CAT Time, Speed and Distance Questions

Let us assume that the distance is D, speed of the train is S and time taken by the train is t. 

t is nothing but $$\frac{D}{S}$$

Statement 1: Had the speed been 6 km per hour more, it would have needed 4 hours less
$$\frac{D}{S+6}=t-4$$

$$\frac{D}{S+6}=\frac{D}{S}-4$$

$$4=\frac{D}{S}-\frac{D}{S+6}$$

$$\frac{S+6-S}{S\left(S+6\right)}=\frac{4}{D}$$

$$\frac{6}{S\left(S+6\right)}=\frac{4}{D}$$

$$D=\frac{2S\left(S+6\right)}{3}$$

Statement 2:  Had the speed been 6 km per hour less, it would have needed 6 hours more
$$\frac{D}{S-6}=t+6$$

$$D\left[\frac{1}{S-6}-\frac{1}{S}\right]=6$$

$$\frac{S-S+6}{S\left(S-6\right)}=6$$

$$D=S\left(S-6\right)$$

Equating the two equations for distance, 
$$S\left(S-6\right)=\frac{2S\left(S+6\right)}{3}$$

$$3S-18=2S+12$$

$$S=30$$

Hence the speed is 30 kmph

We know that the distance $$D=S\left(S-6\right)$$ = 30*24  =720 km

Let's take the scheduled time taken by the bus to be t
From the first statement (bus travelling at 60 kmph), we can get the total distance travelled by bus to 60(t+3.5)

The second scenario gives us that the bus covered two-thirds of the distance in one-third of the time, meaning that the remaining one-third distance was covered in two-thirds of the time, giving us the relation $$\frac{1}{3}st$$ covered in $$\frac{2}{3}t$$ giving the speed to be $$\frac{s}{2}$$ which is given as 40 km/h, thereby giving the usual speed of the bus to be 80 km/hr

Now the first relation we get 60(t+3.5)=80t
Giving us t=10.5 hours

Thus, the bus usually takes 10.5 hours on its journey. 
Staring at 9:00, it will complete the journey at 7:30 pm

Therefore, Option A is the correct answer. 

We can use the formula for when two people moving towards each other and meeting in a straight line. $$t^2=t_1\times\ t_2$$
Where $$t$$ is time taken for them to meet each other. $$t_1$$ is the time taken by person 1 to reach the destination after meeting and 
$$t_2$$ is the time taken by person 2 to reach the destination after meeting

We are told they meet each other in 1 and a half hours, that is 90 minutes. 
And if Sunil takes X minutes to reach A, Anil will take X+75 minutes
Since it is given that, Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A

$$90^2=x\left(x+75\right)$$
$$8100=x^2+75x$$
$$x^2+75x-8100=0$$
$$\dfrac{\left(-75\pm\sqrt{5625+32400}\right)}{2}$$
$$\dfrac{\left(-75+195\right)}{2}$$
$$x=60$$

Total time of travel of Anil is, $$90\ \min\ +60\ \min\ +75\ \min$$
Total of 225 minutes. 
Time in hours will be 3.75 hours. 
Speed is $$\dfrac{45}{3.75}=12\ \dfrac{km}{hr}$$

The given time is 8:48 AM.

Angle made by hours hand w.r.t 12 is 8 * 30 (30 degrees in 1 hour) + 0.5 * 48 (0.5 degree in 1 minute) = 240 + 24 = 264 degrees.

Angle made by minutes hands w.r.t 12 is 48 * 6 = 288 degrees.

=> The angle between them is 288 - 264 = 24 degrees.

This should further increase by 12 degrees (50% of 24)

After m minutes, the further increase in angle = (6 - 0.5)*m = $$\dfrac{11}{2}m=12$$ => m = $$\dfrac{24}{11}$$

It is given that the speed of Ravi is 40 kmph, which is equal to $$\frac{100}{9}$$ m/s. It is also known that the speed of Ashok is 50 kmph, which is equal to $$\frac{125}{9}$$ m/s.

It is known that the distance between Ravi and Ashok is 225 meters, and the relative speed of Ravi and Ashok is $$\frac{125}{9}+\frac{100}{9}=25$$ m/s 

Hence, they will meet each other in $$\frac{225}{25}=9$$ seconds. The distance traveled by Ravi in these 9 seconds is $$\frac{100}{9}\times\ 9=100$$ meters.

Since Vijay was already 54 meters behind Ravi when they were starting, Vijay must travel (100+54) = 154 meters in these 9 seconds.

Hence, the speed of Vijay is $$\frac{154}{9}$$ m/s, which is equal to $$\frac{154}{9}\times\ \frac{18}{5}\ =\frac{308}{5}=61.6$$ kmph.

