GMAT Distance And Speed Questions With Solutions, Download PDF

All the three friends will meet at the starting point again after X seconds, such that X is the LCM of the times taken by J, K and L to complete one round.

=> LCM of 12, 18 and 20 = 180 seconds = 3 minutes.

Hence 3 minutes is the answer.

Let the speed of boat in still water be v and the speed of water be x.

Boat speed while going upstream = v-x = 28

Boat speed while going downstream = v+x = 32

=> 2v = 60 => v = 30

Hence, none of these is the correct answer.

Suppose the required distance be d km/h

Then, d/4-d/5=15min=1/4 h ⇒d 1/20=1/4⇒d=5km

Speed downstream = (15 + 3) kmph = 18 kmph.

Distance traveled =(18×12/60)km = 3.6 km.

Time taken to cover 375 km = (375/75)h = 5h

Number of stoppages = 375/125-1 = 2

Total time to stoppages = (5 × 2 ) min = 10 min

Hence, total time taken = 5 h 10 min

Let the speed of the train be x kmph.
Speed of the man = 10 kmph
Relative speed = (x+10) kmph
$$(x+10)\times\dfrac{4}{60\times60} = \dfrac{400}{1000}$$

$$(x+10) = 360$$
$$x=  350$$
Hence, Speed of the train = 350 kmph.

Let x be the speed of bike and y be the speed of bus.

y=8x

Also, bike covers 186 kilometers of distance in 3 hours

Therefore, x = 186/3 = 62km/hr

Now, y = 8x = 496km/hr

Distance covered by bus in 10 hours with this speed = 4960 km/hr

Let, amit's speed be x and sujit's speed be y. y=.8x
They work for total 8*5= 40 hours
x+y=(576000)/40
1.8x=14400
x=8000

Amit speed is 8000 depressions/hr

Speed of train = x

Relative speed = x + 9(5/18) = x + 2.5

Speed = 210/6 = 35m/s

x + 2.5 = 35 => x = 32.5 m/s = 32.5 * (18/5) = 6.5 * 18 = 117 kmph

Least common multipke of 36,27 and 9 is 108.
Hence Harish, Dilip and Asha will meet at 108 second i.e 1 minute and 48 seconds.
Therefore, the correct option is option A.

Let the speed of boat be b and that of water be w.

Hence, b+w = 32

and b-w = 28

Adding both eqns together, 2b = 60

b=30km/hr

The time of their meeting again can be calciulated by taking LCM of 24, 6 and 14.
LCM of 24, 6 and 14 = 168
Now, 168 seconds = 2 min 48 s.
Option D is correct.

Let the speed of boat rowed by man by x.
A man can row 10.2 km downstream in 18 minutes i.e speed (x+3.5) = (10.2*60)/18
Hence, speed of boat = 30.5 km/hr
Now, relative speed of boat going downwards = 30.5-3.5 = 27km/hr
Now, number of hours required to row upwards = 121.5/27 = 4.5 hours = 4$$\frac{1}{2}$$
Option A is correct.

Distance = Speed $$\times$$ Time

Initial speed = 70 kmph

In first 2 hours, motor covers a distance = $$70 \times 2 = 140~km$$

Speed after 2 hours = 70 + 10 = 80 kmph

In the next 2 hours, it covers a distance = $$80 \times 2 = 160~km$$

Distance covered = 300 km

Now, the speed is 90 kmph

Remaining 45 km is covered in $$\frac{45}{90}$$ = 0.5 hr

Total time taken to cover 345 km = 2 + 2 + 0.5 = 4.5 hr

Let the length of the train be 'x' metres
Length of the platform = 2x
Distance travelled by the train = x + 2x = 3x
Speed of the train = 60 kmph = $$60 \times \frac{5}{18}$$ m/sec
Distance = Speed $$\times$$ Time
$$3x = 60 \times \frac{5}{18} \times 32.4$$
$$x = 180$$
The length of the platform = 2x = 360 m

