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Compound Interest [Using Formula] - Q&A

EXERCISE 3 (A)

1. Find the amount and the compound interest on ₹ 12,000 in 3 years at 5%; interest being compounded annually.
Answer:
Given: Principal (P) = ₹ 12,000, Rate (r) = 5%, Time (n) = 3 years.
Formula: A = P(1 + r/100)ⁿ
Amount (A) = 12000(1 + 5/100)³
= 12000(1 + 1/20)³
= 12000(21/20)³
= 12000 × (9261 / 8000)
= 12 × 9261 / 8
= 1.5 × 9261
= ₹ 13,891.50
Compound Interest (C.I.) = Amount - Principal
= 13,891.50 - 12,000
= ₹ 1,891.50

2. Calculate the amount, if ₹ 15,000 is lent at compound interest for 2 years and the rates for the successive years are 8% p.a. and 10% p.a. respectively.
Answer:
Given: P = ₹ 15,000, r₁ = 8%, r₂ = 10%.
Formula: A = P(1 + r₁/100)(1 + r₂/100)
A = 15000(1 + 8/100)(1 + 10/100)
= 15000(1 + 2/25)(1 + 1/10)
= 15000(27/25)(11/10)
= 1500 × 27 × 11 / 25
= 60 × 27 × 11
= ₹ 17,820

3. Calculate the compound interest accrued on ₹ 6,000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 8% and 10% respectively.
Answer:
Given: P = ₹ 6,000, r₁ = 5%, r₂ = 8%, r₃ = 10%.
Amount (A) = 6000(1 + 5/100)(1 + 8/100)(1 + 10/100)
= 6000(105/100)(108/100)(110/100)
= 6000(21/20)(27/25)(11/10)
= 6000 × 1.05 × 1.08 × 1.1
= 6000 × 1.2474
= ₹ 7,484.40
Compound Interest = A - P
= 7484.40 - 6000
= ₹ 1,484.40

4. What sum of money will amount to ₹ 5,445 in 2 years at 10% per annum compound interest?
Answer:
Given: A = ₹ 5,445, n = 2 years, r = 10%. Let P be the sum.
5445 = P(1 + 10/100)²
5445 = P(11/10)²
5445 = P(121/100)
P = (5445 × 100) / 121
P = 45 × 100
Sum = ₹ 4,500

5. On what sum of money will the compound interest for 2 years at 5 per cent per annum amount to ₹ 768.75?
Answer:
Given: C.I. = ₹ 768.75, n = 2 years, r = 5%.
C.I. = P[(1 + r/100)ⁿ - 1]
768.75 = P[(1 + 5/100)² - 1]
768.75 = P[(21/20)² - 1]
768.75 = P[441/400 - 1]
768.75 = P[41/400]
P = (768.75 × 400) / 41
P = 307500 / 41
P = ₹ 7,500

6. Find the sum on which the compound interest for 3 years at 10% per annum amounts to ₹ 1,655.
Answer:
Given: C.I. = ₹ 1,655, n = 3 years, r = 10%.
1655 = P[(1 + 10/100)³ - 1]
1655 = P[(1.1)³ - 1]
1655 = P[1.331 - 1]
1655 = P(0.331)
P = 1655 / 0.331
P = ₹ 5,000

7. What principal will amount to ₹ 9,856 in two years, if the rates of interest for successive years are 10% and 12% respectively?
Answer:
Given: A = ₹ 9,856, r₁ = 10%, r₂ = 12%.
9856 = P(1 + 10/100)(1 + 12/100)
9856 = P(1.1)(1.12)
9856 = P(1.232)
P = 9856 / 1.232
P = ₹ 8,000

8. On a certain sum, the compound interest in 2 years amounts to ₹ 4,240. If the rates of interest for successive years are 10% and 15% respectively, find the sum.
Answer:
Given: C.I. = ₹ 4,240, r₁ = 10%, r₂ = 15%.
Amount A = P(1 + 10/100)(1 + 15/100) = P(1.1)(1.15) = 1.265P
C.I. = A - P = 1.265P - P = 0.265P
4240 = 0.265P
P = 4240 / 0.265
P = 4240000 / 265
Sum = ₹ 16,000

9. At what rate per cent per annum will ₹ 6,000 amount to ₹ 6,615 in 2 years when interest is compounded annually?
Answer:
Given: P = 6000, A = 6615, n = 2.
A = P(1 + r/100)ⁿ
6615 = 6000(1 + r/100)²
(1 + r/100)² = 6615 / 6000
(1 + r/100)² = 441 / 400
(1 + r/100)² = (21/20)²
Taking square root:
1 + r/100 = 21/20
r/100 = 21/20 - 1 = 1/20
r = 100/20 = 5%

10. At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2 years?
Answer:
Let Principal P = x. Then Amount A = 1.44x. Time n = 2 years.
1.44x = x(1 + r/100)²
1.44 = (1 + r/100)²
(1.2)² = (1 + r/100)²
1.2 = 1 + r/100
0.2 = r/100
r = 20%
Rate = 20% p.a.