The correct option is C

Given the total number of kilometres travelled, including the trip = is 26862 Km, and the duration of the trip is 8 hrs.

If avg. speed of the car during the trip is 's' => the km travelled till just before the trip is 26862 - 8s, which should also be a palindrome.

=> From the options if s = 110 => The reading will be 26862 - 110*8 = 25982 (Not a palindrome)

=> If s = 100 => The reading will be 26862 - 100*8 = 26062 => It is a palindrome.

=> s = 100 is the correct option.

Let us assume the speed of the 1st boat is b, the 2nd boat is s, and the river's speed is r.

Let 'd' be the distance between A and B.

=> d = 2(b+r) and d = 3(b-r)

=> b + r = d/2 and b - r = d/3 => r = d/12 (subtracting both equations).

Now, it is given that

$$\dfrac{d}{s+r}+\dfrac{d}{s-r}=6$$

=> $$\dfrac{d}{s+\dfrac{d}{12}}+\dfrac{d}{s-\dfrac{d}{12}}=6$$

=> $$2ds=6\left(s^2-\dfrac{d^2}{144}\right)$$

=> $$144s^2-48ds-d^2=0$$

Solving the quadratic equation, we get:

$$s=d\left(\dfrac{\left(48+\sqrt{\ 48^2+4\left(144\right)}\right)}{2\times\ 144}\right)$$

$$s=d\left(\dfrac{1}{6}+\dfrac{\sqrt{\ 5}}{12}\right)$$

=> Required value of $$\dfrac{d}{s+r}$$

= $$\dfrac{d}{\dfrac{d}{6}+\dfrac{\sqrt{5}d}{12}+\dfrac{d}{12}}$$

= $$\dfrac{12}{3+\sqrt{\ 5}}=\dfrac{\left(12\right)\left(3-\sqrt{\ 5}\right)}{4}$$

= $$3\left(3-\sqrt{\ 5}\right)$$

Let us assume the speeds of Arvind and Surbhi are 'a' and 's', respectively.

Let us say they meet after 't' hours

=> Arvind travelled s*t distance in 6 hrs and Surbhi travelled a*t in 24 hrs

=> s*t = a*6 and a*t = s*24 => $$t^2=6\times\ 24$$ => t = 12

Given a = 54 => s*12 = 54*6 => s = 27.

=> Total distance between A and B is (s+a)*t = (54+27)*12 = 81*12 = 972 Kms.

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 Car 2 be 'x' kmph and the time taken by the two cars to meet be 't' hours.

In 't' hours, Car 1 travels $$\left(60\ \times\ t\right)\ km$$ while Car 2 travels $$\left(x\ \times\ t\right)\ km$$

It is given that the time taken by Car 1 to travel $$\left(x\ \times\ t\right)\ km$$ is 45 minutes or (3/4) hours. $$\therefore\ \frac{\left(x\ \times\ t\right)}{60}\ =\ \frac{3}{4}\ $$ or $$t=\frac{180}{4x}$$....(i)

Similarly, the time taken by Car 2 to travel $$\left(60\ \times\ t\right)\ km$$ is 20 minutes or (1/3) hours. $$\therefore\ \frac{\left(60\times\ t\right)}{x}=\frac{1}{3}$$ or $$\therefore\ t=\frac{x}{180}$$....(ii)

Equating the values in (i) and (ii), and solving for x:

$$\therefore\ \frac{180}{4x}=\frac{x}{180}\ \ \longrightarrow\ \ \ x\ =90\ kmph$$

Hence, Option B is the correct answer.

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

Car 3 meets car 1 at Q, which is 200 km from A.
Therefore, at the time of their meeting car 1 must have travelled 200 km and car 3 must have travelled 100 km.
As the time is same, ratio of speed of car 1 to speed of car 3 = 2 : 1.

Car 3 meets car 2 at P, which is 100 km from A.
Therefore, at the time of their meeting car 2 must have travelled 100 km and car 3 must have travelled 200 km.
As the time is same, ratio of speed of car 2 to speed of car 3 = 1 : 2.

Speed of car 1 : speed of car 3 = 2 : 1
And speed of car 2 : speed of car 3 = 1 : 2
So, speed of car 1 : speed of car 2 : speed of car 3 = 4 : 1 : 2

Hence, option D is the correct answer. 

Let the distance between A and B be 4x. 
Length of BP is thrice the length of AP.
=> AP = x and BP = 3x

Let the speed of car 1 be s and the speed of car 2 be 0.5s. 
Car 2 reaches P one hour (60 minutes) after Car 1 reaches P.

=> x/s + 60 = 3x/0.5s
x/s + 60 = 6x/s
5x/s = 60
x/s = 12

Time taken by car 1 in reaching P from A = x/s = 12 minutes.
Therefore, 12 is the correct answer.  

