Leaving home at the same time, Amal reaches office at 10:15 am if he travels at 8 km/hr, and at 9:40 am if he travels at 15 km/hr. Leaving home at 9:10 am, at what speed, in km/hr, must he travel so as to reach office exactly at 10 am ?

Let the distance from home to office be d km, and the time he leaves home be t minutes after 9:00 am.

If he travels at 8 km/hr, he reaches at 10:15 am. The time taken is 75 - t minutes.
If he travels at 15 km/hr, he reaches at 9:40 am. The time taken is 40 - t minutes.

Since distance is constant, the ratio of speeds is inversely proportional to the ratio of times:
$\frac{8}{15}=\frac{40-t}{75-t}$
8(75 - t) = 15(40 - t)
600 - 8t = 600 - 15t
This indicates an error in setting up the variable representing time relative to a fixed start. Let's define the duration directly:
Let T be the time in hours taken to reach at 10:15 am (at 8 km/hr).
The difference in arrival times between 10:15 am and 9:40 am is 35 minutes, which is $\frac{35}{60}=\frac{\mathrm{7<!--\; 12</mfrac-->}}{}$ hours.

So, the time taken at 15 km/hr is $T-\frac{7}{12}$ hours.

Using Distance = Speed × Time:
$8T=15(T-\frac{7}{12})$
$8T=15T-\frac{35}{4}$
$7T=\frac{35}{4}\Rightarrow T=\frac{5}{4}\text{hours (75 minutes)}$

This means if he arrives at 10:15 am, he traveled for 75 minutes. Thus, he leaves home at 9:00 am.
The distance d = 8 × 1.25 = 10 km.

If he leaves at 9:10 am and wants to reach at 10:00 am, the time allowed is 50 minutes = $\frac{50}{60}=\frac{5}{6}$ hours.
Required speed = $\frac{\text{Distance}}{\text{Time}}=\frac{10}{5/6}=12\text{km/hr}$.