11. At what rate per cent will a sum of ₹ 4,000 yield ₹ 1,324 as compound interest in 3 years?
Answer:
P = 4000, C.I. = 1324, n = 3.
Amount A = P + C.I. = 4000 + 1324 = 5324.
5324 = 4000(1 + r/100)³
(1 + r/100)³ = 5324 / 4000
(1 + r/100)³ = 1331 / 1000
(1 + r/100)³ = (11/10)³
1 + r/100 = 11/10
r/100 = 1/10
r = 10%
Rate = 10% p.a.

12. A person invests ₹ 5,000 for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to ₹ 6,272. Calculate:
(i) the rate of interest per annum.
(ii) the amount at the end of the third year.
Answer:
(i) P = 5000, Amount after 2 years = 6272.
6272 = 5000(1 + r/100)²
(1 + r/100)² = 6272 / 5000 = 3136 / 2500
(1 + r/100)² = (56/50)²
1 + r/100 = 56/50 = 1.12
r/100 = 0.12 ⇒ r = 12%
(ii) Amount after 3 years can be calculated by applying interest on the 2nd year amount.
A₃ = A₂(1 + r/100)
A₃ = 6272(1 + 12/100)
A₃ = 6272 × 1.12
A₃ = ₹ 7,024.64

13. In how many years will ₹ 7,000 amount to ₹ 9,317 at 10 per cent per annum compound interest?
Answer:
P = 7000, A = 9317, r = 10%.
9317 = 7000(1 + 10/100)ⁿ
9317 / 7000 = (11/10)ⁿ
1.331 = (1.1)ⁿ
(1.1)³ = (1.1)ⁿ
n = 3
Time = 3 years

14. Find the time, in years, in which ₹ 4,000 will produce ₹ 630.50 as compound interest at 5 per cent p.a. interest being compounded annually.
Answer:
P = 4000, C.I. = 630.50, r = 5%.
A = P + C.I. = 4000 + 630.50 = 4630.50.
4630.50 = 4000(1 + 5/100)ⁿ
4630.50 / 4000 = (1.05)ⁿ
463050 / 400000 = (1.05)ⁿ
1.157625 = (1.05)ⁿ
Since 1.05³ = 1.157625
n = 3
Time = 3 years

15. Divide ₹ 28,730 between A and B so that when their shares are lent out at 10 per cent compound interest compounded per year, the amount that A receives in 3 years is the same as what B receives in 5 years.
Answer:
Let A's share be x and B's share be y.
Amount of A in 3 years = x(1 + 10/100)³
Amount of B in 5 years = y(1 + 10/100)⁵
Given: Amount A = Amount B
x(1.1)³ = y(1.1)⁵
x/y = (1.1)⁵ / (1.1)³ = (1.1)² = 1.21 = 121/100
Ratio of shares A : B = 121 : 100
Sum of ratios = 121 + 100 = 221
Total sum = 28730
A's share = (121 / 221) × 28730 = 121 × 130 = ₹ 15,730
B's share = (100 / 221) × 28730 = 100 × 130 = ₹ 13,000

16. A sum of ₹ 44,200 is divided between John and Smith, 12 years and 14 years old respectively, in such a way that if their portions be invested at 10 percent per annum compound interest, they will receive equal amounts on reaching 16 years of age.
(i) What is the share of each out of ₹ 44,200?
(ii) What will each receive, when 16 years old?
Answer:
(i) Let John's share = x, Smith's share = y.
John is 12, time to reach 16 = 4 years.
Smith is 14, time to reach 16 = 2 years.
Amount of John = x(1 + 10/100)⁴
Amount of Smith = y(1 + 10/100)²
x(1.1)⁴ = y(1.1)²
x/y = 1 / (1.1)² = 1 / 1.21 = 100/121
Ratio x : y = 100 : 121. Total parts = 221.
John's share = (100 / 221) × 44200 = 100 × 200 = ₹ 20,000
Smith's share = (121 / 221) × 44200 = 121 × 200 = ₹ 24,200
(ii) Amount received by each (e.g., Smith's amount):
A = 24200(1.1)² = 24200 × 1.21 = ₹ 29,282

17. The simple interest on a certain sum of money at 10% per annum is ₹ 6,000 in 2 years. Find:
(i) the sum.
(ii) the amount due at the end of 3 years and at the same rate of interest compounded annually.
(iii) the compound interest earned in 3 years.
Answer:
(i) S.I. = (P × R × T) / 100
6000 = (P × 10 × 2) / 100
6000 = P / 5 ⇒ P = 30000
Sum = ₹ 30,000
(ii) Amount for 3 years at 10% CI:
A = 30000(1 + 10/100)³ = 30000(1.1)³
A = 30000 × 1.331
A = ₹ 39,930
(iii) C.I. earned in 3 years = A - P
= 39930 - 30000
= ₹ 9,930

18. Find the difference between compound interest and simple interest on ₹ 8,000 in 2 years and at 5% per annum.
Answer:
Method 1: Calculate both.
S.I. = (8000 × 5 × 2) / 100 = 800.
C.I. = 8000[(1 + 5/100)² - 1] = 8000[(1.05)² - 1] = 8000(1.1025 - 1) = 8000(0.1025) = 820.
Difference = 820 - 800 = ₹ 20
Method 2: Difference formula for 2 years = P(R/100)²
Diff = 8000(5/100)² = 8000(1/20)² = 8000(1/400) = 20.
Answer: ₹ 20