Train T starts at 3 PM and train S starts at 4 PM. 
Let the speed of train T be t.
=> Speed of train S = 0.75t.

When the trains meet, train t would have traveled for one more hour than train S. 
Let us assume that the 2 trains meet x hours after 3 PM. Trains S would have traveled for x-1 hours. 

Distance traveled by train T = xt
Distance traveled by train S = (x-1)*0.75t = 0.75xt-0.75t

We know that train T has traveled three fifths of the distance. Therefore, train S should have traveled two-fifths the distance between the 2 cities. 

=> (xt)/(0.75xt-0.75t) = 3/2
2xt = 2.25xt-2.25t
0.25x = 2.25
x = 9 hours. 

Train T takes 9 hours to cover three-fifths the distance. Therefore, to cover the entire distance, train T will take 9*(5/3) = 15 hours. 
Therefore, 15 is the correct answer.

Time taken to cover first 50 km at 100 km/hr = $$\frac{1}{2}$$ hr.
Time taken to cover second 50 km at 50 km/hr = 1 hr.
Time taken to cover last 50 km at 25 km/hr = 2 hr.
When car 2 starts, car 1 has already covered 20 km.
So, time taken by car 1 to reach B after car 1 starts = total time - time required to travel first 20 km
= 3 hr 30 min - 12 min = 3 hr 18 min
Distance travelled by car 2 = (50 + 50 + 45) = 145 km
Distance from B = (150 - 145) km = 5 km

Hence, 5 is the correct answer.

Let the time taken by Partha to cover 60 km be x hours.
Narayan will cover 60 km in x-4 hours. 

Speed of Partha = $$\frac{60}{x}$$
Speed of Narayan = $$\frac{60}{x-4}$$

Partha reaches the mid-point of A and B two hours before Narayan reaches B.

=> $$\dfrac{30}{\frac{60}{x}} + 2 = \dfrac{60}{\frac{60}{(x-4)}}$$

$$\frac{x}{2} + 2 = x-4$$
$$\frac{x+4}{2}=x-4$$
$$x+4=2x-8$$
$$x=12$$

Partha will take 12 hours to cross 60 km.
=> Speed of Partha = $$\frac{60}{12}=5$$ Kmph.
Therefore, option D is the right answer. 

Let 'a' and 'b' be the speed (in km/hr) of cars starting from both A and B respectively. 

If they both move in east direction, then B will catch A if and only if b > a. 

Relative speed of both the cars when they move in east direction = (b - a) km/hr

It takes them 7 hours to meet. i.e. they travel 350 km in 7 hours with a relative speed of (b-a) km/hr. 

Hence, (b - a) = $$\dfrac{350}{7}$$ = 50 km/hr.

By the time A traveled 10 KM, B traveled 9 KM
Hence $$Speed_A : Speed_B = 10:9$$
Similarly $$Speed_B : Speed_C = 10:9$$
Hence $$Speed_A : Speed_B : Speed_C = 100:90:81$$
Hence by the time A traveled 10 KMs , C should have traveled 8.1 KMs
So A beat C by 1.9 KMs = 1900 Mts

We see that the man saves 20 minutes by changing his speed from 12 Km/hr to 15 Km/hr.

Let d be the distance

Hence,

$$\frac{d}{12} - \frac{d}{15} = \frac{1}{3}$$

$$\frac{d}{60} = \frac{1}{3}$$

d = 20 Km.

Let the distance between the home and office be $$2x$$ miles
Time taken for going in the morning = $$\frac{2x}{60}$$ hrs
Time taken for going back in the evening = $$\frac{x}{25} + \frac{x+5}{50}$$. hrs
It is given that he took 30 minutes (0.5 hrs) more in the evening

Hence $$\frac{2x}{60}$$ hrs + 0.5 = $$\frac{x}{25} + \frac{x+5}{50}$$
Solving for x, we get x = 15 miles.
Total distance traveled = 2x + x + x + 5 = 4x + 5 = 65 Miles

Let the speed of the river be $$x$$ and the speed of the boat be $$u$$. Let $$d$$ be the one way distance and $$t$$ be the initial time taken.

