# Time and Distance Questions and Answers updated daily – Aptitude

## Time and Distance Questions

Given:

Speed of the river = 2 km/hr

Speed of the motorboat in upstream = 12 km/hr

**Upstream speed = u - v**

u - v = 12

u - 2 = 12

u = 14 km/hr

Therefore, speed of the motorboat in still water is 14 km/hr.

**Downstream speed = u + v**

= 14 + 2 = 16 km/hr

WKT,

**Time = Distance/Speed**

As per the question, motorboat takes 3 hours more to travel x km in still water than to travel (x – 24) km in downstream

(x/14) - ((x-24)/16) = 3

(8x - 7(x - 24))/112 = 3

8x - 7x + 168 = 336

**x = 168 km.**

Let the sides of the square be

**'a' km.**

WKT, Time = Distance/Speed

Time taken for 1st side = a/300

Time taken for 2nd side = a/600

Time taken for 3rd side = a/900

Time taken for 4th side = a/1200

**Average speed = Total distance/Total time**

Here, Total distance travelled =

**4a**

Therefore, Average speed = 4a/[(a/300) + (a/600) + (a/900) + (a/1200)]

Average speed = 4a/a[(1/300) + (1/600) + (1/900) + (1/1200)]

Average speed = 4/[(12 + 6 + 4 + 3)/3600]

**Average speed = 576 km/hr.**

**'d'.**

WKT,

**Speed = Distance/Time**

Given, first ant cover 8% of distance in 3 hours.

So, the speed of first ant = 8d/(100*3) = 8d/300 feet/hr

Also given, second ant covers 7/120 of distance in 2 hours 30 minutes.

Here, time = 2 hr 30 minutes = 2 (1/2) hr = 5/2 hr

So, speed of second ant = (7d*2)/(120*5) = 7d/300 feet/hr

If the first ant travelled 800 feet and meet the second ant ,

WKT,

**Time = Distance/Speed**

Time taken by first ant = Time taken by second ant

800/ (8d/300) = (d -800)/ (7d/300)

(800*300)/8d = [(d - 800)*300]/7d

800/8 = (d - 800)/7

100 = (d - 800)/7

700 = d - 800

d = 1500 feet

Speed of second ant = 7d/300 = 7(1500)/300 =

**35 feet/hr.**

Given that car can travels

**thrice**the distance than a bike.

The distance travelled by bike in 'x' hrs = 'a' km

Then the distance travelled by car in 'x' hrs = '3a' km

WKT,

**Speed = Distance/Time**

(3a/x) - (a/x) = 20 km/hr

2a/x = 20

a/x = 10 km/hr

In 5 hrs of journey, x = 5

a/5 = 10

a = 50 km

Therefore, total distance travelled by bike = a = 50km

Total distance travelled by car = 3a = 3(50) = 150 km

Required difference in difference between car and bike

= 150 - 50

=

**100 km.**

WKT,

**Distance = Speed * Time**

So, distance travelled by him in 3hrs = 25 * 3 = 75 km

Also given, he travels 2 hours at 40 km/hr.

Therefore, distance travelled by him in 2 hrs = 40*2 = 80 km

Let the total distance travelled be 'x'.

He covered 1/3 of total distance = (75 + 80) = 155 km

1/3 * x = 155

x = 155 *3

**x = 465 km**

Therefore, remaining distance = 465 -155 = 310 km

To cover the remaining 310 km in 5 hours,

WKT,

**Speed = Distance/Time**

**Required speed = 310/5 = 62 km/hr.**

speed of second train = 22 km/hr

Since the two trains are moving in same direction, speed can be subtracted.

