Chapter 14 - Compound Interest
Practice set 14.1
1. Find the amount and the compound interest.
| No. | Principal (₹) | Rate (p.c.p.a.) | Duration (years) |
|---|---|---|---|
1. | 2000 | 5 | 2 |
2. | 5000 | 8 | 3 |
3. | 4000 | 7.5 | 2 |
1.
Given:
Principal = ₹ 2000
Rate of interest = ₹ 5 p.c.p.a.
No of years = 2
To find:
Amount and Compound interest
Solution:
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 2000 \((1 + \large \frac {5}{100})\)²
∴ Amount = 2000 \((1 + \large \frac {1}{20})\)²
∴ Amount = 2000 \((\large \frac {21}{20})\)²
∴ Amount = 2000 × \(\large \frac {21}{20}\) × \(\large \frac {21}{20}\)
∴ Amount = 5 × 21 × 21
∴ Amount = ₹ 2205
Now,
Compound interest = Amount – Principal
∴ Compound interest = 2205 – 2000
∴ Compound interest = ₹ 205
Ans: Amount is ₹ 2205 and compound interest is ₹ 205
2.
Given:
Principal = ₹ 5000
Rate of interest = ₹ 8 p.c.p.a.
No of years = 3
To find:
Amount and Compound interest
Solution:
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 5000 \((1 + \large \frac {8}{100})\)³
∴ Amount = 5000 \((\large \frac {108}{100})\)³
∴ Amount = 5000 × \(\large \frac {108}{100}\) × \(\large \frac {108}{100}\) × \(\large \frac {108}{100}\)
∴ Amount = ₹ 6298.56
Now,
Compound interest = Amount – Principal
∴ Compound interest = 6298.56 – 5009
∴ Compound interest = ₹ 1298.56
Ans: Amount is ₹ 6298.56 and compound interest is ₹ 1298.56
3.
Given:
Principal = ₹ 4000
Rate of interest = ₹ 7.5 p.c.p.a.
No of years = 2
To find:
Amount and Compound interest
Solution:
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 4000 \((1 + \large \frac {7.5}{100})\)²
∴ Amount = 4000 \((1 + \large \frac {75}{1000})\)²
∴ Amount = 4000 \((\large \frac {1075}{1000})\)²
∴ Amount = 4000 × \(\large \frac {1075}{1000}\) × \(\large \frac {1075}{1000}\)
∴ Amount = ₹ 4622.50
Now,
Compound interest = Amount – Principal
∴ Compound interest = 4622.50 – 4000
∴ Compound interest = ₹ 622.50
Ans: Amount is ₹ 4622.50 and compound interest is ₹ 622.50
2. Sameerrao has taken a loan of ₹12500 at a rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?
Given:
Principal = ₹ 12500
Rate of interest = ₹ 12 p.c.p.a.
No of years = 3
To find:
Amount
Solution:
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 12500 \((1 + \large \frac {12}{100})\)²
∴ Amount = 12500 \((\large \frac {112}{100})\)²
∴ Amount = 12500 × \(\large \frac {112}{100}\) × \(\large \frac {112}{100}\) × \(\large \frac {112}{100}\)
∴ Amount = ₹ 17561.60
Ans: Sameerrao should pay ₹ 17561.60 to clear his loan.
3. To start a business Shalaka has taken a loan of ₹8000 at a rate of 10½ p.c.p.a. After two years how much compound interest will she have to pay?
Given:
Principal = ₹ 8000
Rate of interest = ₹ 10.5 p.c.p.a.
No of years = 2
To find:
Compound interest
Solution:
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 8000 \((1 + \large \frac {10.5}{100})\)²
∴ Amount = 8000 \((1 + \large \frac {105}{1000})\)²
∴ Amount = 8000 \((\large \frac {1105}{1000})\)²
∴ Amount = 8000 × \(\large \frac {1105}{1000}\) × \(\large \frac {1105}{1000}\)
∴ Amount = ₹ 9768.20
Now,
Compound interest = Amount – Principal
∴ Compound interest = 9768.20 – 8000
∴ Compound interest = ₹ 1768.20
Ans: Amount is ₹ 9768.20 and compound interest is ₹ 1768.20
Practice set 14.2
1. On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.
Given:
No of workers initially (Principal) = 320
Rate of interest = 25 p.c.p.a.