EXERCISE 3 (B)

1. The difference between simple interest and compound interest on a certain sum is ₹ 54.40 for 2 years at 8 percent per annum. Find the sum.
Answer:
For 2 years, Difference = P(R/100)²
54.40 = P(8/100)²
54.40 = P(64/10000)
P = (54.40 × 10000) / 64
P = 544000 / 64
Sum = ₹ 8,500

2. A sum of money, invested at compound interest, amounts to ₹ 19,360 in 2 years and to ₹ 23,425.60 in 4 years. Find the rate percent and the original sum of money.
Answer:
A₂ = P(1 + r/100)² = 19360 ... (i)
A₄ = P(1 + r/100)⁴ = 23425.60 ... (ii)
Divide (ii) by (i):
(1 + r/100)² = 23425.60 / 19360 = 1.21
1 + r/100 = √1.21 = 1.1
r/100 = 0.1 ⇒ r = 10%
Substitute r in (i):
P(1.1)² = 19360
P(1.21) = 19360
P = 19360 / 1.21
P = ₹ 16,000

3. A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.
Answer:
Let Sum = P. Amount = 3P in 8 years.
3P = P(1 + r/100)⁸ ⇒ 3 = (1 + r/100)⁸
We need Amount = 27P. Let time be T.
27 = (1 + r/100)^T
Since 27 = 3³, we have 3³ = [(1 + r/100)⁸]³
27 = (1 + r/100)²⁴
Comparing powers, T = 24 years

4. On what sum of money will compound interest (payable annually) for 2 years be the same as simple interest on ₹ 9,430 for 10 years, both at the rate of 5 percent per annum?
Answer:
Calculate S.I. first:
S.I. = (9430 × 5 × 10) / 100 = 9430 × 0.5 = 4715.
Now, let Sum for C.I. be P.
C.I. for 2 years at 5% = P[(1.05)² - 1] = P(1.1025 - 1) = 0.1025P.
Given C.I. = S.I.
0.1025P = 4715
P = 4715 / 0.1025
P = ₹ 46,000

5. Kamal and Anand each lent the same sum of money for 2 years at 5% at simple interest and compound interest respectively. Anand received ₹ 15 more than Kamal. Find the amount of money lent by each and the interest received.
Answer:
This is a problem of difference between C.I. and S.I. for 2 years.
Diff = ₹ 15, R = 5%, T = 2 years.
Diff = P(R/100)²
15 = P(5/100)² = P(1/400)
P = 15 × 400 = ₹ 6,000 (Sum lent)
Interest received by Kamal (S.I.) = (6000 × 5 × 2) / 100 = ₹ 600
Interest received by Anand (C.I.) = 600 + 15 = ₹ 615

6. Simple interest on a sum of money for 2 years at 4% is ₹ 450. Find compound interest on the same sum and at the same rate for 2 years.
Answer:
S.I. = (P × 4 × 2) / 100 = 450
8P / 100 = 450 ⇒ P = (45000 / 8) = 5625.
C.I. for 2 years at 4% on 5625:
A = 5625(1 + 4/100)² = 5625(1.04)² = 5625(1.0816) = 6084.
C.I. = 6084 - 5625 = ₹ 459

7. Simple interest on a certain sum of money for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by ₹ 228. Find the sum.
Answer:
Let Sum = P.
S.I. = (P × 4 × 4) / 100 = 0.16P
C.I. (3 yrs, 5%) = P[(1.05)³ - 1] = P(1.157625 - 1) = 0.157625P
Given: S.I. - C.I. = 228
0.16P - 0.157625P = 228
0.002375P = 228
P = 228 / 0.002375
P = ₹ 96,000

8. Compound interest on a certain sum of money at 5% per annum for two years is ₹ 246. Calculate simple interest on the same sum for 3 years at 6% per annum.
Answer:
C.I. = P[(1.05)² - 1] = P(1.1025 - 1) = 0.1025P = 246
P = 246 / 0.1025 = 2400.
Calculate S.I. on 2400 for 3 years at 6%:
S.I. = (2400 × 6 × 3) / 100 = 24 × 18 = ₹ 432

9. A certain sum of money amounts to ₹ 23,400 in 3 years at 10% per annum simple interest. Find the amount of the same sum in 2 years and at 10% p.a. compound interest.
Answer:
Amount (S.I.) = P + (P × 10 × 3)/100 = P + 0.3P = 1.3P
1.3P = 23400
P = 23400 / 1.3 = 18000.
Amount (C.I.) for 2 years at 10%:
A = 18000(1 + 10/100)² = 18000(1.1)² = 18000(1.21)
A = ₹ 21,780

10. Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying ₹ 12,600 at the end of the first year and ₹ 17,640 at the end of the second year. Find the sum borrowed.
Answer:
Let Sum = P.
Amount at end of 1st year = P(1.05).
Balance after paying 12600 = 1.05P - 12600.
Amount at end of 2nd year on this balance = (1.05P - 12600) × 1.05.
This amount clears the debt, so it equals 17640.
(1.05P - 12600) × 1.05 = 17640
1.05P - 12600 = 17640 / 1.05
1.05P - 12600 = 16800
1.05P = 16800 + 12600 = 29400
P = 29400 / 1.05
Sum borrowed = ₹ 28,000

