Average reading rate over equal-page segments at different speeds: A student reads 100 pages in the afternoon at 60 pages/h and then 100 pages in the evening at 40 pages/h. What is her average reading rate (pages per hour) over the 200 pages?

Introduction / Context:
For equal page counts read at different rates, average rate is total pages divided by total time. It mirrors the harmonic-mean style used in speed problems when distances (pages) are equal.

Given Data / Assumptions:

• Afternoon: 100 pages at 60 pages/h
• Evening: 100 pages at 40 pages/h

Concept / Approach:
Total time = time1 + time2. Average rate = total pages / total time.

Step-by-Step Solution:
Time1 = 100 / 60 = 5/3 h ≈ 1.6667 hTime2 = 100 / 40 = 2.5 hTotal pages = 200; Total time ≈ 4.1667 hAverage rate = 200 / 4.1667 ≈ 48 pages/h

Verification / Alternative check:
Harmonic-mean form for equal tasks: v_avg = 2 / (1/60 + 1/40) = 2 / (1/24) = 48, consistent.

Why Other Options Are Wrong:
50 and 60 are simple means; 70 is impossible given a 40 pages/h period; 44 is too low.

Common Pitfalls:
Averaging 60 and 40 directly (→ 50) rather than weighting by time, which is incorrect for equal page counts.

Final Answer:
48