Aptitude::Time and Distance

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Example 1 / 16

How many minutes does Ashwani take to cover a distance of 400 m, if he runs at a speed of 20 km/hr?

Ashwani's speed = 20 km/hr = \(\left(20 \times \frac{5}{18}\right)\) m/sec = \(\frac{50}{9}\) m/sec.
So, Time taken to coer 400 m = \(\left(400 \times \frac{9}{50}\right)\) sec = 72 sec. = \(1\frac{12}{60}\)min. = \(1\frac{1}{5}\) min. Ans.

Formula Used:

Time = \(\left(\frac{Distance}{Speed}\right)\)

x km/hr = \(\left(x \times \frac{5}{18}\right)\) m/sec

first convert speed from km/hr to m/sec. and apply the time formula directly.
Time = \(\left( \frac{400}{20 \times \frac{5}{18}}\right)\) m/sec.

Solution
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Example 2 / 16

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?

Due to stoppages, it covers (54-45)= 9 km less
Time taken to cover 9 km= \(\frac{9}{54}\) * 60 = 10 min Ans.

For stoppage type questions, just subtract the including and excluding stoppage, we will get the distance covered and take out the time by considering the speed of the excluding stoppage.

time taken = \(\frac{distance}{speed}\)

Example 3 / 16

Rishikant, during his journey, travels for 20 minutes at a speed of 30 km/hr, another 30 minutes at a speed of 50 km/hr, and 1 hour at a speed of 50km/hr and 1 hour at a speed of 60 km/hr. What is the average velocity?

for 1st round,
Distance(d1)= speed*time= 30* \(\frac{1}{3}\)= 10 km
For 2nd round,
Distance(d2)= speed*time= 50* \(\frac{1}{2}\)= 25 km
For 3rd round,
Distance(d2)= speed*time= 50* 1= 50 km
For 4th round,
Distance(d4)= speed*time= 60*1= 60 km
So, avg. speed= \(\frac{Total Distance}{total time}\)
Total d = d1+d2+d3+d4
= 10+25+50+60 = 145 km
Total time = \(\frac{1}{3}\)+\(\frac{1}{2}\)+1+1
= \(\frac{17}{6}\)
Average speed= \(\frac{145}{\frac{17}{6}}\) = 51.18 km/hr. Ans.

Avg. speed = \(\frac{Total Distance}{Total Time}\)

Avg. speed= \(\frac{(s1*t1+s2*t2+s3*t3+s4*t4)}{t1+t2+t3+t4}\)
= \(\frac{(10+25+50+60)}{2.83}\)= 51.18 km/hr.

Example 4 / 16

Sambhu beats Kalu by 30 meters or 10 seconds. How much time was taken by Sambhu to complete a race 1200 meters?

speed of Sambhu= \(\frac{d}{t}\)= \(\frac{30}{10}\)m/s = 3 m/s
Time taken by Sambhu to complete 1200 m= \(\frac{1200}{3}\)= 400s

speed = \(\frac{d}{t}\)
Time = \(\frac{d}{s}\)

In these questions, just take out the speed and divide by the distance given.

Example 5 / 16

Walking at \(\frac{3}{4}\) of his normal speed, Abhishek is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and his office is?

Let the normal time be t and speed be s. Now, new speed= \(\frac {3}{4}\)s and new time= t+16
So, since distance is same, then,
S*t=\(\frac {3}{4}\)s* (t+ 16)
On solving, we get, t= 48 minutes Ans.

distance = speed*time

If the change in speed and time is given, then just equate the original speed*time and new speed*time as the distance is the same. Hence, we will get the time.

Example 6 / 16

Rajdhani Express travels 650 km in 5 hr and another 940 km in 10 hr. What is the average speed of the train?

average speed= \(\frac{total distance}{total time}\)
= \(\frac{650+940}{10+5}\)
= \(\frac{1590}{15}\)
= 106kmph Ans.

average speed = \(\frac{total distance}{total time}\)

To find the avg. speed when the different distance is given, just add up total distance and add total time and take out speed as by the formula, average speed= \(\frac{total distance}{total time}\).

Example 7 / 16

What is the time taken by Chandu to cover a distance of 360 km by a motorcycle moving at a speed of 10m/s?

