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# The population of City A for each of three years is shown in the table

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Joined: 02 Sep 2009
Posts: 57155
The population of City A for each of three years is shown in the table  [#permalink]

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03 Nov 2015, 13:46
00:00

Difficulty:

5% (low)

Question Stats:

88% (01:37) correct 13% (02:26) wrong based on 56 sessions

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The population of City A for each of three years is shown in the table above. If the population grew by m percent from 1970 to 1980 and n percent from 1980 to 1990, then m exceeds n by how much?

A. 37.5
B. 20
C. 17.5
D. 15
E. 12.5

Attachment:

2015-11-04_0045.png [ 9.6 KiB | Viewed 1870 times ]

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Re: The population of City A for each of three years is shown in the table  [#permalink]

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03 Nov 2015, 17:10
70 - 80
m = (55-40)/40 = 3/8

80 - 90
n = (66-55)/55 = 1/5

m - n
3/8 - 1/5 = 7/40 = 17,5%
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Re: The population of City A for each of three years is shown in the table  [#permalink]

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03 Nov 2015, 19:29
m = ((55000-40000)/40000) * 100
=(15000/40000)*100
=(3/8)*100
=37.5 %

n=((66000-55000)/55000)*100
= (11000/55000)*100
= (1/5)*100
= 20 %

m-n= 17.5

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Re: The population of City A for each of three years is shown in the table  [#permalink]

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05 Aug 2019, 17:23
Bunuel wrote:

The population of City A for each of three years is shown in the table above. If the population grew by m percent from 1970 to 1980 and n percent from 1980 to 1990, then m exceeds n by how much?

A. 37.5
B. 20
C. 17.5
D. 15
E. 12.5

Attachment:
2015-11-04_0045.png

The percent increase from 1970 to 1980 is:

(55,000 - 40,000)/40,000 x 100

15,000/40,000 x 100

15/40 x 100

37.5

So m = 37.5 percent

Likewise, the percent increase from 1980 to 1990 is:

(66,000 - 55,000)/55,000 x 100

11,000/55,000 x 100

11/55 x 100

20

So n = 20 percent.

Therefore, m - n = 17.5 percent.

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Re: The population of City A for each of three years is shown in the table   [#permalink] 05 Aug 2019, 17:23
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# The population of City A for each of three years is shown in the table

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