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#### Page No 90:

#### Question 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 |

#### Answer:

(1) Here, P = ₹ 2000; R = 5 % ; N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=2000{\left(1+\frac{5}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}=2000{\left(\frac{105}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}=2000{\left(\frac{21}{20}\right)}^{2}\phantom{\rule{0ex}{0ex}}=2205\mathrm{Rupees}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Compound}\mathrm{Interest}\mathrm{after}2\mathrm{years},\phantom{\rule{0ex}{0ex}}\mathrm{I}=\mathrm{Amount}-\mathrm{Principal}\phantom{\rule{0ex}{0ex}}=2205-2000\phantom{\rule{0ex}{0ex}}=205\mathrm{Rupees}$

Hence, Amount = ₹ 2205 and Compound interest = ₹ 205.

(2) Here, P = ₹ 5000; R = 8 % ; N = 3 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=5000{\left(1+\frac{8}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}=5000{\left(\frac{108}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}=5000{\left(\frac{27}{25}\right)}^{3}\phantom{\rule{0ex}{0ex}}=6298.56\mathrm{Rupees}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Compound}\mathrm{Interest}\mathrm{after}3\mathrm{years},\phantom{\rule{0ex}{0ex}}\mathrm{I}=\mathrm{Amount}-\mathrm{Principal}\phantom{\rule{0ex}{0ex}}=6298.56-5000\phantom{\rule{0ex}{0ex}}=1298.56\mathrm{Rupees}$

Hence, Amount = ₹ 6298.56 and Compound interest = ₹ 1298.56

(3) Here, P = ₹ 4000; R = 7.5 % ; N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=4000{\left(1+\frac{7.5}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}=4000{\left(1+\frac{75}{1000}\right)}^{2}\phantom{\rule{0ex}{0ex}}=4000{\left(\frac{1075}{1000}\right)}^{2}\phantom{\rule{0ex}{0ex}}=4000{\left(\frac{43}{40}\right)}^{2}\phantom{\rule{0ex}{0ex}}=4622.50\mathrm{Rupees}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Compound}\mathrm{Interest}\mathrm{after}2\mathrm{years},\phantom{\rule{0ex}{0ex}}\mathrm{I}=\mathrm{Amount}-\mathrm{Principal}\phantom{\rule{0ex}{0ex}}=4622.50-4000\phantom{\rule{0ex}{0ex}}=622.50\mathrm{Rupees}$

Hence, Amount = ₹ 4622.50 and Compound interest = ₹ 622.50

#### Page No 90:

#### Question 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?

#### Answer:

Here, P = ₹ 12500; R = 12 % ; N = 3 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=12500{\left(1+\frac{12}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}=12500{\left(1+\frac{3}{25}\right)}^{3}\phantom{\rule{0ex}{0ex}}=12500{\left(\frac{28}{25}\right)}^{3}\phantom{\rule{0ex}{0ex}}=17561.60\mathrm{Rupees}$

Hence, he should pay an amount of ₹ 17561.60 to clear his loan.

#### Page No 90:

#### Question 3:

To start a business Shalaka has taken a loan of ₹8000 at a rate of 10$\frac{1}{2}$ p.c.p.a. After two years how much compound interest will she have to pay?

#### Answer:

Here, P = ₹ 8000; R = 10$\frac{1}{2}$ % ; N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=8000{\left(1+\frac{10{\displaystyle \frac{1}{2}}}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}=8000{\left(1+\frac{21}{200}\right)}^{2}\phantom{\rule{0ex}{0ex}}=8000{\left(\frac{221}{200}\right)}^{2}\phantom{\rule{0ex}{0ex}}=9768.20\mathrm{Rupees}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Compound}\mathrm{Interest}\mathrm{after}2\mathrm{years},\phantom{\rule{0ex}{0ex}}\mathrm{I}=\mathrm{Amount}-\mathrm{Principal}\phantom{\rule{0ex}{0ex}}=9768.20-8000\phantom{\rule{0ex}{0ex}}=1768.20\mathrm{Rupees}$

Hence, she will have to pay a Compound interest of ₹ 1768.20 after 2 years.

#### Page No 93:

#### Question 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.

#### Answer:

Here, P = Number of workers initially = 320

A = Number of workers after 2 years

R = Rate of increase of number of workers per year = 25 %

N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=320{\left(1+\frac{25}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}=320{\left(1+\frac{1}{4}\right)}^{2}\phantom{\rule{0ex}{0ex}}=320{\left(\frac{5}{4}\right)}^{2}\phantom{\rule{0ex}{0ex}}=500$

Hence, the number of workers after 2 years is 500.

#### Page No 93:

#### Question 2:

A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.

#### Answer:

Here, P = Number of sheeps initially = 200

A = Number of sheeps after 3 years

R = Rate of increase of number of sheeps per year = 8 %

N = 3 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=200{\left(1+\frac{8}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}=200{\left(1+\frac{2}{25}\right)}^{3}\phantom{\rule{0ex}{0ex}}=200{\left(\frac{27}{25}\right)}^{3}\phantom{\rule{0ex}{0ex}}=251.94\phantom{\rule{0ex}{0ex}}=252\left(\mathrm{approx}\right)$

Hence, the number of sheeps after 3 years is 252.

