Arithmetic Chapter 2.1 Uniform Rate Of Increase And Decrease Solved Example Problems
In this chapter, we shall discuss the uniform rate of increase or decrease of some events or objects.
All these objects affect directly or indirectly different events in our daily life or in our society.
Increase or Growth
There are some objects or events such as, the population of any place, the height of trees, the weights of the children, etc, which increases uniformly as time is passed over.
In these cases, we say that there occurs some increment or growth in the respective Helds.
This increment or growth in unit time is known as rate of increase or rate of growth.
WBBSE Solutions for Class 10 Maths
Decrease or depreciation
The values of some objects like machine, value of house, etc diminishes as time passes. This phenomenon of diminishing the values of objects is known as depreciation and the depreciation in unit time is called the rate of decrease or depreciation.
Formulas regarding uniform increase and decrease
Let the present population of a city be P and the rate of uniform increase be R% then the population after n years
= \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)
Also, if the uniform rate of increase in 1st year be R1%, in 2nd year be R2%, in 3rd year be R3%, ………..in n-th year be Rn%, then the population of the city after n years.
= \(P\left(1+\frac{\mathrm{R}_1}{100}\right)\left(1+\frac{\mathrm{R}_2}{100}\right)\left(1+\frac{\mathrm{R}_3}{100}\right) \cdots \cdots \cdots \cdots \cdots\left(1+\frac{\mathrm{R}_n}{100}\right)\)
Similarly, if the present value of an object be ₹P and if the rate of decrease of its value be R% per annum, then the value of the object after n years
= \(\mathrm{P}\left(1-\frac{\mathrm{R}}{100}\right)^n\)
and depreciation = \(₹\left\{P-P\left(1-\frac{\mathrm{R}}{100}\right)^n\right\}\)
Arithmetic Chapter 2.1 Uniform Rate Of Increase And Decrease Multiple Choice Questions
Example 1. In the case of compound interest, the rate of compound interest per annum for every year is
- Equal
- Unequal
- Both equal or unequal
- None of the above
Solution: 2. Unequal
The rate of compound interest per annum for every year is Unequal
ExampIe 2. In the case of compound interest
- The principal remains the same in every year;
- The principal changes every year;
- Every year the principal may remain the same or may change;
- None of the above
Solution: 2. The principal changes every year.
In the case of compound interest The principal changes every year.
Example 3. If the present population of a village be P and the rate of uniform increase in population in every year be 2r%, then the population of the village after n years will be
1. \(\mathrm{P}\left(1+\frac{r}{100}\right)^n\)
2. \(\mathrm{P}\left(1+\frac{r}{50}\right)^n\)
3. \(\mathrm{P}\left(1+\frac{r}{100}\right)^{2 n}\)
4. \(\mathrm{P}\left(1-\frac{r}{100}\right)^n\)
Solution: 2. \(\mathrm{P}\left(1+\frac{r}{50}\right)^n\)
\(\left[because \mathrm{P}\left(1+\frac{2 r}{100}\right)^n=\mathrm{P}\left(1+\frac{r}{50}\right)^n \cdot\right]^{100}\)The population of the village after n years will be \(\mathrm{P}\left(1+\frac{r}{50}\right)^n\)
Example 4. The present value of a machine be ₹2P and every year the value of it decreases by 2r%. Then the value of the machine after 2n years is
1. \(\mathrm{P}\left(1-\frac{r}{100}\right)^n\)
2. \(=2 \mathrm{P}\left(1-\frac{r}{50}\right)^n\)
3. \(₹ \mathrm{P}\left(1-\frac{r}{50}\right)^{2 n}\)
4. \(₹ 2 \mathrm{P}\left(1-\frac{r}{50}\right)^{2 n}\)
Solution: 4. \(₹ 2 \mathrm{P}\left(1-\frac{r}{50}\right)^{2 n}\)
The value of the machine after 2n years is \(₹ 2 \mathrm{P}\left(1-\frac{r}{50}\right)^{2 n}\)
Example 5. If interest is compounded at the interval of 3 months, the amount of the principal ₹a at the rate of compound interest of b % per annum after c years is
1. \(₹ a\left(1-\frac{b}{400}\right)^{4 c}\)
2. \(₹ a\left(1+\frac{b}{100}\right)^{4 c}\)
3. \(₹ a\left(1+\frac{b}{100}\right)^c\)
4. \(₹ a\left(1+\frac{b}{100}\right)^c\)
Solution: 2. \(₹ a\left(1+\frac{b}{100}\right)^{4 c}\)
Arithmetic Chapter 2.1 Uniform Rate Of Increase And Decrease State Whether The Following Statement Is True Or False
“WBBSE Class 10 Uniform Rate of Increase examples”
Example 1. The compound interest of a certain sum of money for a certain period of time at a certain rate of compound interest per annum is less than the simple interest of the same sum of money for the same period of time and at the same rate of interest.
