Test: Heights & Distances - GMAT MCQ

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m,  BAD = 30° ,  BCD = 45°


Distance between the two ships 
= AC = BA + BC 
=100√3+100
=100(√3+1)
= 100(1.73+1) = 100 × 2.73 = 273 m

One of AB, AD and CD must have given.

So, the data is inadequate.

Consider the diagram shown above where QR represents the tree and PQ represents its shadow

We have, QR = PQ
Let angleQPR = θ
i.e., required angle of elevation = 45°

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.

Then, ∠ADC = 30° , ∠ACB = 45°

Let AB = h, BC = x, CD = y



Given that distance y is covered in 8 minutes.
i.e, distance h(√3−1) is covered in 8 minutes.
Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).

Let distance h is covered in t minutes.

since distance is proportional to the time when the speed is constant, we have

Consider the diagram shown above. 

Let AB be the tower. Let C and D be the positions of the boat

Then, ACB = 45° , ADC = 30°, BC = 100 m

CD = (BD - BC) = It is given that the distance CD is covered in 10 seconds.
i.e., the distance 100(√3−1) is covered in 10 seconds.

Consider the diagram shown above. AC represents the tower and DE represents the pole.

Given that AC = 15 m, ADB = 30°, AEC = 60°

Let DE = h
Then, BC = DE = h,
AB = (15-h)    (∵ AC=15 and BC = h),
BD = CE

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m

We have DAC = 30°, DBC = 45°

Let DC = h