Exam practice: Grade 10 Math Unit 3

a.

A place has population P now, and $P_T$ after T years, growing at R% a year. Write the formula for $P_T$ .

Answer

$P_T = P\left(1 + \frac{R}{100}\right)^T$

b.

Chitwan National Park had x tigers in 2079 BS, and their number grows by R% a year. How many tigers will it have after N years?

Answer

$x\left(1 + \frac{R}{100}\right)^N$

c.

The number of Nepalis going abroad for work grows by R₁%, R₂% and R₃% in the first, second and third years. If P went in the year before, write the formula for the number after 3 years.

Answer

$P_3 = P\left(1 + \frac{R_1}{100}\right)\left(1 + \frac{R_2}{100}\right)\left(1 + \frac{R_3}{100}\right)$

a.

If a country’s death rate is lower than its birth rate, does its population increase or decrease?

Answer

It increases (unless more people move away than move in): more people are born each year than die.

b.

A year ago a 25 kg sack of rice cost Rs 1,500. Its price has risen by 10% a year. What does the sack cost now?

Answer

1500 × 1.10 = Rs 1,650.

a.

At the 2021 census a city had 5,18,452 people. If it grows by 4.5% a year, what will its population be after 3 years?

Answer

$518452 \times 1.045^3 = 518452 \times 1.141166125 \approx 591640.3$ : about 5,91,640 people.

b.

A landlord agrees with a businessman to raise the rent by 15% a year. If a shutter’s rent is Rs 10,000 now, what will it be after 3 years?

Answer

$10000 \times 1.15^3 = 10000 \times 1.520875 =$ Rs 15,208.75.

c.

The bacteria that set curd grow by 10% an hour. If there are $4 \times 10^{11}$ at 6 a.m., how many are there 2 hours later?

Answer

$4 \times 10^{11} \times 1.1^2 = 4 \times 10^{11} \times 1.21 =$ $4.84 \times 10^{11}$ .

a.

Two years ago a rural municipality had 28,500 people. If it grows by 2% a year, by how much has its population increased in 2 years?

Answer

$28500\,(1.02^2 - 1) = 28500 \times 0.0404 = 1151.4$ : about 1,151 people.

b.

Three years ago a school’s Grade 10 monthly fee was Rs 6,500. Its rules raise fees by 10% every year. By how much has the fee risen in 3 years?

Answer

$6500\,(1.1^3 - 1) = 6500 \times 0.331 =$ Rs 2,151.50.

c.

A plot of land costs Rs 6,00,000 now. If its price rises by 10% a year, by how much will it rise in 2 years?

Answer

$600000\,(1.1^2 - 1) = 600000 \times 0.21 =$ Rs 1,26,000.

a.

A university has 21,632 students now; 2 years ago it had 20,000. What is the yearly rate of increase?

Answer

$\left(1 + \frac{R}{100}\right)^2 = \frac{21632}{20000} = 1.0816 = 1.04^2$ , so R = 4%.

b.

A rural municipality had 40,000 people at the end of 2018 AD and 44,100 at the end of 2020 AD. Find its yearly growth rate.

Answer

$\left(1 + \frac{R}{100}\right)^2 = \frac{44100}{40000} = \frac{441}{400} = \left(\frac{21}{20}\right)^2$ , so $1 + \frac{R}{100} = 1.05$ : 5%.

c.

Three years ago a liter of cooking oil cost Rs 125; now it costs Rs 216. What is the yearly rate of inflation?

Answer

$\left(1 + \frac{R}{100}\right)^3 = \frac{216}{125} = \left(\frac{6}{5}\right)^3$ , so $1 + \frac{R}{100} = 1.2$ : 20%.

a.

A village has 13,310 people, and its population grows by 10% a year. How many years ago did it have 10,000?

Answer

$\left(\frac{11}{10}\right)^T = \frac{13310}{10000} = \frac{1331}{1000} = \left(\frac{11}{10}\right)^3$ : 3 years ago.

b.

At the start of Baisakh a papaya plant was 4 m tall. If it grows by 4% a month, after how many months will it be 4.3264 m tall?

Answer

$1.04^T = \frac{4.3264}{4} = 1.0816 = 1.04^2$ : 2 months.

a.

Land now costs Rs 19,96,500 an aana, and its price has risen by 10% a year. What was its price per ropani 3 years ago? (1 ropani = 16 aana)

Answer

Per aana: $\frac{1996500}{1.1^3} = \frac{1996500}{1.331} = 1500000$ .
Per ropani: 16 × 1500000 = Rs 2,40,00,000.

b.

Land now costs Rs 2,66,200 an aana, and its price has risen by 10% a year. What was its price per ropani 3 years ago?

Answer

Per aana: $\frac{266200}{1.331} = 200000$ . Per ropani: 16 × 200000 = Rs 32,00,000.

a.

