Mathematics · Chapter 03

53 solved questions

Chapter 3: Growth and Depreciation

Every question in this chapter, answered and explained — step-by-step solutions drawn live from the chapter library across 2 question sections.

Exercise 3.128 Exercise 3.225

Exercise 3.1

1Textbookex3.1-q1-a

Question 1 If the present population of a locality is $P$ , the population after $T$ years is $P_{T}$ and yearly population growth rate is $R %$ , write the formula to find $P_{T}$ .

$P_{T} = P (1 + \frac{R}{100})^{T}$ .

2Textbookex3.1-q1-b

If the number of tigers of Chitwan National Park in 2079 B.S. is $x$ and annual growth rate of tiger is $R %$ , then what will be the number of tigers after $N$ years?

Number of tigers after $N$ years $= x (1 + \frac{R}{100})^{N}$ .

3Textbookex3.1-q1-c

The growth rate of foreign employment from Nepal of the first year, second year and third year are respectively $R_{1} %$ , $R_{2} %$ and $R_{3} %$ . Write the formula to find the number of employees after 3 years.

Number of employees after 3 years $= P (1 + \frac{R _{1}}{100}) (1 + \frac{R _{2}}{100}) (1 + \frac{R _{3}}{100})$ , where $P$ is the initial number of employees.

4Textbookex3.1-q2-a

Question 2 If death rate is less than birth rate of a country, does the population of the country increase or decrease?

If death rate is less than birth rate, more people are born than die, so the population of the country increases.

5Textbookex3.1-q2-b

One year ago, the price of a sack of 25 kg rice was Rs. 1500. As the price has increased 10% p.a., how much does the sack of 25 kg rice cost now?

Initial price $P = Rs. 1, 500$ , $R = 10%$ , $T = 1$ year. $P_{T} = 1, 500 (1 + \frac{10}{100})^{1} = 1, 500 \times 1.10 = Rs. 1, 650$ .

6Textbookex3.1-q3-a

Question 3 According to the census 2021, the population of a city was 5,18,452. If the growth rate was 4.5% p.a., what will be the population of the city after 3 years?

$P = 5, 18, 452$ , $R = 4.5%$ , $T = 3$ years. $P_{T} = 5, 18, 452 (1 + \frac{4.5}{100})^{3} = 5, 18, 452 \times (1.045)^{3} \approx 5, 91, 640$ .

7Textbookex3.1-q3-b

A landlord made an agreement with a businessman to increase the rent 5% per annum. If the rent of a shutter is Rs. 10,000 now, what will be the rent after 3 years?

Flag: taking the rate exactly as transcribed ( $R = 5%$ ) gives $P_{T} = 10, 000 (1.05)^{3} = Rs. 11, 576.25$ , which does not match the textbook's own printed answer of Rs. $15, 208.75$ . That printed figure is reproduced exactly by using $R = 15%$ instead: $P_{T} = 10, 000 (1.15)^{3} = Rs. 15, 208.75$ — so the rate was most likely $15%$ in the original, with the leading "1" lost in scanning. Using $R = 15%$ , $T = 3$ : $P_{T} = 10, 000 (1 + \frac{15}{100})^{3} = 10, 000 \times 1.520875 = Rs. 15, 208.75$ .

8Textbookex3.1-q3-c

The growth rate of a bacteria in curd is 10% per hour. If the number of bacteria at 6 a.m. is $4 \times 1 0^{11}$ , what will be the number of bacteria after 2 hours? Find.

$P = 4 \times 1 0^{11}$ , $R = 10%$ /hour, $T = 2$ hours. $P_{T} = 4 \times 1 0^{11} (1 + \frac{10}{100})^{2} = 4 \times 1 0^{11} \times 1.21 = 4.84 \times 1 0^{11}$ .

9Textbookex3.1-q4-a

Question 4 The population of a rural municipality before 2 years was 28,500. If the population growth rate is 2% p.a., then by how much is the population increased in 2 years?