Given,

$$t = \frac{d}{u - x} + \frac{d}{u + x}$$ ... i

Also,

$$\frac{t}{4} = \frac{d}{2u - x} + \frac{d}{2u + x}$$

$$t = \frac{4d}{2u - x} + \frac{4d}{2u + x}$$ ... ii

Equating both i and ii,

$$\dfrac{d}{u - x}$$+ $$\dfrac{d}{u + x}$$ =  $$\dfrac{4d}{2u - x} + \dfrac{4d}{2u + x}$$

$$\dfrac{2u}{u^2 - x^2} = \dfrac{16u}{4u^2 - x^2}$$

$$4u^2 - x^2 = 8u^2 - 8x^2$$

$$\frac{u^2}{x^2} = \frac{7}{4}$$

$$\frac{u}{x} = \frac{\sqrt{7}}{2}$$

Let the distance traveled by the car be x KMs
Distance traveled by the bike = 168 KMs
Speed of car is double the speed of bike
=> $$\frac{x}{3:40 - 2:00}$$ = $$2 \times \frac{168}{3:40 - 1:00}$$
=> $$\frac{x}{100}$$ = $$2 \times \frac{168}{160}$$
=> x = 210
Hence the distance between A and B is x + 168 = 378 KMs

According to given conditions angle between AC and AB is 30 degrees and between AB and BC is 60 degrees. So the triangle formed is a 30-60-90 triangle. 

Hence, $$\angle\ ABC=60^{\circ\ }$$, $$\angle\ CAB=30^{\circ\ }$$, which implies $$\angle\ ACB=180^{\circ\ }-\left(30^{\circ\ }+60^{\circ\ }\right)=90^{\circ\ }$$

Therefore, The triangle is a right-angled triangle with base = BC, height = AC, and hypotenuse = AB.

We know that $$\cos\ \angle\ ABC\ =\ \frac{\ BC}{AB}=>\ \cos60^{\circ\ }=\frac{BC}{500}\ =>\ 500\times\ \frac{1}{2}=250\ km$$

Similarly, $$\sin\ \angle\ ABC\ =\ \frac{\ AC}{AB}=>\ \sin60^{\circ\ }=\frac{AC}{500}\ =>\ 500\times\ \frac{\sqrt{\ 3}}{2}=250\sqrt{\ 3}\ km$$

So, total time taken by train to travel B to C is is (250/50) = 5 hrs, hence the train reaches at 1 pm. Accordingly, Rahim has to reach C fifteen minutes before i.e. at 12:45 PM.

Time taken by Rahim to travel by car is around $$\ \frac{\ 250\sqrt{\ 3}}{70}=6.19$$ hrs (Around 6.2 hrs = 6 hrs 12 minutes). So, the latest time by which Rahim must leave A and still be able to catch the train is 6:30 am.

Let the speed of the plane be p Kmph.
So the speed of plane from A to B will be 'p+50' and the speed from B to A will be 'p-50'.
We notice that the plane goes from B to A stays there for 1 hr and again come back to B with total time duration 12 hrs.
So we have $$\frac{3000}{p-50} + 1 + \frac{3000}{p+50} = 12$$.
We can clearly see that speed of the plane is 550 which satisfies the above equation.
So for the journey of B to A, the plane takes $$\frac{3000}{550-50} = 6$$ hrs.
So time at B when plane reaches at A is 2 pm .
Hence the time difference between A and B is 1 hr.

Alternatively,
Let speed of flight be s,
Since A is to the east of B, A is ahead of be in time
Let A be ahead of B in time by a hours
Departure from A = 4PM, Arrival at B = 8PM
Travel time = 8 - 4 +a = 4 + a
Since City B is behind city A by 'a' hours, the actual travel time is 'a' hours more than the difference of local times.
Similarly when one travels from B to A, since B is ahead of A by 'a' hrs, actual travel time is 'a' hours less than total
i.e. B->A Travel time = (3PM - 8AM) - a = 7 -a
Total distance travelled = Speed $$\times$$ Time taken ....(1)
From A to B, the wind is favourable / in same direction as flight
Hence from (1), we have 
A->B >>> $$3000 = (s+50)(4+a) => 3000(7-a) = (s+50)(4+a)(7-a)$$ ...(2)
B->A >>> $$3000 = (s-50)(7-a) => 3000(4+a) = (s-50)(7-a)(4+a)$$ ...(3)
(2) - (3) => $$3000(3-2a) = 100(7-a)(4+a) => a^2-63a+62=0 => a=1/62$$
Hence the time difference between A and B is 1 hr.

Let the speed of the plane be p Kmph.

So the speed of plane from A to B will be 'p+50' and the speed from B to A will be 'p-50'.

We notice that the plane goes from B to A stays there for 1 hr and again come back to B with total time duration 12 hrs.

So we have $$\frac{3000}{p-50} + 1 + \frac{3000}{p+50} = 12$$.

On substituting the options, we can clearly see that speed of the plane is 550 which satisfies the above equation.

Let the distance be D.
Time taken by Arun = D/30
Time taken by Barun = D/40
Now, D/40 = D/30 - 2
=> 3D = 4D - 240
=> D = 240
Therefore time taken by Arun to cover 240 km = 240/30 = 8 hr
Time Kiranmala takes to cover 240 km = 240/60 = 4 hr
So, Kiranmala has to start 4 hours after Arun.

