Two friends, Ram and Shyam, start at the same point, at the same time. Ram travels straight north at a speed of 10km/hr, while Shyam travels straight east at twice the speed of Ram. After 15 minutes, Shyam messages Ram that he is just passing by a large telephone tower and after another 15 minutes Ram messages Shyam that he is just passing by an old banyan tree. After some more time has elapsed, Ram and Shyam stop. They stop at the same point of time. If the straight-line distance between Ram and Shyam now is 50 km, how far is Shyam from the banyan tree (in km)? (Assume that Ram and Shyam travel on a flat surface.)
2500=$$\left(5+10t\right)^2+\left(10+20t\right)^2$$ =>Â $$\left(1+2t\right)^2=20$$
$$AB^2=25+100\left(1+2t\right)^2$$ => AB= 45km.
Create a FREE account and get:
- All Quant Formulas and shortcuts PDF
- XAT previous papers with solutions PDF
- XAT Trial Classes for FREE