$P = 28, 500$ , $R = 2%$ , $T = 2$ years. Increase $= P [(1 + \frac{R}{100})^{T} - 1] = 28, 500 [(1.02)^{2} - 1] = 28, 500 \times 0.0404 \approx 1, 152$ .

10Textbookex3.1-q4-b

Monthly fee of grade 10 of a school before 3 years was Rs. 6,500. If fees increase every year by 10% according to the rules and regulation of the school, then by how much has the fee increased in 3 years?

$P = Rs. 6, 500$ , $R = 10%$ , $T = 3$ years. Increase $= 6, 500 [(1.10)^{3} - 1] = 6, 500 \times 0.331 = Rs. 2, 151.50$ .

11Textbookex3.1-q4-c

A land costs Rs. 6,00,000 at present. If the yearly increase rate of the price is 10%, then how much will increase in the price of the land in 2 years? Find.

$P = Rs. 6, 00, 000$ , $R = 10%$ , $T = 2$ years. Increase $= 6, 00, 000 [(1.10)^{2} - 1] = 6, 00, 000 \times 0.21 = Rs. 1, 26, 000$ .

12Textbookex3.1-q5-a

Question 5 The number of students studying in a university at present is 21,632. 2 years ago, the number of students studying in the university was 20,000. What was the yearly increment rate?

$P = 20, 000$ , $P_{T} = 21, 632$ , $T = 2$ years. $P_{T} = P (1 + \frac{R}{100})^{T}$ : $(1 + \frac{R}{100})^{2} = \frac{21 , 632}{20 , 000} = 1.0816$ , so $1 + \frac{R}{100} = 1.04$ , giving $R = 4%$ .

13Textbookex3.1-q5-b

The population of a rural municipality at the end of 2018 A.D. was 40,000. If at the end of 2020 A.D., the population increased to 44,100, then find the yearly population growth rate.

$P = 40, 000$ , $P_{T} = 44, 100$ , $T = 2$ years. $(1 + \frac{R}{100})^{2} = \frac{44 , 100}{40 , 000} = 1.1025$ , so $1 + \frac{R}{100} = 1.05$ , giving $R = 5%$ .

14Textbookex3.1-q5-c

3 years ago, the price per liter of oil was Rs. 125. Now, the price has increased to Rs. 216 per liter, then what is the inflation rate?

$P = 125$ , $P_{T} = 216$ , $T = 3$ years. $(1 + \frac{R}{100})^{3} = \frac{216}{125} = 1.728$ , so $1 + \frac{R}{100} = 1.2$ (since $1. 2^{3} = 1.728$ ), giving $R = 20%$ .

15Textbookex3.1-q6-a

Question 6 The population of a village is 13,310. Whilst the population growth rate of the village is 10% p.a., how many years ago was the population of the village 10,000?

$P_{T} = 13, 310$ , $R = 10%$ , find $T$ such that $P = 10, 000$ . $10, 000 (1.10)^{T} = 13, 310$ , so $(1.10)^{T} = 1.331 = (1.10)^{3}$ , giving $T = 3$ years ago.

16Textbookex3.1-q6-b

At the beginning of Baisakh, the papaya plant was 4 meter high. If the growth rate of the plant is 4% per month, in how many months will the height of the plant be 4.3264 meter?

$P = 4$ , $R = 4%$ /month, $P_{T} = 4.3264$ . $4 (1.04)^{T} = 4.3264$ , so $(1.04)^{T} = \frac{4.3264}{4} = 1.0816 = (1.04)^{2}$ , giving $T = 2$ months.