A patient’s urine sample had $1 \times 10^5$ bacteria at 6 a.m. Tests at 7, 8 and 9 a.m. showed they grew by 3%, 4% and 5% in those hours. How many bacteria were there at 9 a.m.?

Answer

100000 × 1.03 × 1.04 × 1.05 = 112476: $1.12476 \times 10^5$ , about 1,12,476.

b.

Three years ago a city had 1,50,000 people. Its population then grew by 2% in the first year, 4% in the second and 5% in the third. What is it now?

Answer

150000 × 1.02 × 1.04 × 1.05 = 167076: 1,67,076 people.

a.

An insurance company started 3 years ago with 1,000 trained agents. From the start, its rule has been that every group of 5 agents adds 1 new agent each year. How many agents does it have now?

Answer

1 new agent for every 5 is a growth of $\frac{1}{5} = 20\%$ a year.
$1000 \times 1.2^3 = 1000 \times 1.728 =$ 1,728 agents.

b.

A financial institution started at the beginning of 2078 with 200 marketers. Its rule is that every group of 5 marketers adds 1 new marketer each year. How many marketers will it have at the end of 2079?

Answer

From the beginning of 2078 to the end of 2079 is 2 years at 20%: $200 \times 1.2^2 = 200 \times 1.44 =$ 288 marketers.

10.

A rural municipality’s population grows by 10% a year. At the end of 2 years, 5,800 people moved in, and the population then was 30,000. What was the population at the start?

Answer

Before the 5,800 arrived: 30000 − 5800 = 24200.
$P \times 1.1^2 = 24200$ , so $P = \frac{24200}{1.21} =$ 20,000.

11.

Over two years the US dollar rose against the Nepali rupee by 5% a year, and now $\$1 =$ Rs 120. How many rupees was $\$1$ worth 2 years ago?

Answer

$\frac{120}{1.05^2} = \frac{120}{1.1025} \approx 108.84$ : $\$1 =$ Rs 108.84.

12.

Two years ago a village had 31,250 people, and it grows by 6% a year. One year ago, 625 people moved away. What is its population now?

Answer

After 1 year: 31250 × 1.06 = 33125; then 33125 − 625 = 32500.
After the second year: 32500 × 1.06 = 34,450.

13.

Three years ago a district had 3,75,000 people, growing by 2% a year. At the end of the second year 1,480 people moved in and 2,750 died in a natural disaster. What is its population now?

Answer

After 2 years: $375000 \times 1.02^2 = 375000 \times 1.0404 = 390150$ .
Then 390150 + 1480 − 2750 = 388880.
After the third year: 388880 × 1.02 = 396657.6: about 3,96,658.

a.

What was its yearly growth rate?

Answer

The beginning of 2075 to the end of 2077 is 3 years.
$\left(1 + \frac{R}{100}\right)^3 = \frac{665500}{500000} = 1.331 = 1.1^3$ : 10%.

b.

If it keeps growing the same way, what will its population be at the end of 2079?

Answer

Two more years: $665500 \times 1.1^2 = 665500 \times 1.21 =$ 8,05,255.

a.

A thing costs Rs P at first and depreciates by R% a year. Write the formula for its value after T years.

Answer

$P_T = P\left(1 - \frac{R}{100}\right)^T$ (written with V₀: $V_T = V_0\left(1 - \frac{R}{100}\right)^T$ ).

b.

In $V_T = V_0\left(1 - \frac{R}{100}\right)^T$ , the value after T years, what does R stand for?

Answer

The rate of depreciation: the percentage of its value the thing loses every year.

a.

Ram sells a watch that cost Rs 5,000 after 1 year, at 7% depreciation. How much value has it lost?

Answer

$7\% \text{ of } 5000 = 5000 \times \frac{7}{100} =$ Rs 350 (it sells for Rs 4,650).

b.

After 1 year, at 5% depreciation a year, a motorcycle sells for Rs 57,000. What did it cost?

Answer

V₀ × 0.95 = 57000, so $V_0 = \frac{57000}{0.95} =$ Rs 60,000.

a.

What will a cupboard costing Rs 16,800 be worth after 2 years at 15% depreciation a year?

Answer

$16800 \times 0.85^2 = 16800 \times 0.7225 =$ Rs 12,138.

b.

A new 125 cc motorcycle made in India costs Rs 2,50,000 in Nepal. If its value falls by 4% a year, what will it be worth after 3 years of use?

Answer

$250000 \times 0.96^3 = 250000 \times 0.884736 =$ Rs 2,21,184.

a.

Sameer bought a mobile phone for Rs 30,000. Because of a problem at home he has to sell it after 2 years, at 30% depreciation a year. Find the depreciated amount.

Answer

Value after 2 years: $30000 \times 0.7^2 = 14700$ .
Depreciated amount: 30000 − 14700 = Rs 15,300.

b.

A maths teacher bought a laptop for Rs 96,000 for online classes. It depreciates by 15% a year. If he sells it after 3 years, how much value has it lost?

