CMAT Time, Speed and Distance Questions

Let us consider with average speed of 40 km/hr, train reached its destination in time t hrs

with average speed of 30 km/hr, it takes t +$$\frac{36}{60}$$ hrs

Distance will be same in both the cases

40t = 30t + 18

10t = 18

t = $$\frac{18}{10}$$

Distance = 40t = 40$$\times\ \frac{18}{10}$$ = 72 km

Answer is option A.

Now 
When A covers 100 meters 
B covers 90 meters 
so we can say Va :Vb = 100:90 = 10:9   (1) 
Similarly Vb:Vc = 100:95 = 20:19             (2)
Now multiplying (1) and (2)
we get Va : Vc = 200:171
Now therefore when A covers 100 meters
C covers 85.5 meters
And so A beats C by 14.5 meters

Let speed of Raj be v1 and speed of Rahul be v2 
Now Rahul reaches 5 seconds before Raj and beats him by 50m
so Raj covers 50m in 5 seconds
so v1 = 10m/s
Now When Raj covers 950m Rahul covers 1000m (in the same time )
so we get $$\frac{v1}{v2}=\frac{1000}{950}$$
v1 =$$\frac{200}{19}\ =10\ \frac{10}{19}\ $$ m/sec

The distance covered by both Neha and Monica is same .
Now Neha takes 40 minutes to cover the distance which Monica covers in 30 minutes
So we can say their speed ratio = 3:4
Let speed of Neha be 3v and speed of Monica be 4v
Now in 5 minutes Neha would cover 15v of distance
Now Neha is ahead by 15v when Monica starts
so relative distance = 15v
And relative speed = 4v-3v =v
Therefore time to overtake = 15v/v = 15minutes

We will only consider length of bridge 
Now total distance = 100+200 = 300m
Now time taken to cross = 20 seconds 
Therefore speed = 15m/sec = 15(18/5) = 54km/hr

Time taken by X to reach Y's house = 25 minutes and by Y to reach X's house = 50 min

Ratio of time taken = $$1:2$$

$$\because$$ speed is inversely proportional to time, let speed of X = $$2x$$ km/min and speed of Y = $$x$$ km/min and distance between their houses = $$2d$$ km

Using, speed = distance/time

=> $$x=\frac{2d}{50}=\frac{d}{25}$$

=> $$\frac{d}{x}=25$$ -------------(i)

Now, time taken by Y to travel $$d$$ km, i.e. mid way = $$\frac{d}{x}$$

= $$25$$ minutes

Now, time taken by X to travel $$d$$ km, i.e. mid way = $$\frac{d}{2x}$$

= $$12.5$$ minutes

So, the correct answer is $$25-12.5=12.5$$ minutes

=> Ans - (D)

Let Saritha starts from point A and went 15 km to the West from her house to reach B, then turned left and walked 20 km southwards to reach C. She then turned East and walked 25 km to go to D and finally turning left covered 20 km to finally stop at E.

Thus, she is 10 km east of her house.

=> Ans - (D)

Given Speed of surya = 25 km/hr = $$25\times \frac{5}{18} = \frac{125}{18}$$ m/s.

Time taken(T) = Distance Travelled/ Speed

$$\Rightarrow T =\frac{500}{\frac{125}{18}} = 72$$ sec 

Hence the time taken by Mr. Surya is '72 sec' or '1.2 min'. 

Given, speed of a boat in still water (V) =  5 km/hr

Let speed of the stream be V' km/hr.

time taken to go 20km downstream (t) = distance travelled$$\div$$relative speed of the boat 

= $$\frac{20}{V + V'}$$

=$$\frac{20}{5 + V'}$$ hours

time taken to go 20 km upstream (t') =  distance travelled$$\div$$relative speed of the boat 

= $$\frac{20}{V - V'}$$

= $$\frac{20}{5 - V'}$$ hours

Given that t' = 3*t

$$\Rightarrow \frac{20}{5 - V'} = 3\times \frac{20}{5 + V'}$$

$$\Rightarrow 4V' = 10$$

$$\Rightarrow V' = 2.5$$ km/hr.

So the speed of the stream is 2.5 km/hr.

Speed = Distance travelled$$\div$$time taken

So, for the same distance travelled, Speed is inversely proportional to time taken.

$$Speed\propto \frac{1}{time taken}$$

Given ratio of speeds of three cars  is 5 : 4 : 6.

So, ratio of the time taken to travel same distance is $$\frac{1}{5} : \frac{1}{4} : \frac{1}{6}$$

= 12 : 15 : 10

In a clock, 

In one revolution the minute hand covers 60°

$$\Rightarrow$$ 60 minutes = 360°

$$\Rightarrow$$ 1 minute = 6°.

$$\Rightarrow$$ for every minute the minute hand covers 6°.

