on ₹ 8000 for 1 ½ years at 10% p.a. when compounded annually and semi-annually.
Answer :
It is given that
Principal (P) = ₹ 8000
Rate = 10% p.a. or 5% half-yearly
Period = 1 ½ years or 3 half years
Case 1 – When compounded annually
A = P (1 + r/100)n
Substituting the values
= 8000 (1 + 10/100) (1 + 5/100)
By further calculation
= 8000 × 11/10 × 21/20
= ₹ 9240
We know that
CI = A – P
Substituting the values
= 9240 – 8000
= ₹ 1240
Case 2 – When compounded half-yearly
A = P (1 + r/100)n
Substituting the values
= 8000 (1 + 5/100)3
By further calculation
= 8000 × 21/20 × 21/20 × 21/20
= ₹ 9261
We know that
CI = A – P
Substituting the values
= 9261 – 8000
= ₹ 1261
Here the difference between two CI = 1261 – 1240 = ₹ 21
Question 33. A sum of money is lent out at compound interest for two years at 20% p.a., CI being reckoned yearly. If the same sum of money is lent out at compound interest at same rate percent per annum, CI being reckoned half-yearly, it would have fetched ₹ 482 more by way of interest. Calculate the sum of money lent out.
Answer :
It is given hat
Sum = ₹ 100
Rate = 20% p.a. or 10% half-yearly
Period = 2 years or 4 half-years
Case 1 – When the interest is reckoned yearly
A = P (1 + r/100)n
Substituting the values
= 100 (1 + 20/100)2
By further calculation
= 100 × 6/5 × 6/5
= ₹ 144
We know that
CI = A – P
Substituting the values
= 144 – 100
= ₹ 44
Case 2 – When the interest is reckoned half-yearly
A = P (1 + r/100)n
Substituting the values
= 100 (1 + 10/100)4
By further calculation
= 100 × 11/10 × 11/10 × 11/10 × 11/10
= ₹ 146.41
We know that
CI = A – P
Substituting the values
= 146.41 – 100
= ₹ 46.41
So the difference between two CI = 46.41 – 44 = ₹ 2.41
If the difference is ₹ 2.41 then sum = ₹ 100
If the difference is ₹ 482 then sum = (100 × 482)/ 2.41
Multiplying and dividing by 100
= (100 × 482 × 100)/ 241
= ₹ 20000
Question 34. A sum of money amounts to ₹ 13230 in one year and to ₹ 13891.50 in 1 ½ years at compound interest, compounded semi-annually. Find the sum and the rate of interest per annum.
Answer:
It is given that
Amount after one year = ₹ 13230
Amount after 1 ½ years = ₹ 13891.50
So the difference = 13891.50 – 13230 = ₹ 661.50
Here ₹ 661.50 is the interest on ₹ 13230 for ½ years
We know that
Rate = (661.50 × 100 × 2)/ (13230 × 1)
Multiplying and dividing by 100
= (66150 × 100 × 2)/ (13230 × 1 × 100)
= 10% p.a.
Here
A = P (1 + r/100)n
Substituting the values
13891.50 = P (1 + 5/100)3
By further calculation
13891.50 = P × 21/20 × 21/20 × 21/20
So we get
P = 13891.50 × 20/21 × 20/21 × 20/21
P = ₹ 12000
— : End of ML Aggarwal Compound Interest Exe-2.2 Class 9 ICSE Maths Solutions :–
Return to :- ML Aggarawal Maths Solutions for ICSE Class-9
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