The annual decrease in the population of a city was 10%and the present number of inhabitants is 1620. What was the population 2 year hence?

Question

Mudasir · Answer

Anyone solve my question

Elite · Answer

Well if the population is decreasing by 10% each year, and the time span is 2 years, you'd have to do$$2(1620\times0.10)$$ since it's going down by 10 percent (o 0.10), you'd multiply the base population by 0.10 2 times. You will end up with your answer which is 324 inhabitants.

Mudasir · Answer

@elite wrote:

Well if the population is decreasing by 10% each year, and the time span is 2 years, you'd have to do$$2(1620\times0.10)$$ since it's going down by 10 percent (o 0.10), you'd multiply the base population by 0.10 2 times. You will end up with your answer which is 324 inhabitants.

No right answer is 2000 sir. How about is that....

Elite · Answer

How can it be 2000 if the population is decreasing? Your base number of inhabitants was 1620, it can't be higher than that if it's decreasing.

Mudasir · Answer

@elite wrote:

How can it be 2000 if the population is decreasing? Your base number of inhabitants was 1620, it can't be higher than that if it's decreasing.

Elite · Answer

How did you get 2000? I genuinely want to see your method.

Mudasir · Answer

@elite wrote:

How did you get 2000? I genuinely want to see your method.

No sir I can't solve it please help me. Whats your right answer

Elite · Answer

It's saying 2 years hence, meaning 2 years in the future, unless it's asking for 2 years in the past. If it's 2 years in the future it'd be 324 like I said, otherwise, I'll try and solve it for 2 years in the past to see if it matches.

Mudasir · Answer

It's okay sir

Elite · Answer

Yeah I read it again, it's a mixup of words, it should've said henced. The answer is 1944 if it's 2 years in the past. Round that up and you get to 2000. https://www.omnicalculator.com/math/percentage-change On that website, plug in the initial value as 1620 and the change as 20%.

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