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.

Questions and Answers

Q 1 - The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

A - 15√3

B - 7.5

C - 15√2

D - 7.5√2

Answer : D


Height & Distance Solution 1

Let AB be the wall and BC be the ladder.
Then, ∠ACB = 45° and AC = 7.5 m
AC/BC= Cos (45) =1/√2

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?

A - 8√2

B - 30√2

C - 40√2

D - 50√2

Answer : C


Height & Distance Solution 4

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)
=>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

A - 48.28 meters

B - 30.28 meters

C - 24.14 meters

D - 28.14meters

Answer : A


Height & Distance Solution 9

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.

A - 16.32m

B - 17.32 m

C - 18.32 m

D - 19.32m

Answer : B


Height & Distance Solution 12

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 - 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)

A - 135.5m

B - 136.5 m

C - 137.5 m

D - 138.5m

Answer : B


Height & Distance Solution 15

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
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)

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:

A - 7.32 m

B - 8.26 m

C - 8.66 m

D - 8.16 m

Answer : C


Height & Distance Solution 18

Let AB be the step slanted at 30°to the Ground AC.
Then, AB=10m and 

Q 7 - A kite is flying at a tallness of 75 m from the level of ground, joined to a string slanted at 60° to the level. The string's length is:

A - 50 √2 m

B - 50√3 m

C - 50m/√2

D - 50m/√3

Answer : B


Height & Distance Solution 20

Let AB be the kite and AC be the level ground
So that BC - AC.
At that point, ∠BAC=60°and BC=75m. Let AB=x meters.
Presently AB/BC=coses60°=2/ √3
=> x/75=2/√3 =>x=150/√3 =150* √3/3=50 √3m.
∴ Length of the string=50 √3m.

Q 8 - The stature of a tree is 10m. It is twisted by the wind in a manner that its top touches the ground and makes a point of 60 with the ground. At what range from base did the tree get twisted? (√3=1.73)

A - 4.6m

B - 4.8m

C - 5.2m

D - 5.4m

Answer : A


Height & Distance Solution 23

Let AB be the tree bowed at the point C so that part CB takes the position CD.
Then ,CD=CB. Let AC=x meters. At that point, CD = CB= (10-X) m and ∠ADC=60°.
AC/AC=sin60° => x/(10-x) = √3/2
=>2x=10 √3-√3x
=> (2+ √3) x= 10 √3
=>x=10 √3/ (2+ √3)*(2-√3)/(2-√3)=20 √3-30)m
= (20*1.73-30) m=4.6m
=> Required height=4.6m.

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