| Book Name | : Ganit Prakash |
| Subject | : Mathematics |
| Class | : 10 (Madhyamik) |
| Publisher | : Prof Nabanita Chatterjee |
| Chapter Name | : Compound Interest And Uniform Rate Of Increase Or Decrease (6th Chapter) |
Application Questions
Application 1
If I take a loan of ₹ 1400 at 5% compound interest per annum for 2 years, let us write by calculating how much compound interest and total amount I shall pay.
Solution:
Principal (P) = ₹ 1400
Rate (r) = 5 %
Time (n) = 2 yrs
Amount (A) = P × (1 + r/100)n
= 1400 × (1 + 5/100)²
= 1400 × (1.05)²
= 1400 × 1.1025 = ₹1543.50
CI = A – P = 1543.50 – 1400 = ₹143.50
Application 3
Let us write by calculating what is the amount on ₹ 1000 for 2 years at the rate of 5% compound interest per annum. [Let me do it myself]
Solution:
Principal (P) = ₹ 1000
Rate (r) = 5 %
Time (n) = 2 yrs
Amount (A) = \text { ₹ } 1000\left(1+\frac{5}{100}\right)^2
= \text { ₹ } 1000 \times\left(\frac{22}{20}\right)^2
= \text { ₹ } 1000 \times \frac{441}{400}
= \text { ₹ } \frac{2205}{2} = \text { ₹ } 1102.50
Application 4
If ₹ 4000 is invested in a bank at the rate of 5% compound interest per annum, then let us write by calculating the amount after 3 year.
Solution:
Principal (P) = ₹ 4000
Rate (r) = 5 %
Time (n) = 3 yrs
Amount (A) = ₹ 4000\left(1+\frac{5}{100}\right)^3
= ₹ 4000 \times \left(\frac{21}{20}\right)^3
= ₹ 4000 \times \frac{9261}{8000}
= ₹ \frac{37044000}{8000} = ₹4630.50
Compound interest = ₹ (4630.50-4000) = ₹ 630.50.
Application 5
At 5% compound interest per annum, let us find the compound interest on ₹ 10,000 for 3 year [Let me do it myself]
Solution:
Principal (P) = ₹ 10,000
Rate (r) = 5 %
Time (n) = 3 yrs
Amount (A) = ₹ 10000\left(1+\frac{5}{100}\right)^3
= ₹ 10000 \times \left(\frac{21}{20}\right)^3
= ₹ 10000 \times \frac{9261}{8000}
= ₹ \frac{92610000}{8000} = \text{₹ } 11576.25
Compound Interest (CI) = ₹ 11,576.25 – ₹ 10,000
= ₹ 1,576.25
Application 6
If interest is compounded half-yearly, let us write by calculating compounded interest and amount on ₹ 8,000 at the rate of 10% compound interest per annum for 1\frac{1}{2} year
Solution:
Principal (P) = ₹ 8,000
Rate (r) = 10% per annum, compounded half-yearly ⇒ 5% per half-year
Time (n) = 1\frac{1}{2} year = 3 half-years
Amount (A) = ₹ 8000\left(1+\frac{5}{100}\right)^3
= ₹ 8000 \times \left(\frac{21}{20}\right)^3
= ₹ 8000 \times \frac{9261}{8000}
= ₹ \frac{74088000}{8000}
= ₹ 9,261.00
Compound Interest (CI) = ₹ 9,261.00 – ₹ 8,000
= ₹ 1,261.00
Application 7
Let us find compound interest on ₹ 1000 at the rate of 10% compound interest per annum and the interest being compounded at 6 months interval [Let me do it myself]
Solution:
Compound interest on ₹ 1000 for 1 year at 10% per annum compound interest compound semi-annually.
\text { Amount = ₹ } 1000\left(1+\frac{1 / 2}{100}\right)^{2 \times 1}
= \text { ₹ } 1000\left(1+\frac{10}{200}\right)² = \text { ₹ } 1000\left(\frac{21}{20}\right)² = \text { ₹ } 1000 \times \frac{441}{400}
= \text { ₹ } \frac{2205}{2} = ₹ 1102.50
∴ Compound interest = ₹ (1102.50-1000) = ₹ 102.50
Application 8
Let us write by calculating compound interest on ₹ 10,000 at the rate of 8% compound interest per annum for 9 months, compounded at an interval 0f 3 months.
Solution:
Principal (P) = ₹ 10,000
Rate (r) = 8% per annum, compounded quarterly ⇒ 2% per quarter
Time (n) = 9 months = 3 quarters
Amount (A) = ₹ 10000\left(1+\frac{2}{100}\right)^3
= ₹ 10000 \times \left(\frac{51}{50}\right)^3
= ₹ 10000 \times \frac{132651}{125000}
= ₹ \frac{1326510000}{125000}
= ₹ 10,612.08
Compound Interest (CI) = ₹ 10,612.08 – ₹ 10,000 = ₹ 612.08
Application 10
If the rate of compound interest for the first year is 4% and for 2nd year is 5%, let us find the compound interest on ₹ 25000 for 2 year [Let me do myself]
Solution:
Amount after two years on ₹ 25,000 at two rates of 4% for 1st year & 5% for 2nd year
= \text { ₹ } 25,000\left(1+\frac{4}{100}\right) \cdot\left(1+\frac{5}{100}\right)
= \text { ₹ } 25,000 \times \frac{104}{100} \times \frac{105}{100} = ₹ 27300
\therefore \text { Compound interest } = \text { ₹ } 27300-\text { ₹ } 25000
= ₹ 2300
Application 11
I lend ₹ 10,000 at the rate of 4% compound interest per annum for 2 \frac{1}{2} years, let us write by calculating how much total money I shall pay.
