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Aptitude - Height & Distance Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Height & Distance. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 10 minutes for the angle of depression to change from 45° to 60°, how soon after this will the car reach the observation tower?
Answer : B
Explanation
Let AB be the tower and C and D be the two positions of the car. Then,from figure AB/AC=tan 60 =√3 => AB=√3AC AB/AD=tan 45=1 => AB=AD AB=AC+CD CD=AB-AC=√3AC - AC=AC (√3-1) CD = AC (√3-1) =>10 min AC=> ? AC/(AC(√3-1)) x 10=?=10/(√3-1)=13.66=13 min 20 sec(approx)
Q 2 - An aeroplane when 750 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 45° and 30&de; respectively. Approximately, how many meters higher is the one than the other?
Answer : C
Explanation
Let C and D be the position of the aeroplanes. Given that CB = 900 m,∠CAB = 60°,∠DAB = 45° From the right △ ABC, Tan45=CB/AB=>CB=AB From the right △ ADB, Tan30=DB/AB=>DB=ABtan30=CBx(1/√3)=750/√3 CB=CD+DB => Required height CD=CB-DB=750-750/√3=250(3- √3)
Q 3 - A man in a boat is rowing away from a cliff (180 meters high), take 90 seconds to change angle of elevation of the top of cliff from 30° to 45°.The speed of the boat is
Answer : C
Explanation
From right angled triangle ADB, Tan45=AB/AD =>AB=AD=180 From right angled triangle ACB, Tan 30=180/(CD+180) =>CD+180=180√3 =>CD=180(√3-1) Speed =Distance/Time=180(√3-1)/90=2(√3-1) m/sec
Q 4 - A flag staff of 10 meters height stands on a building of 50 meters height. An observer at a height of 60 meters subtends equal angles to the flag staff and the building. The distance of the observer from the top of the flag staff is
Answer : C
Explanation
From the figure tanθ=10/CB tan(2θ)=60/CB=(2tan(θ))/(1-tan(θ)2) =>60/CB=(2tan(θ))/(1-tan(θ)2)=(2(10/CB))/(1-(10/CB)2) =>3/1=1/(1-(10/CB)2) =>3x(1-(10/CB)sup>2)=1 3CB2-300=CB2 2CB2=300=>CB=√150=5√6
Q 5 - Two men are inverse sides of a tower. They gauge the edge of the rise of the highest point of the tower as 30° and 45° respectively. On the off chance that the tallness of the tower is 50 m, discover the separation between the two men. (Take √3=1.732)
Answer : B
Explanation
Let AB be the tower and let C and D be the two's positions men. At that point ∠ACB=30°,∠ADB= 45°and AB= 50 m AC/AB = Cot30°=√3 => AC/50 = √3 =>AC=50√3m AD/AB=cot 45°=1 => AD/50=1 => AD=50M. Separation between the two men =CD= (AC+AD) = (50√3+50) m=50(√3+1) =50(1.73+1)m=(50*2.73)m=136.5m.
Q 6 - The point of height of a tower from a separation 50 m from its foot is 30. The tower's tallness is:
Answer : B
Explanation
Let AB be the tower and AC be the even line such that AC=50 m and ∠ACB=30°. AB/AC=tan 30°=1/√3 =>x/50 = 1/√3 > x=50*1/√3m= 50/√3m. ∴ Height of the tower=50/√3m.
Q 7 - From The highest point of a bluff 90 m high, the edges of Misery of the top and base of a tower are seen to be 30° and 60°. What is the tower's tallness is?
Answer : C
Explanation
Let AB be the precipice and CD be the tower. Draw DE || CA. Then, ∠BDE=30°, ∠BCA=60°and AB= 90m. From right △CAB, we have CA/AB=cost60°=1/√3 => CA/90=1/√3 =>CA=(90*1/√3* √3/√3) =30 √3m. ∴ DE =CE=30/√3m. From right ?DEB, we have BE/DE= tan30°=1/√3 => BE/30 √3=1 √3 =>BE= (30 √3*1 √3) =30m. ∴ CD=AE= (AB-BE) = (90-30) m=60m. Hence, the tower's stature is 60m.
Q 8 - On the level plane, there is a vertical tower with a flagpole on its top. At a point 9m far from the tower, the edges of rise of the top and Base of the flagpole are 60°and 30°respectively.The flagpole's tallness is:
Answer : A
Explanation
Let AB be the tower and BC be the flag pole and let O be the point of observation. Then, A=9m, ∠AOB=30°and ∠AOC=60° AB/OA=tan30°=1 ∠ =>AB/9=1∠ =>AB=(9*1/ √3* √3/√3)= 3 √3m. AC/AO=tan60°=√3 =>AC/9= √3 =>AC= 9√3m. ∴BC= (AC-AB) = (9 √3-3 √3) m=6 √3m. ∴ Height of the flagpole is 6 √m.