SNAP Time, Speed and Distance Questions

Average speed = Total distance/total time

Let us consider total distance =180km

Case 1 :  man covers half of his journey by train at 90 km/hr

               Distance= 90km => time t1=$$\frac{90}{90}$$=1hr

Remaining distance = 90km

Case 2 :  man covers one-third of the remainder by bus at 30 km/hr

                Distance = 30km => time t2=$$\frac{30}{30}$$=1hr

Case 3 : man covers rest by cycle at 10 km/hr

             Distance = 60km => time t3=$$\frac{60}{10}$$=6hr

Therefore Average speed = $$\frac{180}{1+1+6}$$= 22.5km/hr

Since Pradeep and his wife are going in the opposite direction, their relative velocity will be added. 

Their relative velocity = 4.5+3.75 = 8.25 km/hr = 137.5 m/min

Time taken to cover 726 meters = $$\ \frac{\ 726}{137.5}\ =\ 5.28$$ minutes

Total distance covered by horse = Circumference of the circle.

$$2.5 \times 66 = 2\pi \times $$ (radius)

$$2.5 \times 66 = 2 \times \frac{22}{7}\times $$ (radius)

radius $$=\ 26.25$$

Let the total distance covered by the taxi be 100 kms.

For 60% of the distance, i.e. 60 Kms, Speed of the taxi was 40 kmph. Time taken to cover 60 kms= $$\ \frac{\ Dis\tan ce}{Speed}\ =\ \frac{\ 60}{40}=\ \frac{\ 3}{2}\ hrs.$$

For 20% of the distance, i.e. 20 Kms, Speed of the taxi was 30 kmph. Time taken to cover 20 kms= $$\ \frac{\ Dis\tan ce}{Speed}\ =\ \frac{\ 20}{30}=\ \frac{\ 2}{3}\ hrs.$$

For remaining, i.e. 20% of the distance, i.e. 20 Kms, Speed of the taxi was 10 kmph. Time taken to cover 20 kms= $$\ \frac{\ Dis\tan ce}{Speed}\ =\ \frac{\ 20}{10}=\ \ 2\ hrs.$$

So, total distance= 100 kms.

Total time= $$\ \ \frac{\ 3}{2}+\ \frac{\ 2}{3}+2=\ \frac{\ 9+4+12}{6}=\ \frac{25\ }{6}hrs$$

.'. Average speed= $$\ \ \ \frac{\ Total\ Dis\tan ce}{Total\ Time}=\ \frac{\ 100}{\ \frac{\ 25}{6}}=\ \frac{\ 100\cdot6}{25}=4\times\ 6=24\ kmph$$

We have two parameters to look at in this question:

1) Length of the train. Let this be assumed to be a metres.

2) Length of the platform. Let us assume it as b metres.

It is given that speed of the train is 54 km/h Or, 54$$\ \times\ \frac{\ 5}{18}$$= 15 m/s

So, when it passes the man, distance covered will be the length of the train itself= a metres.

So, time taken= $$\frac{Dis\tan ce}{Speed}=\ \ \frac{\ a}{15}$$, which is given to be 20 seconds.

So, $$\frac{\ a}{15}=20$$. And a= 300 metres.

Now, to pass the entire platform, we need to account for both the length of the platform and that of the train.

So, effective distance to be covered= a+b= 300+b metres.

Speed of the train = 15 m/s.

So, time taken= $$\frac{\ 300+b}{15}$$, which is given to be 36 seconds.

So, $$\frac{\ 300+b}{15}=36$$

=>$$\ 300+b=36\times\ 15$$

=>b=540-300

So, b= Length of the train= 240 metres.

$$time\ =\ \frac{dis\tan ce}{speed}$$

$$t1=\ \frac{1500}{S}$$

$$t2=\ \frac{1500}{S+250}$$

Given t1 - t2 = 1/2 hours

So, $$\frac{1500}{S}\ -\frac{1500}{S+250}=\frac{1}{2}$$

On solving S=750kmph

1st case :

Distance travelled by train = 300km and by taxi = 200km

Let the speed of train be x km/hr and that of taxi be y km/hr.

Time taken by train = 300/x and time taken by taxi = 200/y hr

$$\ \frac{\ 300}{x}$$ = $$\ \frac{\ 200}{y}$$

Total time taken = 5hrs 30 mins = 5 1/2 hrs

$$\ \frac{\ 300}{x}$$ +$$\ \frac{\ 200}{y}$$= 5$$\ \frac{\ 1}{2}$$

On solving we get : 600y + 400x = 11xy  ----- Eq (1)

2nd case :

Distance travelled by train = 260 km and by taxi = 240 km

The speed of train and taxi will remain the same.

