Ganit Prakash - Class-X - Uniform Rate of Increase or Decrease
Let us work out 6.2 Population & Depreciation
📘 Let us work out 6.2 Solutions
Growth & DecayGiven:
Present population \((P)\) = 10000
Rate of increase \((r)\) = 3% per year
Time \((n)\) = 2 years
Population after 2 years \((A)\) will be:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ = 10000 \left(1 + \frac{3}{100}\right)^2 \\ = 10000 \times \left(\frac{103}{100}\right)^2 \\ = 10000 \times \frac{103 \times 103}{100 \times 100} \\ = 10609 \end{array}\]Therefore, the population of the village after 2 years will be 10609.
Given:
Present population \((P)\) = 80000000
Rate of increase \((r)\) = 2% per year
Time \((n)\) = 3 years
Population after 3 years \((A)\) will be:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ = 80000000 \left(1 + \frac{2}{100}\right)^3 \\ = 80000000 \times \left(\frac{102}{100}\right)^3 \\ = 80000000 \times \left(\frac{51}{50}\right)^3 \\ = 80000000 \times \frac{51 \times 51 \times 51}{50 \times 50 \times 50} \\ = 80000000 \times \frac{132651}{125000} \\ = 640 \times 132651 \\ = 84896640 \end{array}\]Therefore, the population of the state after 3 years will be 84896640.
Given:
Present price \((P)\) = ₹ 100000
Rate of depreciation \((r)\) = 10% per year
Time \((n)\) = 3 years
Since the price depreciates, the price after 3 years \((A)\) will be:
\[\begin{array}{l} A = P \left(1 - \frac{r}{100}\right)^n \\ = 100000 \left(1 - \frac{10}{100}\right)^3 \\ = 100000 \times \left(\frac{90}{100}\right)^3 \\ = 100000 \times \left(\frac{9}{10}\right)^3 \\ = 100000 \times \frac{9 \times 9 \times 9}{10 \times 10 \times 10} \\ = 100000 \times \frac{729}{1000} \\ = 100 \times 729 \\ = 72900 \end{array}\]Therefore, the price of the machine after 3 years will be ₹ 72900.
Given:
Present number of readmitted students \((A)\) = 3528
Rate of increase \((r)\) = 5% per year
Time \((n)\) = 2 years
Let the number of students readmitted 2 years ago be \(x\).
According to the formula for uniform increase:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ 3528 = x \left(1 + \frac{5}{100}\right)^2 \\ 3528 = x \left(\frac{105}{100}\right)^2 \\ 3528 = x \left(\frac{21}{20}\right)^2 \\ 3528 = x \times \frac{441}{400} \\ \Rightarrow x = \frac{3528 \times 400}{441} \\ \Rightarrow x = 8 \times 400 \\ \Rightarrow x = 3200 \end{array}\]Therefore, the number of students readmitted 2 years ago was 3200.
Given:
Present number of street accidents \((A)\) = 8748
Rate of decrease \((r)\) = 10% per year
Time \((n)\) = 3 years
Let the number of street accidents 3 years ago be \(x\).
According to the formula for uniform decrease:
\[\begin{array}{l} A = P \left(1 - \frac{r}{100}\right)^n \\ 8748 = x \left(1 - \frac{10}{100}\right)^3 \\ 8748 = x \left(\frac{90}{100}\right)^3 \\ 8748 = x \left(\frac{9}{10}\right)^3 \\ 8748 = x \times \frac{729}{1000} \\ \Rightarrow x = \frac{8748 \times 1000}{729} \\ \Rightarrow x = 12 \times 1000 \\ \Rightarrow x = 12000 \end{array}\]Therefore, the number of street accidents 3 years before was 12000.
