Aptitude - Height & Distance Online Quiz


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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 - A man observes the elevation of a balloon to be 45° at a point A .He then walks towards the balloon and at a certain place B finds the elevation to be 60°. He further walks in the direction of the balloon and finds it to be directly over him at a height of 450 m. Distance travelled from A to B is

A - 300√3 m

B - 200√3 m

C - 100√3 m

D - 450√3 m

Answer : A

Explanation

Height & Distance Solution 2

450/BD= tan (60) =>BD =450/√3
450/AD= tan (30) =>AD= 450√3
AD =BD +AB
=>AB=AD-BD= 450√3-450/√3=(450x3-450)/√3=300√3m

Q 2 - When the sun's altitude changes from 45° to 60°, the length of the shadow of a tower decreases by 45m. What is the height of the tower?

A - (45√3)/(√3-1)

B - (45√3)/(√3+1)

C - (45+√3)/(√3-1)

D - (45-√3)/(√3-1)

Answer : A

Explanation

Height & Distance Solution 6

Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 45 m,  ABD = 45°,  ACD = 60°,
Let CD = x, AD = h
From the right   CDA, tan60=h/x
From the right   BDA, tan45=(45+x)/h=>h=45+x
=>h=45+h/√3
=>h(1-1/√3)=45
=>h=45/(1-1/√3)=(45√3)/(√3-1)

Q 3 - From the top of mast head of height 210 meters of a ship, a boat is observed at an angle of depression of 30° then the distance between them is

A - 210√3

B - 210/√3

C - 70√3

D - 105√3

Answer : A

Explanation

Height & Distance Solution 7

From the right angled triangle CAB
Tan(30) =210/X
=>X=210/Tan(30)=210/(1/√3)=210√3

Q 4 - A step inclining toward a vertical divider makes a point of 45 with the even ground. The step's Foot is 3m from the divider. Find Length of the step?

A - 4.23 m

B - 5.23 m

C - 6.23 m

D - 7.23 m

Answer : A

Explanation

Height & Distance Solution 11

Let AB be the step and BC be the divider and let AC be the even ground.
Then, ∠CAB=45 and AC=3m. Let AB= x Meter.
From right △ ACB, we have
AB/AC =sec. 45° = √2 => x/3 = √2
X= 3√2m = (3*1.41) m= 4.23m.
∴ Length of the stepping stool is 4.23 m

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

Explanation

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
=>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 - 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

Explanation

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

Explanation

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

Explanation

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.


aptitude_height_distance.htm

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