Time taken by train = $$\ \frac{\ 260}{x}$$hr and by taxi = $$\ \frac{\ 240}{y}$$ hr
$$\ \frac{\ 260}{x}$$ + $$\ \frac{\ 240}{y}$$ = 5hrs 36 mins

$$\ \frac{\ 260}{x}$$ + $$\ \frac{\ 240}{y}$$ = 5$$\ \frac{\ 3}{5}$$ on solving we get,

1300y + 1200x = 28xy   ---- Eq (2)

Solving 1 and 2 we get:

x=100 km/hr

Speed of the train = 100km/hr

A is the correct answer.

Let the usual speed of the plane be v kmph and the time taken by the plane to cover 1500m = t hrs

To reach the destination on time, the increased speed of the plane = v+250 kmph

Since he started 30 min later, the time taken should be t-$$\ \frac{\ 1}{2}$$ hr

v*t = (v+250)(t-$$\ \frac{\ 1}{2}$$) 

v = (250t-125)2

$$\ \frac{\ 1500}{v+250}=t-\ \frac{\ 1}{2}$$

$$\ \frac{\ 1500}{500t-250+250}=t-\ \frac{\ 1}{2}$$

$$\ \frac{\ 3}{t}=t-\ \frac{\ 1}{2}$$

On solving for t, we get

t= 2 hrs

His original speed = $$\ \frac{\ 1500}{2}$$ = 750 kmph

B is the correct answer.

Let the distance he has to travel to reach the place = x km

and the time given to reach the place = y minutes.

If he walks 4 km in an hour he is 15 minutes late ie if he walks with a speed 4 km/hour (4km/60min)he will take y+15 minutes

distance = time*speed

ie x=$$\ \frac{\ 4}{60}$$*(y+15)

60 x = 4* (y+15)

15 x = y+15    -- Eq (1)

if he walks 6km in an hour he is 10 minutes early ie if he walks with a speed of 6km/hour he will take y-10 minutes.

x=$$\ \frac{\ 6}{60}$$ * (y-10)

10x = (y-10)    -- Eq (2)

On solving Eq 1, 2, we get

y = 60 minutes , x = 5 km

B is the correct answer. 

Let t be the start time

A starts at t, while B starts at (t-2)

Total distance covered = 200 km

As they are travelling in the opposite direction,

Relative distance covered i,e 200=((200/4)*t )+((400/7)*(t-2))

t=176min i.e 2hrs 56min from start (6 am)

Therefore they meet at 8.56 am.

A is the correct answer.

Quantity of water let in by the leak in 1 hr= 15 tonnes

Quantity of water thrown out by the pumps in 1 hr = 12 tonnes

Net quantity of water filled in the ship in 1 hour =15-12=3 tonnes

60 tonnes water is filled in  = 60/3= 20 hours

Required speed = 40/20= 2kmph

Speed of the aeroplane = 756 km/h.

It is given that twice the distance covered by the aeroplane in 9 hrs is equal to the distance covered by the helicopter in 48 hrs.

Distance covered by the aeroplane in 9 hrs = 756*9 = 6804 km.

Speed of the helicopter = $$\ \frac{2\times\ 6804\ }{48}$$ = 283.5

Distance covered by the helicopter in 18 hrs = 283.5*18 = 5103 km.

D is the correct answer.

Speed of the Train = 36 km/hr.
= 36 × 5/18 [Changing into m/s]
= 2 × 5
= 10 m/s.

Time taken to cross the Platform = 20 seconds.
Time taken to cross the man = 10 seconds.
∵ Speed of the Train = Length of the Train/Time to cross the Man

∴ Length of the Train = Speed of the Train × Time to cross the Man.
= 10 × 10
= 100 m.
Sum of the Length of Platform and train = Speed of the Train × Time taken by the Train to cross the Platform.
= 10 × 20
= 200 m.
∵ Length of the Train + Length of the Platform = 200
∴ 100 + Length of the Platform = 200
∴ Length of the Platform = 200 - 100
= 100 m.
∴ Length of the Platform is 100 m.

Let the usual speed and the time taken by the man = 5v, 4t respectively.

If the man walks at $$\frac{4}{5}th$$ of his usual speed, a man reaches office 10 minutes later than usual

i.e the time taken by the man = $$\ \frac{\ 5}{4}$$ of the usual time (Speed and time are inversely proportional to each other)

=5t

5t-4t = 10 minutes.

t = 10 minutes.

