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

[BTGmoderatorDC]

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


OA B

Source: Princeton Review

[swerve]

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


OA B

Source: Princeton Review

\(A \cdot n = S\)

First \(x\) laps: \(51\cdot x\)
Extra lap: \(39 \cdot 1\)
Sum: \(51x + 39\)
Total number \(n\) of laps: \(x + 1\)

\(A = \dfrac{S}{n}\), where \(A = 49\)

\begin{align*}
\dfrac{51x+39}{x+1} &= 49 \\
51x+39 &= 49(x+1) \\
51x+39 &= 49x+49 \\
2x &= 10 \\
x &= 5
\end{align*}

Therefore, B