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Growth

Question 1

The population of a city in 2078 B.S. was 50,000. If the annual population growth rate was 2%, what will be the population in 2080 B.S.? Calculate it.

Solution:

Given information:

  • Initial population of the city (P)=50,000(P) = 50,000
  • Population growth rate (R)=2%(R) = 2\% p.a.
  • Time (T)=2080 B.S.−2078 B.S.=2(T) = 2080 \text{ B.S.} - 2078 \text{ B.S.} = 2 years

Finding the population after 2 years:

We know that, PT=P(1+R100)TP_T = P\left(1 + \frac{R}{100}\right)^T

Substitute the given values:

PT=50,000(1+2100)2P_T = 50,000\left(1 + \frac{2}{100}\right)^2
PT=50,000(102100)2P_T = 50,000\left(\frac{102}{100}\right)^2
PT=50,000(1.02)2P_T = 50,000(1.02)^2
PT=50,000×1.0404=52,020P_T = 50,000 \times 1.0404 = 52,020

Final Answer:

Thus, the population in 2080 B.S. will be 52,020.

Verification:

Population increase = 52,020 - 50,000 = 2,020

This represents a 4.04% increase over 2 years, which is consistent with 2% annual growth compounded.

Note: Population growth follows the same mathematical model as compound interest, where the growth rate is applied to the total population each year.

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Question 2

Two years ago, the price of a sack of 25 kg jeera masino rice was Rs. 1,300. If the inflation rate was 5% p.a., by how much is the price increased now? Find it.

Solution:

Given information:

  • The initial price of jeera masino rice (P)=Rs. 1,300(P) = \text{Rs. } 1,300
  • Increased rate of price (R)=5%(R) = 5\% p.a.
  • Time (T)=2(T) = 2 years
  • Increased price = ?

Finding the price increase:

We know that,

Increased price = P[(1+R100)T−1]P\left[\left(1 + \frac{R}{100}\right)^T - 1\right]

Substitute the given values:

Increased price=Rs. 1,300[(1+5100)2−1]\text{Increased price} = \text{Rs. } 1,300\left[\left(1 + \frac{5}{100}\right)^2 - 1\right]
=Rs. 1,300[(105100)2−1]= \text{Rs. } 1,300\left[\left(\frac{105}{100}\right)^2 - 1\right]
=Rs. 1,300[(1.05)2−1]= \text{Rs. } 1,300\left[(1.05)^2 - 1\right]
=Rs. 1,300[1.1025−1]= \text{Rs. } 1,300[1.1025 - 1]
=Rs. 1,300×0.1025= \text{Rs. } 1,300 \times 0.1025
=Rs. 133.25= \text{Rs. } 133.25

Final Answer:

∴ The increased price of jeera masino rice is Rs. 133.25.

Verification:

  • Original price 2 years ago: Rs. 1,300
  • Current price: Rs. 1,300 + Rs. 133.25 = Rs. 1,433.25
  • Total percentage increase: (133.25/1300) × 100% = 10.25%
  • This matches the compound inflation: (1.05)² - 1 = 0.1025 = 10.25% ✓

Note: Inflation affects prices similarly to compound interest, where the inflation rate is applied to the current price each year, causing prices to grow exponentially over time.

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Question 3

If the price of a photocopy machine increases from Rs. 1,00,000 to Rs. 1,21,000 in 2 years, find the rate of yearly increment.

Solution:

Given information:

  • The initial price of a photocopy machine (P)=Rs. 1,00,000(P) = \text{Rs. } 1,00,000
  • Time (T)=2(T) = 2 years
  • The present price of the photocopy machine (PT)=Rs. 1,21,000(P_T) = \text{Rs. } 1,21,000
  • Rate of price increment (R)=?(R) = ?

Finding the rate of yearly increment:

We know that PT=P(1+R100)TP_T = P\left(1 + \frac{R}{100}\right)^T

Substitute the given values:

Rs. 1,21,000=Rs. 1,00,000(1+R100)2\text{Rs. } 1,21,000 = \text{Rs. } 1,00,000\left(1 + \frac{R}{100}\right)^2

121000100000=(1+R100)2\frac{121000}{100000} = \left(1 + \frac{R}{100}\right)^2

(1110)2=(1+R100)2\left(\frac{11}{10}\right)^2 = \left(1 + \frac{R}{100}\right)^2

Taking square root of both sides:

1110=1+R100\frac{11}{10} = 1 + \frac{R}{100}

1110−1=R100\frac{11}{10} - 1 = \frac{R}{100}

11−1010=R100\frac{11 - 10}{10} = \frac{R}{100}

110×100=R\frac{1}{10} \times 100 = R

R=10%R = 10%

Final Answer:

∴ The rate of yearly increment is 10%.

