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Maths Height and Distance Test.4

**Quiz Instructions:**

- There will be 20 multiple choice question in this online test.
- Answer of the questions will change randomly each time you start this test.
- Practice this test at least 3 times if you want to secure High Marks.
- At the End of the Test you can see your Test score and Rating.

1 / 20

**The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electric pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole ?**

2 / 20

**The angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree, is :________?**

Let AB be the tree and AC be its shadow.

Let ∠ACB = θ.

Then, AC/AB = √3

Cot θ = √3

θ = 30°

Let AB be the tree and AC be its shadow.

Let ∠ACB = θ.

Then, AC/AB = √3

Cot θ = √3

θ = 30°

3 / 20

**From a point P on a level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is :_________?**

Let AB be the tower. Then, ∠APB = 30° and AB = 100 m, AB/AP = tan 30° = 1/√3 AP = (AB X √3)= 100√3 m. = (100 X 1.73) m = 173 m.

Let AB be the tower. Then, ∠APB = 30° and AB = 100 m, AB/AP = tan 30° = 1/√3 AP = (AB X √3)= 100√3 m. = (100 X 1.73) m = 173 m.

4 / 20

**If the height of a pole is 2√3 metres and the length of its shadow is 2 metres, find the angle of elevation of the sun.**

Let AB be the pole and AC be its shadow. Let angle of elevation, ∠ACB = θ. Then, AB = 2√3m, AC = 2 m. tan θ = AB/AC = 2√3/2 = √3 θ = 60°. So,the angle of elevation is 60°

Let AB be the pole and AC be its shadow. Let angle of elevation, ∠ACB = θ. Then, AB = 2√3m, AC = 2 m. tan θ = AB/AC = 2√3/2 = √3 θ = 60°. So,the angle of elevation is 60°

5 / 20

**Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the two ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is :_________?**

Let AB be the lighthouse and C and D be the

positions of the ships. Then,

AB = 100 m, ∠ACB = 300 and ∠ADB = 45°.

AB/AC = tan 30° = 1/√3

AC = AB X √3 = 100√3 m.

AB/AD = tan 45° = 1 ⇒ AD = AB = 100 m.

CD = (AC + AD) = (100√3 + 100) m

= 100 (√3 +1) m = (100 X 2.73) m = 273 m.

Let AB be the lighthouse and C and D be the

positions of the ships. Then,

AB = 100 m, ∠ACB = 300 and ∠ADB = 45°.

AB/AC = tan 30° = 1/√3

AC = AB X √3 = 100√3 m.

AB/AD = tan 45° = 1 ⇒ AD = AB = 100 m.

CD = (AC + AD) = (100√3 + 100) m

= 100 (√3 +1) m = (100 X 2.73) m = 273 m.

6 / 20

**A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the joɡɡer?**

Explanation

With 5 best steps

**1.Data:**

Speed 1=V₁= 9km/h

Distance 1=S₁= 240 m

Distance 2 =S₁= 120 m

Speed 2=V₂= 45km/h

Relative speed(V) = V₂_V₁

Relative speed (V)= 45km/h_9 km/h

Relative speed (V)=36km/h

Relative speed(V)= 36×1000m/3600Sec

Relative speed (V)= 36000m/3600Sec

Relative speed(V)= 10 m/sec

Covered distance(S)= 240m+120m

Covered distance (S)= 360m.

**2. Required:**

Time taken= T=??**3. Formula:**

S= V×T

For T

T = S/V**4. Solution:**

Putting values in formula.

T= 360 m/10 m/Sec

T= 36Sec**5. Result:**

Time taken= T= 36 Sec

Explanation

With 5 best steps

**1.Data:**

Speed 1=V₁= 9km/h

Distance 1=S₁= 240 m

Distance 2 =S₁= 120 m

Speed 2=V₂= 45km/h

Relative speed(V) = V₂_V₁

Relative speed (V)= 45km/h_9 km/h

Relative speed (V)=36km/h

Relative speed(V)= 36×1000m/3600Sec

Relative speed (V)= 36000m/3600Sec

Relative speed(V)= 10 m/sec

Covered distance(S)= 240m+120m

Covered distance (S)= 360m.