Let the time at which they meet be t minutes past 10.

So, distance run by Ram + distance run by Shyam = 10 km

=> (60+t)5/60 [t+60 because he would have traveled for 9 am to 10 am and t minutes more before meeting Shyam]+ (15+t)*10/60 [15+t because he would have traveled from 9:45 to 10:00 and t minutes more]= 10

=> 300+5t+150+10t = 600 => t = 10

So, they meet at 10.10 am

Let the time at which Shyam overtakes Ram be t minutes past 10.
So, distance run by both of them is the same till that moment.
(60+t)5 = (15+t)10 => 300 + 5t = 150 + 10t => 5t = 150 => t = 30 min.
So, at 10.30 am, Shyam overtakes Ram

Let the distance to be travelled be D.

In the first case, D/10 = t

In the second case, D/15 = t-2

=> D/15 = D/10 - 2

=> 2D = 3D - 60

=> D = 60 km and T = 6 hours

Therefore, to get there at noon, he has to travel at 60/5 = 12 km/hr

Let radius be 1 units and p = 3.14 or $$\pi$$ . So circumference is $$2*\pi$$.

According to given condition distance covered in first 1/2 mins = $$\pi$$/2 km, distance covered in next 1 min = $$\pi$$/2 km, distance covered in next 2 mins = $$\pi/2$$ km and finally distance covered in next 4 minutes = $$\pi/2$$ km.

Time taken to cover first round = 1/2 + 1 + 2 + 4 = 7.5 minutes.

Now time taken to cover $$\pi/2$$ is in GP.

For the second round the time taken is = 8+16+32+64 = 120

Ratio = 120/7.5 = 16

The speeds of Karan and Arjun are in the ratio 10:9. Let the speeds be 10s and 9s.
Time taken by Karan to cover 110 m = 110/10s = 11/s
Time taken by Arjun to cover 100 m = 100/9s = 11.11/s
Therefore, Karan reaches the finish line before Arjun. From the options, the only possible answer is d).

The relative speed is 15 km/hr = 15 km/60 min = 0.25 km/min = 250 m/min.
Therefore, one minute before they collide, they are at a distance of 250m.

Let A , B and f,s be the distance traveled and speed of the fastest and the slowest person respectively. Also f=2s so in the given time A=2B. Since the ration of the speeds is 2:1, they will meet at 2-1 points = 1 pont.

Both meet each other for first time at starting point . let b travel distance equal to 1 circumference i.e. 1000m so A=2000m . Both meet after 5 min so speed of slowest is 1000/5=200m/min . So speed of the fastest is 400m/min. So time taken by A to complete race 4000/400 = 10 min

Let the radii of 2 circles be R and r respectively such that R=2*r. Triangle $$ON_2E_1$$ and all the other 3 similar triangles form a right angle at the centre. So, using Pythagoras theorem, the value of chords comes out to be $$\sqrt{\ 5}\times\ \dfrac{R}{2}$$ . Hence, the total distance travelled is $$\sqrt{\ 5}\times\ \dfrac{R}{2}$$ + $$0.5\ \times\ R\ \times\ \pi\ $$. The total time required can be calculated by distance/speed, which comes out to be 3.5 * R. Among options, only 105 is an integral multiple of 3.5.

Let the radius of outer circle be 2R and the centre of both the circles be O.

Triangle $$ON_2E_1$$and all the other 3 similar triangles form a right angle at the centre. 

Let the radius of the inner ring road be R

The radius of outer will be 2R as the circumference of the outer ring road is double that of the inner ring road. 

So, in triangle $$ON_2E_1$$ using Pythagoras theorem the value of chords come out to be $$\sqrt5$$ * R so the total length of the chords 4 * $$\sqrt5$$ * R and circumference is equal to 2 *Pi*2R. The ratio gives option C.

We know that the total time taken is 1.5 hrs. Calculating the individual time taken and the adding and then equating to 1.5.

Let R be the radius of the outer-ring road.

$$\frac{\pi*R}{2*30*\pi} + \frac{\sqrt{5}*R}{2*15*\sqrt{5}}$$ = 1.5 solving we get R=30.

Let the speed of an express train be 4x, normal train be 2x and passenger train be x.
Let the distance between the 2 stations be D. 
Since there is only 1 railway track, train N must reach station B before train S leaves. 
Therefore, D/4x + D/x = 60 
5D/4x = 60
D/x = 48

Train N leaves 20 minutes late. Therefore, the 2 trains must have covered the distance within 40 minutes on this particular day. 
Train N doubles its speed. Therefore, speed of train N will be 8x. Let the new speed of the passenger train be y.