Usual time taken by the man = 4t = 40 minutes

B is the correct answer.

Time taken to meet = $$\ \frac{\text {Distance between them}}{\text{Relative speed}}$$

Relative Speed = 5 +4 = 9 miles per hour.
Time take to cover 81 miles = $$\ \frac{\ 81\ }{9}$$ = 9 hours.
Woman travelled in 9 hours = 9*4 = 36 miles.

1 person :
$$6^2 + 8^2 = OB^2$$
√(36+64) = OB
√(100)= OB
10 = OB...
The distance of the first person from his starting point is 10 miles.
Similarly, the distance of the second person from his starting point is 10 miles.
B is the correct answer

Let us consider A covers 100 m in t sec.

Speed of A = $$\ \frac{100}{t}$$

Since A gave B a headstart of 20m, B covers 80m in (t+5) sec (Given)

Speed of B =  $$\ \frac{80}{t+5}$$

If A gives head start of 40m to B, then the time taken by A to cover 100m is the same as the time taken by B to cover 60m

$$\ \frac{100}{\ \frac{\ 100}{t}}$$ = $$\ \frac{60}{\ \frac{\ 80}{t+5}}$$

n solving for t, we get 

t =  15 sec

Speed of A =  $$\ \frac{100}{15}$$ m/sec

Time taken by A to cover 200m(x) = 

200=time * $$\ \frac{100}{15}$$

x = 30 sec

C is the correct answer.

Total distance = 108+112 = 220m
Total time = 6 seconds 
Now we know that :
Relative speed = Distance / Time
so we get :
220/6 = v1+v2
110/3 =125/9 + v
we get v = $$\ \frac{\ 205}{9}$$m/s =$$=\ \frac{\ 205}{9}\times\ \ \frac{\ 18}{5}=$$ 82km/hr

Let speed of boat be V and speed of stream be U 
so as per case 1 
Downstream i.e V+U = 30/2 =15
And Upstream i.e V-U = 30/6 = 5 
Now adding 
we get 2V =2
V =10

Assume the speed in km/hr be a and b.

Then, the distance travelled after half an hour = $$\ \frac{\ a}{2}$$ and $$\ \frac{\ b}{2}$$

The distance between them = $$\ \ \ \frac{\ 1}{2}\sqrt{\ a^2+b^2}$$  = 17

=> $$\sqrt{\ a^2+b^2}$$ = 34

=> $$a^2+b^2$$ = 1156  .....(1)

After 45 minutes, the difference of the distance travelled = $$\ \frac{\ 3}{4}\left(a-b\right)$$ = 10.5

=> a-b = 14   => $$a^2+b^2-2ab\ $$ = 196.......(2)

From 1 and 2, 2ab = 960 

$$a^2+b^2$$ +2ab = 1156+960= 2116

=> a+b=46

Hence a=30 and b=16

Assuming the cyclist speed without the wind = a  and the wind speed = b

Then $$\ \frac{\ 1km}{a+b}$$ = 3   => a+b = $$\ \frac{\ 1}{3}$$

$$\ \frac{\ 1km}{a-b}$$ = 4  => a-b = $$\ \frac{\ 1}{4}$$

Adding both the equations, we get 2a = $$\ \frac{\ 1}{3}$$ + $$\ \frac{\ 1}{4}$$

=> a = $$\ \frac{\ 7}{24}$$  

Time taken to cover 1 km = $$\ \frac{\ 1\ km}{a}$$ = $$\ \dfrac{\ 1\ km}{\ \frac{\ 7}{24}}$$ = $$\ \frac{\ 24}{\ \ 7}$$ = $$3\frac{3}{7}$$

Since the clock is losing time, the original time will be more than 1:45 PM and options A,B, C are less than 1:45 PM. Hence D is the answer.

Let us assume that the length of each bogy is 1 unit, and the speed of both A and B be s units.

Initially, both A and B had 12 bogies, and ran in opposite direction.

So, using the concept of relative speed, the effective speed of the two trains taken together, when travelling in opposite direction adds up.

Effective distance covered = 12+12 units=24.

Effective speed= 2s units.

Now, using time= $$\ \frac{\ Dis\tan ce}{Speed}$$, we get 2 mins= $$\ \frac{\ 24}{2s}$$

=>120=$$\ \frac{\ 12}{s}$$

.'.s= $$\ \frac{\ 1}{10}$$

In the new scenario, length of B becomes 16 units.

