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# For the past x laps around the track, Steven’s average time per lap

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Math Expert
Joined: 02 Sep 2009
Posts: 44287
For the past x laps around the track, Steven’s average time per lap [#permalink]

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14 Oct 2017, 06:35
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Difficulty:

25% (medium)

Question Stats:

76% (01:10) correct 24% (01:09) wrong based on 45 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
[Reveal] Spoiler: OA

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For the past x laps around the track, Steven’s average time per lap [#permalink]

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14 Oct 2017, 06:41
Bunuel wrote:
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

Total time for $$x$$ laps $$= 51*x$$

total time after additional lap of $$39$$ sec $$= 49*(x+1)$$

therefore $$51x+39=49(1+x)$$

or $$x=5$$

Option B
VP
Joined: 22 May 2016
Posts: 1416
For the past x laps around the track, Steven’s average time per lap [#permalink]

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14 Oct 2017, 06:54
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Bunuel wrote:
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

A* n = S

First x laps: 51 * x
Extra lap : 39 * 1
Sum: 51x + 39
Total number n of laps: x + 1

A = S/n, where A = 49

$$\frac{(51x+39)}{(x+1)} = 49$$

$$(51x+39)= 49(x+1)$$

$$51x+39= 49x+49$$

$$2x = 10$$

$$x = 5$$

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For the past x laps around the track, Steven’s average time per lap   [#permalink] 14 Oct 2017, 06:54
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