Given:
Present production \((P)\) = 406 quintals
Rate of increase \((r)\) = 10% per year
Time \((n)\) = 3 years
Production after 3 years \((A)\) will be:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ = 406 \left(1 + \frac{10}{100}\right)^3 \\ = 406 \left(\frac{110}{100}\right)^3 \\ = 406 \left(\frac{11}{10}\right)^3 \\ = 406 \times \frac{1331}{1000} \\ = \frac{540386}{1000} \\ = 540.386 \end{array}\]Therefore, the production of fishes after 3 years will be 540.386 quintals.
Given:
Present height \((A)\) = 28.8 metre
Rate of increase \((r)\) = 20% per year
Time \((n)\) = 2 years
Let the height of the tree 2 years before be \(x\) metre.
According to the formula for uniform increase:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ 28.8 = x \left(1 + \frac{20}{100}\right)^2 \\ 28.8 = x \left(1 + \frac{1}{5}\right)^2 \\ 28.8 = x \left(\frac{6}{5}\right)^2 \\ 28.8 = x \times \frac{36}{25} \\ x = \frac{28.8 \times 25}{36} \\ x = \frac{288 \times 25}{36 \times 10} \\ x = 8 \times \frac{25}{10} \\ x = \frac{200}{10} = 20 \end{array}\]Therefore, the height of the tree 2 years before was 20 metre.
Given:
Electric bill 3 years ago \((P)\) = ₹ 4000
Rate of reduction \((r)\) = 5% per year
Time \((n)\) = 3 years
Since the expenditure is reduced, the present electric bill \((A)\) will be:
\[\begin{array}{l} A = P \left(1 - \frac{r}{100}\right)^n \\ = 4000 \left(1 - \frac{5}{100}\right)^3 \\ = 4000 \left(\frac{95}{100}\right)^3 \\ = 4000 \left(\frac{19}{20}\right)^3 \\ = 4000 \times \frac{6859}{8000} \\ = \frac{6859}{2} \\ = 3429.50 \end{array}\]Therefore, the family will have to spend ₹ 3429.50 to pay the electric bill in the present year.
Given:
Present weight of Savan babu \((P)\) = 80 kg
Rate of reduction \((r)\) = 10% per year
Time \((n)\) = 3 years
Weight after 3 years \((A)\) will be:
\[\begin{array}{l} A = P \left(1 - \frac{r}{100}\right)^n \\ = 80 \left(1 - \frac{10}{100}\right)^3 \\ = 80 \left(\frac{90}{100}\right)^3 \\ = 80 \left(\frac{9}{10}\right)^3 \\ = 80 \times \frac{729}{1000} \\ = 8 \times \frac{729}{100} \\ = \frac{5832}{100} \\ = 58.32 \end{array}\]Therefore, his weight after 3 years will be 58.32 kg.
Given:
Present number of students \((A)\) = 3993
Rate of increase \((r)\) = 10% per year
Time \((n)\) = 3 years
Let the number of students 3 years before be \(P\).
According to the formula for uniform increase:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ 3993 = P \left(1 + \frac{10}{100}\right)^3 \\ 3993 = P \left(\frac{110}{100}\right)^3 \\ 3993 = P \left(\frac{11}{10}\right)^3 \\ 3993 = P \times \frac{1331}{1000} \\ P = \frac{3993 \times 1000}{1331} \\ P = 3 \times 1000 \\ P = 3000 \end{array}\]Therefore, the sum of the number of students 3 years before was 3000.
Given:
Number of farmers 3 years ago \((P)\) = 3000
Rate of decrease \((r)\) = 20% per year
Time \((n)\) = 3 years
Since the number is decreasing, the present number of farmers \((A)\) will be:
\[\begin{array}{l} A = P \left(1 - \frac{r}{100}\right)^n \\ A = 3000 \left(1 - \frac{20}{100}\right)^3 \\ A = 3000 \left(1 - \frac{1}{5}\right)^3 \\ A = 3000 \left(\frac{4}{5}\right)^3 \\ A = 3000 \times \frac{64}{125} \\ A = 24 \times 64 \\ A = 1536 \end{array}\]Therefore, the number of such farmers in that village now is 1536.