Solution:
Principal (P) = ₹ 10,000
Rate (r) = 4% per annum
Time (n) = 2 \frac{1}{2} years = 2 full years + 6 months
Step 1: Amount after 2 years
Amount (A₁) =
= ₹ 10000\left(1+\frac{4}{100}\right)^2
= ₹ 10000 \times \left(\frac{26}{25}\right)^2
= ₹ 10000 \times \frac{676}{625}
= ₹ \frac{6760000}{625}
= ₹ 10,816
Step 2: Interest for remaining 6 months (simple interest on ₹ 10,816)
Simple Interest = \frac{10816 \times 4 \times \frac{1}{2}}{100}
= \frac{21632}{100}
= ₹ 216.32
Step 3: Total Amount after 2 \frac{1}{2} years
= ₹ 10,816 + ₹ 216.32 = ₹ 11,032.32
Application 12
Let us find the amount of ₹ 30,000, at the rate of 6% compound interest per annum for 2 \frac{1}{2} year [Let me do it myself]
Solution:
Principal (P) = ₹ 30,000
Rate (r) = 6% per annum
Time (n) = 2 \frac{1}{2} years = 2 full years + 6 months
Step 1: Amount after 2 years
Amount (A₁) = ₹ 30000\left(1+\frac{6}{100}\right)^2
= ₹ 30000 \times \left(\frac{53}{50}\right)^2
= ₹ 30000 \times \frac{2809}{2500}
= ₹ \frac{84270000}{2500}
= ₹ 33,708
Step 2: Interest for remaining 6 months (simple interest on ₹ 33,708)
Simple Interest = \frac{33708 \times 6 \times \frac{1}{2}}{100}
= \frac{101124}{100}
= ₹ 1,011.24
Step 3: Total Amount after 2 \frac{1}{2} years
= ₹ 33,708 + ₹ 1,011.24 = ₹ 34,719.24
Application 14
Let us write by calculating what sum of money will amount ₹ 3528 after 2 years at the rate of 5% compound interest per annum. [Let me do it myself]
Solution:
Amount (A) = ₹ 3528
Rate (r) = 5% per annum
Time (n) = 2 years
Let the Principal (P) = ?
Step 1: Use the formula
A = P \left(1+\frac{r}{100}\right)^n
3528 = P \left(1+\frac{5}{100}\right)^2
3528 = P \times \left(\frac{21}{20}\right)^2
3528 = P \times \frac{441}{400}
Step 2: Solve for P
P = \frac{3528 \times 400}{441}
P = \frac{1411200}{441}
= ₹ 3,200
Application 19
The simple interest and compound interest of a certain sum of money for 2 years are ₹ 840 and ₹ 869.40 respectively. Let us calculate that sum of money and the rate of interest. [Let me do it myself]
Solution:
Simple interest for 2 years = ₹ 840
∴ Simple interest for 1 year = ₹ \frac{840}{2} = ₹ 420.
Difference between Compound interest & Simple interest for 2 years = ₹ (869.40-840) = ₹ 29.40.
∴ Interest on ₹ 420 for 1 year = ₹ 29.40
∴ Interest on ₹ 100 for 1 year = ₹ \frac{29.40}{420} × 100. = ₹ 7.
∴ Rate of interest = 7%.
Principal = ₹ \frac{420 \times 100}{7} = ₹ 6,000
Application 21
Let us calculate at what rate of compound interest, ₹ 5,000 will amount to ₹ 5832 in 2 year [Let me do it myself]
Solution:
Principal (P) = ₹ 5,000
Amount (A) = ₹ 5,832
Time (n) = 2 years
Rate (r) = ?
Step 1: Use the compound interest formula
A = P \left(1+\frac{r}{100}\right)^n
5832 = 5000\left(1+\frac{r}{100}\right)^2
Divide both sides by 5000:
\left(1+\frac{r}{100}\right)^2 = \frac{5832}{5000} = 1.1664Take square root of both sides:
1+\frac{r}{100} = \sqrt{1.1664} = 1.08
Now solve for r:
\frac{r}{100} = 1.08 - 1 = 0.08or, r = 0.08 \times 100 = 8%
Application 23
Let us write by calculating in how many years ₹ 5000 will compound interest at the rate of 10% per annum amount to ₹ 6050.
Solution:
Principal (P) = ₹ 5,000
Amount (A) = ₹ 6,050
Rate (r) = 10% per annum
Time (n) = ?
∴ Amount (A) = P\left(1+\frac{r}{100}\right)^n
or, 6050 = 5000\left(1+\frac{10}{100}\right)^n
\frac{6050}{5000} = \left(\frac{11}{10}\right)^n
or, \left(\frac{11}{10}\right)^2 = \left(\frac{1.1}{10}\right)^n
∴ n = 2.
∴ Required Time = 2 years
Let Us See By Calculating 6.1
Question 1
I have ₹ 5000 in my hand. I deposited that money in a bank at the rate of 8.5% compound interest per annum for two year Let us write by calculating how much money I shall get at the end of 3 year
Solution:
Principal (P) = ₹ 5,000
Rate (r) = 8.5% per annum
Time (n) = 3 years
∴ Amount A = P\left(1+\frac{R}{100}\right)^n
= \text { ₹ } 5,000\left(1+\frac{8.5}{100}\right)^2
= \text { ₹ } 5,000\left(\frac{1085}{1000}\right)^2
= ₹ \frac{5000 \times 1085 \times 1085}{1000 \times 1000}</p> <p dir="auto">= ₹ \frac{5886125}{1000}
= ₹ 5886.125
Question 2
Let us calculate the amount on ₹ 5000 at the rate of 8% compound interest per annum for 3 year
Solution:
Principal (P) = ₹ 5,000
Rate (r) = 8% per annum
Time (n) = 3 years
∴ Amount (A) = P \left(1+\frac{R}{100}\right)^n
= ₹ 5,000\left(1+\frac{8}{100}\right)^3
= ₹ 5,000\left(\frac{108}{100}\right)^3
= ₹ \frac{5000 \times 108 \times 108 \times 108}{100 \times 100 \times 100}
= ₹ \frac{629856000}{1000000}
= ₹ 6,298.56
Question 3
Goutam babu borrowed ₹ 2000 at the rate of 6% compound interest per annum for 2 year Let us write by calculating how much compound interest at the end of 3 years he will pay.
