Two runners, P and Q, competed in a race of 400 m. P ran at

Race distance = 400m
P ran at a constant speed of 5 strides in every 2 seconds covering 10m in that span.
Speed of P = 10m/2secs = 5 m/sec
Time taken for P to finish the race = 400m / 5m/s = 80secs
So, given that the time taken by P=Q, then Q took 80secs yo complete the race as well.
After running for 1 minute, Q increased its speed by 1 stride per second
Let stride in 60 seconds = x
Strides in (80-60) seconds = x+1

Stride in 20 secs = x+1 stride per sec

Q covers 4m in every stride.
The total distance = [Initial speed and time] + [final speed and time]
$$400m=4\left[60x+20\left(x+1\right)\right]m$$
Divide through by 4, we have
$$100m=80x\ +\ 20m$$ $$80x=80m\ $$
$$x=1\ \frac{stride}{\sec}$$
Initial speed of Q = 1 stride per sec covering 4m in that span
$$=4ms^{-1}$$
Conversion;
4m = 4 * 1/1000 km
1sec = 1 * 1/3600 hr
$$\frac{4m}{s}\cdot\frac{3600s}{1hr}\cdot\frac{1km}{1000m}=\frac{14400km}{1000hr}=14.4kph$$

Option E is the correct answer.

Thanks

BTGmoderatorDC wrote:Two runners, P and Q, competed in a race of 400 m. P ran at a constant speed of 5 strides in every 2 seconds, covering 10 m in that span. Q started at a constant speed. After running for 1 minute, Q increased its speed by 1 stride per second and finally finished the race at the same time with P. What is Q’s initial speed in kph if Q covered 4 m in every stride?

A. 10 kph
B. 11.5 kph
C. 12.8 kph
D. 13.6 kph
E. 14.4 kph

OA E

Source: e-GMAT

Velocity of P = 10/2 = 5m/s
The total time taken by P = 400/5 = 80 s.

Now, out of 80 sec, Q has travelled at a constant speed of a fro 60 sec. Therefore, total distance travelled by Q in 60 Sec = 60a. In the remaining 20 seconds for every 1 second the velocity of Q increases by 4m/s.

400 = 60a + 20a + 80
a = 4m/sec = 14.4 kph