Speed, Distance, and Time Problems

Let original speed be v and distance to breakdown be x. The equations derived from the two scenarios (30km total) lead to v = 10 km/hr and x = 10 km.

Total distance to cover = 2000 m + 500 m = 2500 m = 2.5 km. Speed = 36 km/hr = 10 m/s. Time = distance / speed = 2500 / 10 = 250 seconds. 250 seconds = 4 minutes and 10 seconds.

Letting the going speed be $v$, the returning speed is $v+10$. The time equation $150/v - 150/(v+10) = 2.5$ simplifies to $v^2 + 10v - 600 = 0$, yielding $v = 20$ km/hr going and 30 km/hr returning. Distractors either list incorrect speed combinations or physically impossible negative speeds.

$$speed=\frac{distance}{time}$$
$$time=\frac{distance}{speed}$$
$$time=\frac{300}{60}h$$
$$time=5 hr$$

Time taken to walk 17km=5hrs.

Time taken to walk $1km=\dfrac{5}{17}hrs.$

Time taken to walk $20.4km=\dfrac{5}{17} \times20.4$

$=6 hrs.$

Circumference = 2 * pi * r = 2 * 3.14 * 50 = 314 m. Total distance for 20 rounds = 20 * 314 = 6280 m = 6.28 km. Speed = 12 km/hr. Time = 6.28 / 12 hours = 0.5233 hours = 31.4 minutes. 32 minutes is the closest approximation.

Let the actual distance travelled be x km.

Then, $\dfrac{x}{10}=\dfrac{x+20}{14}$

$14x=10x+200$

$4x=200$

$x=50 km$

Distance covered by B in 6 sec $= \dfrac{(1000}{300}×6)=20 m$

$Speed=\dfrac{distance}{time}=\dfrac{150}{5}$
$=30 km/hr$