Answer

Value: $96000 \times 0.85^3 = 96000 \times 0.614125 = 58956$ .
Lost: 96000 − 58956 = Rs 37,044.

a.

Three years ago, 4 ropani of land in the hills was bought for Rs 12,50,000. Now it can be sold for only Rs 1,60,000 a ropani. By what percentage a year has its price depreciated?

Answer

Now: 4 × 160000 = 640000.
$\left(1 - \frac{R}{100}\right)^3 = \frac{640000}{1250000} = 0.512 = 0.8^3$ , so $1 - \frac{R}{100} = 0.8$ : 20% a year.

b.

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

Answer

$\left(1 - \frac{R}{100}\right)^3 = \frac{625}{5000} = \frac{1}{8} = \left(\frac{1}{2}\right)^3$ , so $1 - \frac{R}{100} = \frac{1}{2}$ : 50% a year.

a.

A printing press bought for Rs 4,00,000 some years ago has lost 30% of its value every year and is now worth Rs 1,96,000. How many years ago was it bought?

Answer

$0.7^T = \frac{196000}{400000} = 0.49 = 0.7^2$ : 2 years ago.

b.

A car costs Rs 8,00,000. If its value falls by 10% a year, after how many years will it be worth Rs 5,83,200?

Answer

$0.9^T = \frac{583200}{800000} = 0.729 = 0.9^3$ : 3 years.

a.

Over 2 years the US dollar lost 5% of its value against the rupee each year, and now $\$1 =$ Rs 125. How many rupees was $\$1$ worth 2 years ago?

Answer

$V_0 \times 0.95^2 = 125$ , so $V_0 = \frac{125}{0.9025} \approx 138.50$ : $\$1 =$ Rs 138.50.

b.

Three years ago a group of young people back from work abroad started a cow farm with Rs 2,80,000. Because of political instability, the farm’s value has fallen by 5% a year. What is it worth now?

Answer

$280000 \times 0.95^3 = 280000 \times 0.857375 =$ Rs 2,40,065.

a.

A photocopy machine costs Rs 5,00,000 now. It depreciates by 15% in the first year, 10% in the second and 5% in the third. What will it be worth after 3 years?

Answer

500000 × 0.85 × 0.90 × 0.95 = Rs 3,63,375.

b.

A lamination machine was sold for Rs 24,168 after depreciating by 4% in the first year and 5% in the second. What did it cost 2 years ago?

Answer

V₀ × 0.96 × 0.95 = 24168, so $V_0 = \frac{24168}{0.912} =$ Rs 26,500.

9.

A businessman bought a heavy truck for Rs 48,00,000 and earned Rs 6,80,000 with it in 2 years. It depreciates by 10% a year, and he sells it after 2 years. Find his profit or loss.

Answer

Value after 2 years: $4800000 \times 0.9^2 = 3888000$ .
Received: 3888000 + 680000 = 4568000, less than the Rs 48,00,000 paid.
A loss of Rs 2,32,000.

10.

A bus owner bought a bus for Rs 16,00,000 and ran it on the Kathmandu–Baglung route for 3 years, earning only Rs 5,10,000. Its value depreciates by 5% a year, and he sells it after 3 years. Find his profit or loss.

Answer

Value after 3 years: $1600000 \times 0.95^3 = 1600000 \times 0.857375 = 1371800$ .
Received: 1371800 + 510000 = 1881800. Paid 16,00,000.
A profit of Rs 2,81,800.

11.

A finance company’s shares on the Nepal Stock Exchange fell in price by 10% a year for 2 years. Your neighbor sold all his shares for Rs 28,350. If he had bought them 2 years ago at Rs 100 a share, how many shares had he bought?

Answer

$V_0 = \frac{28350}{0.9^2} = \frac{28350}{0.81} = 35000$ . At Rs 100 a share: 350 shares.

a.

What were his shares worth 2 years ago?

Answer

$\frac{710775}{0.81} =$ Rs 8,77,500.

b.

He bought them in the IPO, 2 years ago, at Rs 100 a share. How many shares did he buy?

Answer

$\frac{877500}{100} =$ 8,775 shares.

a.

Find the rate of depreciation.

Answer

The third year multiplies by $1 - \frac{R}{100} = \frac{8192}{10240} = 0.8$ : 20%.

b.

What was it worth at first?

Answer

$V_0 \times 0.8^2 = 10240$ , so $V_0 = \frac{10240}{0.64} =$ Rs 16,000.

a.

What is the rate of depreciation?

Answer

$1 - \frac{R}{100} = \frac{514425}{541500} = 0.95$ : 5%.

b.

What was it worth at first?

Answer

$V_0 = \frac{541500}{0.95^2} = \frac{541500}{0.9025} =$ Rs 6,00,000.

Exam practice

Unit 3 · Growth and Depreciation

Exercise 3.1

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Exercise 3.2

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14