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To the nearest percent, by what percent did the population o
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22 May 2017, 01:05
22 May 2017, 01:05
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Question Stats:
27% (02:33) correct
72% (02:06) wrong based on 44 sessions
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#GREpracticequestion To the nearest percent, by what percent did the population of the United States increase from 1994 to 1998.jpg [ 57.93 KiB | Viewed 8660 times ]
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%
Re: To the nearest percent, by what percent did the population o
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24 Sep 2017, 20:52
24 Sep 2017, 20:52
1
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Re: To the nearest percent, by what percent did the population o
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25 Sep 2017, 02:38
25 Sep 2017, 02:38
Expert Reply
Please post your answer as text and your resoning. We solve together.
Do not post a screen.
Thank you
Do not post a screen.
Thank you
Re: To the nearest percent, by what percent did the population o
[#permalink]
01 Jul 2018, 10:05
01 Jul 2018, 10:05
Carcass wrote:
Please post your answer as text and your resoning. We solve together.
Do not post a screen.
Thank you
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?
Re: To the nearest percent, by what percent did the population o
[#permalink]
01 Jul 2018, 10:32
01 Jul 2018, 10:32
1
Expert Reply
The solution posted is partial, missing the last key sentence.
Ask for further assistance.
Regards
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 \(3.8%\).
Ask for further assistance.
Regards
Re: To the nearest percent, by what percent did the population o
[#permalink]
02 Jul 2018, 00:17
02 Jul 2018, 00:17
Carcass wrote:
The solution posted is partial, missing the last key sentence.
Ask for further assistance.
Regards
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 \(3.8%\).
Ask for further assistance.
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.
Re: To the nearest percent, by what percent did the population o
[#permalink]
02 Jul 2018, 15:25
02 Jul 2018, 15:25
Expert Reply
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.
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.
Re: To the nearest percent, by what percent did the population o
[#permalink]
07 Aug 2018, 00:34
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?
Re: To the nearest percent, by what percent did the population o
[#permalink]
07 Aug 2018, 01:05
07 Aug 2018, 01:05
Expert Reply
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
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
Re: To the nearest percent, by what percent did the population o
[#permalink]
09 Aug 2019, 19:22
09 Aug 2019, 19:22
carcass please the explain the question using other solution other than that is mentioned above
Re: To the nearest percent, by what percent did the population o
[#permalink]
10 Aug 2019, 11:54
10 Aug 2019, 11:54
Expert Reply
This is NOT a good question at a closer look.
The graph mentions of offenses in general and then the explanation talks about vehicles and cars which is not the same thing, strictly consider.
Practice with tons of questions on the board on the hard side yet OFFICIAL question. Do not waste your time here
Data Interpretation HARD questions
Regards
The graph mentions of offenses in general and then the explanation talks about vehicles and cars which is not the same thing, strictly consider.
Practice with tons of questions on the board on the hard side yet OFFICIAL question. Do not waste your time here
Data Interpretation HARD questions
Regards
Re: To the nearest percent, by what percent did the population o
[#permalink]
16 Aug 2019, 07:11
16 Aug 2019, 07:11
1
Is there missing any info ?
I think, enough data is not exist here.
I think, enough data is not exist here.
Re: To the nearest percent, by what percent did the population o
[#permalink]
09 Dec 2020, 07:48
09 Dec 2020, 07:48
3
Percentage of cases in 94 = (Number of cases in 94)/(Population in 94)*100 - (1)
Percentage of cases in 98 = (Number of cases in 98)/(Population in 98)*100 - (2)
Percentage of cases in 98 = (100-22.4)/100 * (Percentage of cases in 94) - (3)
Number of cases in 98 = (100-19.4)/100 * (Number of cases in 94) - (4)
Substituting (3) and (4) in (2)
77.6/100 * (Percentage of cases in 94) = 80.6/100 * (Number of cases in 94)/(Population in 98)*100 - (5)
Divide (5) by (1) to get the following
77.6 = 80.6 * (Population in 94)/(Population in 98)
(Population in 98)/(Population in 94) = 1.03866
Subtract 1 from both sides
(Population in 98)/(Population in 94)-1= 1.03866-1
[(Population in 98)-(Population in 94)]/(Population in 94) = 0.03866
So the percentage change is 0.03866*100 = 3.866% = approximately 4% (D)
It took me way too long to come up with the logic of how to solve this problem!
Percentage of cases in 98 = (Number of cases in 98)/(Population in 98)*100 - (2)
Percentage of cases in 98 = (100-22.4)/100 * (Percentage of cases in 94) - (3)
Number of cases in 98 = (100-19.4)/100 * (Number of cases in 94) - (4)
Substituting (3) and (4) in (2)
77.6/100 * (Percentage of cases in 94) = 80.6/100 * (Number of cases in 94)/(Population in 98)*100 - (5)
Divide (5) by (1) to get the following
77.6 = 80.6 * (Population in 94)/(Population in 98)
(Population in 98)/(Population in 94) = 1.03866
Subtract 1 from both sides
(Population in 98)/(Population in 94)-1= 1.03866-1
[(Population in 98)-(Population in 94)]/(Population in 94) = 0.03866
So the percentage change is 0.03866*100 = 3.866% = approximately 4% (D)
It took me way too long to come up with the logic of how to solve this problem!
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