# Aptitude::Time and Distance

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Example 1 / 16

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.

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

NA

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.

Time = \(\left( \frac{400}{20 \times \frac{5}{18}}\right)\) m/sec.

Example 2 / 16

Due to stoppages, it covers (54-45)= 9 km less

Time taken to cover 9 km= \(\frac{9}{54}\) * 60 = 10 min

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.

NA

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

Example 3 / 16

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.

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 = \(\frac{Total Distance}{Total Time}\)

NA

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.

= \(\frac{(10+25+50+60)}{2.83}\)= 51.18 km/hr.

Example 4 / 16

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

Time taken by Sambhu to complete 1200 m= \(\frac{1200}{3}\)= 400s

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

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

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

NA

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

Example 5 / 16

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

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

NA

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

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

= \(\frac{650+940}{10+5}\)

= \(\frac{1590}{15}\)

= 106kmph

= \(\frac{650+940}{10+5}\)

= \(\frac{1590}{15}\)

= 106kmph

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

NA

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

Given,

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

Time = \(\frac{distance}{speed}\)

=\(\frac{360}{36}\)

= 10 hr.

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

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

NA

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

Example 8 / 16

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.

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.

NA

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

Example 9 / 16

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.

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

NA

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.

Time = \(\frac{x}{s1}\)- \(\frac{x}{s2}\)

Time difference will be given, hence x is obtained.

Example 10 / 16

Due to stoppages, it covers (60-20)= 40 km less

Time taken to cover 40 km= \(\frac{40}{60}\) * 60 = 40 min/hr

Time taken to cover 40 km= \(\frac{40}{60}\) * 60 = 40 min/hr

**Ans.**
time taken = \(\frac{distance}{speed}\)

NA

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

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

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

NA

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

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

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

NA

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

Example 13 / 16

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

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\(\left(\frac{s1*s2}{s1+s2}\right)\)

NA

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

Example 14 / 16

\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.

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

NA

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

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.

.

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

NA

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

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.

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

NA

D=sum of the total distance

Srinivasan S1 year ago

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