EXERCISE 3 (C)

1. If the interest is compounded half-yearly, calculate the amount when principal is ₹ 7,400; the rate of interest is 5% per annum and the duration is one year.
Answer:
P = 7400, R = 5% p.a., n = 1 year.
Compounded half-yearly: Rate = 5/2 = 2.5%, Time = 1 × 2 = 2 half-years.
A = 7400(1 + 2.5/100)² = 7400(1.025)²
A = 7400 × 1.050625
A = ₹ 7,774.63

2. Find the difference between the compound interest compounded yearly and half-yearly on ₹ 10,000 for 18 months at 10% per annum.
Answer:
Time = 1.5 years.
Case 1: Yearly Compounding.
A = P(1 + 10/100)¹(1 + (10/2)/100) [1 year full + 1/2 year simple]
A = 10000(1.1)(1.05) = 11550.
C.I. = 1550.
Case 2: Half-yearly Compounding.
Rate = 5% per half-year. Time = 3 half-years.
A = 10000(1.05)³ = 10000(1.157625) = 11576.25.
C.I. = 1576.25.
Difference = 1576.25 - 1550 = ₹ 26.25

3. A man borrowed ₹ 16,000 for 3 years under the following terms: 20% simple interest for the first 2 years. 20% C.I. for the remaining one year on the amount due after 2 years, the interest being compounded half-yearly. Find the total amount to be paid at the end of three years.
Answer:
Step 1: S.I. for 2 years.
S.I. = (16000 × 20 × 2) / 100 = 6400.
Amount due after 2 years = 16000 + 6400 = 22400.
Step 2: C.I. for 3rd year (1 year duration) on 22400, 20% p.a. compounded half-yearly.
Principal = 22400. Rate = 10% per half-year. Time = 2 half-years.
Amount = 22400(1 + 10/100)² = 22400(1.1)²
= 22400 × 1.21
= ₹ 27,104

4. What sum of money will amount to ₹ 27,783 in one and a half years at 10% per annum compounded half yearly?
Answer:
A = 27783, n = 1.5 years (3 half-years), r = 10% p.a. (5% semi-annual).
27783 = P(1 + 5/100)³
27783 = P(1.05)³
27783 = P(1.157625)
P = 27783 / 1.157625
Sum = ₹ 24,000

5. Ashok invests a certain sum of money at 20% per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half-yearly. If Geeta gets ₹ 33 more than Ashok in 18 months, calculate the money invested.
Answer:
Let Sum = P. Rate = 20%. Time = 1.5 years.
Ashok (Yearly): A_A = P(1 + 20/100)¹(1 + 10/100) = P(1.2)(1.1) = 1.32P.
Geeta (Half-yearly): Rate = 10% per half-year. Time = 3 half-years.
A_G = P(1 + 10/100)³ = P(1.1)³ = 1.331P.
Difference = A_G - A_A = 1.331P - 1.32P = 0.011P.
0.011P = 33
P = 33 / 0.011 = 3000.
Money invested = ₹ 3,000

6. At what rate of interrest per annum will a sum of ₹ 62,500 earn a compound interest of ₹ 5,100 in one year? The interest is to be compounded half-yearly.
Answer:
P = 62500, C.I. = 5100, A = 67600. Time = 1 year (2 half-years). Let rate be r% p.a.
A = P(1 + r/200)²
67600 = 62500(1 + r/200)²
67600 / 62500 = (1 + r/200)²
676 / 625 = (1 + r/200)²
(26 / 25)² = (1 + r/200)²
26/25 = 1 + r/200
1.04 = 1 + r/200
0.04 = r/200
r = 8
Rate = 8% p.a.

7. In what time will ₹ 1,500 yield ₹ 496.50 as compound interest at 20% per year compounded half-yearly?
Answer:
P = 1500, C.I. = 496.50, A = 1996.50. Rate = 20% p.a. (10% semi-annual).
Let time be n half-years.
1996.50 = 1500(1 + 10/100)ⁿ
1996.5 / 1500 = (1.1)ⁿ
1.331 = (1.1)ⁿ
(1.1)³ = (1.1)ⁿ
n = 3 half-years.
Time = 1.5 years

8. Calcualte the C.I. on ₹ 3,500 at 6% per annum for 3 years, the interest being compounded half-yearly. (Given 1.03⁶ = 1.194052)
Answer:
P = 3500, R = 6% p.a., n = 3 years.
Half-yearly: r = 3%, time = 6 half-years.
A = 3500(1.03)⁶
A = 3500 × 1.194052
A = 4179.182
C.I. = 4179.182 - 3500
C.I. = ₹ 679.18

9. Find the difference between compound interest and simple interest on ₹ 12,000 and in 1½ years at 10% p.a. compounded yearly.
Answer:
S.I. = (12000 × 10 × 1.5) / 100 = 1800.
C.I. (Yearly compounding for 1.5 yrs):
A = 12000(1 + 10/100)¹(1 + 5/100) = 12000(1.1)(1.05) = 13860.
C.I. = 13860 - 12000 = 1860.
Difference = 1860 - 1800 = ₹ 60