=>

**Relative speed**= Speed of 1st train - Speed of 2nd train

= (40 - 22) km/hr

= 18 km/hr

--> Converting into m/sec

= 18 x (5/18) m/sec

= 5 m/sec

=>

**Relative speed = 5 m/sec**

Given, Time taken to cross each other = 1 minute = 60 seconds

Length of first train = 125 meter

Let the length of other train = X meter

Then,

**Relative speed = Sum of the Lengths of trains / Time taken in crossing**

--> 5 = (X + 125)/60

--> 5 * 60 = (X + 125)

--> 300 = (X + 125)

--> 300 - 125 = X

---> X = 175 meter

Therefore, the length of second train = 175 meter

Speed = 54 kmph

= 54 * (5/18) m/sec

=15 m/sec

When a train crosses a pole, it covers its own length

Required time = (150/15)

= 10 seconds

Time taken to pass a pole = 10 seconds.

WKT, Time = Distance/Speed

Accoroding to the question,

(240/x) - (240/(x + 3)) = 4

240[(1/x) - (1/(x + 3))] = 4

240[(x + 3 - x)/x(x + 3)] = 4

[3/x(x + 3)] = 4/240

x(x + 3) = (240*3)/4

x

^{2}+ 3x = 180

x

^{2}+ 3x - 180 = 0

x

^{2}+ 15x -12x -180 = 0

x(x + 15) - 12(x + 15) = 0

(x + 15) (x - 12) = 0

x = -15, 12

x = 12km/hr

Therefore, the original speed of Prabhat = 12 km/hr.

WKT, Time = Distance/ Speed

Total time = (24/6) + (24/8) + (24/12)

= 4 + 3 + 2 = 9 hrs

Average speed = Total distance/Total time

= 72/9 = 8 km/hr.

After one and half an hour train A reached 59*(3/2) = 88.5 km

Remaining distance = 677 – 88.5 = 588.5 km

Both the trains cross each other in,

= 588.5/(59 + 48) = 588.5/107

= 5.5 hours = 5 hours 30 min

Both the trains cross each other after 5 hours 30 min, that is,

= 2 + 5. 30 = 7.30 PM

Downstream Time (T) = 66 min

---> converting time in minutes to hour

---> Downstream Time (T) = (66/60) hour

W.K.T:

**Speed = Distance/Time**

**Speed Downstream**= (55/66) * 60 =

**50 km/hr**

Given, Speed of the boat in still water : Speed of the stream = 4: 1

---> Speed of the boat in still water = 4x

---> Speed of the stream = x

W.K.T:

**Speed Downstream = (Speed of the boat in still water + Speed of stream)**

---> Speed Downstream = 4x + x = 50

---> 5x = 50

---> x = 10

Thus, Speed of the boat in still water = 4x = 4 * 10 = 40 km/hr

and, Speed of the stream = x = 10 km/hr

Now,

**Speed upstream = (Speed of the boat in still water - Speed of the stream)**

--->

**Speed upstream**= 40 - 10 =

**30 km/hr**

Given, Upstream distance = 72 km

**Time taken by the boat to cover upstream distance**= Upstream distance/ Speed upstream

= 72/30

= 2 (2/5) hr

= 2 hour [(2/5) * 60 mins]

=

**2 hour 24 mins**

Total distance = S*T = 50*3 = 150 Km

Speed of the bike = D/T = 150/2 = 75 Km/hr

Let's take the fixed charges = Rs. x (for the first 5 km)

And the additional charges = Rs. y /km

As per the question,

Charge for distance 10 km = Rs. 350

So for charge for first 5 km is Rs.x and additional charge for next 5 km is Rs. 5y which can be written as follows,

x + 5y = 350 ....(i)

Charge for distance 25 km = Rs. 800

So for charge for first 5 km is Rs.x and additional charge for next 20 km is Rs. 20y which can be written as follows,

x + 20y = 800 ...(ii)

On solving eqn. (i) and (ii), we get

x = 200, y = 30

Therefore, charge for a distance of 30 km = Charge for first 5 km is Rs.x and additional charges for next 25 km is Rs. 25y

= x + 25y

= 200 + 25(30) = Rs. 950

The ratio of length of Tejas express, Shatabdi express and Rajdhani express is 6 : 7 : 8.