**2. Required:**

Time taken= T=??**3. Formula:**

S= V×T

For T

T = S/V**4. Solution:**

Putting values in formula.

T= 360 m/10 m/Sec

T= 36Sec**5. Result:**

Time taken= T= 36 Sec

7 / 20

**Two trains are moving in opposite directions 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is_________?**

Explained with 5 Steps.

**1.Data:**

Speed 1 = 60 km/hr

Speed 2= 90 km/hr

Relative speed= (60+90)km/hr

Relative speed= 150 km/hr

Relative speed= 150×1000m/3600 Sec

Relative speed= 150,000m/3600 Sec

Relative speed= 1500m/36Sec

Relative speed= 42m/Sec

Distance 1= 1.10km

Distance 2= 0.9km

Total distance= (1.10+0.9) km

D(t)= 2 km => 2×1000m

D(t) = 2000m

**2 Required:**

Time=T=???**3. Formula:**

S = V×T

For T

T= S/V**4. Solution:**

Putting value in formulas.

T= 2000m/ 42 m/sec

T= 48Sec**5. Result:**

Time =T= **48 Sec**

8 / 20

**A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of**

Explanation with best method.

5 Steps

**1.Data:**

Train station platform= 36Sec

Standing platform = 20Sec

Speed of train = 54km/hr

**2.Required:**

Length of platform= x=???**3.Formula:**

X+length of train/Train station platform**4.Solution:** first for terms

Speed = 54 ×1000m/3600sec = 15 m/Sec.

Length of the train = (15 m/Sec × 20Sec = 300 m.

Let the length of the platform be x metres.

Now, x + 300m /36Sec= 15 m/Sec

x+300m= 15m/Sec×36Sec

x + 300m = 540m

x= 540m-300m

x = 240 m.

5.Result:

Length of platform=x=**240m**

9 / 20

**A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?**

Explanation:

Best 5 steps methods.

1.Data:

Distance= d1= 240m

Time =t= 24 Sec

Distance =d2=650

2.Required:

Time =T=??

3. Formula: D=S×T

d1+d2= S× T

Now for T=??

T= d1+d2/S

4.Solution: first for speed

S=d1/t=240m/24Sec = 10 m/Sec

Now for time

T =d1+d2/S

Putting values in T formula.

T= (240 m+ 650 m) /10 m/Sec

T= 890 m/10 m/Sec

T = 89 Sec

5.Result:

T= 89 Seconds

Explanation:

Best 5 steps methods.

1.Data:

Distance= d1= 240m

Time =t= 24 Sec

Distance =d2=650

2.Required:

Time =T=??

3. Formula: D=S×T

d1+d2= S× T

Now for T=??

T= d1+d2/S

4.Solution: first for speed

S=d1/t=240m/24Sec = 10 m/Sec

Now for time

T =d1+d2/S

Putting values in T formula.

T= (240 m+ 650 m) /10 m/Sec

T= 890 m/10 m/Sec

T = 89 Sec

5.Result:

T= 89 Seconds

10 / 20

**A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?**

11 / 20

**A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.**

12 / 20

**A man covers a distance of 20 km in 3.5 hours partly by running and partly by walking. If he walks at 4 kmph and runs at 10 kmph, find the distance he covers by walking.**

Let the distance covered by walking be x km

the distance covered by running be (20-x) km Time for walking + time for running = 3.5 hours (x/4)+[(20-x)/10]=3.5 Solving,

5x+40-2x = 70

3x=30

x=10km

Let the distance covered by walking be x km

the distance covered by running be (20-x) km Time for walking + time for running = 3.5 hours (x/4)+[(20-x)/10]=3.5 Solving,