10. Find the difference between compound interest and simple interest on ₹ 12,000 and in 1½ years at 10% compounded half-yealry.
Answer:
S.I. remains same = ₹ 1800.
C.I. (Half-yearly): Rate = 5%, n = 3.
A = 12000(1.05)³ = 12000(1.157625) = 13891.50.
C.I. = 1891.50.
Difference = 1891.50 - 1800 = ₹ 91.50

EXERCISE 3 (D)

1. The cost of a machine is supposed to depreciate each year by 12% of its value at the beginning of the year. If the machine is valued at ₹ 44,000 at the beginning of 2008, find its value :
(i) at the end of 2009.
(ii) at the beginning of 2007.
Answer:
Present Value (Start 2008) = 44000. Dep rate = 12%.
(i) End of 2009 is 2 years later.
Value = 44000(1 - 12/100)² = 44000(0.88)² = 44000 × 0.7744 = ₹ 34,073.60
(ii) Beginning of 2007 is 1 year before 2008.
Let Value in 2007 be P.
P(1 - 12/100) = 44000
P(0.88) = 44000
P = 44000 / 0.88 = ₹ 50,000

2. The value of an article decreased for two years at the rate of 10% per year and then in the third year it increased by 10%. Find the original value of the article, if its value at the end of 3 years is ₹ 40,095.
Answer:
Let Original Value = P.
A = P(1 - 10/100)²(1 + 10/100)
40095 = P(0.9)²(1.1)
40095 = P(0.81)(1.1)
40095 = P(0.891)
P = 40095 / 0.891
Original Value = ₹ 45,000

3. According to a census taken towards the end of the year 2009, the population of a rural town was found to be 64,000. The census authority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2009 did the population of this town reach 74,088?
Answer:
P = 64000, A = 74088, r = 5%.
74088 = 64000(1 + 5/100)ⁿ
74088 / 64000 = (1.05)ⁿ
9261 / 8000 = (1.05)ⁿ
(21/20)³ = (1.05)ⁿ
(1.05)³ = (1.05)ⁿ
n = 3 years

4. The population of a town decreased by 12% during 1998 and then increased by 8% during 1999. Find the population of the town, at the beginning of 1998, if at the end of 1999 its population was 2,85,120.
Answer:
Let Population at start of 1998 = P.
P(1 - 12/100)(1 + 8/100) = 285120
P(0.88)(1.08) = 285120
P(0.9504) = 285120
P = 285120 / 0.9504
P = 3,00,000

5. A sum of money, invested at compound interest, amounts to ₹ 16,500 in 1 year and to ₹ 19,965 in 3 years. Find the rate per cent and the original sum of money invested.
Answer:
A₁ = P(1 + r/100) = 16500
A₃ = P(1 + r/100)³ = 19965
Divide A₃ by A₁:
(1 + r/100)² = 19965 / 16500 = 1.21
1 + r/100 = 1.1 ⇒ r = 10%
P(1.1) = 16500
P = 16500 / 1.1 = ₹ 15,000

6. The difference between C.I. and S.I. on ₹ 7,500 for two years is ₹ 12 at the same rate of interest per annum. Find the rate of interest.
Answer:
Diff = P(r/100)²
12 = 7500(r² / 10000)
12 = 3r² / 4
r² = 16
r = 4%

7. A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 10 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.
Answer:
(1 + r/100)¹⁰ = 3
We need 27 times, i.e., 3³.
(1 + r/100)ⁿ = 27 = 3³ = [(1 + r/100)¹⁰]³
(1 + r/100)ⁿ = (1 + r/100)³⁰
n = 30 years

8. Mr. Sharma borrowed a certain sum of money at 10% per annum compounded annually. If by paying ₹ 19,360 at the end of the second year and ₹ 31,944 at the end of the third year he clears the debt; find the sum borrowed by him.
Answer:
Let Sum = P. Rate = 10%.
Amount at end of 2nd year (before payment) = P(1.1)² = 1.21P.
He pays 19360. Remaining Balance = 1.21P - 19360.
Interest accrues on this balance for the 3rd year.
Amount at end of 3rd year = (1.21P - 19360) × 1.1.
This equals final payment 31944.
1.331P - 21296 = 31944
1.331P = 53240
P = 53240 / 1.331
P = ₹ 40,000

9. The difference between compound interest for a year payable half-yearly and 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:
Let Sum = P.
S.I. (1 yr, 10%) = 0.1P.
C.I. (1 yr, 10%, half-yearly) = P(1 + 5/100)² - P = P(1.025) - P = 0.1025P.
Diff = 0.1025P - 0.1P = 0.0025P.
0.0025P = 15
P = 15 / 0.0025
Sum = ₹ 6,000

10. The ages of Pramod and Rohit are 16 years and 18 years respectively. In what ratio must they invest money at 5% p.a. compounded yearly so that both get the same sum on attaining the age of 25 years?
Answer:
Pramod needs to wait 25 - 16 = 9 years.
Rohit needs to wait 25 - 18 = 7 years.
Let Pramod invest x and Rohit invest y.
x(1.05)⁹ = y(1.05)⁷
x/y = (1.05)⁷ / (1.05)⁹ = 1 / (1.05)²
x/y = 1 / 1.1025 = 10000 / 11025 = 400 / 441
Ratio = 400 : 441

