9. The population of a town increased at the rate of 5% per year consecutively for 2 years. Then, due to an epidemic in the third year the population decreased at the rate of 1.5% for the next 3 years. What is the current population of that town, if it was 23,45,000, 5 years ago? abhi tak nahi pta chla
Answers
Step-by-step explanation:
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Solution
FORMULA TO BE IMPLEMENTED
Suppose present population = P
1. If population increased by r % per year. Then the population after n years will be
= \displaystyle \: P {(1 + \frac{r}{100} \: ) }^{n}=P(1+
100
r
)
n
2. If population decreased by r % per year. Then the population after n years will be
= \displaystyle \: P {(1 - \frac{r}{100} \: ) }^{n}=P(1−
100
r
)
n
EVALUATION
The population before 5 years ago = 2345000
Now for the first two years the population of the town increased at rate of 5 % per year consecutively.
So after 2 years population will be
= \displaystyle \: 2345000 \times {(1 + \frac{5}{100} \: ) }^{2}=2345000×(1+
100
5
)
2
= \displaystyle \: 2345000 \times {( \frac{105}{100} \: ) }^{2}=2345000×(
100
105
)
2
= \displaystyle \: 2345000 \times {( \frac{21}{20} \: ) }^{2}=2345000×(
20
21
)
2
= \displaystyle \: 2345000 \times \frac{441}{400}=2345000×
400
441
= 258536.25=258536.25
Again for the first next 3 years the population of the town decreased at rate of 1.5 % per year consecutively.
5 % per year consecutively. So after 3 years population will be
= \displaystyle \: 258536.25 \times {(1 - \frac{1.5}{100} \: ) }^{3}=258536.25×(1−
100
1.5
)
3
= \displaystyle \: 258536.25 \times {( \frac{98.5}{100} \: ) }^{3}=258536.25×(
100
98.5
)
3
= \displaystyle \: 258536.25 \times {(1 - \frac{1.5}{100} \: ) }^{3}=258536.25×(1−
100
1.5
)
3
= 247075.76=247075.76