Solution: False
Example 2. In the case of compound interest the quantity of principal gradually increases.
Solution: True
Example 3. In the case of compound interest, the rate of interest changes every year.
Solution: False
Example 4. In the case of compound interest, the same quantity of interest is obtained in every year.
Solution: False
Example 5. In the case of compound interest the obtained interest is proportional to the period of time.
Solution: True
Arithmetic Chapter 2.1 Uniform Rate Of Increase And Decrease Fill In The Blanks
Example 1. If the increase of any object occurs at a certain rate with respect to a certain period of time, then it is called _______ increase.
Solution: Uniform
Example 2. If the decrease of any object occurs at a certain rate with respect to certain period of time, then it is called uniform ______
Solution: Decrease
Example 3. Compound interest = _______ – principal
Solution: Amount
Example 4. If the rate of compound interest decreases, then the quantity of compound interest also ______
Solution: Decreases
Example 5. For the 1st year, the simple and compound interest are the ______
Solution: Same
Arithmetic Chapter 2.1 Uniform Rate Of Increase And Decrease Short Answer Type Questions
“Solved problems on uniform rate of increase for Class 10”
Example 1. At a certain rate of compound interest, if a sum of money becomes double in n years, then find the period of time in which it will become 4 times.
Solution:
Given:
At a certain rate of compound interest, if a sum of money becomes double in n years
Let ₹x becomes double in nyears at the rate of compound interest r% per annum.
∴ \(x\left(1+\frac{r}{100}\right)^n=2 x\)
⇒ \(\left(1+\frac{r}{100}\right)^n=2 \) …….(1)
Also, let ₹ x become 4 times in t years at the same rate of interest r% per annum.
\(\begin{aligned}& x \times\left(1+\frac{r}{100}\right)^t=4 x \\
& \Rightarrow\left(1+\frac{r}{100}\right)^t=4 \\
& \Rightarrow\left(1+\frac{r}{100}\right)^t=2^2 \\
& \Rightarrow\left(1+\frac{r}{100}\right)^{t^*}=\left\{\left(1+\frac{r}{100}\right)^n\right\}^2 \quad\left[because \text { by }(1), 2=\left(1+\frac{r}{100}\right)^n\right] \\
& \Rightarrow\left(1+\frac{r}{100}\right)^t=\left(1+\frac{r}{100}\right)^{2 n} \quad
\end{aligned}\)
Hence, in 2n years the given principal becomes 4 times.
Example 2. The value of a machine being decreased in n years at the rate of r% per year becomes ₹V. Find the value of the machine before n years
Solution:
Given:
The value of a machine being decreased in n years at the rate of r% per year becomes ₹V.
Let the value of the machine before n years was ₹x
As per question, \(x\left(1-\frac{r}{100}\right)^n=\mathrm{V}\)
⇒ \(x=\frac{\mathrm{V}}{\left(1-\frac{r}{100}\right)^n}=\mathrm{V}\left(1-\frac{r}{100}\right)^{-n}\)
Hence, before n yeras, the value of the machine was ₹\(\mathrm{V}\left(1-\frac{r}{100}\right)^{-n}\)
Example 3. At the end of 2nd and 3rd year, the compound interest of a sum of money is ₹880 and ₹968 respectively. What will be the rate of interest per annum?