No of years = 2
To find:
No of workers after 2 years (Amount)
Solution:
No of workers after 2 years (Amount) = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 320 \((1 + \large \frac {25}{100})\)²
∴ Amount = 320 \((\large \frac {125}{100})\)²
∴ Amount = 320 × \(\large \frac {125}{100}\) × \(\large \frac {125}{100}\)
∴ Amount = 500
Ans: There will be 500 workers after 2 years.
2. A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in the number of sheeps is 8% every year.
Given:
No of sheeps initially (Principal) = 200
Rate of interest = 8 p.c.p.a.
No of years = 3
To find:
No of sheeps after 3 years (Amount)
Solution:
No of sheeps after 3 years (Amount) = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 200 \((1 + \large \frac {8}{100})\)³
∴ Amount = 200 \((\large \frac {108}{100})\)³
∴ Amount = 200 × \(\large \frac {108}{100}\) × \(\large \frac {108}{100}\) × \(\large \frac {108}{100}\)
∴ Amount = 251.94 ≈ 252 sheeps
Ans: No of sheeps after 3 years is 252.
3. In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.
Given:
No of trees initially (Principal) = 40000
Rate of interest = 5 p.c.p.a.
No of years = 3
To find:
No of trees after 3 years (Amount)
Solution:
No of trees after 3 years (Amount) = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 40000 \((1 + \large \frac {5}{100})\)³
∴ Amount = 40000 \((\large \frac {105}{100})\)³
∴ Amount = 40000 × \(\large \frac {105}{100}\) × \(\large \frac {105}{100}\)
∴ Amount = 46306 trees
Ans: There will be 46,306 trees
P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 40000 \((1 + \large \frac {5}{100})\)³
4. The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year, find the depreciation in price of the machine after two years.
Given:
Principal = ₹ 2,50,000
Rate of interest = –10 p.c.p.a. (depreciation)
No of years = 2
To find:
Depreciation in price after 2 years
Solution:
We know that,
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 2,50,000 \([1 + \large (\frac {– 10}{100})]\)²
∴ Amount = 2,50,000 \((1 \,– \large \frac {1}{10})\)²
∴ Amount = 2,50,000 \((\large \frac {9}{10})\)²
∴ Amount = 2,50,000 × \(\large \frac {9}{10}\) × \(\large \frac {9}{10}\)
∴ Amount = ₹ 2,02,500
Decrease in price = Principal – Amount
∴ Decrease in price = 2,50,000 – 2,02,500
∴ Decrease in price = ₹ 47,500
Ans: Price of the machine will reduce by 47,500 in two years.
5. Find the compound interest if the amount of a certain principal after two years is ₹4036.80 at the rate of 16 p.c.p.a.
Given:
Amount = ₹4036.80
Rate of interest = 16 p.c.p.a.
No of years = 2
To find:
Compound interest
Solution:
We know that,
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ 4036.80 = P \((1 + \large \frac {16}{100})\)²
∴ 4036.80 = P \((\large \frac {116}{100})\)²
∴ 4036.80 = P × \(\large \frac {116}{100}\) × \(\large \frac {116}{100}\)
∴ \(\large \frac {4036.80\, ×\, 100\, ×\, 100}{116\, ×\, 116}\) = P
∴ P = ₹ 3000
Now,
Compound interest = Amount – Principal
∴ Compound interest = 4036.80 – 3000
∴ Compound interest = ₹ 1036.80
Ans: Compound interest is ₹ 1036.80
6. A loan of ₹15000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.
Given:
Principal = ₹ 15,000
Rate of interest = 12 p.c.p.a
No of years = 3
To find:
Amount to settle the loan
Solution:
We know that,
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 15,000 \((1 + \large \frac {12}{100})\)³
∴ Amount = 15,000 \((\large \frac {112}{100})\)³
∴ Amount = 15,000 × \(\large \frac {112}{100}\) × \(\large \frac {112}{100}\) × \(\large \frac {112}{100}\)
∴ Amount = ₹ 21,073.92
Ans: Amount to settle the loan is ₹ 21,073.92
7. A principal amounts to ₹13924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.
Given:
Amount = ₹13924
Rate of interest = 18 p.c.p.a.