Let the distance between A and B be 'd' km and time taken be 't' hrs.
In both the cases, distance travelled is the same.
Distance = Speed $$\times$$ Time
Case (i) Time taken = 48 min late = (t + 0.8) hrs
$$d = 5 \times (t + 0.8)$$
Case (ii) Time taken = 15 min early = (t - 0.25) hrs
$$d = 8 \times (t - 0.25)$$
On equating them, we get
5t + 4 = 8t - 2
t = 2
Distance, d = 5 (2 + 0.8) = 14 km

Let the speed of Ram be 'v' kmph
Time taken by Shyam be 't' hrs
Distance = Speed $$\times$$ Time
60 = v (t + 1)
After doubling the speed,
60 = 2v(t - 0.5)
Simplifying them, we get
v (t + 1) = 2v (t - 0.5)
t = 2
v = 60/3 = 20 kmph

Let the speed of boat in still water and speed of river be B km/hr and R km/hr respectively

Speed of boat in downstream = (B+R)km/hr

Speed of boat in upstream = (B-R) km/hr

It is given that $$\frac{B-R}{B+R} = \frac{3}{4}$$

we get , B = 7 R

Now it is given that boat covers 70 km in 3.5 hours downstream

so ,

70 = (B+R)3.5

B+R = 20

R = 2.5 km/hr

B = 17.5 km/hr

Let the speed of boat in still water and speed of river be B km/hr and R km/hr respectively .

Speed of boat in downstream = B+ R km/hr

Speed of boat in upstream = B-R km/hr = $$\frac{7.5}{1.5}$$= 5km/hr

Let the same distance travelled in upstream and downstream be D km

So ,

$$\frac{D}{B-R}$$ =2× $$\frac{D}{B+R}$$

B+R = 10 km/hr

Let the speed of boat in still water and speed of river be B km/hr and R km/hr respectively

speed of boat in downstream = (B+R) km/hr

speed of boat in upstream = (B-R) km/hr

Distance covered = 16 km

it is given that tie taken ti move 16 km downstream is 2 hours and for upstream it is 4 hours

16 = (B+R) 2 .............(1)

16 = (B-R) 4...............(2)

from equation 1 and 2

B = 6 km/hr

distance travelled by car which started from point A from 11 am to 1 pm is = 36 ×2 = 72 km

Now it is given that total distance is 592 km

Distance left to be covered = 592 - 72 = 520 km

Speed of car started from B = 44 km/hr

Soeed of car started from A = 36km/hr

Relative speed = 80 km/hr

Time taken = 520/80 = 6.5 hours

So the two cars will meet at 7.30 pm

let the speed of faster train be F m/s

And slower train be S m/s

When they move in same direction relative speed = (F -S) m/s

Time taken = 14 sec

When they move in opposite direction = (F+S) m/s

Time taken in this case = 182 sec

In both cases they are covering same distance

So , (F+S )× 14 = (F-S ) × 182

12 F = 14 S

F = $$\frac{7}{6}$$ S

so faster train is 1/6 more than the slower one which means it is 16.67% faster than the slower train.

Distance travelled by boat = $$x$$ km

Let speed of current = $$y$$ km/h

Speed of boat = 12 km/h

=> Downstream speed = $$(12 + y)$$ km/h and upstream speed = $$(12 - y)$$ km/h

Acc. to ques, time taken by a boat to travel; `x' km upstream is twice the time taken by the same boat to travel `x' km downstream

=> $$\frac{x}{12 - y} = 2 \times \frac{x}{12 + y}$$

=> $$\frac{1}{12 - y} = \frac{2}{12 + y}$$

=> $$12 + y = 24 - 2y$$

=> $$y + 2y = 3y = 24 - 12$$

=> $$y = \frac{12}{3} = 4$$ km/h

Le the speed of boat be B km/hr and speed of river be R km/hr

Speed of boat in downstream = (B + R) km/hr

Speed of boat in upstream = (B- R)km/hr

It is given that $$\frac{B+R}{B-R}$$ = $$\frac{7}{5}$$

B = 6R...........(1)

Distance travelled downstream = 63 km

Time taken = 3 hr

63 = (B+R)3

B+R = 21 ..........(2)

From equation 1 and 2

R = 18 km/hr

as it is given that train which starts at M at 6 am has a speed of 63 km/hr So distance covered till 7 am = 63 km

Now the remaining distance between M and N is = 287 km

Now a train also starts from N towards M with a speed of 77 km/hr .

