\begin{tabular}{l} On the first day of Keenan's speed-reading class, the students tested their current reading \\ rate. Keenan predicted that his rate would be 200 words per minute, but his prediction was \\ \( 25 \% \) faster than his actual rate. What was Keenan's actual reading rate? \\ \( \qquad \) words per minute \\ \hline Submit \end{tabular}
Solución de tutoría real
Responder
Solución
¡Inicia sesión para desbloquear respuestas gratis!
Una plataforma de aprendizaje en la que confían millones de estudiantes y profesores reales.
Revisado y aprobado por el equipo de tutoría de UpStudy
Extra Insights
Keenan predicted his reading rate to be 200 words per minute, which he thought was \( 25\% \) faster than his actual rate. To find his actual rate, we can represent it as \( x \). According to the problem, we can set up the equation: \[ x + 0.25x = 200 \] This simplifies to: \[ 1.25x = 200 \] Now, dividing both sides by \( 1.25 \): \[ x = \frac{200}{1.25} \] \[ x = 160 \] So, Keenan's actual reading rate is \( 160 \) words per minute.