\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}

Question

UpStudy AI · Accepted Answer

To find Keenan's actual reading rate, we can follow these steps:

1. **Identify the known values:**
   - Keenan predicted his reading rate to be $$ P = 200 $$ words per minute.
   - His prediction was $$ 25\% $$ faster than his actual rate.

2. **Define the actual reading rate:**
   - Let $$ A $$ be Keenan's actual reading rate in words per minute.

3. **Set up the relationship:**
   - Since his prediction is $$ 25\% $$ faster than his actual rate, we can express this relationship mathematically:
   $$
   P = A + 0.25A
   $$
   This simplifies to:
   $$
   P = 1.25A
   $$

4. **Substitute the known value of $$ P $$:**
   $$
   200 = 1.25A
   $$

5. **Solve for $$ A $$:**
   To find $$ A $$, we can rearrange the equation:
   $$
   A = \frac{200}{1.25}
   $$

Now, let's calculate $$ A $$.

$$
A = \frac{200}{1.25} = 160
$$

Thus, Keenan's actual reading rate is $$ 160 $$ words per minute.

Final answer:
$$
\text{Keenan's actual reading rate is } 160 \text{ words per minute.}
$$
Keenan's actual reading rate is 160 words per minute.

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

faster than his actual rate. What was Keenan's actual reading rate?

words per minute

Submit

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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 faster than his actual rate. What was Keenan's actual reading rate? words per minute Submit