5x+40-2x = 70

3x=30

x=10km

13 / 20

**Rayan crosses a 400m bridge in 3 minutes. What is speed?**

He crosses 0.4 km in 3/60th of an hour. Speed = 0.4/(1/20) = 8 kmph

He crosses 0.4 km in 3/60th of an hour. Speed = 0.4/(1/20) = 8 kmph

14 / 20

**Two bikes start at the same time from two destination 300 km apart and travel towards each other. If they cross each other at a distance of 130 km from one of the destination, what is the ratio of their speeds?**

15 / 20

**How much time does a train 125 metres long running at 60 km/hr take to pass a pole?**

60 km/hr = 60 * 5/18 = 16.67 m/s

Speed = distance/time; v = d/t

16.67 = 125/t

t = 7.5s

60 km/hr = 60 * 5/18 = 16.67 m/s

Speed = distance/time; v = d/t

16.67 = 125/t

t = 7.5s

16 / 20

**Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?**

New speed = 6/7 of usual speed

Speed and time are inversely proportional.

Hence new time = 7/6 of usual time

Hence, 7/6 of usual time – usual time = 12 minutes

=> 1/6 of usual time = 12 minutes

=> usual time = 12 x 6 = 72 minutes

= 1 hour 12 minutes

New speed = 6/7 of usual speed

Speed and time are inversely proportional.

Hence new time = 7/6 of usual time

Hence, 7/6 of usual time – usual time = 12 minutes

=> 1/6 of usual time = 12 minutes

=> usual time = 12 x 6 = 72 minutes

= 1 hour 12 minutes

17 / 20

**Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?**

speed of the bus excluding stoppages = 54 kmph

speed of the bus including stoppages = 45 kmph

Loss in speed when including stoppages = 54 – 45 = 9kmph

=> In 1 hour, bus covers 9 km less due to stoppages

Hence, time that the bus stop per hour = time taken to cover 9 km

=distance/speed=9/54 hour=1/6 hour = 60/6 min=10 min

speed of the bus excluding stoppages = 54 kmph

speed of the bus including stoppages = 45 kmph

Loss in speed when including stoppages = 54 – 45 = 9kmph

=> In 1 hour, bus covers 9 km less due to stoppages

Hence, time that the bus stop per hour = time taken to cover 9 km

=distance/speed=9/54 hour=1/6 hour = 60/6 min=10 min

18 / 20

**There are 21 poles with a constant distance between each pole. A car takes 30 seconds to reach the 16th pole. How long does it take to reach the last pole?**

The first pole is crossed at the zeroth second. The timer starts after the car crosses the first pole. Let the distance between each pole be x m. To reach the 16th pole, the distance travelled by the car is 15x. The speed of the car is 15x/30 = x/2 m/s To reach the 21st pole, it has to travel a distance of 20x. Time taken

= 20x÷(x/2)

= 40 seconds

The first pole is crossed at the zeroth second. The timer starts after the car crosses the first pole. Let the distance between each pole be x m. To reach the 16th pole, the distance travelled by the car is 15x. The speed of the car is 15x/30 = x/2 m/s To reach the 21st pole, it has to travel a distance of 20x. Time taken

= 20x÷(x/2)

= 40 seconds

19 / 20

**A person walking at 5/6th of his usual speed is 40 minutes late to his office. What is his usual travel time to his office?**

Let his usual speed be x kmph, the distance to his office be y km and his usual travel time be t hrs. y = xt = (5/6)x * (t+(40/60)) Solving the equation,

t = 3.33 hrs = 3 hrs 20 minutes

Let his usual speed be x kmph, the distance to his office be y km and his usual travel time be t hrs. y = xt = (5/6)x * (t+(40/60)) Solving the equation,

t = 3.33 hrs = 3 hrs 20 minutes

20 / 20

**A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?**