Speed Time and Distance Quiz 1

Solution: Let the distance be x km

Travelling at 10 kmph, Arun will reach point A at 2 pm.
Travelling at 15 kmph, Arun will reach point A 12 noon.

Therefore, time taken when travelling at 10 km - time taken when travelling at 15 km =2 hours

\frac{x}{10}_{x}{15}=2
=> 3x-2x=2 \times 30
=> x=60

Time needed if travelled at 10 kmph = \frac{60}{10}=6 hours

Therefore, to reach 1 pm, his travelling time must be (6−1)=5 hours.

Hence, required speed = \frac{60}{5}=12 kmph

Solution: Franz Kafka needs to cover 6 km in 45 minutes

Given that he covers one-half of the distance in two-thirds of the total time
=> he covers half of 6 km in two-thirds of 45 minutes
=> He covers 3 km in 30 minutes

Now he needs to cover the remaining 3 km in remaining 15 minutes
Distance =3 km

Time =15 minutes = \frac{1}{4}

Required Speed= \frac{3}{\frac{1}{4}}

=12 kmph

Solution: If two bodies A and B move from each other's starting point in opposite directions, they reach their destinations after having met, then their speeds S_{a} & S_{b} are given by,

\frac{S_{a}}{S_{b}}=\frac{^{\sqrt{t_{b}}}}{^{\sqrt{t_{a}}}}

where t is the time taken by them to cover the distance.

S_{b}=\frac{25\times 4}{5}

S_{b}=20 kmph

Solution: Let the distance be x km. Then,
( Time taken to walk x km) + (Time taken to ride x km) =37 min.
( Time taken to walk 2x km ) + (Time taken to ride 2x km )= 74 min.
But, the Time taken to walk 2x km = 55 min.
Time taken to ride 2x km = (74-55)min =19 min.

Solution: Original speed is 60 miles per hour.
If the speed is reduced by 10 miles per hour, then the new speed = 50 miles per hour
Distance to be covered is 300 miles.
The formula to find time is
Time = Distance / Speed
Time taken to cover 300 miles distance at the speed of 50 miles per hour is
= 300 / 50 = 6 hours
So, if the person reduces his speed by 10 miles per hour, he will take 6 hours to cover 300 miles.

Solution: Ratio of speeds = 4 : 3 : 5.
Since time is inversely proportional to speed.
Ratio of times taken = \frac{1}{3}:\frac[1}{4}:\frac{1}{5}
=15:20:12

Solution:
Relative speed = (2 + 3) = 5 rounds per hour.

So, they cross each other 5 times in an hour and 2 times in half an hour.

Hence, they cross each other 7 times before 9.30 a.m.

Solution: Clearly, the two will meet when they are 726 m apart.
To be (4.5 + 3.75) = 8.25 km apart, they take 1 hour.
To be 726 m apart, they take \frac{100}{825} \times \frac{726}{1000}
\frac{242}{2750} \times 60 min
=5.28 min

Solution: Let the distance be x km. Then,
(Time taken to walk x km) + (Time taken to ride x km) = \frac{23}{4}
(Time taken to walk 2x km) + (Time taken to ride 2x km) = \frac{23}{2}
But, time taken to ride 2x km = \frac{15}{4}
Time taken to walk 2x km= \frac{23}{2}-\frac{15}{4}
= \frac{31}{4} hrs
=7 hrs 45 mins

Solution: When Kevin Feige meets David Heyman at R, Kevin Feige has walked the distance PQ + QR and David Heyman the distance PR.
That is, both of them have together walked twice the distance from P to Q, i.e. 42 km.
Now, the speed of David Heyman and Kevin Feige are in the ratio 3 : 4, and they have walked 42 km.
Hence, the distance PR travelled by David Heyman= \frac{3}{7} of 42 km = 18 km.