Given:
Present price \((P)\) = ₹ 180000
Rate of decrease \((r)\) = 10% per year
Time \((n)\) = 3 years
Since the price decreases, the price after 3 years \((A)\) will be:
\[\begin{array}{l} A = P \left(1 - \frac{r}{100}\right)^n \\ A = 180000 \left(1 - \frac{10}{100}\right)^3 \\ A = 180000 \left(\frac{90}{100}\right)^3 \\ A = 180000 \left(\frac{9}{10}\right)^3 \\ A = 180000 \times \frac{729}{1000} \\ A = 180 \times 729 \\ A = 131220 \end{array}\]Therefore, the price of the machine after 3 years will be ₹ 131220.
Given:
Present number of families without electricity \((P)\) = 1200
Since electricity is provided to 75% of such families every year, the number of families *without* electricity decreases by 75%.
Rate of decrease \((r)\) = 75% per year
Time \((n)\) = 2 years
Number of families without electricity after 2 years \((A)\) will be:
\[\begin{array}{l} A = P \left(1 - \frac{r}{100}\right)^n \\ A = 1200 \left(1 - \frac{75}{100}\right)^2 \\ A = 1200 \left(\frac{25}{100}\right)^2 \\ A = 1200 \left(\frac{1}{4}\right)^2 \\ A = 1200 \times \frac{1}{16} \\ A = 75 \end{array}\]Therefore, the number of families without electricity after 2 years will be 75.
Given:
Number of users 3 years ago \((P)\) = 80000
Rate of decrease \((r)\) = 25% per year
Time \((n)\) = 3 years
The present number of users \((A)\) will be:
\[\begin{array}{l} A = P \left(1 - \frac{r}{100}\right)^n \\ A = 80000 \left(1 - \frac{25}{100}\right)^3 \\ A = 80000 \left(\frac{75}{100}\right)^3 \\ A = 80000 \left(\frac{3}{4}\right)^3 \\ A = 80000 \times \frac{27}{64} \\ A = 1250 \times 27 \\ A = 33750 \end{array}\]Therefore, the number of users of cold drinks in the present year is 33750.
Given:
Present number of smokers \((A)\) = 33750
Rate of decrease \((r)\) = \(6\frac{1}{4}\% = \frac{25}{4}\%\) per year
Time \((n)\) = 3 years
Let the number of smokers 3 years before be \(P\).
According to the formula for uniform decrease:
\[\begin{array}{l} A = P \left(1 - \frac{r}{100}\right)^n \\ 33750 = P \left(1 - \frac{25/4}{100}\right)^3 \\ 33750 = P \left(1 - \frac{25}{400}\right)^3 \\ 33750 = P \left(1 - \frac{1}{16}\right)^3 \\ 33750 = P \left(\frac{15}{16}\right)^3 \\ 33750 = P \times \frac{3375}{4096} \\ P = \frac{33750 \times 4096}{3375} \\ P = 10 \times 4096 \\ P = 40960 \end{array}\]Therefore, the number of smokers in that city 3 years before was 40960.
📘 16. Very short answer type questions (V.S.A.)
(A) Multiple Choice QuestionsExplanation:
In compound interest, the rate of interest can be the same for all years (e.g., \(10\%\) every year) or it can be different for successive years (e.g., \(R_1\%\) for the first year, \(R_2\%\) for the second year, etc.). Thus, it can be both equal or unequal.
Correct Option: (c) both equal or unequal
Explanation:
In compound interest, the interest accrued during a period is added to the principal. Therefore, the amount at the end of the first year becomes the principal for the second year. Hence, the principal changes every year.