EXERCISE 3 (E)

1. Simple interest on a sum of money for 2 years at 4% is ₹ 450. Find compound interest on the same sum and at the same rate for 1 year, if the interest is reckoned half-yearly.
Answer:
S.I. = (P × 4 × 2) / 100 = 0.08P = 450.
P = 450 / 0.08 = 5625.
C.I. (1 yr, 4%, half-yearly): Rate = 2%, n = 2.
A = 5625(1.02)² = 5625(1.0404) = 5852.25.
C.I. = 5852.25 - 5625 = ₹ 227.25

2. Find the compound interest to the nearest rupee on ₹ 10,800 for 2½ years at 10% per annum.
Answer:
P = 10800, n = 2.5 yrs, r = 10%.
A = 10800(1 + 10/100)²(1 + 5/100)
A = 10800(1.21)(1.05)
A = 10800(1.2705)
A = 13721.40
C.I. = 13721.40 - 10800 = 2921.40
Nearest rupee = ₹ 2,921

3. The value of a machine, purchased two years ago, depreciates at the annual rate of 10%. If its present value is ₹ 97,200, find:
(i) its value after 2 years.
(ii) its value when it was purchased.
Answer:
Present Value (P) = 97200.
(i) Value after 2 years = P(1 - 10/100)² = 97200(0.9)² = 97200(0.81) = ₹ 78,732
(ii) Let Original Purchase Value be V.
V(1 - 10/100)² = 97200
V(0.81) = 97200
V = 97200 / 0.81 = ₹ 1,20,000

4. Anuj and Rajesh each lent the same sum of money for 2 years at 8% simple interest and compound interest respectively. Rajesh received ₹ 64 more than Anuj. Find the money lent by each and interest received.
Answer:
Difference between C.I. and S.I. for 2 years is ₹ 64. Rate = 8%.
Diff = P(r/100)²
64 = P(8/100)² = P(64/10000)
P = 640000 / 64 = ₹ 10,000 (Sum lent)
Interest Anuj (S.I.) = (10000 × 8 × 2) / 100 = ₹ 1,600
Interest Rajesh (C.I.) = 1600 + 64 = ₹ 1,664

5. Calculate the sum of money on which the compound interest (payable annually) for 2 years be four times the simple interest on ₹ 4,715 for 5 years, both at the rate of 5 per cent per annum.
Answer:
S.I. on 4715 = (4715 × 5 × 5) / 100 = 4715 × 0.25 = 1178.75.
C.I. required = 4 × S.I. = 4 × 1178.75 = 4715.
Let P be the sum for C.I. (2 years, 5%).
C.I. = P[(1.05)² - 1] = 0.1025P.
0.1025P = 4715.
P = 4715 / 0.1025
P = ₹ 46,000

6. A sum of money was invested for 3 years, interest being compounded annually. The rates for successive years were 10%, 15% and 18% respectively. If the compound interest for the second year amounted to ₹ 4,950, find the sum invested.
Answer:
Let sum = P.
Amount at end of 1st year = P(1.10) = 1.1P.
Interest for 2nd year = Amount(1st yr) × Rate(2nd yr) / 100
Interest 2nd year = 1.1P × 15 / 100 = 0.165P.
0.165P = 4950.
P = 4950 / 0.165
Sum = ₹ 30,000

7. A sum of money is invested at 10% per annum compounded half-yearly. If the difference of amounts at the end of 6 months and 12 months is ₹ 189, find the sum of money invested.
Answer:
Rate = 5% per half-year.
Amount at 6 months (1 half-year) = P(1.05).
Amount at 12 months (2 half-years) = P(1.05)².
Difference = P(1.05)² - P(1.05) = P(1.05)[1.05 - 1] = P(1.05)(0.05) = 0.0525P.
0.0525P = 189.
P = 189 / 0.0525
Sum = ₹ 3,600

8. Rohit borrows ₹ 86,000 from Arun for two years at 5% per annum simple interest. He immediately lends out this money to Akshay at 5% compound interest compounded annually for the same period. Calculate Rohit's profit in the transaction at the end of two years.
Answer:
Profit is the difference between C.I. earned and S.I. paid.
Diff = P(r/100)²
Profit = 86000(5/100)² = 86000(1/400)
Profit = 860 / 4 = ₹ 215

9. The simple interest on a certain sum of money for 3 years at 5% per annum is ₹ 1,200. Find the amount due and the compound interest on this sum of money at the same rate and after 2 years, interest is reckoned annually.
Answer:
S.I. = 1200, T=3, R=5.
1200 = (P × 5 × 3) / 100 = 0.15P.
P = 1200 / 0.15 = 8000.
Calculate C.I. for 2 years on 8000 at 5%.
A = 8000(1.05)² = 8000(1.1025) = ₹ 8,820 (Amount)
C.I. = 8820 - 8000 = ₹ 820

10. Nikita invests ₹ 6,000 for two years at a certain rate of interest compounded annually. At the end of first year it amounts to ₹ 6,720. Calculate:
(a) the rate of interest.
(b) the amount at the end of the second year.
Answer:
(a) Interest for 1st year = 6720 - 6000 = 720.
Rate = (720 × 100) / (6000 × 1) = 12%
(b) Amount end of 2nd year = A₁(1 + r/100) = 6720(1.12)
Amount = ₹ 7,526.40

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Quick Review Flashcards - Click to flip and test your knowledge!

