Question

During one year, the population of a town increased by 10% but during the next year, it diminished by 10%.
If at the end of the second year the population was 99000 what was the population at the beginning.
(Answer :- 1,00,000)



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Answer 1

Answer:

Let the population of a town is 100

Then 5% increased population after first year =100+5=105

So, 5% decreased population after second year = = 105−( 1005×105)=105−5.25=99.75

But, the population after second year is 9975

Then, the population at the beginning of the year is

99.75100 ×9975= 10000

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Answer 2

Given:

• During one year, the population of a town increased by 10%
• but during the next year it diminished by 10%.
• If at the end of the second year the population was 99000

To Find:

• What was the population in the beginning

Solution:

◐ Here, we have said that the population increases 10% in the first year and decreased by 10% the next year

and that we have to find the Original population.

❒ Let the original population be X

According to the question:

• the population increases by 10 % in the first year

Now, let's find the increase in population:

$\longrightarrow \sf \: x + (x \times \cancel\frac{10}{100} ) \\ \\ \longrightarrow \sf \: x + \frac{x}{10} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \frac{10x + x}{10} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \orange{\sf{ \frac{11x}{10} } \bigstar} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bigstar \: { \blue{ \underline{ \frak{Hence, \: the \: population \: after \: 1 \: year \: was \: \frac{11x}{10} }}}}$

Now, let's find the population decreased in the second year:

$\longrightarrow \sf \: \frac{11x}{10} - ( \frac{11x}{10} \times \frac{10}{100} ) \\ \\ \\ \longrightarrow \sf \frac{11x}{10} - \frac{11x}{100} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \: \frac{110x - 11x}{100} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \orange {\sf{ \: \frac{99x}{100} \bigstar}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

According to condition 2

• We know, the population 2nd year was 99000

${ : \implies} \sf \: \frac{99x}{100} = 99000 \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf99x = 99000 \times 100 \\ \\ \\ { : \implies} \sf \: 99 x = 9900000 \: \: \: \: \\ \\ \\{ : \implies} \sf \: x = \cancel \frac{9900000}{99} \: \: \: \: \: \: \: \\ \\ \\{ : \implies} \orange{ \boxed{\bf \: x = 100000 \bigstar}} \: \:$

$\bigstar{ \blue{ \underline{ \frak{Hence, the \: population \: in \: the \: begeinng \: year \: was \: 100000}}}}$

More to know:

To Increase a Quantity

• by a percentage find the percentage of the quality and add in to the original quantity

To decrease a quantity

• by percentage in the similar way find the percentage but subtract it from the original quantity.

$\bigstar \sf \: percentage \: change = \Big( ( \frac{actual \: change}{original \: quantity} ) \times 100 \Big)\%$