17Textbookex3.1-q7-a

Question 7 The price of a plot of land is Rs. 15,97,200 per aana. If the rate of increase in the price is 10% p.a., what was the price of the land per ropani before 3 years? Find it. (1 ropani = 16 aana)

This is identical to Example 5 of this section: $P_{T} = 15, 97, 200$ , $R = 10%$ , $T = 3$ years. $P = \frac{15 , 97 , 200}{( 1.10 ) ^{3}} = \frac{15 , 97 , 200}{1.331} = Rs. 12, 00, 000$ per aana, 3 years ago. Per ropani $= 16 \times 12, 00, 000 = Rs. 1, 92, 00, 000$ .

18Textbookex3.1-q7-b

The price of a land is Rs. 2,66,200 per aana. If the rate of increase in price is 10% p.a., what was the price of the land per ropani before 3 years? Find it.

$P_{T} = 2, 66, 200$ , $R = 10%$ , $T = 3$ years. $P = \frac{2 , 66 , 200}{( 1.10 ) ^{3}} = \frac{2 , 66 , 200}{1.331} = Rs. 2, 00, 000$ per aana, 3 years ago. Per ropani $= 16 \times 2, 00, 000 = Rs. 32, 00, 000$ .

19Textbookex3.1-q8-a

Question 8 When urine of a patient was tested in a laboratory at 6 a.m., it was found that the number of bacteria was $1 \times 1 0^{5}$ . After that, it was tested at 7 a.m., 8 a.m. and 9 a.m. and found that the increase rate per hour was 3%, 4% and 5% respectively. Find the total number of bacteria at 9 a.m.

$P = 1 \times 1 0^{5}$ , $R_{1} = 3%$ , $R_{2} = 4%$ , $R_{3} = 5%$ . $P_{3} = 1 \times 1 0^{5} \times 1.03 \times 1.04 \times 1.05 = 1.12476 \times 1 0^{5} \approx 1.1248 \times 1 0^{5}$ .

20Textbookex3.1-q8-b

3 years ago, the population of a city was 1,50,000. In the following 3 years, the population increased by 2% in the first year, 4% in the second year and 5% in the third year. Find the present population of the city.

$P = 1, 50, 000$ , $R_{1} = 2%$ , $R_{2} = 4%$ , $R_{3} = 5%$ . $P_{3} = 1, 50, 000 \times 1.02 \times 1.04 \times 1.05 = 1, 67, 076$ (very close to the printed answer of $1, 67, 070$ , the small difference being rounding in the source).

21Textbookex3.1-q9-a

Question 9 In an insurance company established 3 years ago, 1000 agents were trained by all the branches throughout the country. Customers should be increased according as the market competition. So that, from the very beginning, a policy "a group of every 5 agents should add 1 more agent every year" was implemented and the number of agents has been increased. How many agents are there in the company now?

"A group of every 5 agents adds 1 more every year" means a growth rate of $\frac{1}{5} \times 100% = 20%$ per year, compounded. $P = 1000$ , $R = 20%$ , $T = 3$ years. $P_{T} = 1000 (1.20)^{3} = 1000 \times 1.728 = 1, 728$ .

22Textbookex3.1-q9-b

A financial institution established at the beginning of 2078 has 200 marketers. After the expansion of market of the institution, a policy of a group of every 5 marketers should add 1 more new marketer every year was implemented and the number of marketers has been increased. Find how many marketers will be there at the end of 2079?

$R = 20%$ as in (a). $P = 200$ , $T = 2$ years (start of 2078 to end of 2079). $P_{T} = 200 (1.20)^{2} = 200 \times 1.44 = 288$ .

23Textbookex3.1-q10

The population of a rural municipality is increasing by 10% every year. At the end of 2 years, the population reached to 30,000. If 5,800 migrated to the place finally, find the initial population.

Let the initial population $= P$ . After 2 years of $10%$ growth, the population (before migration) $= P (1.10)^{2} = 1.21 P$ . Adding the $5, 800$ who migrated in gives the final population of $30, 000$ : $1.21 P + 5, 800 = 30, 000$ , so $1.21 P = 24, 200$ , giving $P = \frac{24 , 200}{1.21} = 20, 000$ .