D/8x  + D/y = 40
48/8 + D/y = 40
D/y = 34.

Speed of super fast train = D/8x = 6
Speed of passenger train = D/y = 34

Ratio of the speeds = 6/34 = 3/17.
The ratio is approximately equal to 1:6. Therefore, option D is the right answer.

Let the distance between gutter 1 and A be x and between gutter 1 and 2 be y.

Hence, x + y + 2y + x = 20 => 2x+3y=20

Also x = 30kmph * 5/60 = 2.5km

Hence, y = 5km

After the ambulance doubles its speed it goes at 60kmph i.e. 1km per min. Hence, time taken for the rest of the journey = 15*2 + 2.5 = 32.5

It takes 1 min to load and unload the patient.

Hence, total time = 5 + 32.5 + 1 = 38.5 mins

So, the doctor would get 1.5 min to attend to the patient.

Let the length of the tunnel be x and distance of the train to entrance A be y. Let the speeds of train and cat be t and c respectively.

Hence, when the cat runs 3x/8, the train covers y.

=> (3x/8)/c = y/t --- (1)

When the cat runs 5x/8 to the other end, the train covers x+y

=>(5x/8)/c = (x+y)/t ---(2)

Taking ratio of (1) to (2)

3/5 = y/(x+y) => 3x = 2y ---(3)

Substituting (3) in (1)

(2y/8)/c = y/t

=> t = 4c

Hence the ratio t:c is 4:1.

$$12/(R - S) = T$$
$$12/(R + S) = T - 6$$
$$12/(2R - S) = t$$
$$12/(2R + S) = t - 1$$
=> $$12/(R - S) - 12/(R + S) = 6$$ and $$12/(2R - S) - 12/(2R + S) = 1$$
=> $$12R + 12S - 12R + 12S = 6R^2 - 6S^2$$ and $$24R + 12S - 24R + 12S = 4R^2 - S^2$$
=> $$24S = 6R^2 - 6S^2 and 24S = 4R^2 - S^2$$
=> $$6R^2 - 6S^2 = 4R^2 - S^2$$
=> $$2R^2 = 5S^2$$
=> $$24S = 10S^2 - S^2 = 9S^2$$
=> $$S = 24/9 = 8/3$$

Let the number of steps on the escalator be x.

So, by the time Shyama covered 25 steps, the escalator moved 'x-25' steps.

Hence, the ratio of speeds of Shyama and escalator = 25:(x-25)

Similarly, the ratio of speeds of Vyom and escalator = 20:(x-20)

But the ratio is 3:2

Ratio of speeds of Shyama and Vyom = 25(x-20)/20*(x-25) = 3/2

=> 10(x-20) = 12(x-25)

=> 2x = 100 => x = 50

Distance between A-B , A-C, C-B is 180, 120 and 60 km respectively.

Let x be the distance from A where the 2 trains meet.

According to given condition we have

$$\frac{x}{70}=\frac{60}{50} + \frac{1}{4} + \frac{120-x}{50}$$.

Solving the equation we get x around 112 km.

Let x be the required distance.

Let a,b,c be speed of the A,B and C respectively.

From the given conditions we have, 

$$\frac{a}{b}=\frac{x}{x-12}$$ and $$\frac{a}{c}=\frac{x}{x-18}$$ and $$\frac{b}{c}=\frac{x}{x-8}$$ . From first 2 equations we can deduce $$\frac{b}{c}=\frac{x-12}{x-18}$$. 

$$\frac{b}{c}=\frac{x-12}{x-18} = \frac{x}{x-8}$$

x = 48 satisfy the equation.

If the truck is being driven at 70 kmph, it takes 1.3 liters of diesel to travel 19.5 km.
Therefore, with 10 liters of diesel, the truck can travel 10/1.3 * 19.5 km = 150 km.

At the beginning, the robot gets the message from both A and D. So it will cated A and D in the same trip.
Distance travelled= 40*2=80m
Time taken=8 sec
At the beginning of 9th second,it recieves message from both B and C and hence caters to them in the same trip.
Distance travelled=30*2=60m

Total distance travelled= (60+80)= 140m

Since the machine recieves the message from A and D, it will cater to them in the same trip.In return journey,it will go to the point which is 10 m away from E.
Distance travelled=40+20=60m
Now it recieves the message from B and C and caters them in the single journey.In return journey,it will go to origin.
Distance travelled=40+20=60m

Total distance travelled=60m+60m=120m

The distance between Ahmedabad and Baroda is 100 Km
Navjivan express starts at 6:30 am at 50 Km/hr and Howrah expresses starts at 7:00 am at 40 Km/hr.
Distance covered by Navjivan express in 30 minutes (by 7 am) is 25 Km/hr.

