Tag: speed math

Questions Related to speed math

A cyclist, after riding a certain distance, stopped for half an hour to repair his bicycle, after which he completes the whole journey of 30km at half speed in 5 hours. If the breakdown had occurred 10 km farther off, he would have done the whole journey is 4 hours. Find where the breakdown occurred and his original speed.

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  1. 5km/hr,10km

  2. 10km/hr,10km

  3. 10km/hr,5km

  4. 5km/hr,5km

Show answer
Correct Option: B

Ramu walks at 3kms per hour while his brother Somu walks at 4kms per hour. They both started to school, which is 2kms away from their house at 7:00 A.M. Then, which of these is are correct?
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  1. Somu reaches school at 7:20 AM

  2. Somu reaches school 10 mins earlier than Ramu

  3. Ramu reaches school at 7:30 AM

  4. All the above

Show answer
Correct Option: B
Explanation:

Ramu =3 kmph Somu =4 kmph Both starts at 7 AM Distance =2km

Time taken by Ramu ⇒S=v×t

2=3×t

t=23hr=23×60

=40min

Time taken by Somu ⇒2=4×t

t=12hr=30 min.

∴ Somu reaches school 10 mins earlier than Ramu.

A goods train of length 500 m transverse a 2-km-long tunnel at a constant speed of 36 km/hour. The time needed for the train to completely pass through the tunnel is
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  1. 3 minutes 20 seconds

  2. 4 minutes 10 seconds

  3. 4 minutes 40 seconds

  4. 6 minutes 10 seconds

Show answer
Correct Option: B

The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10km/hr more than the speed of going, what was the speed per hour in each direction?
ime speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

  1. going speed =−30km/hr, returning=−20km/hr

  2. going speed =20km/hr, returning=30km/hr

  3. going speed =30km/hr, returning=40km/hr

  4. going speed =30km/hr, returning=20km/hr

Show answer
Correct Option: B

The present ages of Reena and Usha ar 24 and 36 yr respectively. What was the ratio between the ages of Usha and Reena respectively 8 yr ago?
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  1. 7:4

  2. 4:7

  3. 11:8

  4. 8:11

Show answer
Correct Option: A
Explanation:

Age of Reena after 8yr=24−8=16yr
Age of Usha After 8yr=36−8=28
So, the ratio will be 1628=47

A train is running at a uniform speed of 60km/h. How much time it will take to cover a distance of 300km?
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  1. 5 h

  2. 4.5 h

  3. 6 h

  4. 7 h

Show answer
Correct Option: A
Explanation:

$$speed=\frac{distance}{time}$$ $$time=\frac{distance}{speed}$$ $$time=\frac{300}{60}h$$ $$time=5 hr$$

If a man walks 17 km in 5 hours, how long(in hrs) would he take to walk 20.4 km
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  1. 6

  2. 5

  3. 8

  4. None of these

Show answer
Correct Option: A
Explanation:

Time taken to walk 17km=5hrs.
Time taken to walk $1km=\dfrac{5}{17}hrs.$
Time taken to walk $20.4km=\dfrac{5}{17} \times20.4$

$=6 hrs.$

A man runs round a circular field of radius 50m at the speed of 12km/hr. What is the time taken by the man to take twenty rounds of the field?
time speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

  1. 30 min

  2. 32 min.

  3. 34 min.

  4. None of these

Show answer
Correct Option: B

Two boys P and Q walk at 6 km/hr and 6 km/hr respectively. If Q covers a certain distance in 10 minutes less than P. What is the time taken by Q to cover the same distance (in minutes)?

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  1. 40

  2. 60

  3. 30

  4. 70

Show answer
Correct Option: A

A can run 1 km in 4 min 54 sec and B in 5 min. How many metres start can A can give B in a km race so that the race may end in a dead heat ?
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  1. 16 m

  2. 18 m

  3. 20 m

  4. 22 m

Show answer
Correct Option: C
Explanation:

Distance covered by B in 6 sec $= \dfrac{(1000}{300}×6)=20 m$