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# For the past x laps around a track, Stevens avergae time per

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CEO
Joined: 21 Jan 2007
Posts: 2694
Location: New York City
For the past x laps around a track, Stevens avergae time per  [#permalink]

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18 May 2007, 12:42
For the past x laps around a track, Stevens avergae time per lap was 51 seconds. If a lap of 39 seconds would reduce his avg time per lap to 49 seconds, what is the value of x?

Princeton Review uses the "plug the answer choices in and see which one fits" which i dont like. is there an algebraic wa to solve this quickly?

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VP
Joined: 08 Jun 2005
Posts: 1139

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18 May 2007, 12:47
1
[51x+39]/[x+1] = 49

[51x+39]/[x+1] = 49

51x+39 = 49x+49

2x=10

x=5

Director
Joined: 26 Feb 2006
Posts: 883

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18 May 2007, 13:34
1
bmwhype2 wrote:
For the past x laps around a track, Stevens avergae time per lap was 51 seconds. If a lap of 39 seconds would reduce his avg time per lap to 49 seconds, what is the value of x?

Princeton Review uses the "plug the answer choices in and see which one fits" which i dont like. is there an algebraic wa to solve this quickly?

total time = t

t/x = 51
t = 51x

(t+39)/(x+1) = 49
t = 49x+10

51x = 49x +10
x = 5
CEO
Joined: 21 Jan 2007
Posts: 2694
Location: New York City

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18 May 2007, 14:50
1
1
Himalayan wrote:
bmwhype2 wrote:
For the past x laps around a track, Stevens avergae time per lap was 51 seconds. If a lap of 39 seconds would reduce his avg time per lap to 49 seconds, what is the value of x?

Princeton Review uses the "plug the answer choices in and see which one fits" which i dont like. is there an algebraic wa to solve this quickly?

total time = t

t/x = 51
t = 51x

(t+39)/(x+1) = 49
t = 49x+10

51x = 49x +10
x = 5

thanks

sum/n = avg

let t= sum

t/x=51
t=51x

( t+39 ) / (x + 1) = 49
cross multiply. T + 39 = 49x + 49
T = 49x +10

51x = 49x + 10
2x=10
x=5
Non-Human User
Joined: 09 Sep 2013
Posts: 7326
Re: For the past x laps around a track, Stevens avergae time per  [#permalink]

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14 Oct 2017, 05:26
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