Solution:
Principal (P) = ₹ 2,000
Rate (r) = 6% per annum
Time (n) = 3 years
∴ Amount (A) = P \left(1+\frac{R}{100}\right)^n
= ₹ 2,000\left(1+\frac{6}{100}\right)^3
= ₹ 2,000\left(\frac{106}{100}\right)^3
= ₹ \frac{2000 \times 106 \times 106 \times 106}{100 \times 100 \times 100}
= ₹ \frac{2383280000}{1000000}
= ₹ 2,383.28
Compound Interest = ₹ 2,383.28 – ₹ 2,000 = ₹ 383.28
Question 4
Let us write by calculating the amount on ₹ 30,000 at the rate of 9% compound interest per annum for 3 year
Solution:
Principal (P) = ₹ 30,000
Rate (r) = 9% per annum
Time (n) = 3 years
∴ Amount A = P
\left(1+\frac{R}{100}\right)^n= ₹ 30,000\left(1+\frac{9}{100}\right)^3
= ₹ 30,000\left(\frac{109}{100}\right)^3
= ₹ \frac{30000 \times 109 \times 109 \times 109}{100 \times 100 \times 100}
= ₹ 38850.87
Question 5
Let us write by calculating the amount on ₹ 80,000 for 2 \frac{1}{2} years at the rate of 5 % compound interest per annum.
Solution:
Principal (P) = ₹ 80,000
Rate (r) = 5% per annum
Time (n) = 2 \frac{1}{2} years = 2 full years + 6 months
Step 1: Amount after 2 years
∴ Amount A₁ = P
\left(1+\frac{R}{100}\right)^2= ₹ 80,000\left(1+\frac{5}{100}\right)^2
= ₹ 80,000\left(\frac{105}{100}\right)^2
= ₹ \frac{80000 \times 105 \times 105}{100 \times 100}
= ₹ \frac{882000000}{10000}
= ₹ 88,200
Step 2: Interest for remaining 6 months (simple interest on ₹ 88,200)
Simple Interest = \frac{88200 \times 5 \times \frac{1}{2}}{100}
= \frac{220500}{100}
= ₹ 2,205
Step 3: Total Amount after 2 \frac{1}{2} years
= ₹ 88,200 + ₹ 2,205 = ₹ 90,405
Question 6
Chandadavi borrowed some money for 2 years in the compound interest at the rate of 8% per annum. Let us calculate, if the compound interest is ₹ 2496, then how much money she had lended.
Solution:
Compound Interest (CI) = ₹ 2,496
Rate (r) = 8% per annum
Time (n) = 2 years
Principal (P) = ?
Step 1: Use the amount formula
Let the amount be A, and principal be P
∴ A = P \left(1+\frac{R}{100}\right)^n
= P \left(1+\frac{8}{100}\right)^2
= P \left(\frac{108}{100}\right)^2 = P \times \frac{11664}{10000}
Step 2: Use relation
Compound Interest = Amount – Principal
So: CI = A - P = P \left( \frac{11664}{10000} - 1 \right) = P \times \frac{1664}{10000}
P \times \frac{1664}{10000} = 2496
P = \frac{2496 \times 10000}{1664}
P = \frac{24960000}{1664} = 15000
Question 7
Let us write by calculating the principal which becomes ₹ 2648 after getting 8% compound interest per annum for 3 year.
Solution:
Amount (A) = ₹ 2,648
Rate (r) = 8% per annum
Time (n) = 3 years
Principal (P) = ₹ x
According to the problem,
x\left\{\left(1+\frac{10}{100}\right)^3-1\right\} = 2648
or, x\left\{\left(\frac{11}{10}\right)^3-1\right\} = 2648
or, x\left(\frac{1331-1000}{1000}\right) = 2648
or, x × 331 = 2648 × 1000
or, x = \frac{2648 × 1000}{331}
∴ x = 8000
∴ Required principal = ₹ 8,000.
Question 8
Rahaman chacha deposited some money in a cooperative bank at the rate of 9% compound interest and he received amount ₹ 29702.50 after 2 year Let us calculate how much money Rahaman chacha had deposited in cooperative bank.
Solution:
Amount (A) = ₹ 29,702.50
Rate (r) = 9% per annum
Time (n) = 2 years
Principal (P) = ?
Step 1: Use the compound amount formula
∴ A = P \left(1+\frac{R}{100}\right)^n
= P \left(1+\frac{9}{100}\right)^2
= P \left(\frac{109}{100}\right)^2 = P \times \frac{11881}{10000}
So, 29702.50 = P \times \frac{11881}{10000}
Step 2: Solve for P
P = \frac{29702.50 \times 10000}{11881}
P = \frac{297025000}{11881} = 25000
Question 9
Let us write by calculating what sum of money at the rate of 8% compound interest per annum for 3 years will amount to ₹ 31492.80.
Solution:
Amount (A) = ₹ 31,492.80
Rate (r) = 8% per annum
Time (n) = 3 years
Principal (P) = ?
According to the problem,
x \left(1+\frac{8}{100}\right)^3 = 31492.80
or, x × \left(\frac{108}{100}\right)^3 = 31492.80
or, x = \frac{31492.80 × 100 × 100}{108 × 108} = 25000
∴ Principal = ₹ 25000.
Question 10
Let us calculate the difference between the compound interest and simple interest on ₹ 12,000 for 2 years, at 7.5% interest per annum.