No of years = 2
To find:
Principal
Solution:
We know that,
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ 13924 = P \((1 + \large \frac {18}{100})\)²
∴ 13924 = P \((\large \frac {118}{100})\)²
∴ 13924 = P × \(\large \frac {118}{100}\) × \(\large \frac {118}{100}\)
∴ \(\large \frac {13924\, ×\, 100\, ×\, 100}{118\, ×\, 118}\) = P
∴ P = ₹ 10,000
Ans: Principal is ₹ 10,000
8. The population of a suburb is 16000. Find the rate of increase in the population if the population after two years is 17640.
Given:
Present Population (Principal) = 16000
No of years = 2
Population after 2 years (Amount) = 17640
To find:
Rate of increase
Solution:
We know that,
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ 17640 = 16000 \((1 + \large \frac {R}{100})\)²
∴ \(\large \frac {17640}{16000}\) = \((1 + \large \frac {R}{100})\)²
∴ 1.1025 = \((1 + \large \frac {R}{100})\)²
∴ \(\large \frac {11025}{10000}\) = \((1 + \large \frac {R}{100})\)²
∴ \( \sqrt {\frac {11025}{10000}}\) = \( \sqrt {(1 + \frac {R}{100})²}\) …[By Taking square root of both sides]
∴ \(\large \frac {105}{100}\) = \(1 + \large \frac {R}{100}\)
∴ \(\large \frac {105}{100}\) – 1 = \(\large \frac {R}{100}\)
∴ \(\large \frac {105\, –\, 100}{100}\) = \(\large \frac {R}{100}\)
∴ \(\large \frac {5}{100}\) = \(\large \frac {R}{100}\)
∴ \(\large \frac {5\, ×\, 100}{100}\) = R
∴ R = 5
Ans: Rate of increase in population is 5 p.c.p.a.
9. In how many years ₹700 will amount to ₹847 at a compound interest rate of 10 p.c.p.a.
Given:
Principal = ₹ 700
Amount = ₹ 847
Rate of interest = 10 p.c.p.a
To find:
No. of years
Solution:
We know that,
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ 847 = 700 \((1 + \large \frac {10}{100})^N\)
∴ \(\large \frac {847}{700}\) = \((1 + \large \frac {1}{10})^N\)
∴ 1.21 = \((\large \frac {11}{10})^N\)
∴ \(\large \frac {121}{100}\) = \((\large \frac {11}{10})^N\)
∴ \((\large \frac {11}{10})^2\) = \((\large \frac {11}{10})^N\)
By Comparing the equations, we get,
N = 2
Ans: 2 years will be required.
10. Find the difference between simple interest and compound interest on ₹20000 at 8 p.c.p.a for 2 years.
Given:
Principal = ₹ 20000
No. of years = 2
Rate of interest = 8 p.c.p.a.
To find:
Difference between Compound interest and Simple interest
Solution:
We know that,
Simple Interest = \(\frac {P\, ×\, N\, ×\, R }{100}\)
∴ Simple Interest = \(\large \frac {20000\, ×\, 2\, ×\, 8}{100}\)
∴ Simple Interest = 200 × 2 × 8
∴ Simple Interest = ₹ 3200
Now,
Amount = P\( \left(1+\large \frac R{100}\right)^N\)
∴ Amount = 20000 \((1 + \large \frac {8}{100})\)²
∴ Amount = 20000 \((\large \frac {108}{100})\)²
∴ Amount = 20000 × \(\large \frac {108}{100}\) × \(\large \frac {108}{100}\)
∴ Amount = 2 × 108 × 108
∴ Amount = ₹ 23,328
Compound interest = Amount – Principal
∴ Compound interest = 23,328 – 20000
∴ Compound interest = ₹ 3,328
And,
Difference between Compound interest and Simple interest = 3328 – 3200
∴ Difference between Compound interest and Simple interest = ₹ 128
Ans: Difference between Compound interest and Simple interest is ₹ 128