24Textbookex3.1-q11

After continuous inflation of American dollar at the rate of 5% p.a., in two years it becomes $1 = R s .120. B e f or e 2 y e a r s, h o w m u c h N e p a l esec u r r e n cy w a se q u a l t o$ 1?

$P_{T} = Rs. 120$ , $R = 5%$ , $T = 2$ years. $P = \frac{P _{T}}{( 1 + R /100 ) ^{T}} = \frac{120}{( 1.05 ) ^{2}} = \frac{120}{1.1025} \approx Rs. 108.84$ . So 2 years ago, $1 was equal to about Rs. $108.84$ .

25Textbookex3.1-q12

The population of a village before 2 years was 31,250. The population growth rate is 6% p.a. One year ago, 625 migrated to other places. Find the number of present population of the village.

Population 2 years ago $P = 31, 250$ , $R = 6%$ . After 1 year of growth: $31, 250 (1.06) = 33, 125$ . One year ago, $625$ migrated out: $33, 125 - 625 = 32, 500$ . Growing this for 1 more year to reach the present: $32, 500 (1.06) = 34, 450$ . So the present population is $34, 450$ .

26Textbookex3.1-q13

A district had 3,75,000 population before 3 years. If 1480 migrated to the village at the end of 2 years and 2,750 died due to natural disaster and yearly population growth rate is 2%, then find the number of present population of the district.

Population 3 years ago $P = 3, 75, 000$ , $R = 2%$ . After 2 years of growth: $3, 75, 000 (1.02)^{2} = 3, 90, 150$ . At the end of year 2, add migrants and subtract deaths: $3, 90, 150 + 1, 480 - 2, 750 = 3, 88, 880$ . Growing this for the third year: $3, 88, 880 (1.02) = 3, 96, 657.6 \approx 3, 96, 658$ . So the present population of the district is $3, 96, 658$ .

27Textbookex3.1-q14-a

At the beginning of 2075 B.S., the population of a metropolitan city was 5,00,000. At the end of 2077 B.S., the population of the city was 6,65,500. What was the population growth rate?

From beginning of 2075 to end of 2077 is $T = 3$ years. $P = 5, 00, 000$ , $P_{T} = 6, 65, 500$ . $(1 + R /100)^{3} = \frac{6 , 65 , 500}{5 , 00 , 000} = 1.331 = (1.10)^{3}$ , so $R = 10%$ .

28Textbookex3.1-q14-b

If the population increases in the same way, what will be the population of the city at the end of 2079?

From end of 2077 to end of 2079 is $2$ more years. $P = 6, 65, 500$ , $R = 10%$ , $T = 2$ . $P_{T} = 6, 65, 500 (1.10)^{2} = 6, 65, 500 \times 1.21 = 8, 05, 255$ .

Exercise 3.2

1Textbookex3.2-q1-a

Question 1 If the initial price of a good is Rs. $P$ and the rate of depreciation is $R %$ p.a., then write the formula to find the price of the good after $T$ years.

$V_{T} = P (1 - \frac{R}{100})^{T}$ .

2Textbookex3.2-q1-b

What does ' $R$ ' in $V_{T} = V_{0} (1 - \frac{R}{100})^{T}$ represent?

$R$ represents the yearly (or per-period) rate of depreciation — the percentage by which the good's value decreases each year.

3Textbookex3.2-q2-a

Question 2 If Ram sold a watch which cost Rs. 5,000 after 1 year at 7% depreciation, then what will be the depreciated price?

$V_{0} = Rs. 5, 000$ , $R = 7%$ , $T = 1$ year. The depreciated (lost) amount $= V_{0} \times \frac{R}{100} = 5, 000 \times 0.07 = Rs. 350$ .

4Textbookex3.2-q2-b

A motorcycle is sold for Rs. 57,000 in 1 year after depreciating at the rate of 5% p.a. At what price was the motorcycle purchased for?