Solution:
Given:
At the end of 2nd and 3rd year, the compound interest of a sum of money is ₹880 and ₹968 respectively.
Here the interest of ₹880 in 1 year = ₹(968 – 880) = ₹88.
Let the rate of interest be r% per annum.
∴ \(880\left(1+\frac{r}{100}\right)-880\)=88
or, \(880\left(1+\frac{r}{100}-1\right)\)=88
or, \(880 \times \frac{r}{100}\)=88
or, \(r=\frac{88 \times 100}{880}\)=10
Hence, the rate of interest is 10% per annum.
“Chapter 2.1 Uniform Rate of Decrease solutions WBBSE”
Example 4. What is the equivalent rate of compound interest per annum if the half-yearly compound interest be 10%?
Solution: Let the principal be ₹x.
∴ The amount of ₹x in 1 year at the rate of compound interest 10% compounded half-yearly
= \(₹ x \times\left(1+\frac{\frac{10}{2}}{100}\right)^{2 \times 1}\)
= \(₹ x \times\left(1+\frac{1}{20}\right)^2\)
= \(₹ x \times\left(\frac{21}{20}\right)^2\)
=\(₹ \frac{441 x}{440}\)
Let the equivalent rate be r%.
∴ \(x\left(1+\frac{r}{100}\right)^1=\frac{441 x}{400}\)
⇒ \(\frac{r}{100}=\frac{441}{400}-1\)
⇒ \(\frac{r}{100}=\frac{441-400}{400}\)
⇒ \(\frac{r}{100}=\frac{41}{400}\)
Hence, the required equivalent rate = 10.25% per annum.
Arithmetic Chapter 2.1 Uniform Rate Of Increase And Decrease Short Answer Type Questions Long Answer Type Questions
Example 1. The present population of the town is 16000. If the rate of increase of population be 5% per annum, then what will be the population of the town after 2 years?
Solution:
Given:
The present population of the town is 16000. If the rate of increase of population be 5% per annum
The present population = 16000
Rate of increase of population = 5% per annum.
Period of time = 2 years.
∴ The population of the town after 2 years
= \(16000 \times\left(1+\frac{5}{100}\right)^2\)
= \(16000 \times\left(1+\frac{1}{20}\right)^2\)
= \(16000 \times\left(\frac{21}{20}\right)^2\)
= \(16000 \times \frac{21 \times 21}{20 \times 20}\)
= 17640.
Hence, the population of the town after 2 years = 17640.
Example 2. The rate of increase of the population of a state is 2% per annum. If the present population of a state be 80000000, then what will be the population of the state after 3 years?
Solution:
Given:
The rate of increase of the population of a state is 2% per annum. If the present population of a state be 80000000
The present population of the state = 80000000.
The rate of increase of population = 2% per annum.
Time = 3 years.
∴ After 3 years the population of that state will be
= \(80000000 \times\left(1+\frac{2}{100}\right)^3\)
= \(80000000 \times\left(1+\frac{1}{50}\right)^3\)
= \(80000000 \times\left(\frac{51}{50}\right)^3\)
= \(80000000 \times \frac{51 \times 51 \times 51}{50 \times 50 \times 50}\)
Hence, the required population will be 84896640.
“Class 10 Maths exercises on uniform increase and decrease”
Example 3. The value of a machine of factory decreases by 10% per year. If the present value of the machine be ₹100000, then what will be its value of it after 3 years?
Solution:
Given:
The value of a machine of factory decreases by 10% per year. If the present value of the machine be ₹100000
The present value of the machine = ₹100000.
The rate of decrease = 10 % per year
Time = 3 years.