Question

In the context of compound interest, what is the 'amount' ($A$)?

Answer

The sum of the principal and the interest accumulated over a specific period.

Question

How is compound interest ($C.I.$) fundamentally defined as a process?

Answer

As a repeated simple interest computation where the principal grows in each conversion period.

Question

What makes manual computation of compound interest 'quite tedious' as the number of conversion periods increases?

Answer

The requirement to calculate simple interest repeatedly for every year or half-year.

Question

What is the standard formula for the amount ($A$) when interest is compounded yearly?

Answer

$A = P(1 + \frac{r}{100})^n$

Question

In the compound interest formula, what does the variable $P$ represent?

Answer

The principal, which is the initial sum of money invested or borrowed.

Question

In the compound interest formula, what does the variable $r$ represent?

Answer

The rate of interest compounded yearly.

Question

In the compound interest formula, what does the variable $n$ represent?

Answer

The number of years (or conversion periods).

Question

Formula: Calculate Compound Interest ($C.I.$) using Amount ($A$) and Principal ($P$).

Answer

$C.I. = A - P$

Question

What is the direct formula to calculate Compound Interest ($C.I.$) without finding the amount first?

Answer

$C.I. = P[(1 + \frac{r}{100})^n - 1]$

Question

Formula: What is the amount ($A$) when rates for successive years are different ($r_1, r_2, r_3$)?

Answer

$A = P(1 + \frac{r_1}{100})(1 + \frac{r_2}{100})(1 + \frac{r_3}{100})...$

Question

How is the principal ($P$) calculated if the amount ($A$), rate ($r$), and time ($n$) are known?

Answer

By rearranging the formula to $P = \frac{A}{(1 + \frac{r}{100})^n}$.

Question

To find the rate percent ($r$) when $A$ and $P$ are known, what is the first algebraic step?

Answer

Divide the amount by the principal to isolate the term $(1 + \frac{r}{100})^n$.

Question

If $\frac{A}{P} = (\frac{21}{20})^3$ and the time is $3$ years, what is the value of $(1 + \frac{r}{100})$?

Answer

$\frac{21}{20}$

Question

When finding the number of years ($n$), if $(\frac{11}{10})^n = (\frac{11}{10})^3$, what is the value of $n$?

Answer

$3$ years.

Question

When interest is compounded half-yearly, how is the annual rate ($r$) adjusted in the formula?

Answer

The rate percent is divided by $2$ (i.e., $\frac{r}{2}$).

Question

When interest is compounded half-yearly, how is the number of years ($n$) adjusted in the formula?

Answer

The number of years is multiplied by $2$ (i.e., $n \times 2$).

Question

What is the formula for Amount ($A$) when interest is compounded half-yearly?

Answer

$A = P(1 + \frac{r}{2 \times 100})^{n \times 2}$

Question

Although not in the I.C.S.E. syllabus, what is the formula for Amount ($A$) when interest is compounded quarterly?

Answer

$A = P(1 + \frac{r}{4 \times 100})^{n \times 4}$

Question

How is the amount calculated when the time is not an exact number of years (e.g., $2 \frac{1}{2}$ years) and interest is compounded yearly?

Answer

Calculate the amount for the full years, then use that as the principal for the remaining fractional year.

Question

Formula: Amount ($A$) for $2 \frac{1}{2}$ years at $10\%$ interest compounded yearly.

Answer

$A = P(1 + \frac{10}{100})^2 \times (1 + \frac{10}{2 \times 100})^1$

Question

What is the conversion period for interest compounded half-yearly?

Answer

Six months.

Question

In growth problems, what formula is used to find production after $n$ years given an initial production and growth rate $r$?

Answer

Production after $n$ years = Initial production $\times (1 + \frac{r}{100})^n$

Question

Concept: Depreciation

Answer

Definition: The reduction in the value of an asset (like machinery) over time due to wear and tear or age.

Question

What is the formula for the value of a machine after $n$ years if it depreciates at $r\%$ every year?

Answer

Value after $n$ years = Present value $\times (1 - \frac{r}{100})^n$

Question

How is the present value of a machine calculated if its value $n$ years ago and depreciation rate are known?

Answer

Present value = Value $n$ years ago $\times (1 - \frac{r}{100})^n$

Question

What is 'scrap value' in the context of depreciation?

Answer

The reduced value of an asset at the time it is sold or discarded after several years of use.

Question

Formula: Population after $n$ years given present population ($P$) and growth rate ($r$).

Answer

Population after $n$ years = $P(1 + \frac{r}{100})^n$

Question

How do you find the population $n$ years ago if the present population and growth rate $r$ are given?

