CAT Time Speed Distance Questions

M - First meeting point

Let the speeds of trains A and B be 'a' and 'b', respectively.

$$\dfrac{x}{a}=\ \dfrac{\ D-x}{b}$$

It is given,

$$\dfrac{D}{a}=10$$ and $$\dfrac{x}{b}=9$$

$$\dfrac{x}{\dfrac{D}{10}}=\ \dfrac{\ D-x}{\dfrac{x}{9}}$$

$$\dfrac{10x}{D}=\ \dfrac{\ 9D-9x}{x}$$

$$10x^2=\ \ 9D^2-9Dx$$

$$10x^2+9Dx-9D^2=\ 0$$

Solving, we get $$x=\dfrac{3D}{5}$$

$$\dfrac{x}{b}=9$$

$$\dfrac{3D}{b\times5}=9$$

$$\dfrac{D}{b}=15$$ 

The total time taken by train B to travel from station Y to station X is 15 minutes.

The answer is option B

Let the speeds of two ships be 'x' and 'x+6' km per hour

Distance covered in 2 hours will be 2x and 2x+12

It is given,

$$\left(2x\right)^2+\left(2x+12\right)^2=60^2$$

$$\left(x\right)^2+\left(x+6\right)^2=30^2$$

$$2x^2+12x+36=900$$

$$x^2+6x+18=450$$

$$x^2+6x-432=0$$

Solving, we get x = 18

The speed of slower ship is 18 kmph

The answer is option C.

Both the cars take 1.5 hrs to meet when they travel towards each other.

It is given, speed of slower car is 60 km/hr

Therefore, distance covered by slower car before they meet = 60*1.5 = 90 km

The answer is option B.

Let the speed of Moody be 'x' steps/sec and that of the escalator be 'y' steps/sec.

In 30 seconds, Moody will finish riding the escalator when going in the same direction.

Thus, total steps = 30(x+y)

If Moody's speed becomes twice, the time becomes 20 seconds.

Thus, total steps = 20(2x+y)

Or    30x + 30y = 40x + 20y

Or    x = y

So, total steps = 60y.

Time taken by only escalator= 60y/y = 60s.

Let S be the slower ship and F be the faster ship.

It is given that when S travelled 8 km, the positions of ships with the port is forming a right triangle.

Since one of the angles is 60(since one vertex is still part of the equilateral triangle),

the other two vertexes will have angles of 30 and 90.

The distance between O and S = 24 - 8 = 16

In triangle OFS, $$\cos60^0\ =\ \frac{OF}{OS}$$

Thus, OF = 8.

Thus in the time, S covered 8 km, F will cover 24 - 8 = 16 km.

Thus, the ratio of their speeds is 2:1,

Thus, when F covers 24 km, S will cover 12 km.

The correct option is B.

Speed of the faster train = $$\frac{160}{12}=\frac{40}{3}\ $$ m/s

Speed of the slower train = $$\frac{40}{3}-\left(6\times\ \frac{5}{18}\right)=\frac{35}{3}$$ m/s

Sum of speeds (when the trains travel towards each other) = $$\frac{40}{3}+\frac{35}{3}=25$$ m/s

Let the slower train be $$x$$ metres long; then: $$\frac{160+x}{25}=14$$

On solving, $$x=190\ m$$

Considering the length of train A = La, length of train B = Lb.

The speed of train A be 5*x, speed of train B be 3*x.

From the information provided :

The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other.

In this case, train A traveled a distance equivalent to the length of train B which is Lb at a speed of 5*x+3*x = 8*x because both the trains are traveling in the opposite direction.

Hence (8*x)*(46) = Lb.

In the information provided :

It took another 69 seconds for the rear ends of the trains to cross each other.

In the next 69 seconds 

The train B traveled a distance equivalent to the length of train A in this 69 seconds.

Hence (8*x)*(69) = La.

La/Lb = 69/46 = 3/2 = 3 : 2

Considering the distance travelled by Mira in one minute = M,

The distance traveled by Amal in one minute = A.

Given if they walk in the opposite direction it takes 3 minutes for both of them to meet. Hence 3*(A+M) = C. (1)

C is the circumference of the circle.

