# TANCET 2008 Quant Question 10 : Speed, Distance - Geometry

## Question

A plane flying north at 500 kmph passes over a city at 12 noon. A plane flying east at the same attitude passes over the same city at 12.30 pm. The plane is flying east at 400 kmph. To the nearest hundred km, how far apart are the two planes at 2 pm?

- 600 km
- 1000 km
- 1100 km
- 1200 km
- 1300 km

**Correct Answer:** 1200 km. Choice (4)

#### Explanatory Answers

The plane flying north would have traveled for 2 hours at 2 PM.

It travels at 500 kmph.

Hence, the plane flying north would have covered 1000 kms from the city at 2 PM.

The plane flying east would have traveled for 1.5 hours at 2 PM.

It travels at 400 kmph.

Hence, the plane flying east would have covered 1.5*400 = 600 kms from the city at 2 PM.

The two planes are flying at right angles to each other.

The shortest distance between the two planes is the length of the hypotenuse of the right triangle whose legs are the distances covered by the two planes.

Hence, distance =

= 1166 km

The nearest 100 km distance between the two planes = 1200 kms.

Level of difficulty : Moderate to difficult

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