So, at 7 am, the distance between the two trains is 75 Kms and they are travelling towards each other a relative speed of 50+40=90 Km/hr.

So, time taken them to meet is 75/90*60 = 50 minutes.

As, Mr. Shah realizes the problem after thirty minutes, time left to avoid collision is 50-30 = 20 minutes

The function of the speed of the train = 42 - k$$\sqrt{n}$$ where n is the number of compartments and k is a constant.

42 - k$$\sqrt{9}$$ = 24

=> 3k = 18 => k = 6

=> Function of speed = 42 - 6$$\sqrt{n}$$

Speed is 0 when 42 - 6$$\sqrt{n}$$ = 0

=> 42 = 6$$\sqrt{n}$$

=> n = 49

=> So, with a positive speed, the train can carry 48 compartments.

Total distance = 35*2 + 45*2 = 160 km

Distance traveled by Aditi at each constant speed value = $$\frac{160}{3}$$

Total petrol consumed by Aditi = $$\frac{160/3}{16}$$ + $$\frac{160/3}{24}$$ + $$\frac{160/3}{16}$$ = $$\frac{10}{3}$$ + $$\frac{20}{9}$$ + $$\frac{10}{3}$$ = 6.67 + 2.22 = 8.9 litres

To minimize the petrol consumption, he must travel at 40kmph constantly through out the 160 km.

Total petrol consumed = $$\frac{160}{24}$$ = 6.67 litres.

Y takes the direct route AC and travels at 45 km per hour on segment AD. Y's speed on segment DC is 55 km per hour.

AD=DC =a(D is the midpoint on the road connecting A and C.)

The time taken by Y to reach D= a/45

Further, the time taken by Y to reach C= a/55

Y has travelled equal distances are constant speed.

=> Average speed = $$\frac{2a}{\frac{a}{45}+\frac{a}{55}}$$=$$\frac{2*45*55}{100}$$ = 49.5 kmph

$$BC^2$$ + 10000 = $$AC^2$$.

$$\frac{BC+100}{61.875}$$ = $$\frac{AC}{49.5}$$ => $$\frac{BC+100}{AC}$$ = 1.25 = > BC = 1.25AC - 100

On solving these equations, we get AC as 105 and BC a 31.

=> Y traveled 105 km.

$$BC^2$$ + 10000 = $$AC^2$$.

$$\frac{BC+100}{61.875}$$ = $$\frac{AC}{49.5}$$ => $$\frac{BC+100}{AC}$$ = 1.25 = > BC = 1.25AC - 100

On solving these equations, we get AC as 105 and BC a 31.

Let the point B be (0,0) => A = (0, -100) and C = (31,0)

D = Mid point of AC = (15.5, -50)

BD = $$\sqrt{15.5^2+50^2}$$ = 52.5 (approximately).

In the first hour, they cover 6km. In the second hour they cover 6.5km. In the third hour, they cover 7km and so on.

Finally, they cover 72km in 9 hours.

Distance covered by A in 9 hours = 4*9 = 36km

=> They meet mid-way between A and B.

The first wheel completes a revolution in $$\frac{60}{60}=1$$ second
The second wheel completes a revolution in $$\frac{60}{36}=1\frac{2}{3}$$ second
The third wheel completes a revolution in $$\frac{60}{24}=2\frac{1}{2}$$ second

The three wheels touch the ground simultaneously at time which are multiples of the above three times.
Hence, the required number is $$LCM(1,\frac{5}{3},\frac{5}{2}) = 5$$ seconds.

So, the correct option is option (c)

The total time taken for both the journeys is the same. So, a) is incorrect. If he goes with the same speed both ways, he would've been at the same point at noon on both days. So, d) is incorrect. If he does not travel with the same speed in both the directions, he need not be at the same spot at noon. So, option b) is incorrect. Option c) is the correct answer.

Let's say length of express train = $$x$$
So length of goods train = $$2x$$
Total length travelled by express train = $$3x = ((80-40) \times \frac{5}{18}) \times 54 $$ (Where $$(80-40) \times \frac{5}{18})$$ = relative velocity of express train w.r.t. goods train in meter/sec.)
So $$x =200$$ meter.
Now crossing a plateform of length 400 m., total length travelled by train = 600 m=$$t\times(80 \times \frac{5}{18})$$
$$t = 27$$ sec.

Time taken to catch the thief = $$\frac{d}{v}$$  (Where d is relative distance achieved in 15 min. = 15km. and v is relative velocity of poiliceman i.e. = 65-60 = 5 kmph)
So time taken = 3hr.
Hence he will catch thief at 3:15 pm

As relative velocity of other poilceman will be 0 , hence distance remain same as starting i.e. = 15 km.