$V_{T} = Rs. 57, 000$ , $R = 5%$ , $T = 1$ year. $V_{T} = V_{0} (1 - 0.05)$ : $57, 000 = V_{0} \times 0.95$ , so $V_{0} = \frac{57 , 000}{0.95} = Rs. 60, 000$ .

5Textbookex3.2-q3-a

Question 3 What will be the price of a cupboard of cost Rs. 16,800 after 2 years at the rate of 15% depreciation p.a.?

$V_{0} = Rs. 16, 800$ , $R = 15%$ , $T = 2$ years. $V_{T} = 16, 800 (1 - 0.15)^{2} = 16, 800 \times (0.85)^{2} = 16, 800 \times 0.7225 = Rs. 12, 138$ .

6Textbookex3.2-q3-b

The present price of a motorcycle of efficiency 125 c.c. produced in India is Rs. 2,50,000. If its cost depreciates every year at the rate of 4% p.a., what will be the price of the motorcycle after using it for 3 years? Find it.

$V_{0} = Rs. 2, 50, 000$ , $R = 4%$ , $T = 3$ years. $V_{T} = 2, 50, 000 (1 - 0.04)^{3} = 2, 50, 000 \times (0.96)^{3} = 2, 50, 000 \times 0.884736 = Rs. 2, 21, 184$ .

7Textbookex3.2-q4-a

Question 4 Sameer has bought a mobile phone for Rs. 30,000. Due to his household problem he has to sell it after using 2 years at 30% rate of depreciation. Find the depreciated amount.

$V_{0} = Rs. 30, 000$ , $R = 30%$ , $T = 2$ years. Depreciated amount $= V_{0} [1 - (1 - 0.30)^{2}] = 30, 000 [1 - 0.49] = 30, 000 \times 0.51 = Rs. 15, 300$ .

8Textbookex3.2-q4-b

As the laptop is convenient for online class, a mathematics teacher purchased a laptop for Rs. 96,000. It is depreciated at the rate of 15% every year. If he sells it after using it for 3 years, how much money is depreciated? Find it.

$V_{0} = Rs. 96, 000$ , $R = 15%$ , $T = 3$ years. Depreciated amount $= V_{0} [1 - (1 - 0.15)^{3}] = 96, 000 [1 - 0.614125] = 96, 000 \times 0.385875 = Rs. 37, 044$ .

9Textbookex3.2-q5-a

Question 5 3 years ago, a plot of land of 4 ropani in hilly region was purchased for Rs. 12,50,000. It can be sold now for Rs. 1,60,000 per ropani. By what percent per annum is the price of land depreciated? Calculate it.

Price per ropani 3 years ago $= \frac{12 , 50 , 000}{4} = Rs. 3, 12, 500$ . Present price per ropani $= Rs. 1, 60, 000$ . $V_{T} = V_{0} (1 - R /100)^{T}$ : $(1 - \frac{R}{100})^{3} = \frac{1 , 60 , 000}{3 , 12 , 500} = 0.512 = (0.8)^{3}$ , so $1 - \frac{R}{100} = 0.8$ , giving $R = 20%$ p.a.

10Textbookex3.2-q5-b

A man bought a watch for Rs. 5,000 and he sold it after using it for 3 years for Rs. 625 only. Find the annual rate of depreciation.

$V_{0} = Rs. 5, 000$ , $V_{T} = Rs. 625$ , $T = 3$ years. $(1 - \frac{R}{100})^{3} = \frac{625}{5 , 000} = 0.125 = (0.5)^{3}$ , so $1 - \frac{R}{100} = 0.5$ , giving $R = 50%$ p.a.

11Textbookex3.2-q6-a

Question 6 A press machine was purchased for Rs. 4,00,000 before some years and it is depreciated to Rs. 1,96,000 by reducing the price every year by 30%. How many years ago was the machine purchased?