Correct Option: (b) Principal changes in each year
Explanation:
Initial population \(= p\)
Rate of increase \(= 2r\%\) per year
Time \(= n\) years
Using the uniform increase formula \(A = P\left(1 + \frac{R}{100}\right)^t\) :
\[\begin{array}{l} \text{Population after } n \text{ years} = p \left(1 + \frac{2r}{100}\right)^n \\ = p \left(1 + \frac{r}{50}\right)^n \end{array}\]Correct Option: (b) \(p \left(1 + \frac{r}{50}\right)^n\)
Explanation:
Present price \((P)\) = ₹ \(2p\)
Rate of decrease \((R)\) = \(2r\%\) per year
Time \((t)\) = \(2n\) years
Using the uniform decrease formula \(A = P\left(1 - \frac{R}{100}\right)^t\) :
\[\begin{array}{l} \text{Price after } 2n \text{ years} = 2p \left(1 - \frac{2r}{100}\right)^{2n} \\ = 2p \left(1 - \frac{r}{50}\right)^{2n} \end{array}\]Correct Option: (d) ₹ \(2p \left(1 - \frac{r}{50}\right)^{2n}\)
Explanation:
Principal \((P)\) = ₹ 100
Amount \((A)\) = ₹ 121
Time \((n)\) = 2 years
Let the rate of interest be \(r\%\) per annum.
According to the formula \(A = P\left(1 + \frac{r}{100}\right)^n\) :
\[\begin{array}{l} 121 = 100 \left(1 + \frac{r}{100}\right)^2 \\ \Rightarrow \frac{121}{100} = \left(1 + \frac{r}{100}\right)^2 \\ \Rightarrow \left(\frac{11}{10}\right)^2 = \left(1 + \frac{r}{100}\right)^2 \end{array}\]Taking square root on both sides:
\[\begin{array}{l} 1 + \frac{r}{100} = \frac{11}{10} \\ \Rightarrow \frac{r}{100} = \frac{11}{10} - 1 \\ \Rightarrow \frac{r}{100} = \frac{1}{10} \\ \Rightarrow r = \frac{100}{10} = 10 \end{array}\]Correct Option: (a) 10%
📘 16. Very short answer type questions (V.S.A.)
(B) True or FalseExplanation:
For the first year, compound interest and simple interest are equal. For any subsequent year (more than 1 year), compound interest is always greater than simple interest because interest is calculated on the accumulated principal (principal + previous interest).
Answer: False
Explanation:
This is the fundamental definition of compound interest. At fixed intervals (yearly, half-yearly, etc.), the accumulated interest is added back to the principal, and this new amount becomes the principal for the next time interval.
Answer: True
📘 16. Very short answer type questions (V.S.A.)
(C) Fill in the blanksExplanation:
For the first year, the principal remains the same for both simple and compound interest. Therefore, the interest earned in the first year is identical for both methods.
Answer: equal
Explanation:
When a quantity increases at a constant percentage or fixed rate over equal intervals of time, it is referred to as a uniform rate of growth (or increase).
Answer: uniform rate
Explanation:
When a quantity decreases at a constant percentage or fixed rate over equal intervals of time, it is referred to as a uniform rate of decrease (or decay/depreciation).
Answer: decrease
📘 17. Short answer (S.A.)
Compound InterestGiven:
Principal \( (P) \) = ₹ 400
Amount \( (A) \) = ₹ 441
Time \( (n) \) = 2 years
Let the rate of compound interest be \( r\% \) per annum.
We know the formula for amount in compound interest:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ 441 = 400 \left(1 + \frac{r}{100}\right)^2 \\ \Rightarrow \frac{441}{400} = \left(1 + \frac{r}{100}\right)^2 \\ \Rightarrow \left(\frac{21}{20}\right)^2 = \left(1 + \frac{r}{100}\right)^2 \end{array}\]Taking square root on both sides:
\[\begin{array}{l} 1 + \frac{r}{100} = \frac{21}{20} \\ \Rightarrow \frac{r}{100} = \frac{21}{20} - 1 \\ \Rightarrow \frac{r}{100} = \frac{21 - 20}{20} \\ \Rightarrow \frac{r}{100} = \frac{1}{20} \\ \Rightarrow r = \frac{100}{20} \\ \Rightarrow r = 5 \end{array}\]Therefore, the rate of compound interest is 5% per annum.
