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To the nearest percent, by what percent did the population o

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To the nearest percent, by what percent did the population o [#permalink]  22 May 2017, 01:05
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Question Stats:

44% (01:46) correct 55% (02:15) wrong based on 9 sessions

To the nearest percent, by what percent did the population of the United States increase from 1994 to 1998?

A) 1%

B) 2%

C) 3%

D) 4%

E) 5%
[Reveal] Spoiler: OA

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Re: To the nearest percent, by what percent did the population o [#permalink]  24 Sep 2017, 20:52
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Moderator
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Kudos [?]: 983 [0], given: 4518

Re: To the nearest percent, by what percent did the population o [#permalink]  25 Sep 2017, 02:38
Expert's post

Do not post a screen.

Thank you
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Re: To the nearest percent, by what percent did the population o [#permalink]  01 Jul 2018, 10:05
Carcass wrote:

Do not post a screen.

Thank you

I agree Carcass, but why he assumed it was 1000 stolen vehicles? And why he assumed that the population must be increased?
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Re: To the nearest percent, by what percent did the population o [#permalink]  01 Jul 2018, 10:32
Expert's post
The solution posted is partial, missing the last key sentence.

Quote:
Simplify the situation by assuming that in 1994 the population was 100,000 and there were 1000 vehicles stolen.The number of thefts per 100,000 inhabitants decreased 22.4% from 1000 to 776. So if there were 776 vehicles stolen for every 100,000 inhabitants, and 806 cars were stolen, the number of inhabitants must have increased. To know by how much, solve the proportion: $$\frac{776}{100,000}$$ = $$\frac{806}{x}$$ Cross-multiplying, we get $$776x = 80,600,000$$. So, $$x = 103,800$$. Then for every 100,000 inhabitants in 1994, there were 103,800 in 1998, an increase of $$3.8%$$.

Regards
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Re: To the nearest percent, by what percent did the population o [#permalink]  02 Jul 2018, 00:17
Carcass wrote:
The solution posted is partial, missing the last key sentence.

Quote:
Simplify the situation by assuming that in 1994 the population was 100,000 and there were 1000 vehicles stolen. The number of thefts per 100,000 inhabitants decreased 22.4% from 1000 to 776. So if there were 776 vehicles stolen for every 100,000 inhabitants, and 806 cars were stolen, the number of inhabitants must have increased. To know by how much, solve the proportion: $$\frac{776}{100,000}$$ = $$\frac{806}{x}$$ Cross-multiplying, we get $$776x = 80,600,000$$. So, $$x = 103,800$$. Then for every 100,000 inhabitants in 1994, there were 103,800 in 1998, an increase of $$3.8%$$.

Regards

I have got the general paradigm for solving this question as you can assume the initial number of stolen cars to be any number from 100 to 1 billion, but you will always get the same ratio always. But the logic of combining everything altogether and assuming that it increased is still vague to me.
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Re: To the nearest percent, by what percent did the population o [#permalink]  02 Jul 2018, 15:25
Expert's post
I do not think honestly this is a good question.

Actually, you can perform the difference between 22.4 -19.4 = 3

However, considering that the population had a major increase during those years than the theft of the vehicle, the difference is not 3 but slightly more.

D is the answer. But is not a very clear question.
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Re: To the nearest percent, by what percent did the population o [#permalink]  07 Aug 2018, 00:34
Is the question incomplete? There are no numbers to being with. How are assuming how many cars were stolen? How do we know 806 cars were stolen? i am so lost. Is this from an actual test?
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Re: To the nearest percent, by what percent did the population o [#permalink]  07 Aug 2018, 01:05
Expert's post
You didnt read carefully the OE above.

It says that alike the previous question in the book you have to compute a % decrease.

Read now above, I have fixed it.

Hope now is more clear to you.

Regards
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