Relative speed = 140 km/hr

Time taken to cover the remaining 287 km is = $$\frac{287}{140}$$ = 2hours 3 min

And hence both trains will meet at 9.03 am

Dheeraj, Sonu and Deepak start running around a circular stadium and complete one round in 14 seconds. 8 seconds and 15 seconds respectively.

The time they will meet again = L.C.M of 14seconds,8seconds,15seconds

= 840 seconds

= 14 minutes

To calculate the speed of the train,we need to know the length of the platform.

Hence the speed of the train cannot be determined.

Distance covered by Hike = 420 -36 = 384 km
Time taken by Hike = $$\frac{420}{70}$$ = 6 hrs
Speed of Hike = $$\frac{384}{6}$$ = 64 kmph

Speed S=$$\frac{\text{distance}}{\text{time}}$$

Speed of train= 430/5= 86

Speed of bike= 86/2= 43

Time to travel 301km= 301/43= 7hrs

At 7am the distance between the two trains will be 726kms.
Since the trains are moving in opposite directions, the relative speed is the sum of individual speeds.
The relative speed is 121kmph. and the distance is 726 kms.
The time taken to cover 726kms at a speed of 121kmph is nothing but the time taken by the two trains to meet each other.
Time taken to meet = 726/121 = 6 hours
Hence the trains would meet at 1:00 pm.
Hence Option A is the correct answer.

Let length of train A = $$l$$ metres

and speed of train A = $$x$$ m/s

Using, $$time = \frac{distance}{speed}$$

=> $$\frac{l}{x} = 18$$

=> $$x = \frac{l}{18}$$

Length of platform = $$157.5$$ metres

=> $$\frac{157.5 + l}{x} = 39$$

Substituting value of $$x$$, we get :

=> $$\frac{157.5 + l}{39} = \frac{l}{18}$$

=> $$13l = 6l + 945$$

=> $$13l - 6l = 7l = 945$$

=> $$l = \frac{945}{7} = 135$$ metres

$$\therefore$$ Length of train B

= $$(\frac{135}{2}) + (157.5 \times 2)$$

= $$67.5 + 315 = 382.5$$ metres

Let speed of stream = $$x$$ kmph

=> Speed of boat downstream = $$(6 + x)$$ kmph

Speed upstream = $$(6 - x)$$ kmph

Using, $$time = \frac{distance}{speed}$$

=> $$\frac{36}{6 - x} - \frac{36}{6 + x} = 8$$

=> $$\frac{1}{6 - x} - \frac{1}{6 + x} = \frac{8}{36}$$

=> $$\frac{(6 + x) - (6 - x)}{(6 + x) (6 - x)} = \frac{2}{9}$$

=> $$\frac{2x}{36 - x^2} = \frac{2}{9}$$

=> $$\frac{x}{36 - x^2} = \frac{1}{9}$$

=> $$36 - x^2 = 9x$$

=> $$x^2 + 9x - 36 = 0$$

=> $$x^2 + 12x - 3x - 36 = 0$$

=> $$x (x + 12) - 3 (x + 12) = 0$$

=> $$x = 3 , -12$$

$$\because$$ Speed cant be negative, => $$x = 3$$ kmph

Let speed of current = $$x$$ km/h

=> Speed of boat in still water = $$x + (\frac{500}{100} \times x)$$

= $$x + 5x = 6x$$ km/h

=> Speed of the boat downstream = $$6x + x = 7x$$ km/h

Speed of boat upstream = $$6x - x = 5x$$ km/h

$$\therefore$$ Required ratio = $$\frac{7x}{5x}$$

= $$7 : 5$$

Let length of train = $$x$$ m

=> Length of platform = $$2x$$ m

Using, $$time = \frac{distance}{speed}$$

While travelling at $$s$$ m/s, time taken

=> $$\frac{x + 2x}{s} = 21$$

=> $$x = 7s$$ --------------Eqn(1)