Solution:
Principal (P) = ₹ 12,000
Rate (r) = 7.5% per annum
Time (n) = 2 years
Simple Interest (SI) = \frac{12000 \times 7.5 \times 2}{100} = \frac{180000}{100}
= ₹ 1,800
Amount (A) = ₹ 12000 \left(1 + \frac{7.5}{100} \right)^2
= ₹ 12000 \left( \frac{107.5}{100} \right)^2
= ₹ 12000 \times \frac{11556.25}{10000}</p> <p>= \frac{138675000}{10000}
= ₹ 13,867.50
Compound Interest (CI) = ₹ 13,867.50 – ₹ 12,000 = ₹ 1,867.50
Difference = ₹ 1,867.50 – ₹ 1,800 = ₹ 67.50
Question 11
Let us write by calculating the difference between compound interest and simple interest of ₹ 10,000 for 3 years at 5% per annum.
Solution:
Principal (P) = ₹ 10,000
Rate (r) = 5% per annum
Time (n) = 3 years
Simple Interest (SI) = \frac{10000 \times 5 \times 3}{100} = \frac{150000}{100}
= ₹ 1,500
Amount (A) = ₹ 10000 \left(1 + \frac{5}{100} \right)^3
= ₹ 10000 \left( \frac{105}{100} \right)^3
= ₹ 10000 \times \frac{1157625}{1000000}
= ₹ \frac{11576250000}{1000000}
= ₹ 11,576.25
Compound Interest (CI) = ₹ 11,576.25 – ₹ 10,000 = ₹ 1,576.25
Difference = CI - SI
= ₹ 1,576.25 – ₹ 1,500
= ₹ 76.25
Question 12
Let us write by calculating the sum of money, if the difference between compound interest and simple interest for 2 years at the rate of 9% interest per annum is ₹ 129.60.
Solution:
Rate (r) = 9% per annum
Time (n) = 2 years
Difference between CI and SI = ₹ 129.60
Let the principal be P
Difference = P \times \left( \frac{R}{100} \right)^2
or, ₹ 129.60 = P \times \left( \frac{9}{100} \right)^2
or, 129.60 = P \times \frac{81}{10000}
or, P = \frac{129.60 \times 10000}{81}
or, P = \frac{1296000}{81} = 16000
Principal (P) = ₹ 16,000
Question 13
Let us write by calculating the sum of money if the difference between compound interest and simple interest for 3 years becomes ₹ 930 at the rate of 10% interest per annum.
Solution:
Let the principal = ₹ x
∴ C.I. on ₹ x for 3 years at the rate of 10%
= ₹ x\left\{\left(1+\frac{10}{100}\right)^3-1\right\} = x\left\{\left(\frac{11}{10}\right)^3-1\right\}
= ₹ x ×\left(\frac{331}{1000}-1\right)
= ₹ x ×\frac{331}{1000} = \text{ ₹ }\frac{331 x}{1000}
Again, the S.I on ₹ x for 3 years at the rate of 10%
= ₹ x × \frac{10}{100} × 3 = \text{ ₹ } \frac{30 x}{100} = ₹ \frac{3 x}{10}
According to the problem,
₹ \left(\frac{331 x}{1000}-\frac{3 x}{10}\right) = ₹ 930
or, \frac{331 x-300 x}{1000} = 930
or, \frac{31 x}{1000} = 930
\therefore x = \frac{930 × 1000}{31} = 30,000
∴ Required principal = ₹ 30,000
Question 14
If the rates of compound interest for the first and the second year are 7% and 8% respectively, let us write by calculating compound interest on ₹ 6000 for 2 year
Solution:
After 2 years amount on ₹ 6,000 at the rate of 7% for 1st year & 8% for 2nd year respectively
= ₹ 6000 ×\left(1+\frac{7}{100}\right)\left(1+\frac{8}{100}\right)
= ₹ 6000 × \frac{107}{100} × \frac{108}{100} = ₹ \frac{69336}{10}
= ₹ 6933.6
∴ Compound interest = ₹ (6933.6-6000)
= ₹ 933.6
Question 15
If the rate of compound interest for the first and second year are 5% and 6% respectively, let us calculate the compound interest on ₹ 5000 for 2 year
Solution:
After 2 years amount on ₹ 5000 at the rate of 5% for the 1st year & 6% for the 2nd year respectively
= ₹ 5,000 ×\left(1+\frac{5}{100}\right) ×\left(1+\frac{6}{100}\right)
= ₹ 5,000 × \frac{105}{100} × \frac{106}{100} = ₹ 5565
∴ Comound interest = ₹ (5565-5000) = ₹ 565
Question 16
If simple interest on a certain sum of money for 1 year is ₹ 50 and compound interest for 2 years is ₹ 102, let us write by calculating the sum of money and the rate of interest.
Solution:
Simple Interest (SI) for 1 year = ₹ 50
Compound Interest (CI) for 2 years = ₹ 102
Let the principal be P and rate be R%
∴ Interest on ₹ 50 for 1 year = ₹ (102-100) = ₹ 2
∴ Interest on ₹ 100 for 1 year = \frac{2}{50} × 100 = ₹ 4.
∴ Rate of interest = 4%
∴ Principal = \frac{S . I × 100}{R × t}
= \frac{50 × 100}{4 × 1} = ₹ 1250.
Question 17
If simple interest and compound interest of a certain sum of money for two years are ₹ 8400 and ₹ 8652, then let us write by calculating the sum of money and the rate of interest.
Solution:
Simple Interest (SI) for 2 years = ₹ 8,400
Compound Interest (CI) for 2 years = ₹ 8,652
Difference = ₹ 8652 – ₹ 8400 = ₹ 252
This ₹ 252 is the interest on the 1st year’s interest (i.e., on ₹ 8400 ÷ 2 = ₹ 4200) for 1 year.
Let the rate be R%
Use SI formula on ₹ 4200 for 1 year:
\frac{4200 \times R}{100} = 252
or, 4200R = 25200
or, R = \frac{25200}{4200} = 6%
Now use SI formula to find Principal:
SI = \frac{P \times R \times T}{100}
or, 8400 = \frac{P \times 6 \times 2}{100}
or, P = \frac{8400 \times 100}{12} = 70000
Principal = ₹ 70,000
Rate = 6% per annum
Question 18
Let us calculate compound interest on ₹ 6000 for 1 year at the rate of 8% compound interest per annum compounded at the interval of 6 months.