∴ The value of machine will b after 3 years
= \(₹ 100000\left(1-\frac{10}{100}\right)^3\)
= \(₹ 100000\left(1-\frac{1}{10}\right)^3=₹ 100000 \times\left(\frac{9}{10}\right)^3\)
= \(₹ 100000 \times \frac{9 \times 9 \times 9}{10 \times 10 \times 10}\)
= ₹72900
The value of machine will b after 3 years = ₹72900
Hence, the value of the machine after 3 years will be ₹72900.
Example 4. As a result of Sarba Siksha Abhiyan, the students leaving the school before completion, are re-admitted, so the number of students in a year is increased by 5% in comparison to the previous year. If, the number of such re-admitted students in a district be 3528 in the present year. Find the number of students re-admitted 2 years before in this manner.
Solution:
Given:
As a result of Sarba Siksha Abhiyan, the students leaving the school before completion, are re-admitted, so the number of students in a year is increased by 5% in comparison to the previous year. If, the number of such re-admitted students in a district be 3528 in the present year.
Let the number of re-admitted students be x.
Rate of increase of re-admission = 5% per annum.
∴ At the present year the number of students re-admitted
= \(x \times\left(1+\frac{5}{100}\right)^2=x \times\left(1+\frac{1}{20}\right)^2\)
= \(x \times\left(\frac{21}{20}\right)^2=\frac{441 x}{400}\)
As per the question, \(\frac{441x}{400}\) = 3528
or, \(x=\frac{3528 \times 400}{441}\)
or, x = 3200
Hence, the required number of students re-admitted before 2 years = 3200.
“WBBSE Class 10 Maths solved examples for uniform rates”
Example 5. The present population of a town is 576000. If the rate of increase of population be 6 \(\frac{2}{3}\) % per annum, then what was the population of the town before 2 years?
Solution:
Given:
The present population of a town is 576000. If the rate of increase of population be 6 \(\frac{2}{3}\) % per annum,
Let the population before 2 years was x.
The rate of increase of population =6 \(\frac{2}{3}\) % per annum. = \(\frac{20}{3}\) % per annum.
∴ The population of the town in present year
= \(x \times\left(1+\frac{\frac{20}{3}}{100}\right)^2\)
= \(x \times\left(1+\frac{1}{15}\right)^2 \)
= \(x \times\left(\frac{16}{15}\right)^2\)
= \(x \times \frac{16 \times 16}{15 \times 15}\)
As per question \(x \times \frac{16 \times 16}{15 \times 15}\) = 576000
⇒ \(x=\frac{576000 \times 15 \times 15}{16 \times 16}\)
= 506250
The population of the town in present year = 506250
Hence, the population of the town before 2 years was 506250.
Example 6. Through the publicity of the road-safety program, street accidents in the Purulia district are decreased by 10% in comparison to the previous year. If the number of street accidents in this year be 8748, then find the number of street accidents 3 years before in the district.
Solution:
Given:
Through the publicity of the road-safety program, street accidents in the Purulia district are decreased by 10% in comparison to the previous year. If the number of street accidents in this year be 8748
Let before 3 years the number of road accidents was x.
Rate of decrease of accidents = 10% per annum.
∴ The number of road-accidents in the present year
=\(x \times\left(1-\frac{10}{100}\right)^3\)
= \(x \times\left(1-\frac{1}{10}\right)^3\)
= \(x \times\left(\frac{9}{10}\right)^3\)
As per the question,
= \(x \times\left(\frac{9}{10}\right)^3=8748\)
⇒ \(x \times \frac{9 \times 9 \times 9}{10 \times 10 \times 10}=8748\)
⇒ \(x=\frac{8748 \times 10 \times 10 \times 10}{9 \times 9 \times 9}\)
⇒ x = 12000
Hence, the heights of the tree was 20 meters before 2 years.
“Understanding uniform rate of increase in Class 10 Maths”
Example 7. The height of a tree increases by 20% per annum. If the present height of the tree be 28.8 metres, then what was the height of the tree before 2 years?