Also, $$\frac{x + 2x}{s + 5} = 18$$

=> $$3x = 18 (s + 5)$$

Using eqn(1), we get :

=> $$3 \times 7s = 18s + 90$$

=> $$21s - 18s = 3s = 90$$

=> $$s = \frac{90}{3} = 30$$ m/s

Length of train A = 220 m

=> Speed of train A = $$\frac{220 + 340}{32}$$

= $$\frac{560}{32} = 17.5$$ m/s

Speed of train A : Speed of train B = 7 : 9

=> Speed of train B = $$\frac{9}{7} \times 17.5$$

= $$9 \times 2.5 = 22.5$$ m/s

Length of train B = 270 m

$$\therefore$$ Time taken by train B to cross the pole = $$\frac{270}{22.5}$$

= 12 sec

Let the length of train = $$l$$ metre

$$\because speed = \frac{distance}{time}$$

=> $$\frac{l + 50}{14} = \frac{l}{10}$$

=> $$10l + 500 = 14l$$

=> $$14l - 10l = 4l = 500$$

=> $$l = \frac{500}{4} = 125$$ metre

$$\therefore$$ Speed of train = $$\frac{125}{10} = 12.5$$ m/s

= $$12.5 \times \frac{18}{5} = 2.5 \times 18$$

= $$45$$ km/hr

Let distance between AB = $$5x$$ and BC = $$7x$$

Downstream speed = $$\frac{35}{2.5} = 14$$ km/h

Let upstream speed = $$y$$ km/h

Now, $$time = \frac{distance}{speed}$$

=> $$\frac{5x}{y} = \frac{7x}{14}$$

=> $$\frac{5}{y} = \frac{1}{2}$$

=> $$y = 5 \times 2 = 10$$ km/h

Now, speed of stream = $$\frac{1}{2}$$ (downstream - upstream)

= $$\frac{1}{2} (14 - 10)$$

= $$\frac{4}{2} = 2$$ km/h

$$\because$$ The cars are moving in opposite direction

$$\therefore$$ Relative speeds = 34.5 + 41.5 = 76 km/hr

=> Required time  = $$\frac{684}{76} = 9$$ hrs

Shaurya's speed = $$\frac{18}{1}$$ = 18 kmph

=> Speed of Pratyusha = $$18 \times \frac{7}{3}$$

= 42 kmph

Average speed = $$\frac{(39 + 25) km}{(45 + 35) min}$$

= $$\frac{64 km}{80 min}$$

= $$\frac{64 \times 60}{80}$$ kmph

= $$48$$ kmph

Let the distance of the station from the house of the person = $$x$$ km

Difference of time = 8 + 7 = 15 minutes = $$\frac{1}{4}$$ hr

Since, Time = $$\frac{Distance}{Speed}$$

$$\therefore \frac{x}{6} - \frac{x}{10} = \frac{1}{4}$$

=> $$\frac{10x - 6x}{60} = \frac{1}{4}$$

=> $$\frac{2x}{30} = \frac{1}{4}$$

=> $$x = \frac{15}{4} = 3\frac{3}{4}$$ km

Speed of boat in still water

= $$\frac{1}{2}$$ (downstream + upstream)

