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 - 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

Explanation

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
BC=7.5√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

Explanation

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)
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 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

A - 2(√3) m/sec

B - 3(√3-1) m/sec

C - 2(√3-1) m/sec

D - (√3-1) m/sec

Answer : C

Explanation

Height & Distance Solution 8

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 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 - Consider vertical shaft 6 m high has a shadow of length 2√3m, discover the angle of elevation of the sun?

A - 60°

B - 30°

C - 40°

D - 50°

Answer : A

Explanation

Height & Distance Solution 13

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:

A - 50√3m

B - 50/√3m

C - 23√3m

D - 100m/√3

Answer : B

Explanation

Height & Distance Solution 16

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 - The point of the height of a stepping stool inclining toward a divider is 60°and the step's foot is 7.5 m far from the divider. The stepping stool's length is

A - 15 m

B - 14.86 m

C - 15.64 m

D - 15.8 m

Answer : A

Explanation

Height & Distance Solution 19

Let AB be the step inclining toward the divider CB.
Let AC be the flat such that AC=7.5M
What's more, ∠CAB=60°
∴ AB/AC=sec60°=2 => AB/7.5m=2 => AB=15m.
∴ length of the stepping stool is 15m.

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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