For the past x laps around the track, Stevens average time per lap wa

Question

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

Sr1994 · Answer

\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

divyadna · Answer

51x+39=49(x+1)= 49x+49
2x=10
x=5
Ans option B

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phambrenda · Answer

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!

bumpbot · Answer

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For the past x laps around the track, Stevens average time per lap wa

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For the past x laps around the track, Stevens average time per lap wa

Prep Toolkit

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