Similarly, it is mentioned that if both of them walk in the same direction Amal completes 3 more rounds than Mira :

Hence 45*(A-M) = 3C. (2)

Multiplying (1)*15 we have :

45A + 45M = 15C.

45A - 45M = 3C.

Adding the two we have A = $$\frac{18C}{90}$$

Subtracting the two M = $$\frac{12C}{90}$$

Since Mira travels $$\frac{12C}{90}$$ in one minute, in one hour she travels :$$\frac{12C}{90}\cdot60\ =\ 8C$$

Hence a total of 8 rounds.

Alternatively, 

Let the length of track be L 
and velocity of Mira be a and Amal be b 
Now when they meet after 45 minutes Amal completes 3 more rounds than Mira
so we can say they met for the 3rd time moving in the same direction
so we can say they met for the first time after 15 minutes
So we know Time to meet = Relative distance /Relative velocity
so we get $$\frac{15}{60}=\frac{L}{a-b}$$     (1)
Now When they move in opposite direction
They meet after 3 minutes
so we get $$\frac{3}{60}=\frac{L}{a+b}$$    (2)
Dividing (1) and (2)
we get $$\frac{\left(a+b\right)}{\left(a-b\right)}=5$$
or 4a =6b
or a = 3b/2
Now substituting in (1)
we get :
$$\frac{L}{b}\times\ 2=\ \frac{15}{60}$$
so $$\frac{L}{b}\ =\frac{1}{8}$$
So we can say 1 round is covered in $$\frac{1}{8}$$ hours
so in 1-hour total rounds covered = 8.

Let the distance from A to B be "x", then the distance from B to C will be 3x. Now the speed of Train 2 is double of Train 1. Let the speed of Train 1 be "v", then the speed of Train 2 will be "2v" while travelling from A to B.

Time taken by Train 1 = (x/v)

Time taken by Train 2 = (x/2v)

Now from B to C distance is "3x" and the speed of Train 2 is (v) and the speed of Train 1 is (2v).

Time taken by Train 1 = 3x/2v

Time taken by Train 2 = 3x/v

Total time taken by Train 1 = x/v(1+(3/2)) = (5/2)(x/v)

Total time taken by Train 2 = x/v(3+(1/2))= (7/2)(x/v)

Ratio of time taken = $$\frac{5}{\frac{2}{\frac{7}{2}}}=\frac{5}{7}$$

Now car A beats car B by 45km. Let the speed of car A be v(a) and speed of car B be v(b).

$$\frac{v\left(a\right)}{v\left(b\right)}=\frac{m}{m-45}$$ .....(1)where '"m" is the entire distance of the race track.

Moreover $$\frac{v\left(b\right)}{v\left(c\right)}=\frac{m}{m-50}$$.......(2)

and finally $$\frac{v\left(a\right)}{v\left(c\right)}=\frac{m}{m-90}$$......(3)

Multiplying (1) and (2) we get (3). $$\frac{m}{m-90}=\frac{m}{m-45}\left(\frac{m}{m-50}\right)$$

Solving we get m=450 which is the length of the entire race track

Let distance = d

Given, $$\frac{d}{35}-\frac{d}{40}=\frac{6}{60}$$

=> d = 28km

The actual time  taken to travel 28km = 28/40 = 7/10 hours = 42 min.

Given time taken to travel 58/3 km = 1/3 *42 = 14 min.

Then a break of 8 min.

To reach on time, he should cover remaining 28/3 km in 20 min => Speed = $$\frac{\left(\frac{28}{3}\right)}{\frac{20}{60}}=28\ $$ km/hr

Let the speed of Ram be v(r) and the speed of Rahim be v(h) respectively. Let them meet after time "t" from the beginning.

Hence Ram will cover v(r)(t) during that time and Rahim will cover v(h)t respectively.

Now after meeting Ram reaches his destination in 1 min i.e. Ram covered v(h)t in 1 minute or v(r)(1)= v(h)(t)

Similarly Rahim reaches his destination in 4 min i.e. Rahim covered v(r)t in 4 minutes or v(h)(4)= v(r)(t)

Dividing both the equations we get $$\frac{v\left(r\right)}{4v\left(h\right)}=\frac{v\left(h\right)}{v\left(r\right)}\ or\ \frac{v\left(r\right)}{v\left(h\right)}=2$$ Hence the ratio is 2.