Let the length of Tejas express = 6x meter

Length of Shatabdi express = 7x meter

Length of Rajdhani express = 8x meter

Given that, Tejas express crossed stationary Shatabdi express = 52/3 sec.

Speed of Tejas express = Length of Tejas and Shatabdi/Time

108*(5/18) = (6x + 7x)/(52/3)

13x = (108 *5 *52)/(18*3)

13x = 520

x = 40 meters

Therefore, length of Tejas express = 6*40 = 240 meters

Length of Rajdhani express = 8*40 = 320 meters

Relative speed Tejas and Rajdhani = 144 - 108 = 36 km/hr

= 36 *(5/18) = 10 m/s

Required time to cross each other(Tejas and Rajdhani) = Length of Tejas and Rajdhani/Relative speed

= (240 + 320)/10

= 560/10

= 56 sec.

If Hiten covers 1 km, then in the same time Priyank covers only 875 mt.

If Vikash covers 1 km in (x + 10) sec, then Priyank covers 1 km in (x + 25) sec

Thus in x seconds Priyank covers the distance of 875 mt.

So the speed of Priyank = 875/x

WKT, Time = Distance/Speed

(x + 25) = 1000/(875/x)

x = 175

Time taken by Hiten to complete 1 km = 175 seconds

Time taken by Vikash to complete 1 km = (175 + 10) = 185 seconds

Required ratio of Hiten and Vikash = 175 : 185

= 35 : 37

= (3/60)*120 = 6 km

Distance travelled by Govinda in 3 minutes = (3/60)*80 = 4 km

Hence, Salman will have to travel totally 10 km distance before catching up with Govinda.

Relative speed of Salman to that of Govinda after taking U turn = 120 - 80 = 40 kmph

Therefore, time taken by Salman after taking U turn = Distance/Relative speed

= 10/40 = 15 minutes

Total time taken by Salman to meet Govinda = time taken for U turn + time taken after U turn

= 3 + 15 = 18 minutes

And length of second train be L.

Speed of first train = 48 km/hr

Speed of second train = 42 km/hr

WKT, Speed = Distance/Time

(48 + 42)5/18 = (2L + L)/12

(90*5)/18 = 3L/12

L = (25*12)/3

L = 100 m

Therefore, the length of first train = 2*100 = 200 m

Let the length of platform = x

Distance of first train and the platform = 200 + x

Speed of first train = 48 km/hr

First train passes a railway platform in 45 seconds, So

(48 * 5)/18 = (200 + x)/45

600 = 200 + x

x = 400 m

Therefore, length of the platform is 400 m.

Speed of the Sai = 50 km/hr

Total distance = 710 km

WKT, Distance = Speed * Time

Distance covered by Sai in 3 hours = 50 * 3 = 150 km

Remaining distance = 710 – 150 = 560 km

Relative speed = Speed of sai + Speed of Raj

= 50 + 30 = 80 km/hr

They both meet after = Remaining distance/Relative speed

= 560/80 = 7 hours

Now, Sai covers the total distance in = 7 + 3 = 10 hours

Distance covered by Sai in 10 hours = 50 * 10 = 500 km

Distance between C and A = 500 km.

**Common Explanation:**

The step measure of 3 persons:

P1= 40cm

P2=42cm

P3=45cm

To find the

**minimum distance, L.C.M**should be calculated

L.C.M = 40,42, 45

= 2223357

=2520 cm

Converting

**cm to m,**

2520cm = 25.2m (1m = 100cm)

The minimum distance each should walk to cover the same distance = 25.2m

**Common Explanation:**

Let the speed of the slower bull = x

The faster bull is 3 times faster than the slower bull

Therefore, the speed of the faster bull = 3x

The total distance between the bulls:

**Distance = Speed*Time**

Total distance between the bulls = (3x + x)*(10/3) = 40x/3 km

Time required by the slower bull:

Time = Distance/Speed

Time = 40x/3 / x

Slower Bulls time required = 40/3 hours