Let the principal be \( P \) and the rate of interest be \( r\% \) per annum.
According to the first condition, the sum doubles in \( n \) years. Therefore, Amount \( (A) = 2P \).
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ 2P = P \left(1 + \frac{r}{100}\right)^n \\ \Rightarrow 2 = \left(1 + \frac{r}{100}\right)^n \quad \dots \text{(i)} \end{array}\]Let the sum become four times (\( 4P \)) in \( t \) years.
\[\begin{array}{l} 4P = P \left(1 + \frac{r}{100}\right)^t \\ \Rightarrow 4 = \left(1 + \frac{r}{100}\right)^t \\ \Rightarrow 2^2 = \left(1 + \frac{r}{100}\right)^t \end{array}\]Substituting the value of 2 from equation (i):
\[\begin{array}{l} \left\{ \left(1 + \frac{r}{100}\right)^n \right\}^2 = \left(1 + \frac{r}{100}\right)^t \\ \Rightarrow \left(1 + \frac{r}{100}\right)^{2n} = \left(1 + \frac{r}{100}\right)^t \end{array}\]Comparing the powers, we get:
\[\begin{array}{l} t = 2n \end{array}\]Therefore, the sum will become four times in 2n years.
Note: Based on standard mathematical interpretation of such problems, "compound interest... becomes ₹ 615" implies the accumulated compound interest is ₹ 615.
Given:
Compound Interest \( (CI) \) = ₹ 615
Rate of interest \( (r) \) = 5% per annum
Time \( (n) \) = 2 years
Let the principal be \( P \).
Formula for Compound Interest:
\[\begin{array}{l} CI = P \left\{ \left(1 + \frac{r}{100}\right)^n - 1 \right\} \\ 615 = P \left\{ \left(1 + \frac{5}{100}\right)^2 - 1 \right\} \\ 615 = P \left\{ \left(\frac{105}{100}\right)^2 - 1 \right\} \\ 615 = P \left\{ \left(\frac{21}{20}\right)^2 - 1 \right\} \\ 615 = P \left\{ \frac{441}{400} - 1 \right\} \\ 615 = P \left( \frac{441 - 400}{400} \right) \\ 615 = P \times \frac{41}{400} \\ \Rightarrow P = \frac{615 \times 400}{41} \\ \Rightarrow P = 15 \times 400 \\ \Rightarrow P = 6000 \end{array}\]Therefore, the required principal is ₹ 6000.
Let the price of the machine \( n \) years before (i.e., its initial price) be \( P \).
Rate of depreciation = \( r\% \) per annum.
Time = \( n \) years.
Price of the machine after \( n \) years \( (A) \) = ₹ \( v \).
According to the formula for uniform decrease:
\[\begin{array}{l} A = P \left(1 - \frac{r}{100}\right)^n \\ v = P \left(1 - \frac{r}{100}\right)^n \\ \Rightarrow P = \frac{v}{\left(1 - \frac{r}{100}\right)^n} \end{array}\]Therefore, the price of the machine \( n \) years before was ₹ \( \frac{v}{\left(1 - \frac{r}{100}\right)^n} \).
Let the population \( n \) years before (i.e., the initial population) be \( P_0 \).
Rate of increase = \( r\% \) per year.
Time = \( n \) years.
Population after \( n \) years \( (A) \) = \( p \).
According to the formula for uniform increase:
\[\begin{array}{l} A = P_0 \left(1 + \frac{r}{100}\right)^n \\ p = P_0 \left(1 + \frac{r}{100}\right)^n \\ \Rightarrow P_0 = \frac{p}{\left(1 + \frac{r}{100}\right)^n} \end{array}\]Therefore, the population that was \( n \) years before was \( \frac{p}{\left(1 + \frac{r}{100}\right)^n} \).
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