Solution:
Principal (P) = ₹ 6,000
Rate (r) = 8% per annum compounded half-yearly ⇒ 4% per half-year
Time (n) = 1 year = 2 half-years
Amount (A) = \text{₹ } 6000 \left(1 + \frac{4}{100} \right)^2
= ₹ 6000 \left( \frac{104}{100} \right)^2
= ₹ 6000 \times \frac{10816}{10000}
= ₹ \frac{64896000}{10000}
= ₹ 6,489.60
Compound Interest (CI) = ₹ 6,489.60 – ₹ 6,000
= ₹ 489.60
Question 19
Let us write by calculating compound interest on ₹ 6250 for 9 months at the rate of 10% compound interest per annum, compounded at the interval of 3 months.
Solution:
Principal (P) = ₹ 6,250
Rate (r) = 10% per annum compounded quarterly ⇒ 2.5% per quarter
Time (n) = 9 months = 3 quarters
Amount (A) = \text{₹ } 6250 \left(1 + \frac{2.5}{100} \right)^3
= ₹ 6250 \left( \frac{102.5}{100} \right)^3
= ₹ 6250 \times \frac{1076890625}{1000000000}
= ₹ \frac{6730566406250}{1000000000}
= ₹ 6,730.57
Compound Interest (CI) = ₹ 6,730.57 – ₹ 6,250
= ₹ 480.57
Question 20
Let us write by calculating at what rate of interest per annum ₹ 60000 will amount to ₹ 69984 in 2 year
Solution:
Principal (P) = ₹ 60,000
Amount (A) = ₹ 69,984
Time (n) = 2 years
Rate (r) = ?
A = P \left(1 + \frac{R}{100} \right)^2
69984 = 60000 \left(1 + \frac{R}{100} \right)^2
or, \left(1 + \frac{R}{100} \right)^2 = \frac{69984}{60000} = 1.1664
or, 1 + \frac{R}{100} = \sqrt{1.1664} = 1.08
or, \frac{R}{100} = 0.08
or, R = 8%
Rate of interest = 8% per annum
Question 21
Let us calculate in how many years ₹ 40000 will amount to ₹ 46656 at the rate of 8% compound interest per annum.
Solution:
Principal (P) = ₹ 40,000
Amount (A) = ₹ 46,656
Rate (r) = 8% per annum
Time (n) = ?
A = P \left(1 + \frac{R}{100} \right)^n
46656 = 40000 \left(1 + \frac{8}{100} \right)^n
or, \left( \frac{108}{100} \right)^n = \frac{46656}{40000} = 1.1664
or, \left( \frac{27}{25} \right)^n = \frac{2916}{2500}
or, \left( \frac{27}{25} \right)^3 = \frac{19683}{15625} \approx 1.26 \quad \text{(too high)}
\left( \frac{27}{25} \right)^2 = \frac{729}{625} = 1.1664or, n = 2
∴ Time = 2 years
Question 22
Let us write by calculating at what rate of compound interest per annum, the amount on ₹ 10,000 for 2 years is ₹ 12100.
Solution:
Principal (P) = ₹ 10,000
Amount (A) = ₹ 12,100
Time (n) = 2 years
Rate (r) = ?
A = P \left(1 + \frac{R}{100} \right)^2
12100 = 10000 \left(1 + \frac{R}{100} \right)^2
or, \left(1 + \frac{R}{100} \right)^2 = \frac{12100}{10000} = 1.21
or, 1 + \frac{R}{100} = \sqrt{1.21} = 1.1
or, \frac{R}{100} = 0.1
R = 10%
Rate of interest = 10% per annum
Question 23
Let us calculate in how many years ₹ 50000 will amount to ₹ 60500 at the rate of 10% compound interest per annum.
Solution:
Principal (P) = ₹ 50,000
Amount (A) = ₹ 60,500
Rate (r) = 10% per annum
Time (n) = ?
A = P \left(1 + \frac{R}{100} \right)^n
60500 = 50000 \left(1 + \frac{10}{100} \right)^n
or, \left( \frac{11}{10} \right)^n = \frac{60500}{50000} = 1.21
\left( \frac{11}{10} \right)^2 = \frac{121}{100} = 1.21
n = 2
∴ Time = 2 years
Question 24
Let us write by calculating in how many years ₹ 30,000 will amount to ₹ 399300 at the rate of 10% compound interest per annum.
Solution:
Principal (P) = ₹ 30,000
Amount (A) = ₹ 3,99,300
Rate (r) = 10% per annum
Time (n) = ?
\therefore P\left(1+\frac{R}{100}\right)^{n} = A
₹ 300000\left(1+\frac{10}{100}\right)^{n} = ₹ 399300
or, \left(1+\frac{1}{10}\right)^{n} = \frac{399300}{300000}
or, \left(\frac{11}{10}\right)^{n} = \left(\frac{11}{10}\right)^{3} \quad \therefore n = 3
∴ Time = 3 year
Question 25
Let us calculate the compound interest and amount on ₹ 1600 for 1 \frac{1}{2} years at the rate of 10% compound interest per annum, compounded at an interval of 6 months.