Verification:

  • Original price: Rs. 1,00,000
  • After 2 years at 10% annual growth: Rs. 1,00,000 × (1.10)² = Rs. 1,00,000 × 1.21 = Rs. 1,21,000 ✓
  • Total increase: Rs. 21,000 over 2 years
  • Annual compound growth rate: 10%

Note: This problem demonstrates how to find an unknown growth rate when we know the initial value, final value, and time period. The solution involves working backwards from the compound growth formula.

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Question 4

The number of students of a basic school is 500 now. In how many years will the number be 720 if the number of students increases by 20% p.a. every year?

Solution:

Given information:

  • Initial number of students (P)=500(P) = 500
  • Increase rate (R)=20%(R) = 20\% p.a.
  • Time (T)=T(T) = T years
  • The number of students after T years (PT)=720(P_T) = 720

Finding the time required:

We know that, PT=P(1+R100)TP_T = P\left(1 + \frac{R}{100}\right)^T

Substitute the given values:

720=500(1+20100)T720 = 500\left(1 + \frac{20}{100}\right)^T

720500=(1+20100)T\frac{720}{500} = \left(1 + \frac{20}{100}\right)^T

720500=(120100)T\frac{720}{500} = \left(\frac{120}{100}\right)^T

3625=(65)T\frac{36}{25} = \left(\frac{6}{5}\right)^T

Simplifying the left side:

3625=6252=(65)2\frac{36}{25} = \frac{6^2}{5^2} = \left(\frac{6}{5}\right)^2

Therefore:

(65)2=(65)T\left(\frac{6}{5}\right)^2 = \left(\frac{6}{5}\right)^T

Since the bases are equal, the indices must be equal:

⇒T=2\Rightarrow T = 2

Final Answer:

Therefore, after 2 years, the number of students will be 720.

Verification:

  • Initial students: 500
  • After 1 year: 500 × 1.20 = 600 students
  • After 2 years: 600 × 1.20 = 720 students ✓
  • Alternative: 500 × (1.20)² = 500 × 1.44 = 720 ✓

Note: This problem demonstrates solving for time in compound growth situations. The key insight is recognizing when we can express both sides of the equation with the same base, allowing us to equate the exponents.

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Question 5

The price of a plot of land is Rs. 15,97,200 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. (1 ropani = 16 aana)

Solution:

Given information:

  • The price of a plot of land per aana (PT)=Rs. 15,97,200(P_T) = \text{Rs. } 15,97,200
  • Time (T)=3(T) = 3 years
  • Rate of increase in price (R)=10%(R) = 10\% p.a.
  • Price of the land per aana before 3 years (P)=?(P) = ?

Finding the original price per aana:

We know that, PT=P(1+R100)TP_T = P\left(1 + \frac{R}{100}\right)^T

Substitute the given values:

15,97,200=P(1+10100)315,97,200 = P\left(1 + \frac{10}{100}\right)^3

15,97,200=P(110100)315,97,200 = P\left(\frac{110}{100}\right)^3

15,97,200=P(1.10)315,97,200 = P(1.10)^3

15,97,200=P×1.33115,97,200 = P \times 1.331

P=15,97,2001.331P = \frac{15,97,200}{1.331}

P=Rs. 12,00,000P = \text{Rs. } 12,00,000

Converting to price per ropani:

Thus, the price of land per aana (P)=Rs. 12,00,000(P) = \text{Rs. } 12,00,000

We know that, 16 aana = 1 ropani

∴ The price of land per ropani = 16×Rs. 12,00,000=Rs. 1,92,00,00016 \times \text{Rs. } 12,00,000 = \text{Rs. } 1,92,00,000

Final Answer:

The price of land per ropani before 3 years was Rs. 1,92,00,000.

Verification:

  • Original price per aana: Rs. 12,00,000
  • After 3 years at 10% growth: Rs. 12,00,000 × (1.10)³ = Rs. 12,00,000 × 1.331 = Rs. 15,97,200 ✓
  • Original price per ropani: Rs. 12,00,000 × 16 = Rs. 1,92,00,000
  • Current price per ropani: Rs. 15,97,200 × 16 = Rs. 2,55,55,200

Note: This problem demonstrates finding the original principal amount when we know the final amount, rate, and time. It involves working backwards from the compound growth formula and also requires unit conversion from aana to ropani.

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Question 6

The number of SEE appeared students from a district in 2076 B.S. was 50,000. If in the coming 3 years, the number increased by 5%, 6% and 4% respectively, find how many students will appear SEE in 2079 B.S.

Solution:

Given information:

  • The number of SEE appeared students before 3 years (P)=50,000(P) = 50,000
  • Time (T)=3(T) = 3 years
  • Rate of increase in the first year (R1)=5%(R_1) = 5\% p.a.
  • Rate of increase in the second year (R2)=6%(R_2) = 6\% p.a.
  • Rate of increase in the third year (R3)=4%(R_3) = 4\% p.a.
  • Number of SEE appeared students after 3 years (PT)=?(P_T) = ?

