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Of the families in City X in 1994, 40 percent owned a person

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Of the families in City X in 1994, 40 percent owned a person [#permalink] New post 09 Jun 2007, 19:33
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74% (02:32) correct 26% (02:10) wrong based on 245 sessions
Of the families in City X in 1994, 40 percent owned a personal computer. The number of families in City X owning a computer in 1998 was 30 percent greater than it was in 1994, and the total number of families in City X was 4 percent greater in 1998 than it was in 1994. what percent of the families in City X owned a personal computer in 1998?

A. 50%
B. 52%
C. 56%
D. 70%
E. 74%
[Reveal] Spoiler: OA

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Last edited by Bunuel on 31 Jul 2013, 10:05, edited 2 times in total.
Added the OA.
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 [#permalink] New post 09 Jun 2007, 19:46
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The answer is A - 50%

Say a 100 families existed in 1994 then the number of families owning a computer in 1994 - 40
Number of families owning computer in 1998 = 40 * 130/100 = 52
Number of families in 1998 = 104

The percentage = 52/104 * 100 = 50%.
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Re: Q37: Of the families in City X in 1994, 40 percent owned a [#permalink] New post 31 Jul 2013, 09:07
jet1445 wrote:
Q37:
Of the families in City X in 1994, 40 percent owned a personal computer. The number of families in City X owning a computer in 1998 was 30 percent greater than it was in 1994, and the total number of families in City X was 4 percent greater in 1998 than it was in 1994. What percent of the families in City X owned a personal computer in 1998?

A. 50%
B. 52%
C. 56%
D. 70%
E. 74%


1994 = 40 families (40% of total)
1998 = 40 F * 30% = 12+40 = 52 families.

Population in 1994 = 100; 1998 = 104

==> 52/104 = 1/2 = 50%
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Re: Of the families in City X in 1994, 40 percent owned a person [#permalink] New post 11 Oct 2014, 04:17
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Re: Of the families in City X in 1994, 40 percent owned a person [#permalink] New post 15 Oct 2014, 23:20
Year .................. Family .................. Computer

1994 .................. 100 (Assume) ............. 40

1998 .................... 104 ......................... \(40 + 40 * \frac{30}{100} = 52\)

Answer = A\(= \frac{52}{104} * 100 = 50%\)
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Of the families in City X in 1994, 40 percent owned a person [#permalink] New post 09 May 2015, 06:53
clear algebra

(1.3*0.4x/1.04x)*100=50%

A
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Re: Of the families in City X in 1994, 40 percent owned a person [#permalink] New post 31 May 2015, 11:30
Using Smart Numbers is a good technique for this problem.

After calculating the percentage increase of families owning PC is 52, when you see the population go up, the percent owned in 1998 has to inversely go down.

Only answer A is below 52.
Re: Of the families in City X in 1994, 40 percent owned a person   [#permalink] 31 May 2015, 11:30
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