Answer

Present population = Population $n$ years ago $\times (1 + \frac{r}{100})^n$

Question

In population problems, which variable in the standard $A = P(1 + \frac{r}{100})^n$ formula represents the population at the earlier point in time?

Answer

The principal ($P$).

Question

In population problems, which variable in the standard $A = P(1 + \frac{r}{100})^n$ formula represents the population at the later point in time?

Answer

The amount ($A$).

Question

When calculating the difference between $C.I.$ and $S.I.$ for $2$ years, what is the standard formula for Simple Interest ($S.I.$)?

Answer

$S.I. = \frac{P \times r \times 2}{100}$

Question

Concept: Conversion Periods

Answer

Definition: The fixed intervals of time (e.g., year, half-year) after which interest is added to the principal.

Question

Under what condition does a sum of money double itself in $n$ years in a compound interest scenario?

Answer

When $(1 + \frac{r}{100})^n = 2$.

Question

If the rate of interest is $20\%$ per annum, what is the rate used for half-yearly compounding calculations?

Answer

$10\%$ per half-year.

Question

In a $2 \frac{1}{2}$ year period, how many half-yearly conversion periods are there?

Answer

$5$

Question

If a sum of money is borrowed and repaid in two yearly instalments, how is the total sum borrowed calculated?

Answer

By finding the present value (Principal) for each instalment separately and adding them together.

Question

Why is the interest for the second year in compound interest higher than the interest for the first year?

Answer

Because the interest from the first year is added to the principal, increasing the sum on which interest is calculated.

Question

Formula: Find the rate ($r$) when the amount in $1$ year and amount in $2$ years are given (yearly compounding).

Answer

$r = \frac{\text{Amount in 2nd year} - \text{Amount in 1st year}}{\text{Amount in 1st year}} \times 100$

Question

In growth problems, which formula applies to 'the rapid growth of plants' or 'inflation'?

Answer

The standard compound interest formula: $A = P(1 + \frac{r}{100})^n$.

Question

If the question asks for 'the rate of interest per annum' and the calculation for a half-year yields $5\%$, what is the final answer?

Answer

$10\%$ per annum ($5\% \times 2$).

Question

True or False: The principal remains constant throughout the entire duration in compound interest.

Answer

False; the principal increases after every conversion period.

Question

When comparing yearly and half-yearly compounding for the same rate and time, which yields a higher amount?

Answer

Half-yearly compounding.

Question

What is the value of $(1 + \frac{10}{100})^2$ expressed as a decimal?

Answer

$1.21$

Question

If $(1 + \frac{r}{100})^n = 1.44$ for $n=2$, what is the value of $1 + \frac{r}{100}$?

Answer

$1.2$ (since $\sqrt{1.44} = 1.2$).

Question

In the formula for depreciation, why is a minus sign used inside the bracket?

Answer

To represent the decrease in value over time.

Question

When $n = 1 \frac{1}{2}$ years and interest is compounded half-yearly, what value of $n$ is used in the power of the formula?

Answer

$3$ (conversion periods).

Question

How do you calculate the $C.I.$ earned in the $3^{rd}$ year only?

Answer

Subtract the Amount at the end of $2$ years from the Amount at the end of $3$ years ($A_3 - A_2$).

Question

If an asset depreciates by $10\%$ in year 1 and $12\%$ in year 2, what is the formula for final value $V$?

Answer

$V = P(1 - \frac{10}{100})(1 - \frac{12}{100})$

Question

In the equation $\frac{9261}{8000} = (1 + \frac{r}{100})^3$, what is the cubical root of $\frac{9261}{8000}$?

Answer

$\frac{21}{20}$

Question

If a sum of money doubles in $5$ years, how many years will it take to become eight times itself ($2^3$) at the same rate?

Answer

$15$ years ($5 \times 3$).

Question

What is the relationship between Amount ($A$) and Principal ($P$) for the first year if interest is compounded annually?

Answer

They are the same as in simple interest for one year.

Question

In the 'Direct Method' formula for $C.I.$, what does the term $(1 + \frac{r}{100})^n - 1$ represent?

Answer

The total interest earned on a principal of $1$ unit.

Question

How is the 'rate of growth' usually expressed in population or industrial problems?

Answer

As a percentage ($r\%$) per year.

Question

If $18,000$ amounts to $23,805$ in $2$ years, what is the total compound interest earned?

Answer

$5,805$ ($23,805 - 18,000$).

Question

Cloze: In compound interest, the interest for the first year becomes _____ for the second year.

Answer

Part of the principal

Question

Formula: Calculate simple interest ($I$) to find principal ($P$) when $I$, $R$, and $T$ are known.

Answer

$P = \frac{I \times 100}{R \times T}$

Question

If interest is reckoned half-yearly, what is the amount formula for $n=1$ year?

Answer

$A = P(1 + \frac{r}{200})^2$

Question

What is the result of $1,600(1 + \frac{20}{100})^2$?

Answer

$2,304$

Question

To find the rate $r$ from $(1 + \frac{r}{100}) = \frac{11}{10}$, what is the final percentage?

Answer

$10\%$

Question

When solving for $n$ in $(\frac{21}{20})^n = \frac{9261}{8000}$, what power is $9261$ of $21$?

Answer

The $3^{rd}$ power (cube).