Finding the number of students after 3 years:

When growth rates are different for different years:

PT=P(1+R1100)(1+R2100)(1+R3100)P_T = P\left(1 + \frac{R_1}{100}\right)\left(1 + \frac{R_2}{100}\right)\left(1 + \frac{R_3}{100}\right)

Substitute the given values:

PT=50,000(1+5100)(1+6100)(1+4100)P_T = 50,000\left(1 + \frac{5}{100}\right)\left(1 + \frac{6}{100}\right)\left(1 + \frac{4}{100}\right)
PT=50,000(105100)(106100)(104100)P_T = 50,000\left(\frac{105}{100}\right)\left(\frac{106}{100}\right)\left(\frac{104}{100}\right)
PT=50,000(1.05)(1.06)(1.04)P_T = 50,000(1.05)(1.06)(1.04)
PT=50,000×1.15752P_T = 50,000 \times 1.15752
PT=57,876P_T = 57,876

Final Answer:

Therefore, 57,876 students will appear in the SEE in 2079 B.S.

Step-by-step verification:

  • 2076 B.S.: 50,000 students (initial)
  • 2077 B.S.: 50,000 × 1.05 = 52,500 students
  • 2078 B.S.: 52,500 × 1.06 = 55,650 students
  • 2079 B.S.: 55,650 × 1.04 = 57,876 students ✓

Note: This problem demonstrates compound growth with variable rates for different years, which is common in real-world scenarios where growth rates fluctuate annually due to various factors like policy changes, economic conditions, or demographic shifts.

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Question 7

The population of a municipality in 2078 B.S. was 1,00,000. In 2079 B.S., 8000 migrated there from other places and 500 died due to several circumstances. If the population increase rate is 2% p.a. every year, what will be the population in 2081 B.S.? Find it.

Solution:

Case I: Finding population in 2079 B.S.

Given information:

  • Population in 2078 B.S. (P)=1,00,000(P) = 1,00,000
  • Population increase rate (R)=2%(R) = 2\% p.a.
  • Time (T)=1(T) = 1 year
  • In-migration population (Min)=8,000(M_{in}) = 8,000
  • Death population (D)=500(D) = 500

Finding the population in 2079 B.S.:

Since the population increases by 2% every year:

PT=P(1+R100)TP_T = P\left(1 + \frac{R}{100}\right)^T
PT=1,00,000(1+2100)1P_T = 1,00,000\left(1 + \frac{2}{100}\right)^1
PT=1,00,000(102100)1P_T = 1,00,000\left(\frac{102}{100}\right)^1
PT=1,00,000×1.02=1,02,000P_T = 1,00,000 \times 1.02 = 1,02,000

Now, the final population in 2079 B.S.:

Final population = Natural growth + Migration - Deaths

Population in 2079 B.S.=1,02,000+8,000−500=1,09,500\text{Population in 2079 B.S.} = 1,02,000 + 8,000 - 500 = 1,09,500

Case II: Finding population in 2081 B.S.

Given information:

  • Population in 2079 B.S. (P)=1,09,500(P) = 1,09,500
  • Time (T)=2(T) = 2 years (from 2079 to 2081)
  • Population increase rate (R)=2%(R) = 2\% p.a.

Finding the population in 2081 B.S.:

PT=P(1+R100)TP_T = P\left(1 + \frac{R}{100}\right)^T
PT=1,09,500(1+2100)2P_T = 1,09,500\left(1 + \frac{2}{100}\right)^2
PT=1,09,500(102100)2P_T = 1,09,500\left(\frac{102}{100}\right)^2
PT=1,09,500×(1.02)2P_T = 1,09,500 \times (1.02)^2
PT=1,09,500×1.0404=1,13,924P_T = 1,09,500 \times 1.0404 = 1,13,924

Final Answer:

Therefore, the population of the municipality in 2081 B.S. will be 1,13,924.

Step-by-step verification:

  • 2078 B.S.: 1,00,000 (initial population)
  • 2079 B.S.:
    • Natural growth: 1,00,000 × 1.02 = 1,02,000
    • Add migration: 1,02,000 + 8,000 = 1,10,000
    • Subtract deaths: 1,10,000 - 500 = 1,09,500
  • 2080 B.S.: 1,09,500 × 1.02 = 1,11,690
  • 2081 B.S.: 1,11,690 × 1.02 = 1,13,924 ✓

Note: This problem demonstrates how compound growth applies to real population dynamics, including factors like migration and mortality. The migration and deaths are one-time events in 2079 B.S., while the 2% growth rate continues to apply annually to the adjusted population base.

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