Similarly In one revolution the hour hand covers 12 hours.

$$\Rightarrow$$ 12 hours = 360°

$$\Rightarrow$$ 1 hour = 30°.

$$\Rightarrow$$ for every hour, the hour hand covers 30°.

So at 5.20 P.M. the hour hand must have covered 5 * 30° = 150° + 20*1/2 = 160

and minute hand must have covered 20 * 6° = 120°.

So the angle created by two hands of the clock when the time is 5.20 P.M. = 160 - 120 = 40°.

Given speed of Ankush = 3 rounds per hour = 1  round per 20 min

Whereas speed of Babulal = 5 rounds per hour = 1 round per 12 min.

Let us consider Ankush as stationery at the starting point.

The relative speed of Babulal with respect to Ankush = 3 rounds per hour + 5 rounds per hour = 8 rounds per hour

This implies, In one hour Babulal crosses Ankush 8 times.

So, starting from 9.00 am to 10.30 am i.e., 1.5 hours or 90 min, Babulal crosses Ankush 12 times.

$$Average  speed = Total  distance  covered\div Total  time  taken$$.

Total distance covered = 12 + 36 + 32 = 80 km.

Total time taken = $$12\div 6 + 36\div 9 + 32\div 4 = 2 + 4 + 8 = 14 hrs$$.

$$\Rightarrow Average  speed = 80\div 14 = 5.71 km/hr$$.

Let the speed of the boat in still water be V and speed of the stream be V'.

Relative speed of boat in upstream = V - V', as water stream flows against the direction of boat.

whereas Relative speed of boat in downstream = V + V', as water stream flows in the direction of boat.

Case (1)

Given Total time taken = Time taken during upstream + Time taken during downstream = 8 hours.

$$\Rightarrow \frac{24}{V - V'} + \frac{72}{V + V'} = 8$$

$$\Rightarrow 3 [\frac{1}{V - V'} + \frac{3}{V + V'}] =1$$.............(1)

$$\Rightarrow 3 [4V - 2V'] = V^2 - V'^2$$......................(2)

Case (2)

Given Total time taken = Time taken during upstream + Time taken during downstream = 14 hours

$$\Rightarrow \frac{48}{V - V'} + \frac{108}{V + V'} = 14$$

$$\Rightarrow 6 [\frac{4}{V - V'} + \frac{9}{V + V'}] = 7$$

$$\Rightarrow 6[13V - 5V'] = 7 [V^2 - V'^2]$$...................(3)

Dividing equation (3) by (2), we get

$$\frac{2 [13V - 5V']}{4V - 2V'} = 7$$

$$\Rightarrow 26V - 10V' = 28V - 14V'$$

$$\Rightarrow V = 2V'$$

Substituting this value in equation (1) we get,

$$\Rightarrow 3 [\frac{1}{V'} + \frac{1}{V'}] =1$$

$$\Rightarrow V' = 6$$

$$\Rightarrow V = 12$$

Hence, Speed of the boat in still water = 12 km/h.

and Speed of the stream = 6 km/h.

Average  Speed = Total  distance  covered $$\div$$ Total  time  taken

Total distance travelled = 50 + 50 + 30 = 130 km.

Total time taken = Time taken to travel first 50 km + Time taken to travel next 50 km + Time taken to travel next 30 km = $$40 + 50\div 2 + 30\div 1 = 95$$ minutes = $$\frac{95}{60}$$ hours.

$$\Rightarrow$$ Average Speed = $$130\div \frac{95}{60} = 82.1$$ kmph

First wheel makes 60 revolutions in 1 minute

$$\Rightarrow$$ It makes 60 revolutions in 60 seconds

$$\Rightarrow$$ It makes 1 revolution in 1 second.

This implies, after every 1 second the certain point at which the wheel started its revolution reaches its initial position.

Similarly, Second wheel and Third wheel makes 36 and 24 revolutions in 1 minute respectively.

$$\Rightarrow$$ Second and Third wheel makes 1 revolution in $$\frac{5}{3} and \frac{5}{2}$$ seconds respectively.

So for all the multiples of $$\frac{5}{3} and \frac{5}{2}$$ seconds the certain point of second wheel and third wheel reaches its initial position respectively.

After LCM {1, $$\frac{5}{3}, \frac{5}{2}$$} seconds all the three wheels will come together in the same position.

LCM of fractions = LCM of numerators/ HCF of denominators 

$$\Rightarrow$$ LCM {1,$$\frac{5}{3}, \frac{5}{2}$$} = LCM {1,5,5}$$\div$$HCF {1,3,2} = 5$$\div$$1 = 5.

Hence, after 5 seconds all the wheels will come again together in the same position.