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 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?

A - 10 min 30 sec

B - 13 min 20 sec

C - 15 min 10 sec

D - 16 min 30 sec

Answer : B

Explanation

Height & Distance Solution 3

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 - 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 - 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

Explanation

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 - 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

A - 2√6

B - 3√6

C - 5√6

D - √6

Answer : C

Explanation

Height & Distance Solution 10

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 - 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 - 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 - 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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