Solution:
Principal (P) = ₹ 1,600
Rate (r) = 10% per annum, compounded half-yearly ⇒ 5% per half-year
Time (n) = 1 \frac{1}{2} years = 3 half-years
Amount (A) = \text{₹ }1600 \left(1 + \frac{5}{100} \right)^3
= ₹ 1600 \left( \frac{105}{100} \right)^3
= ₹ 1600 \times \frac{1157625}{1000000}
= ₹ \frac{1852200000}{1000000}
= ₹ 1,852.20
Compound Interest (CI) = ₹ 1,852.20 – ₹ 1,600 = ₹ 252.20
Application Questions
Application 24
At present, 4000 students have been taking training from this training centre. In last 2 years it has been decided that the facility to get a chance for training progreamme in this centre will be increased by 5% in comparison to its previous year. Let us see by calculating how many students will get chance to join this training programme at the end of 2 year
Solution:
Present number of students = 4,000
Rate of increase = 5% per year
Time = 2 years
Number of students after 2 years = 4000 \left(1 + \frac{5}{100} \right)^2
= 4000 \left( \frac{105}{100} \right)^2
= 4000 \times \frac{11025}{10000}
= \frac{44100000}{10000}
= ₹ 4,410
Number of students after 2 years = 4,410
Application 27
The price of a motor car is ₹ 3 lakhs. If the price of the car depreciates at the rate of 30% every year, let us write by calculating the price of the car after 3 year [Let me do it myself]
Solution:
Present price of the car = ₹ 3,00,000
Rate of depreciation = 30% per annum
Time = 3 years
Price after 3 years = 300000 \left(1 - \frac{30}{100} \right)^3
= 300000 \left( \frac{70}{100} \right)^3
= 300000 \times \frac{343000}{1000000}
= \frac{102900000000}{1000000}
= ₹ 1,02,900
Price of the car after 3 years = ₹ 1,02,900
Application 29
At present the population of a city is 576000; if it is being increased at the rate of 6 \frac{2}{3}% every year, let us calculate its population 2 years ago. [Let me do it myself]
Solution:
Present population = 5,76,000
Rate of increase = 6 \frac{2}{3}% = \frac{20}{3}%
Time = 2 years
Let population 2 years ago = P
Then, present population = P \left(1 + \frac{20}{300} \right)^2 = P \left( \frac{16}{15} \right)^2
or, 576000 = P \times \frac{256}{225}
or, P = \frac{576000 \times 225}{256} = \frac{129600000}{256} = 5,06,250
Population 2 years ago = ₹ 5,06,250
Let Us Work Out - 6.2
Question 1
At present the population of the village of Pahalanpur is 10000; if population is being increased at the rate of 3% every year, let us write by calculating its population after 2 year
Solution:
Present population = 10,000
Rate of increase = 3% per annum
Time = 2 years
Population after 2 years = 10000 \left(1 + \frac{3}{100} \right)^2
= 10000 \left( \frac{103}{100} \right)^2
= 10000 \times \frac{10609}{10000}
= \frac{106090000}{10000}
= ₹ 10,609
Population after 2 years = 10,609
Question 2
Rate of increase in population of a state is 2% in a year. The present population s 80000000; let us calculate the population of the state after 3 year
Solution:
Present population = 8,00,00,000
Rate of increase = 2% per annum
Time = 3 years
Population after 3 years = 80000000 \left(1 + \frac{2}{100} \right)^3
= 80000000 \left( \frac{102}{100} \right)^3
= 80000000 \times \frac{1061208}{1000000}
= \frac{84996640000000}{1000000}
= ₹ 84,996,640
Population after 3 years = 8,49,96,640
Question 3
The price of a machine in a leather factory depreciates at the rate of 10% every lear. If the present price of the machine be ₹ 100000, let us calculate what will be he price of that machine after 3 year
Solution:
Present price of the machine = ₹ 1,00,000
Rate of depreciation = 10% per annum
Time = 3 years
Price after 3 years = 100000 \left(1 - \frac{10}{100} \right)^3
= 100000 \left( \frac{90}{100} \right)^3
= 100000 \times \frac{729}{1000}
= \frac{72900000}{1000}
= ₹ 72,900
Price of the machine after 3 years = ₹ 72,900
Question 4
As a result of Sarba Siksha Abhiyan, the students leaving the school before completion, the students are readmitted, so the number of students in a year increased by 5% in comparison to its previous year. If the number of such readmitted students in a district be 3528 in the present year, let us write by calculating, the number of students readmitted 2 years before in this manner.
Solution:
Present number of students = 3,528
Rate of increase = 5% per year
Time = 2 years
Let the number of students 2 years ago = P
Then, present number = P \left(1 + \frac{5}{100} \right)^2 = P \left( \frac{105}{100} \right)^2
= P \times \frac{11025}{10000}
or, 3528 = P \times \frac{11025}{10000}
P = \frac{3528 \times 10000}{11025}
= \frac{35280000}{11025} = 3,200
Number of students readmitted 2 years ago = 3,200
Question 5
Through the publicity of road-safety programme, the street accidents in Purulia district are decreased by 10% in comparison to its previous year. If the number of street accidents in this year be 8748, let us write by calculating the number of street accidents 3 years before in the district.
Solution:
Present number of accidents = 8,748
Rate of decrease = 10% per annum
Time = 3 years
Let the number of accidents 3 years ago = P
Then, present number = P \left(1 - \frac{10}{100} \right)^3 = P \left( \frac{90}{100} \right)^3
= P \times \frac{729}{1000}
or, 8748 = P \times \frac{729}{1000}
P = \frac{8748 \times 1000}{729}
= \frac{8748000}{729}
= 12,000
Number of street accidents 3 years ago = 12,000
Question 6
A cooperative society of fishermen implemented such an improved plan for the production of fishes that the production in a year will be increased 10% in comparison son to its previous year. In the present year if the cooperative society can produce 400 quintals of fishes, let us write by calculating cooperative society can produce after 3 year.
Solution:
Present production = 400 quintals
Rate of increase = 10% per annum
Time = 3 years
Production after 3 years = 400 \left(1 + \frac{10}{100} \right)^3
= 400 \left( \frac{110}{100} \right)^3
= 400 \times \frac{1331}{1000}
= \frac{532400}{1000}
= ₹ 532.4 quintals
Production after 3 years = 532.4 quintals
Question 7
The height of a tree increases at the rate of 20% every year. If the present height of the tree is 28.8 metre, let us calculate the height of the tree 2 years before.