Given,
distance = 360km, speed = 10m/s= 10*\(\frac{18}{5}\)kmph= 36kmph
Time = \(\frac{distance}{speed}\)
=\(\frac{360}{36}\)
= 10 hr. Ans.

1. Time= \(\frac{distance}{speed}\)
2.To convert x m/s to kmph= x* \(\frac{18}{5}\)kmph
3. To convert y kmph to m/s= y*\(\frac{5}{18}\)m/s

It is based on the simple direct formula, Time= \(\frac{distance}{speed}\)

Example 8 / 16

Alok walks to a viewpoint and returns to the starting point by his car and thus takes a total time of 6 hrs 45 minutes. He would have gained 2 hours by driving both ways. How long would it have taken for him to walk both ways?

Total time is taken by the car to cover both ways= 6.45-2= 4.45 hours.
The time taken to cover one way driving= \(\frac{4.45}{2}\)= 2.225 hrs.
Time is taken by walking to a viewpoint = 6.45-2.225= 4.225 hrs.
The time is taken for him to walk both the way is= 2*4.225= 8.45 hours= 8 hours 45 mins. Ans.

just subtract the two way to one way.

to take out the time, just subtract the given one way from two ways.

Example 9 / 16

Ram and Bharat travel the same distance at the rate of 6km per hour and 10 km per hour respectively. If Ram takes 30 minutes longer than Bharat, the distance traveled by each is:

Let the distance be 'x' km.
Now, time = \(\frac{d}{s}\)
Time = \(\frac{x}{6}\)-\(\frac{x}{10}\) = \(\frac{1}{2}\)
On solving, we get, x = 7.5 km. Ans.

time = \(\frac{d}{s}\)

when we have to take out the distance, for different speeds, let s1 and s2, then,
Time = \(\frac{x}{s1}\)- \(\frac{x}{s2}\)
Time difference will be given, hence x is obtained.

Example 10 / 16

Without stoppage, a train travels a certain distance with an average speed of 60 km/hr, and with stoppage, it covers the same distance with an average speed of 40 km/hr. On an average, how many minutes per hour does the train stops during the journey?

Due to stoppages, it covers (60-20)= 40 km less
Time taken to cover 40 km= \(\frac{40}{60}\) * 60 = 40 min/hr Ans.

time taken = \(\frac{distance}{speed}\)

For stoppage type questions, just subtract the including and excluding stoppage, we will get the distance covered and take out the time by considering the speed of the excluding stoppage.

Example 11 / 16

Walking at \(\frac{3}{4}\) of his normal speed, a man takes 2 \(\frac{1}{2}\) hours more than the normal time. Find the normal time.

Let the normal time be t and speed be s.
Now, new speed = \(\frac{3}{4}\)s and new time = t + 2 \(\frac{1}{2}\).
So, since distance is same, then,
S*t=\(\frac{3}{4}\)s* (t+ 2 \(\frac{1}{2}\))
On solving, we get, t= 7.5 hours Ans.

distance= speed*time

if the change in speed and time is given, then just equate the original speed*time and new speed*time as the distance is same. Hence, we will get the time.

Example 12 / 16

Two trains for Mumbai leave Delhi at 6:00 a.m. and 6:45 a.m. and travel at 100 kmph and 136 kmph respectively. How many kilometers from Delhi will the two trains be together?

Difference in time of departure between two trains= 45 mins.= \(\frac{3}{4}\)hour.
Let the distance be x km from Delhi where the two trains will be together.
Time is taken to cover x km with speed 136kmph be t hour.
Time taken to cover x km with speed 100 kmph be \(\left({{\text{t}+\frac{3}{4}\right)\)
Now, 100* \(\left({\frac{{4{\text{t}} + 3}}{4}}\right)\) = 136t
=> 25(4t+3) = 136t
=> 100t+75 = 136t
=> 36t = 75
t = \( \frac{{75}}{{36}}\) = 2.08hours
Then, distance x km = 136 * 2.084= 283.33 km Ans.

time= \(\frac{d}{s}\)

first take difference in time, and then use the formula of time.

Example 13 / 16

The Sinhagad Express left Pune at noon sharp. Two hours later, the Deccan Queen started from Pune in the same direction. The Deccan Queen overtook the Sinhagad Express at 8pm. Find the average speed of the two trains over this journey if the sum of their average speeds is 70km/hr.

