 # GMAT Quant: Time, Speed & Distance Problems

Here is a shortcut to memorize the formulas for Time, Speed and Distance. $Distance = Time \times Speed$ $Time = \frac{Distance}{Speed}$ $Speed = \frac{Distance}{Time}$ $Average \; Speed = \frac{Total \; Distance}{Total \; Time}$

## Shortcut Tips & Formulas to Solve Time, Speed & Distance Problems Faster

1. Km/hr to m/sec Conversion
$x \; Km/hr = \left ( x \times \frac{5}{18} \right ) m/sec$

1. m/sec to km/hr Conversion
$x \; m/sec = \left ( x \times \frac{18}{5} \right ) km/hr$

1. If the ratio of the speeds of A and B is a : b, then the ratio of the

the time taken by them to cover the same distance is –

$\frac{1}{a} : \frac{1}{b} \; or \; b:a$

1. If a train covers a certain distance x km/hr and an equal distance at y km/hr. Then,

Average Speed during the whole journey is $\left ( \frac{2xy}{x + y} \right ) \; km/hr$

Example 1 – A train traveling at 108 kmph crosses a platform in 30 seconds and a man standing on the platform in 20 seconds. What is the length of the platform in meters?

Extra time taken by the train to cross the platform = 30 – 20 = 10 sec

$Length \; of \; the \; platform = speed \; of \; train \times extra\; time \; taken \; to \; cross \; the \; platform.$ $Length \; of \; the \; platform = 108 \; km/hr \times 10 \; sec$ $Length \; of \; the \; platform = 108 \; km/hr \times 10 \; sec$Convert 108 km/hr into m/sec$108 \; km/ph = \frac{5}{18} \times 108 = 30 \; m/sec$

Convert 108 km/hr into m/sec$108 \; km/ph = \frac{5}{18} \times 108 = 30 \; m/sec$ $108 \; km/ph = \frac{5}{18} \times 108 = 30 \; m/sec$Therefore, length of the platform = $30 \; m/s \times 10 \; sec = 300 \; meters$

Therefore, length of the platform = $30 \; m/s \times 10 \; sec = 300 \; meters$

Example 2 – Karen travels the first 4 hours of his journey at 75 mph speed and the remaining 5 hours at 35 mph speed. What is the average speed of Jim’s travel in mph?

Total distance travelled and total time taken –

$Average \; Speed = \frac{Total \; Distance}{Total \; Time}$

Total distance traveled by Karen = Distance covered in the first 4 hours + Distance covered in the next 5 hours.

Distance covered in the first 4 hours $= 4 \times 75 = 300 \; miles$

Distance covered in the first 5 hours $= 5 \times 35 = 175 \; miles$

Therefore, total distance traveled = 300 + 175 = 475 miles.

Total time taken = 4 + 5 = 9 hours.

Therefore,

$Average \; Speed = \frac{Total \; Distance}{Total \; Time}$ $= \frac{475}{9}$ $= 52.77 \; m/h$

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