Solution:
Present height of the tree = 28.8 metres
Rate of increase = 20% per annum
Time = 2 years
Let the height of the tree 2 years ago = H
Then, present height = H \left(1 + \frac{20}{100} \right)^2 = H \left( \frac{120}{100} \right)^2
= H \times \frac{144}{100} = H \times \frac{36}{25}
or, 28.8 = H \times \frac{36}{25}
or, H = \frac{28.8 \times 25}{36}
= \frac{720}{36} = 20
Height of the tree 2 years ago = 20 metres
Question 8
Three years before from today a family had planned to reduce the expenditure of electric bill by 5% in comparison to its previous year. 3 years before, that family had p spend ₹ 4000 in a year for electric bill. Let us write by calculating how much amount the family will have to spend to pay the electric bill in the present year.
Solution:
Expenditure 3 years ago = ₹ 4,000
Rate of decrease = 5% per annum
Time = 3 years
Expenditure in the present year = 4000 \left(1 - \frac{5}{100} \right)^3
= 4000 \left( \frac{95}{100} \right)^3
= 4000 \times \frac{857375}{1000000}
= \frac{3429500000}{1000000}
= ₹ 3,429.50
Present year’s electric bill expenditure = ₹ 3,429.50
Question 9
The weight of Savan Babu is 80 kg. In order to reduce his weight, he started regular morning walks. He decided to reduce his weight every year by 10%. Let us write by calculating his weight after 3 year.
Solution:
Present weight = 80 kg
Rate of decrease = 10% per annum
Time = 3 years
Weight after 3 years = 80 \left(1 - \frac{10}{100} \right)^3
= 80 \left( \frac{90}{100} \right)^3
= 80 \times \frac{729}{1000}
= \frac{58320}{1000}
= 58.32 kg
Weight after 3 years = 58.32 kg
Question 10
At present the sum of the number of students in all M.S.K in a district is 399. If number of students increased in a year was 10% of its previous year, let us calculate the sum of the number of students 3 years before in all the M.S.K in the district.
Solution:
Present number of students = 399
Rate of increase = 10% per annum
Time = 3 years
Let the number of students 3 years ago = P
Then, present number = P \left(1 + \frac{10}{100} \right)^3 = P \left( \frac{110}{100} \right)^3
= P \times \frac{1331}{1000}
399 = P \times \frac{1331}{1000}
P = \frac{399 \times 1000}{1331}
= \frac{399000}{1331} = 300
Number of students 3 years ago = 300
Question 11
As the farmers are becoming more alert of the harmful effects of applying only the chemical fertilisers and insecticides in agricultural lands, the number of farmers using fertilisers and insecticides in the village of Rasulpur decreases by 20% in a year in comparison to its previous year. Three years before, the number of such farmers was 3000; let us calculate the number of such farmers in that village now.
Solution:
Number of farmers 3 years ago = 3,000
Rate of decrease = 20% per annum
Time = 3 years
Number of farmers now = 3000 \left(1 - \frac{20}{100} \right)^3
= 3000 \left( \frac{80}{100} \right)^3
= 3000 \times \frac{512}{1000}
= \frac{1536000}{1000} = 1536
Number of such farmers now = 1,536
Question 12
The price of a machine of a factory is ₹ 18000. The price of that machine decreases by 10% in each year. Let us calculate its price after 3 year.
Solution:
Present price of the machine = ₹ 18,000
Rate of depreciation = 10% per annum
Time = 3 years
Price after 3 years = 18000 \left(1 - \frac{10}{100} \right)^3
= 18000 \left( \frac{90}{100} \right)^3
= 18000 \times \frac{729}{1000}
= \frac{13122000}{1000}
= 13122
Price of the machine after 3 years = ₹ 13,122
Question 13
For the families having no electricity in their house, a Panchayat samity village Bakultala accepted a plan to offer electricity connections. 1200 families this village have no electric connection in their house. In comparison to its previous year, it is possible to arrange electricity every year for 75% of the families having no electricity, let us write by calculating the number of families without electricity after 2 year
Solution:
Initial number of families without electricity = 1,200
Rate of electrification = 75% per year
⇒ 25% remain without electricity each year
Time = 2 years
Number of families without electricity after 2 years = 1200 \left(1 - \frac{75}{100} \right)^2 = 1200 \left( \frac{25}{100} \right)^2
= 1200 \times \frac{625}{10000}
= \frac{750000}{10000}
= 75
Number of families without electricity after 2 years = 75
Question 14
As a result of continuous publicity on harmful reactions to the use of cold drinks filled, bottles, the number of users of cold drinks has decreased by 25% every year in comparison to the previous year. 3 years before the number of users of cold drinks in a town was 80000. Let us write by calculating the number of users of cold drinks in the present year.
Solution:
Number of users 3 years ago = 80,000
Rate of decrease = 25% per annum
Time = 3 years
Number of users in the present year = 80000 \left(1 - \frac{25}{100} \right)^3
= 80000 \left( \frac{75}{100} \right)^3
= 80000 \times \frac{421875}{1000000}
= \frac{33750000000}{1000000}
= 33750
Number of users in the present year = 33,750
Question 15
As a result of publicity on smoking, the number of smokers is decreased by 6 \frac{1}{4}% every year in comparison to its previous year. If the number of smokers at present in a city is 33750, let us write by calculating the number of smokers in that city 3 years before.
Solution:
Present number of smokers = 33,750
Rate of decrease = 6 \frac{1}{4}% = \frac{25}{4}%
Time = 3 years
Let the number of smokers 3 years ago = P
Then, present number = P \left(1 - \frac{25}{400} \right)^3 = P \left( \frac{375}{400} \right)^3
= P \times \left( \frac{15}{16} \right)^3 = P \times \frac{3375}{4096}
or, 33750 = P \times \frac{3375}{4096}
or, P = \frac{33750 \times 4096}{3375}
= \frac{138240000}{3375} = 40,960
Number of smokers 3 years ago = 40,960
Multiple Choice Questions
Question 1
In case of compound interest, the rate of compound interest per annum is-
- Equal
- Unequal
- Both equal or unequal
- None of these
Solution:
(a) Equal
Explanation:
In compound interest, the rate per annum can be either:
- Equal: When the rate remains the same each year, such as 10% per annum compounded annually.