Time taken by Sinhagad express: time is taken by Deccan queen = 8:6
Since, we know that speed is inversely proportional to time, so,
Speed of Sinhagad express: speed of Deccan queen= 6:8
Total speed= 70kmph.
Speed of sinhagad express(s1)= 70* \(\frac{3}{7}\)= 30 kmph
Speed of deccan queen(s2)= 70* \(\frac{3}{7}\)= 40 kmph
Average speed of both trains= \(\frac{Total Distance}{Total Time}\)
= 2D\(\left(\frac{s1*s2}{s1+s2}\right)\)
= \(\frac{(2*40*30)}{40+30}\) [let D=1km]
= 34.28kmph Ans.

when distance is same, and speeds are given, let s1 nand s2,then average speed is = 2D\(\left(\frac{s1*s2}{s1+s2}\right)\)

when distance is same, formula used is 2D\(\left(\frac{s1*s2}{s1+s2}\right)\)

Example 14 / 16

Two trains, Calcutta Mail and Bombay Mail, start at the same time from stations Kolkata and Mumbai respectively towards each other. After passing each other, they take 12 hours and 3 hours to reach Mumbai and Kolkata respectively. If the Calcutta Mail is moving at the speed of 48 km/h, the speed of the Bombay Mail is:

\Lets assume the speed of the Bombay Mail = x kmph
Then by formula,
\(\frac{speed of Calcutta}{ mailspeed of Bombay mail}\) = \(\sqrt{\frac{3}{12}}\)
so, \(\frac{48}{x}\) = \(\sqrt{\frac{3}{12}}\)
=> X = 96kmph.Ans.

\(\frac{s1}{s2}\)=\(\sqrt{\frac{t2}{t1}}\)

when the time to reach the destination is given after passing of two trains to each other, let t1 and t2 be the time and speed be s1 and s2 then, \(\frac{s1}{s2}\)=\(\sqrt{\frac{t2}{t1}}\)

Example 15 / 16

Lonavala and Khandala are two stations 600 km apart. A train starts from Lonavala and moves towards Khandala at the rate of 25 km/h. After two hours, another train starts from Khandala at a rate of 35 km/h. How far from Lonavala will they cross each other?

Distance between the Two Stations = 600 km.
In the First Case, (Train From Lonavala)
Speed = 25 km/hr.
In first two hours,
∴ Time = 2 hrs.
∵ Distance = Speed × Time.
∴ Distance = 25 × 2
Distance = 50 km.
∴ Distance covered by the Train after starting from the Lonavala before the train from Khandala starts = 50 km.
Now, Train from the Khandala also starts, thus the distance between the two trains = 600 - 50
= 550 km.
Relative speed = Speed of train which starts from Lonavala + Speed of the train which starts from the Khandala
= 25 + 35
= 60 km/hr.
Now, Time at which these two trains meet after traveling = Distance between these Train/Relative Speed
= 550/60
= 55/6 hrs.
Now, In 55/6 hrs, Distance covered by the First Train after the train from the khandala starts = (55/6) × 25
= 1375/6 km.
= 229.167 km.
∴ The distance at which the Both train meets from Lonavala = Distance traveled by train from Lonavala in first two hours + Distance traveled by it after another train from the Khandala starts.
= 229.167 + 50
= 279.167 km.
∴ Both the trains meet after a distance of 279.167 km from the Lonavala.
.Ans.

distance = speed * time

take out the distance travelled by the first train and then subtract from the actual distance, then take out the distance at which they meet.

Example 16 / 16

Shyam’s house, his office, and his gym are all equidistant from each other. The distance between any 2 of them is 4km. Shyam starts walking from his gym in a direction parallel to the road connecting his office and his house and stops when he reaches a point directly east of his office. He then reverses direction and walks till he reaches a point directly south of his office. The total distance walked by Shyam is?

it is said that Shyam starts walking from his gym in a direction parallel to the road connecting his office and his house. The path traversed is P1, G, P2.
Shyam stops when he reaches a point directly east of his office.he then reverses direction and walks till he reaches a point directly south of his office.
Therefore, the total distance walked by shyam will be sum of the distance of (G to P1, P1 to G and G to P2) = 4 + 4 + 4 = 12 km.Ans.

distance = speed * time

D=sum of the total distance
Shyams-house-his-office-and-his-gym-are-all-equal-distance.png

Aptitude::Time and Distance

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