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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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24% (02:27) correct
75% (02:07) wrong based on 62 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 12412 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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09 Dec 2020, 07:48
09 Dec 2020, 07:48
4
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!
Re: To the nearest percent, by what percent did the population o
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01 Jul 2018, 10:32
01 Jul 2018, 10:32
2
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Expert Reply
The solution posted is partial, missing the last key sentence.
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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