- Unequal: When the rate changes from year to year, such as 8% for the first year, 9% for the second year, and so on.
Hence, compound interest calculations can be done with both constant or varying rates depending on the terms of the investment or loan.
Question 2
In case of compound interest
- The principal remains unchanged each year
- Principal changes in each year
- Principal may be equal or unequal in each year
Solution:
(c) Principal changes in each year
Explanation:
- In compound interest, interest is added to the principal at the end of each compounding period.
- This means the principal for the next year includes the previous year's interest.
- As a result, the principal increases with time — it does not remain constant.
Question 3
At present the population of a village is p and if the rate of increase of po tion per year be 2r%, the population will be after n year
- p\left(1+\frac{r}{100}\right)^{n}
- p\left(1+\frac{r}{50}\right)^{n}
- p\left(1+\frac{r}{50}\right)^{2 n}
- p\left(1-\frac{r}{50}\right)^{n}
Solution:
(b) P \left(1+\frac{r}{50}\right)^{n}
Question 4
Present price of a machine is ₹ 2 p and if price of the machine decreases b 2r% in each year, the price of machine will be
- ₹ p\left(1-\frac{r}{50}\right)^{n}
- ₹ 2 p\left(1-\frac{r}{50}\right)^{n}
- ₹ p\left(1-\frac{r}{100}\right)^{2 n}
- ₹ 2 p\left(1-\frac{r}{100}\right)^{2 n}
Solution:
(d) ₹ 2 p\left(1-\frac{r}{100}\right)^{2 n}
Question 5
A person deposited ₹ 100 in a bank and got the amount ₹ 121 for two years
- 10 %
- 20 %
- 5 %
- 10 \frac{1}{2} %
Solution:
(a) 10%
Explanation:
Use the compound interest formula:
A = P \left(1 + \frac{R}{100} \right)^n
Given:
A = ₹121,\quad P = ₹100,\quad n = 2
Substitute values:
121 = 100 \left(1 + \frac{R}{100} \right)^2
or, \left(1 + \frac{R}{100} \right)^2 = \frac{121}{100} = 1.21
or, 1 + \frac{R}{100} = \sqrt{1.21} = 1.1
or, \frac{R}{100} = 0.1
R = 10%
Let us write true or false for the following statements:
Question 1
The compound interest will be always less than simple interest for some money at fixed rate of interest for fixed time.
Solution:
False
Explanation
Compound Interest is never less than Simple Interest for fixed rate and time. It is either equal (for 1 year) or greater (for more than 1 year).
Question 2
In case of compound interest, interest is to be added to the principal at the fixed time interval, i.e., the amount of principal increases continuously.
Solution:
True
Let us fill in the blanks:
Question 1
The compound interest and simple interest for one year at the fixed rate of interest on fixed sum of money are ___.
Solution:
Equal.
Question 2
If some things are increased by fixed rate with respect to time, that is ___.
Solution:
Same rate.
Question 3
If some things are decreased by fixed rate with respect to time, this is uniform rate of ___.
Solution:
Decrease
Short answer (S.A.):
Question 1
Let us write the rate of compound interest per annum, so that the amount on ₹ 400 for 2 years becomes ₹ 441.
Solution:
Principal (P) = ₹ 400
Amount (A) = ₹ 441
Time (n) = 2 years
Rate (r) = ?
A = P \left(1 + \frac{R}{100} \right)^n
441 = 400 \left(1 + \frac{R}{100} \right)^2
or, \left(1 + \frac{R}{100} \right)^2 = \frac{441}{400} = \left( \frac{21}{20}\right)^2
or, 1 + \frac{R}{100} = \frac{21}{20}
or, \frac{R}{100} = \frac{21}{20} - 1
or, \frac{R}{100} = \frac{1}{20}
or, R = \frac{1}{20} × 100
or, R = 5%
Rate of compound interest = 5% per annum
Question 2
If a sum of money doubles itself at compound interest in n years, let us write in how many years it will become four times.
Solution:
Let the principal = ₹ p & rate of C.I = R
\therefore p\left(1+\frac{R}{100}\right)^{n} = 2p
or, \left(1+\frac{R}{100}\right)^{n} = 2
Now in 't' years, amount will be 4p
\therefore\left(1+\frac{R}{100}\right)^{t}= 4p
\left(1+\frac{R}{100}\right)^{t} = 4 = 2^{2} = \left(1+\frac{R}{100}\right)^{2 n}
∴ Time = 2n.
Question 3
Let us calculate the principal that at the rate of 5% compound interest per annum becomes ₹ 615 after two year
Solution:
Let the principal = ₹ p.
\therefore P\left\{\left(1+\frac{5}{100}\right)^{2}-1\right\} = 615
or, P × \left\{\left(\frac{21}{20}\right)^{2}-1\right\} = 615
or, P × \left(\frac{441-400}{400}\right) = 615
\therefore P = \frac{615 × 400}{41} = 15 × 400
P = ₹ 6000
Question 4
The price of a machine depreciates at the rate of r% per annum, let us find the price of the machine that was n years before.
Solution:
Present price of the machine = P
Rate of depreciation = r% per annum
Time = n years
Price n years ago = A
Formula:
To find the price n years ago, we reverse the depreciation process:
P = A \left(1 - \frac{r}{100} \right)^nSo,
A = \frac{P}{\left(1 - \frac{r}{100} \right)^n}
Therefore, the price of the machine n years ago is:
A = \frac{P}{\left(1 - \frac{r}{100} \right)^n}
Question 5
If the rate of increase in population is r% per year, the population after n years p; let us find the population that was n years before.
Solution:
n years before, no. of population was = \frac{P}{\left(1+\frac{r}{100}\right)^{n}}
= P ×\left(1+\frac{r}{100}\right)^{-n}