To complete one round Ram takes 100m/15kmph and Rahim takes 20m/5kmph

They meet for the first time after L.C.M of (100m/15kmph , 20m/5kmph) = 100m/5kmph=20m/kmph.

Distance traveled by Ram =20m/kmph * 15kmph =300m.

So, he must have ran 300/100=3 rounds.

Note:

CAT gave both 2 and 3 as correct answers because of the word 'before'. 

The distance travelled by A between 9:00 Am and 10:30 Am is 3/2*40 =60 km.

Now they are separated by 30 km

Let the time taken to meet =t 

Distance travelled by A in time t + Distance travelled by B in time t = 30

40t + 20t =30 => t=1/2 hour

Hence they meet at 11:00 AM

The difference in the time take to traverse the same distance $$'d'$$ at two different speeds is 35 minutes. Equating this: $$\frac{d}{8}-\frac{d}{15}\ =\ \frac{35}{60}$$

On solving, we obtain $$d = 10 kms$$. Let $$x kmph$$ be the speed at which Amal needs to travel to reach the office in 50 minutes; then

$$\frac{10}{x}=\frac{50}{60}\ or\ x\ =\ 12\ kmph$$.Hence, Option B is the correct answer.

Let the total distance be 'D' km and the speed of the train be 's' kmph. The time taken to cover D at speed d is 't' hours. Based on the information: on equating the distance, we get $$s\ \times\ t\ =\ \frac{s}{3}\times\ \left(t+\frac{1}{2}\right)$$ 
On solving we acquire the value of $$\ t\ =\frac{1}{4}$$ or 15 mins. We understand that during the return journey, the first 5 minutes are spent traveling at speed 's' {distance traveled in terms of s = $$\ \frac{s}{12}$$}. Remaining distance in terms of 's' = $$\ \frac{s}{4}-\frac{s}{12}\ =\frac{s}{6}$$

The rest 4 minutes of stoppage added to this initial 5 minutes amounts to a total of 9 minutes. The train has to complete the rest of the journey in $$15 - 9 = 6 mins$$ or {1/10 hours}. Thus, let 'x' kmph be the new value of speed. Based on the above, we get $$\frac{s}{\frac{6}{x}}\ =\frac{1}{10}\ or\ x\ =\frac{10s}{6}$$

Since the increase in speed needs to be calculated: $$\frac{\left(\frac{10s}{6}\ -s\right)}{s}\times\ 100\ =\frac{200}{3}\approx\ 67\%$$ increase.

Hence, Option C is the correct answer.

Anil and Sunil will meet at a first point after LCM ( $$\frac{3}{15},\frac{3}{10}$$) = 3/5 hr

In the mean time, distance travelled by ravi = 8 * 3/5 = 4.8 km

Let the length of the train be $$ l kms$$ and speed be $$ s kmph$$. Base on the two scenarios presented, we obtain:

$$\frac{l}{s-2}=\frac{90}{3600}$$....(i) and $$\frac{l}{s-4}=\frac{100}{3600}$$...(ii)

On dividing (ii) by (i) and simplifying we acquire the value of $$s$$ as $$22 kmph$$. Substituting this value in (i), we have $$l=\frac{90}{3600}\times\ 20\ kms$$ {keeping it in km and hours for convenience}

Since we need to find $$\frac{l}{s}$$, let this be equal to $$x$$. Then, $$x\ =\ 90\times\frac{20}{22}\ =81.81\ \approx\ 82\ \sec onds\ $$

Hence, Option B is the correct choice.

Let the speed of cars be a and b and the distance =d

Minimum time taken by 1st car = 6 hours,

For maximum difference in time taken by both of them, car 1 has to start at 10:00 AM and car 2 has to start at 11:00 AM. 

Hence, car 2 will take 5 hours. 

Hence a= $$\ \frac{\ d}{6}$$ and b = $$\ \frac{\ d}{5}$$

Hence the speed of car 2 will exceed the speed of car 1 by $$\ \dfrac{\ \ \frac{\ d}{5}-\ \frac{\ d}{6}}{\ \frac{\ d}{6}}\times\ 100$$ = $$\ \dfrac{\ \ \frac{\ d}{30}}{\ \frac{\ d}{6}}\times\ 100$$ = 20

When A and B met for the first time at 10:00 AM, A covered 60% of the track.