= $$\frac{1}{2} (13 + 9) = \frac{22}{2}$$

= $$11$$ kmph

Let speed of the train = $$20x$$ m/s

Now, 75% of the speed = $$\frac{75}{100} \times 20x = 15x$$ m/s

Length of train = 150 m

Time taken to cross the pole = 12 sec

Using, $$speed = \frac{distance}{time}$$

=> $$15x = \frac{150}{12}$$

=> $$x = \frac{10}{12} = \frac{5}{6}$$

Length of platform = $$l$$ m

Acc. to ques, => $$20x = \frac{150 + l}{24}$$

=> $$20 \times \frac{5}{6} = \frac{150 + l}{24}$$

=> $$\frac{50}{3} = \frac{150 + l}{24}$$

=> $$150 + l = 400$$

=> $$l = 400 - 150 = 250$$ m

Let speed of current = $$x$$ km/hr

=> Speed of boat in still water = $$8x$$ km/hr

Acc. to ques, => $$\frac{63}{9x} + \frac{63}{7x} = 8$$

=> $$\frac{7}{x} + \frac{9}{x} = 8$$

=> $$\frac{16}{x} = 8$$

=> $$x = \frac{16}{8} = 2$$

$$\therefore$$ Speed of boat in still water = $$8 \times 2 = 16$$ km/hr

Let usual speed of the train = $$10x$$ m/s

Now, 50% of the speed = $$\frac{50}{100} \times 10x = 5x$$ m/s

Length of train = $$l$$ m

Time taken to cross the pole = 20 sec

Using, $$speed = \frac{distance}{time}$$

=> $$5x = \frac{l}{20}$$

=> $$x = \frac{l}{100}$$

Length of platform = 300 m

Acc. to ques, => $$10x = \frac{300 + l}{25}$$

=> $$10 \times \frac{l}{100} = \frac{300 + l}{25}$$

=> $$\frac{l}{10} = \frac{300 + l}{25}$$

=> $$25l = 3000 + 10l$$

=> $$25l - 10l = 15l = 3000$$

=> $$l = \frac{3000}{15} = 200$$ m

Let speed of current = $$x$$ km/hr

=> Speed of boat in still water = $$6x$$ km/hr

Acc. to ques, => $$\frac{105}{7x} + \frac{105}{5x} = 12$$

=> $$\frac{15}{x} + \frac{21}{x} = 12$$

=> $$\frac{36}{x} = 12$$

=> $$x = \frac{36}{12} = 3$$

$$\therefore$$ Speed of boat in still water = $$6 \times 3 = 18$$ km/hr

Let speed of the train = $$10x$$ m/s

Length of train = $$l$$ m

Time taken to cross the pole = 6 sec

Using, $$speed = \frac{distance}{time}$$

=> $$10x = \frac{l}{6}$$

=> $$x = \frac{l}{60}$$

Now, 60% of the speed = $$\frac{60}{100} \times 10x = 6x$$ m/s

Length of platform = 240 m

Acc. to ques, => $$6x = \frac{240 + l}{15}$$

=> $$6 \times \frac{l}{60} = \frac{240 + l}{15}$$

=> $$\frac{l}{10} = \frac{240 + l}{15}$$

=> $$15l = 2400 + 10l$$

=> $$15l - 10l = 5l = 2400$$

=> $$l = \frac{2400}{5} = 480$$ m

Let speed of boat in still water = $$x$$ km/hr

=> Speed of current = $$y$$ km/hr

Let distance travelled = $$d$$ km

Acc. to ques, => $$2.2 (\frac{d}{x + y}) = \frac{d}{x - y}$$

=> $$2.2x - 2.2y = x + y$$

=> $$2.2x - x = y + 2.2y$$

=> $$3x = 8y$$ -------------(i)

Also, the man takes 2 hrs 30 mins in travelling 55 km downstream.

=> $$\frac{55}{x + y} = 2 + \frac{1}{2}$$

=> $$\frac{55}{x + y} = \frac{5}{2}$$

=> $$x + y = 22$$

Multiplying both sides by 8, and using eqn(i), we get :

=> $$8x + 3x = 22 \times 8$$

=> $$x = \frac{22 \times 8}{11} = 16$$ km/hr

Length of first train = 240 m and second train = 270 m

Total length of the two trains = 240 + 270 = 510 m

Speed of first train = 60 kmph and second train = 48 kmph

Since, the trains are moving in opposite direction, thus relative speed = 60 + 48 = 108 kmph

= $$(108 \times \frac{5}{18})$$ m/s = $$30$$ m/s

Let time taken = $$t$$ seconds

Using, time = distance/speed

=> $$t=\frac{510}{30}=17$$ seconds

=> Ans - (A)

Time taken by boat to travel from point A to B in still water = 2 hours

To find the upstream speed, we definitely need the speed of stream, thus statement (iii) is mandatory.

Also, the distance between points A and B or the speed of boat in still water is needed.

Thus, statements (i) and (iii) are required to find the upstream speed of the boat.

=> Ans - (D)