- Aptitude - Home
- Aptitude - Overview
- Quantitative Aptitude
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 - Consider vertical shaft 6 m high has a shadow of length 2√3m, discover the angle of elevation of the sun?
Answer : A
Explanation
Let AB be the building and AC be its shadow. Then, AB= 6m and AC= 2√3m. Let ∠ACB= θ Then tan θ = AB/AC= 6/2√3m =√3= >θ =60° Point of rise of the sun is 60°
Q 6 - A 10 m long stepping stool is put against a divider. It is slanted at a point of 30°to the ground. The separation of the stepping stool's foot from the divider is:
Answer : C
Explanation
Let AB be the step slanted at 30°to the Ground AC. Then, AB=10m and
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 - The point of dejection of two boats from the highest point of a beacon are 45°and 30°towards east. On the off chance that the boat are 200 m separated, the light's stature hours is :( Take √3=1.73)
Answer : D
Explanation
Let AB be the beacon and C and D be the positions the boats such that CD=200m. ∠ABC=45°and ∠ADB=30°. AB/AC =tan45°=1=>AB=AC=x m. Presently AB/AD =tan30°=1/√3 h/h+200=1 √3?√3h=h+200 => (√3-1) h=200 => h=200/√3-1)*(√3+1/ (√3+1) =100*(√3+1) =100(1.73+1) =273m.