Ranking and Direction Test
A person starts from a point A and travels 3 km eastwards to B and then turns left and travels thrice that distance to reach C. he again turns left and travels five times the distance he covered between A and B and reaches his destination D. the shortest distance between the starting point and the destination is
|E||None of these|
Clearly, AB = 3 km; BC = 3AB = (3 × 3) km = 9 km; CD = 5 AB = (5 × 3) km = 15 km.
Draw AE _ CD. Then, CE = AB = 3 km and AE = BC = 9 km.
DE = (CD – CE) = (15 – 3) km = 12 km. In ∆AED, AD2 = AE2 + DE2 AD = √92 + 122 km = √225 km = 15 km. So, required distance = AD = 15 km.