#### Page No 93:

#### Question 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.

#### Answer:

Here, P = Number of trees initially = 40,000

A = Number of trees after 3 years

R = Rate of increase of number of trees per year = 5 %

N = 3 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=40000{\left(1+\frac{5}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}=40000{\left(1+\frac{1}{20}\right)}^{3}\phantom{\rule{0ex}{0ex}}=40000{\left(\frac{21}{20}\right)}^{3}\phantom{\rule{0ex}{0ex}}=46305$

Hence, the expected number of trees after 3 years is 46,305.

#### Page No 93:

#### Question 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.

#### Answer:

Here, P = Cost price of the machine = 2,50,000

A = Cost price after 2 years

I = Depreciation in price after 2 years

R = Rate of depreciation = 10 %

N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=250000{\left(1+\frac{-10}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}=250000{\left(1-\frac{1}{10}\right)}^{2}\phantom{\rule{0ex}{0ex}}=250000{\left(\frac{9}{10}\right)}^{2}\phantom{\rule{0ex}{0ex}}=202500$

Also,

I = P − A

= 250000 − 202500

= 47500

Hence, the depreciation in price of the machine after two years is Rs 47,500.

#### Page No 93:

#### Question 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.

#### Answer:

Here, P = Principal

A = ₹ 4036.80

I = Compound Interest

R = 16 %

N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}\Rightarrow 4036.80=\mathrm{P}{\left(1+\frac{16}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4036.80=\mathrm{P}{\left(1+\frac{4}{25}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4036.80=\mathrm{P}{\left(\frac{29}{25}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=\frac{4036.80\times 25\times 25}{29\times 29}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=3000$

Also,

I = A − P

= 4036.80 − 3000

= 1036.80

Hence, the compound interest is ₹ 1036.80

#### Page No 93:

#### Question 6:

A loan of ₹15000 was taken on compound interest. If the rate of compound interest s 12 p.c.p.a. find the amount to settle the loan after 3 years.

#### Answer:

Here, P = Principal = ₹ 15000

A = Amount

R = 12 %

N = 3 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=15000{\left(1+\frac{12}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}=15000{\left(1+\frac{3}{25}\right)}^{3}\phantom{\rule{0ex}{0ex}}=15000{\left(\frac{28}{25}\right)}^{3}\phantom{\rule{0ex}{0ex}}=21073.92$

Hence, the amount is ₹ 21073.92

#### Page No 93:

#### Question 7:

A principal amounts to ₹13924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.

#### Answer:

Here, P = Principal

A = ₹ 13924

R = 18 %

N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}\Rightarrow 13924=\mathrm{P}{\left(1+\frac{18}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 13924=\mathrm{P}{\left(1+\frac{9}{50}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 13924=\mathrm{P}{\left(\frac{59}{50}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=\frac{13924\times 50\times 50}{59\times 59}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=10000$

Hence, the principal is ₹ 10000.

#### Page No 93:

#### Question 8:

The population of a suburb is 16000. Find the rate of increase in the population if the population after two years is 17640.

#### Answer:

Here, P = Population of a suburb = 16000

A = Population after two years =17640

R = R %

N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}\Rightarrow 17640=16000{\left(1+\frac{\mathrm{R}}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{17640}{16000}={\left(1+\frac{\mathrm{R}}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{441}{400}={\left(1+\frac{\mathrm{R}}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{21}{20}\right)}^{2}={\left(1+\frac{\mathrm{R}}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 1+\frac{\mathrm{R}}{100}=\frac{21}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{R}}{100}=\frac{21}{20}-1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{R}}{100}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{R}=5$

Hence, the rate of increase in the population is 5 p.c.p.a.

#### Page No 93:

#### Question 9:

In how many years ₹700 will amount to ₹847 at a compound interest rate of 10 p.c.p.a.

#### Answer:

Here, P = Principal = ₹ 700

A = Amount = ₹ 847

R = 10 %

N = N years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}\Rightarrow 847=700{\left(1+\frac{10}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{847}{700}={\left(1+\frac{1}{10}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{121}{100}={\left(\frac{11}{10}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{11}{10}\right)}^{2}={\left(\frac{11}{10}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{N}=2$

Hence, the number of years is 2 years.

#### Page No 93:

#### Question 10:

Find the difference between simple interest and compound interest on ₹20000 at 8 p.c.p.a.

#### Answer:

**Disclaimer:** *In the question "Time" is not given. So the question is solved taking time as 2 years, because simple interest and compound interest will be same for one year.*

Here, P = Principal = ₹ 20000

R = 8 %

N = 2 years

$\mathrm{Simple}\mathrm{interest}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{N}}{100}\phantom{\rule{0ex}{0ex}}=\frac{20000\times 8\times 2}{100}\phantom{\rule{0ex}{0ex}}=3200\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}=20000{\left(1+\frac{8}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}=20000{\left(1+\frac{2}{25}\right)}^{2}\phantom{\rule{0ex}{0ex}}=20000{\left(\frac{27}{25}\right)}^{2}\phantom{\rule{0ex}{0ex}}=23328\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Compound}\mathrm{interest}=23328-20000\phantom{\rule{0ex}{0ex}}=3328\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Compound}\mathrm{interest}-\mathrm{Simple}\mathrm{interest}=3328-3000\phantom{\rule{0ex}{0ex}}=128$

Hence, the difference between simple interest and compound interest is ₹ 128.

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