$V_{0} = Rs. 4, 00, 000$ , $R = 30%$ , $V_{T} = Rs. 1, 96, 000$ . $(1 - 0.30)^{T} = \frac{1 , 96 , 000}{4 , 00 , 000} = 0.49 = (0.7)^{2}$ , so $T = 2$ years ago.

12Textbookex3.2-q6-b

A car costs Rs. 8,00,000. If its price reduces every year by 10%, in how many years will its price be Rs. 5,83,200?

$V_{0} = Rs. 8, 00, 000$ , $R = 10%$ , $V_{T} = Rs. 5, 83, 200$ . $(0.9)^{T} = \frac{5 , 83 , 200}{8 , 00 , 000} = 0.729 = (0.9)^{3}$ , so $T = 3$ years.

13Textbookex3.2-q7-a

Question 7 After continuous devaluation of the American dollar in 2 years by 5% p.a., American dollar $1 = R s .125 n o w, h o w m u c h N e p a l i r u p ees w a se q u a l t o$ 1 before 2 years? Find it.

$V_{T} = Rs. 125$ , $R = 5%$ , $T = 2$ years. $V_{0} = \frac{V _{T}}{( 1 - 0.05 ) ^{2}} = \frac{125}{( 0.95 ) ^{2}} = \frac{125}{0.9025} \approx Rs. 138.50$ .

14Textbookex3.2-q7-b

3 years ago, a small group of youths returned back from foreign employment and started a cow farm investing Rs. 2,80,000. Due to political instability of state, if the price of the farm depreciates by 5% p.a. then how much does the farm cost now?

$V_{0} = Rs. 2, 80, 000$ , $R = 5%$ , $T = 3$ years. $V_{T} = 2, 80, 000 (1 - 0.05)^{3} = 2, 80, 000 \times (0.95)^{3} = 2, 80, 000 \times 0.857375 = Rs. 2, 40, 065$ .

15Textbookex3.2-q8-a

Question 8 A photocopy machine costs Rs. 5,00,000 now. If the machine depreciates by 15% in the first year and similarly depreciates by 10% and 5% in the second and third year, then what will be the price of the machine in 3 years?

$V_{0} = Rs. 5, 00, 000$ , $R_{1} = 15%$ , $R_{2} = 10%$ , $R_{3} = 5%$ . $V_{3} = 5, 00, 000 (1 - 0.15) (1 - 0.10) (1 - 0.05) = 5, 00, 000 \times 0.85 \times 0.90 \times 0.95 = Rs. 3, 63, 375$ .

16Textbookex3.2-q8-b

A lamination machine is sold for Rs. 24,168 after depreciating it by 4% and 5% in the first and second year respectively. Find what the price of the machine was before 2 years.

$V_{T} = Rs. 24, 168$ , $R_{1} = 4%$ , $R_{2} = 5%$ . $V_{0} (1 - 0.04) (1 - 0.05) = 24, 168$ : $V_{0} \times 0.96 \times 0.95 = 24, 168$ , so $V_{0} = \frac{24 , 168}{0.912} = Rs. 26, 500$ .

17Textbookex3.2-q9

An entrepreneur purchased a heavy truck by investing Rs. 48,00,000. He earned Rs. 6,80,000 in 2 years. If it is depreciated by 10% every year and sold in 2 years, then find his loss or profit.

$V_{0} = Rs. 48, 00, 000$ , $R = 10%$ , $T = 2$ years. Depreciated (resale) value $= 48, 00, 000 (1 - 0.10)^{2} = 48, 00, 000 \times 0.81 = Rs. 38, 88, 000$ . Total amount recovered $=$ resale value $+$ earnings $= 38, 88, 000 + 6, 80, 000 = Rs. 45, 68, 000$ . Compared to the original investment of Rs. $48, 00, 000$ : $45, 68, 000 - 48, 00, 000 = - Rs. 2, 32, 000$ , i.e. a loss of Rs. $2, 32, 000$ .

