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Bunuel
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For the past x laps around the track, Stevens average time per lap wa 17 Mar 2022, 06:45
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Difficulty:

15% (low)

Question Stats:

82% (01:28) correct 18% (02:05) wrong based on 49 sessions

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For the past x laps around the track, Steven’s average time per lap was 51 seconds. If a lap of 39 seconds would reduce his average time per lap to 49 seconds, what is the value of x?

(A) 2
(B) 5
(C) 6
(D) 10
(E) 12
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B
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Sr1994
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For the past x laps around the track, Stevens average time per lap wa 17 Mar 2022, 08:21
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\(\frac{t}{x}\) = 51

t = 51x

\(\frac{t+39}{x+1}\) = 49

51x+39 = 49x+49
51x-49x = 49-39
2x = 10
x = 5

Answer B
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For the past x laps around the track, Stevens average time per lap wa 17 Mar 2022, 08:32
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51x+39=49(x+1)= 49x+49
2x=10
x=5
Ans option B

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For the past x laps around the track, Stevens average time per lap wa 01 Jul 2022, 07:26
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You can approach this problem two ways

1) Algebra (examples shown above
2) or as a weighted average:

- so you have 51 (x laps) and then 39 for 1 lap so

39 ------ 51
1 lap -----x laps

and this gets you to a weighted average of 49. you can see that 39 is 10 from 49 and 51 is 2 from x

so the ratio is 1:5 and notice that 49 is closer to 51, so there's more laps for 51 than 39.

so the ratio is 5 <- bigger number / (1+5)

Hope this helps! I solved both ways and feel like the weighted average is a little bit quicker!
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