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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 - A vertical pole fixed to the ground is divided in the ratio 1:4 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 16 m away from the base of the pole, what is the height of the pole?
Answer : C
Explanation
Let CB be the pole and point D divides it such that BD : DC = 1 : 4 = X:4X Given that AB = 16 m Let the the two parts subtend equal angles at point A such that CAD = BAD = Θ =>tan Θ=X/16 =>X=16 tan ( Θ) ------ (1) =>tan( Θ+ Θ)=4X/16 =>16 tan (2 Θ)=4X =>16(2tan ( Θ))/(1-tan ( Θ)2)=4X ------ (2) From eqn 1 & 2 2X/(1-tan ( Θ)2)=4X (X=16tan Θ) 1/(1-(X/16)2)=2 1-(X/16)2=1/2=>162- X2=162/2=>X2=128 =>X=8√2 =>Height of pole BC = X+4X=5X=40√2
Q 3 - A straight tree is broken due to thunder storm. The broken part is bent in such a way that the peak touches the ground at an angle elevation of 45°. The distance of peak of tree (where it touches the root of the tree is 20 m. Then the height of the tree is
Answer : A
Explanation
Let the total length of the tree be X+Y meters From the figure tan 45=X/20 =>X=20 cos 45 = 20/Y =>Y=20/cos 45 =20√2 X+Y=20+20radic;2=20+2x10x1.414 =48.28 meters
Q 4 - The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.
Answer : B
Explanation
Let AB be the building and AC be its shadow. Then, AC=20m and ∠ACB=60°.Let AB= x m. Presently AB/AC=tan 60°=√3=>x/10=√3 =>x=10√3m= (10*1.732) m=17.32m. ∴ Height of the building is 17.32m.
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 - 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 - From a point on a scaffold over the waterway, the edge of dejection of the banks on inverse sides of the waterway is 30°and 45°respectively. In the event that the scaffold is at tallness of 2.5m from the banks, find the width of the Stream. (Take √3=1.732)
Answer : B
Explanation
Let and B be two point on the banks on inverse sides of the stream. Let P be a point on the scaffold at stature of 2.5m. Let PQ-AB. PQ=2.5m.∠BAP=30°and ∠ABP=45°. QB/PQ=cot45°=1 => QB/2.5=1 => QB=2.5m. AQ/PQ =cot30°=√3 => AQ/2.5= √3 => AQ= (2.5)√3m. Width of the stream =AB= (AQ+QB)=2.5(√3+1) 5/2(1.732+1) m=6.83m.