Solution:
Given:
The height of a tree increases by 20% per annum. If the present height of the tree be 28.8 metres
Let the height of the tree was x metres before 2 years.
The rate of increase of height = 20% per annum.
∴ The present height of the tree
= \(x\left(1+\frac{20}{100}\right)^2\) metres.
= \(x \times\left(\frac{6}{5}\right)^2 \text { metres }=\frac{36 x}{25} \text { metres. }\)
As per question, \(\frac{36x}{25}\) = 28.8
or, \(x=\frac{28 \cdot 8 \times 25}{36}\)=20
Hence, the height of the tree was 20 metres before 2 years
Example 8. The weight of Sovanbabu is 80 kg. In order to reduce his weight, he started a regular morning walk. He decided to reduce his weight every year by 10%. Calculate the weight of Sovanbabu after 3 years.
Solution:
Given:
The weight of Sovanbabu is 80 kg. In order to reduce his weight, he started a regular morning walk. He decided to reduce his weight every year by 10%.
The present weight of Sovanbabu = 80 kg.
Rate of decrease in weight per year = 10%.
∴ The weight of Sovanbabu will be after 3 years
= \(80 \times\left(1-\frac{10}{100}\right)^3 \mathrm{~kg}\)
= \(80 \times\left(\frac{9}{10}\right)^3 \mathrm{~kg}\)
= \(80 \times \frac{9 \times 9 \times 9}{10 \times 10 \times 10} \mathrm{~kg}\)
=58.32 kg.
Hence, after 3 years the required weight of Savanbabu will be 58-32 kg.
Example 9. At present, the sum of the number of students in all M. S. K. in a district is 3993. If the number of students increased in a year was 10% of its previous year, calculate the sum of the number of students 3 years before in all the M. S. K. in the district.
Solution:
Given:
At present, the sum of the number of students in all M. S. K. in a district is 3993. If the number of students increased in a year was 10% of its previous year
Let the number of students before 3 years was x.
The rate of increase of students per year = 10%.
∴ The number of students in the present year
= \(x \times\left(1+\frac{10}{100}\right)^3\)
= \(x \times\left(1+\frac{1}{10}\right)^3\)
= \(x \times\left(\frac{11}{10}\right)^3\)
As per question, \(x \times\left(\frac{11}{10}\right)^3\) = 3993
or, \(x=\frac{3993 \times 10 \times 10 \times 10}{11 \times 11 \times 11}\)
or, x = 3000
Hence, the required number of students was 3000 before three years.
“Step-by-step solutions for uniform rate problems Class 10”
Example 10. The present value of a machine of a factory is ₹180000. If the value of the machine decreases at the rate of 10% per annum, then what will be the value of the machine after 3 years?
Solution:
Given:
The present value of a machine of a factory is ₹180000. If the value of the machine decreases at the rate of 10% per annum
The present value of the machine = ₹180000.
Rate of decrease of the value = 10% per annum.
∴ The value og the machine will be after 3 years
= \(₹ 180000 \times\left(1-\frac{1}{10}\right)^3\)
= \(₹ 180000\left(\frac{10-1}{10}\right)^3\)
= \(₹ 180000 \times \frac{9 \times 9 \times 9}{10 \times 10 \times 10}\) =₹ 131220 .
Hence, the required value of the machine will be ₹131220 after 3 years.
Example 11. For the families having no electricity in their houses, a Panchayat samity of village Bakultala accepted a plan to offer electric connections. 1200 families in this village have no electric connection in their houses. In comparison to the previous year, it is possible to arrange electricity every year for 75% of the families having no electricity. Find the number of families without electricity after 2 years.
Solution:
Given:
For the families having no electricity in their houses, a Panchayat samity of village Bakultala accepted a plan to offer electric connections. 1200 families in this village have no electric connection in their houses. In comparison to the previous year, it is possible to arrange electricity every year for 75% of the families having no electricity.
At present the no. of families having no electric connections = 1200.