So, Effective distance covered = 12+16 units=28 units.

Effective speed= 2s units.

New time= $$\ \frac{\ 28}{2s}=\ \frac{\ 14}{s}$$=140.

So, increase in time in the new case= 140-120 seconds= 20 Seconds- Option D.

First person: For every 2 steps he takes, the escalator moves by 3 steps at the same time.
Hence, 5 steps are covered at a time.
This is repeated 24 times(120/5).
Hence, first person takes = 24 *2 = 48 steps
Second person: For every 3 steps he takes, the escalator moves by 5 steps at the same time.
Hence, 8 steps are covered at a time.
This is repeated for 15 times(120/8)
Hence, second person takes = 3 * 15 = 45 steps
Total number of steps = 48 + 45 = 93

At t=0, speed was 40 kmph

At t=180, speed becomes 70 kmph

Therefore, increase in speed= $$\ \frac{\ 70-40}{40}.100=\ \frac{\ 30}{40}.100=75\%\ \ \ $$

Each of the options given assumes the speed of the train to vary linearly wrt time.

So, let us assume the speed of the train at time t be given by 40+ at, because at t=0, the speed was already 40 kmph.

When we put x=30, we get,

45= 40+ 30a

=> 5= 30a Or, a= $$\ \frac{\ 1}{6}\ \ \ $$

So, Speed= 40+ $$\ \frac{\ t}{6}\ \ \ $$

We will check again to verify the results. 

When t=60, Speed= 40+ $$\ \frac{\ 60}{6}\ \ \ $$= 50 kmph, which matches with the value given in the table.

So, Option D is correct

$$2\ \frac{\ 1}{2}\ $$ hours after the beginning of the time period means 120 +$$\left(2\ \cdot60\right)+\left(\frac{\ 1}{2}\cdot60\right)=\ 120+30=150\ \ \ $$ minutes.

From the table, we can infer that after 150 minutes, Speed of the train was 65 km/hour

Let them meet at time "t"

Floor of David at time t = 11+57t

Floor of Albert at time t = 51- 63t

When they meet their floor will be same. 11+57t = 51 - 63t

t = $$\dfrac{1}{3}$$ minutes.

Floor at which they meet = 11 + 57 *$$\dfrac{1}{3}$$ = 11+19 = 30

Let speed of crew= 'a' mph

Speed of current = 'b' mph

Downstream, the speeds will be added 

Upstream, the speeds will be subtracted

a+b = $$\frac{\left(10\right)}{\frac{5}{6}}$$ = 12 miles/hr

a-b = $$\frac{\left(12\right)}{\frac{3}{2}}$$ = 8 miles/hr

Solving the above two equations, we get a = 10 mph and b as 2 mph. 

Therefore, Option B is the correct answer. 

At 7 the hour hand is  $$\frac{7}{12}\cdot360$$ = 210 degrees ahead of the minute hand.

Let after x minutes they are at 180 degrees with each other (since they are exact opposite)

Total angle covered by hour hand = 210 + 0.5x ( as hour hand moves at 0.5 degrees/minute)

Total angle covered by minute hand = 6x ( speed of minute hand is 6 degrees/minute)

As per the question

    Angle by hour hand - Angle by minute hand = 180

(210 + 0.5x) - 6x = 180

210 - 5.5x = 180

5.5x = 30

x  = 5.45 minutes

The closest answer is 5.5 minutes

Assume the distance to be $$x$$ km.

When he travels at the speed of 2.5 km/hr he is 6 minutes late.

When he travels at the speed of 3.5 km/hr he is 6 minutes early

$$\frac{\ x}{2.5}$$ - $$\frac{\ x}{3.5}$$ = $$\frac{\ 12}{60}$$

$$\frac{\ 2x}{5}-\frac{\ 2x}{7}\ =\frac{\ 1}{5}$$

$$\frac{\ 4x}{35}=\frac{\ 1}{5}$$

$$\ x=\frac{\ 7}{4}=\ 1\frac{\ 3}{4}$$ km

Let x be the distance traveled by water

Distance by foot is $$\ \frac{\ 4x}{7}$$

Distance by horseback is $$\ \frac{\ 2x}{5}$$

Total distance = x + $$\ \frac{\ 4x}{7}$$ + $$\ \frac{\ 2x}{5}$$ = 3036

Hence x = 1540

$$\ \frac{\ 4x}{7}$$ = 880

$$\ \frac{\ 2x}{5}$$ = 616

Option A is correct