18Textbookex3.2-q10

A bus owner bought a bus for Rs. 16,00,000 and conducted it in the Kathmandu-Baglung route for 3 years. He earned Rs. 5,10,000 only. If the value depreciates every year by 5% and he sells it in 3 years, then find his loss or profit.

$V_{0} = Rs. 16, 00, 000$ , $R = 5%$ , $T = 3$ years. Resale value $= 16, 00, 000 (1 - 0.05)^{3} = 16, 00, 000 \times 0.857375 = Rs. 13, 71, 800$ . Total amount recovered $=$ resale value $+$ earnings $= 13, 71, 800 + 5, 10, 000 = Rs. 18, 81, 800$ . Profit $= 18, 81, 800 - 16, 00, 000 = Rs. 2, 81, 800$ .

19Textbookex3.2-q11

The price of a finance company listed in Nepal share market is falling down its price by 10% in 2 years. Your neighbour having the shares of the finance company sold his shares for Rs. 28,350. How many shares were sold by the company before 2 years at the rate of Rs. 100 per share?

Note: "falling by 10% in 2 years" is read as a compounding annual rate of $R = 10%$ p.a. (this reading reproduces the textbook's published answer). $V_{T} = Rs. 28, 350$ , $R = 10%$ , $T = 2$ years. $V_{0} = \frac{V _{T}}{( 1 - 0.10 ) ^{2}} = \frac{28 , 350}{0.81} = Rs. 35, 000$ . Since shares were Rs. $100$ each 2 years ago, number of shares $= \frac{35 , 000}{100} = 350$ .

20Textbookex3.2-q12-a

The price of the share of a hydropower company is reducing by 10% p.a. If Sashi sells all his shares now and has the present value Rs. 7,10,775, then What was the price of his shares before 2 years?

$V_{T} = Rs. 7, 10, 775$ , $R = 10%$ , $T = 2$ years. $V_{0} = \frac{V _{T}}{( 1 - 0.10 ) ^{2}} = \frac{7 , 10 , 775}{0.81} = Rs. 8, 77, 500$ .

21Textbookex3.2-q12-b

He purchased the shares at the rate of Rs. 100 per share in IPO, how many shares did he buy before 2 years?

Number of shares $= \frac{8 , 77 , 500}{100} = 8, 775$ .

22Textbookex3.2-q13-a

In certain rate of annual depreciation, the price of a good will be Rs. 10,240 and Rs. 8,192 in 2 years and 3 years respectively. Then, Find the rate of depreciation.

Let $V_{0} (1 - R /100)^{2} = 10, 240$ ...(i) and $V_{0} (1 - R /100)^{3} = 8, 192$ ...(ii). Dividing (ii) by (i): $1 - \frac{R}{100} = \frac{8 , 192}{10 , 240} = 0.8$ , giving $R = 20%$ .

23Textbookex3.2-q13-b

What was the initial price of the good? Find it.

From (i): $V_{0} = \frac{10 , 240}{( 0.8 ) ^{2}} = \frac{10 , 240}{0.64} = Rs. 16, 000$ .

24Textbookex3.2-q14-a

In a certain rate of yearly depreciation, the price of an article will be Rs. 5,41,500 and Rs. 5,14,425 in 2 and 3 years respectively. Then What is the rate of depreciation?

$V_{0} (1 - R /100)^{2} = 5, 41, 500$ ...(i); $V_{0} (1 - R /100)^{3} = 5, 14, 425$ ...(ii). Dividing (ii) by (i): $1 - \frac{R}{100} = \frac{5 , 14 , 425}{5 , 41 , 500} = 0.95$ , giving $R = 5%$ .

25Textbookex3.2-q14-b

What was the initial price of the article? Find it.

From (i): $V_{0} = \frac{5 , 41 , 500}{( 0.95 ) ^{2}} = \frac{5 , 41 , 500}{0.9025} = Rs. 6, 00, 000$ .

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