Every year, the electric connection increases = 75%.
∴ Every year the no.of families having no electric connections decreases = 100 % – 75% = 25%.
∴ After 2 years the number of families having no electric connections will be
\(1200\times\left(1-\frac{25}{100}\right)^2\)= \(1200 \times\left(1-\frac{1}{4}\right)^2\)
= \(1200 \times\left(\frac{3}{4}\right)^2\)
= \(1200 \times \frac{3 \times 3}{4 \times 4}\)=675
Hence, the number of families without electricity after 2 years will be 675.
Example 12. As a result of continuous publicity on harmful reactions in the use of cold drinks filled bottles, the number of users of cold drinks is decreased by 25% every year in comparison to the previous year. 3 years before number of users of cold drinks in a town was 80000. Find the number of users of cold drinks in the present year.
Solution:
Given:
As a result of continuous publicity on harmful reactions in the use of cold drinks filled bottles, the number of users of cold drinks is decreased by 25% every year in comparison to the previous year. 3 years before number of users of cold drinks in a town was 80000.
The number of users of cold drinks before 3 years is 80000.
Every year this number decreases by 25%
∴ At present this number will be
\(80000 \times\left(1-\frac{25}{100}\right)^3\)= \(80000 \times\left(1-\frac{1}{4}\right)^3\)
= \(80000 \times\left(\frac{3}{4}\right)^3\)
= \(80000 \times \frac{3 \times 3 \times 3}{4 \times 4 \times 4}\)
= 33750
Hence, the required number of users of cold drinks in the present year is 33750.
“WBBSE Maths Chapter 2.1 solved exercises”
Example 13. As a result of publicity on smoking, the number of smoker is decreased by 6 \(\frac{1}{4}\) % every year in comparison to the previous year. If the number of smokers at present in a city is 33750, find the number of smokers in that city 3 years before.
Solution:
Given:
As a result of publicity on smoking, the number of smoker is decreased by 6 \(\frac{1}{4}\) % every year in comparison to the previous year. If the number of smokers at present in a city is 33750,
At present the number of smokers in the city = 33750.
Every year the number of smokers decreases by 6\(\frac{1}{4}\)% = \(\frac{25}{4}\) %
Let the number of smokers before 3 years was x.
∴ At present the number of smokers
\(x \times\left(1-\frac{25}{4 \times 100}\right)^3\)=\( x \times\left(1-\frac{1}{16}\right)^3\)
= \( x \times\left(\frac{15}{16}\right)^3\)
As per question,
\(x \times\left(\frac{15}{16}\right)^3=33750\)⇒ \(x \times \frac{15 \times 15 \times 15}{16 \times 16 \times 16}=33750 \)
⇒ \(x=\frac{33750 \times 16 \times 16 \times 16}{15 \times 15 \times 15}\)
⇒ x = 40960
Hence, the number smokers before 3 years was 40960.
Example 14. The population of a village is 20000. If the rate of birth be 4% per annum and the rate of death be 2% per annum, then find the population of the village after 2 years.
Solution:
Given:
The population of a village is 20000. If the rate of birth be 4% per annum and the rate of death be 2% per annum
Rate of birth = 4% per annum.
Rate of death = 2% per annum.
∴ Rate of growth of population = (4% – 2%) per annum
= 2% per annum
The present population of the village is 20000.
∴ The population of the village after 2 years will be
\(20000 \times\left(1+\frac{2}{100}\right)^2\)= \(20000 \times\left(1+\frac{1}{50}\right)^2\)
= \(20000 \times\left(\frac{51}{50}\right)^2\)
= \(20000 \times \frac{51 \times 51}{50 \times 50}\)
Hence, the required population of the village after 2 years will be 20808.
“Uniform rate of increase and decrease practice problems WBBSE”
Example 15. The population of a town was 160000 before 3 years. If the rate of growth of population in these three years be 3%, 2.5% and 5% respectively, then find the population of the town at present.