So B must have covered 40% of the track.

It is given that A returns to P at 10:12 AM i.e A covers 40% of the track in 12 minutes

60% of the track in 18 minutes

B covers 40% of track when A covers 60% of the track.

B covers 40% of the track in 18 minutes.

B will cover the rest 60% in 27 minutes, hence it will return to B at 10:27 AM

Distance covered by A in 1 revolution = 2$$\pi\ $$*30 = 60$$\pi\ $$

Distance covered by B in 1 revolution = 2$$\pi\ $$*40 = 80$$\pi\ $$

Now, (5000+n)60$$\pi\ $$ = 80$$\pi\ $$n

=> 15000= 4n-3n   =>n=15000

Then distance travelled by B = 15000*80$$\pi\ $$ cm = 12$$\pi\ $$ km

Hence, the speed = $$\ \frac{\ 12\pi\times\ 60\ }{45}$$ = 16$$\pi\ $$

Assuming the length of race course = x and the speed of three horses be a,b and c respectively.

Hence, $$\ \frac{\ x}{a}=\ \frac{\ x-11}{b}$$......(1)

and $$\ \frac{\ x}{a}=\ \frac{\ x-90}{c}$$......(2)

Also, $$\ \frac{\ x}{b}=\ \frac{\ x-80}{c}$$......(3)

From 1 and 2, we get, $$\ \frac{\ x-11}{b}=\ \frac{\ x-90}{c}$$ .....(4)

Dividing (3) by (4), we get, $$\ \frac{\ x-11}{x}=\ \frac{\ x-90}{x-80}$$  

=> (x-11)(x-80)=x(x-90)

=> 91x-90x=880  => x=880

Assume the total distance between A and B as d and time taken by Amal = t

Since Amal travelled $$\frac{1}{3}^{rd}$$ of his total journey time in different speeds
d = $$\ \frac{\ t}{3}\times\ 10+\ \frac{\ t}{3}\times\ 20+\frac{\ t}{3}\times\ 30\ \ =\ 20t$$

$$\text{Total time taken by Bimal} = \ \frac{d_1}{s_1}+\frac{d_2}{s_2}+\frac{d_3}{s_3}$$

$$=\ \frac{20t}{3}\times\ \frac{1}{10}+\frac{20t}{3}\times\ \frac{1}{20}+\frac{20t}{3}\times\ \frac{1}{30}\ \ =\frac{20t\left(6+3+2\right)}{3\ \times30}\ =\frac{11}{9}t$$

Hence, the ratio of time taken by Bimal to time taken by Amal = $$\frac{\frac{11t}{9}}{t}=\frac{11}{9}$$
Therefore, Bimal will exceed Amal's time by $$\ \ \ \frac{\ \ \frac{\ 11t}{9}-t}{t}\times\ 100 = 22.22%$$

It is given that starting from 10:01 am, every minute a motorcycle leaves A and moves towards B.

Forty-five such motorcycles reach B by 11 am.

It means that the forty-fifth motorcycle starts at 10:45 AM at A and reaches B by 11:00 AM i.e 15 minutes.

Since the speed of all the motorcycles is the same, all the motorcycles will take the same duration i.e 15 minutes.

If the cyclist doubles the speed, then he will reach B by 10:30 AM. (Since if the speed is doubled, time is reduced by half)

Since each motorcycle takes 15 minutes to reach B, 15 motorcycles would have reached B by the time the cyclist reaches B

Speed of John = 6kmph 

Speed of Mary = 7.5 kmph 

Lengths of tracks A and B = 325 m

Let the length of track A be a, then the length of track B = 325-a

9 rounds of John on track A = 5 rounds of Mary on track B

$$\ \frac{\ 9\times\ a}{6\ \times\ \ \frac{\ 5}{18}}\ =\ \ \frac{\ 5\cdot\left(325-a\right)}{7.5\times\ \ \frac{\ 5}{18}}$$

On solving we get , 13a=1300

a=100

The length of track A = 100m, track B = 225m

Mary makes one round of track A = $$\ \frac{\ 100}{7.5\times\ \ \frac{\ 5}{18}}$$

= 48 sec