Solution:
Given:
The population of a town was 160000 before 3 years. If the rate of growth of population in these three years be 3%, 2.5% and 5% respectively
The population of the town was 160000 before 3 years.
∴ before 2 years, the population of the town was
\(160000 \times\left(1+\frac{3}{100}\right)^1\)= \(160000 \times \frac{103}{100}\)=164800
Also, before 1 year, the population of the town was
= \(164800 \times\left(\frac{102 \cdot 5}{100}\right)^1\) = 168920
At last, the population in the present year
= \(168920 \times\left(1+\frac{5}{100}\right)^1\)
= \(168920 \times\left(1+\frac{1}{20}\right)\)
=\(168920 \times \frac{21}{20}\)=177366 .
Hence, the population of the town at present is 177366.
Example 16. The present value of a car is ₹360000. If the value of the car decreases at the rate of 10% in the first year and in the next every year the rate of decrease is 20% per year. Then find the value of the car after 3 years.
Solution:
Given:
The present value of a car is ₹360000. If the value of the car decreases at the rate of 10% in the first year and in the next every year the rate of decrease is 20% per year.
The present value of the car = ₹360000.
The value of the car will be after one year
= \(₹ 360000 \times\left(1-\frac{10}{100}\right)^1\)
= \(₹ 360000 \times\left(1-\frac{1}{10}\right)\)
= \(₹ 360000 \times \frac{9}{10}\) = ₹324000
Hence, the value of the car will be ₹207360 after 3 years.
Example 17. This year the number of blood donners in a hospital is 24000. If the number of blood donners increases by 5% in every 6 months, then in how much time will the number of blood donners be 27783?
Solution:
Given:
This year the number of blood donners in a hospital is 24000. If the number of blood donners increases by 5% in every 6 months
This year the number of blood donners is 24000.
The number of blood donners increases by 5%in every 6 months; i.e., 10% in every year.
Let the required time be t years to reach the number of blood donners from 24000 to 27783.
As per question,
\(24000 \times\left(1+\frac{\frac{10}{2}}{100}\right)^{2 t}\)=27783
⇒ \(24000 \times\left(1+\frac{1}{20}\right)^{2 t}\)=27783
⇒ \(\left(\frac{20+1}{20}\right)^{2 t}\)
= \(\frac{27783}{24000}\)
⇒ \(\left(\frac{21}{20}\right)^{2 t}=\frac{27783}{24000}=\frac{9261}{8000}=\left(\frac{21}{20}\right)^3\)
⇒ \(\left(\frac{21}{20}\right)^{2 t}=\left(\frac{21}{20}\right)^3\)
⇒ t=3
⇒ \(t=\frac{3}{2}=1 \frac{1}{2}\)
Hence, after 1\(\frac{1}{2}\) years the number of blood donners will be 27783.
Example 18. The number of bicycles produced in the year 2012 and in the year 2015 are 80000 and 92610 respectively. Then find the rate of increase of production in percentage per annum.
Solution:
Given:
The number of bicycles produced in the year 2012 and in the year 2015 are 80000 and 92610 respectively.
The number of bicycles produced in 2012 is 80000.
The number of bicycles produced in 2015 is 92610.
Let the rate of increase of production in every year be r% and the time is 3 years.
As per question, \(8000 \times\left(1+\frac{r}{100}\right)^3\)
\(\begin{aligned}& \left(1+\frac{r}{100}\right)^3=\frac{92610}{80000} \\
& \Rightarrow\left(1+\frac{r}{100}\right)^3=\left(\frac{21}{20}\right)^3 \\
& \Rightarrow 1+\frac{r}{100}=\frac{21}{20} \\
& \Rightarrow \frac{r}{100}=\frac{21}{20}-1 \\
& \Rightarrow \frac{r}{100}=\frac{21-20}{20} \\
& \Rightarrow \frac{r}{100}=\frac{1}{20} \Rightarrow r=5 .
\end{aligned}\)
Hence, the rate of increase of the production of bi-cycles is 5% per annum.