Anshuman travels at a minimum speed by car over A, so the time taken = $$\frac{2}{40}$$*60= 3 min
Anshuman travels stretch B at the fastest speed, so the time taken = $$\frac{2}{50}$$*60= 12/5 min
So in order to break the record, he has to cover the three stretches in less than 10 min
10-(3+$$\frac{12}{5}$$) = $$\frac{23}{5}$$
So the time taken to cover stretch C should be less than $$\frac{23}{5}$$ min.
Let x kmph be the speed at which he covers stretch C
$$\frac{2}{x}$$*60 < $$\frac{23}{5}$$
x > 26.08 kmph but the maximum speed at which he can cover stretch C is 20 kmph.
Hence it is not possible for him to break his previous record.
C is the correct answer.

Total time taken by Mr. Hare to complete the race = 50% more of (10 min.) = 15 min.= $$\frac{1}{4}$$ hr.
Or $$\frac{1}{4} = \frac{2}{40} + \frac{2}{40} + \frac{2}{v}$$ ( i.e. complete time for strech A,B and C Where v is velocity at strech C)
Or  v= 13.3 km./hr

Average speed to cover complete race = 20 km/hr = $$\frac{2+2+2}{t_1 + t_2 + t_3}$$ (Where $$t_1$$ is time taken to cover the distance A, $$t_2$$ is time taken to cover the distance B and $$t_3$$ is time taken to cover the distance C)

So total time =$$ t_1 + t_2 + t_3 = 18$$ min.

Avg. speed for first two streches = $$\frac{4}{t_1 + t_2}$$

Avg. speed for last strech = $$\frac{2}{t_3}$$

Given: $$\frac{4}{t_1 + t_2} = 4 \times \frac{2}{t_3}$$

Or $$2t_1 + 2t_2 = t_3$$

Or $$t_3 = 12$$ min.

So $$V_3 = \frac{2}{12} \times 60$$ = 10 km/hr

X leaves Frankfurt at 6 PM and reaches Boston at 10 AM, which is 6 AM in Frankfurt => 12-hour journey

Leaves Boston at 12 PM ad reaches India at 1 AM, which is 11 PM in Boston => 11-hour journey

=> Total time = 12 + 2 + 11 = 25 hours

Average speed = 180 mph

=> Distance = 25 * 180 = 4500 miles

X leaves Frankfurt at 6 PM and reaches Boston at 10 AM, which is 6 AM in Frankfurt => 12-hour journey

Leaves Boston at 12 PM ad reaches India at 1 AM, which is 11 PM in Boston => 11-hour journey

=> Total time = 12 + 2 + 11 = 25 hours

Return journey is 1 hour lesser => 25 - 1 = 24 hours.

Distance is not known to find the average speed.

=> Data insufficient

Akshay can complete 1600 - 128 = 1472 m and Bhairav can completer 1600 m in the same time.

Bhairav can complete 100 m and Chinmay can complete 96 m in the same time.

=> Bhairav can complete 1600 m and Chinmay can complete 1536 m in the same time.

=> Akshay can complete 1472 m and Chinmay can complete 1536 m in the same time.

1.5 miles => 2400 m

Distance travelled by Akshay by the time Chinmay completes 1.5 miles = $$\frac{1472}{1536}*2400$$ = 2300 m

=> Akshay lost by 100 m, which is $$\frac{1}{16}th$$ of a mile.

$$\frac{\frac{3x}{5}}{3a} + \frac{\frac{2x}{5}}{2b} = \frac{2x}{5c}$$

(Where x is distance between A and B; $$\frac{\frac{3x}{5}}{3a}$$ = time taken to cover the distance with speed 3a ; $$\frac{\frac{2x}{5}}{2b}$$ = time taken to cover the distance with speed 2b; $$\frac{2x}{5c}$$ = time taken to cover the distance x from B to A then return.)

$$\frac{\frac{3x}{5}}{3a} + \frac{\frac{2x}{5}}{2b} = \frac{2x}{5c}$$

Or $$\frac{1}{a} + \frac{1}{b} = \frac{2}{c}$$

Total time taken to reach at D:

$$\frac{12}{x} + x + \frac{12}{2x} + 2x + \frac{12}{4x} = 16$$

Or $$3x^2 - 16x + 21 = 0$$

From the options we can see that only, $$x$$ = 3hr satisfies the equation. Thus, A is the right choice.

In a normal watch, the minute hand crosses the hour's hand after every 1 hour 5 minutes and 27 seconds.
So, the third time the hour's hand crosses the minute's hand is after 3 hours 16 minutes and 21 seconds.

In this watch, the time taken for this to happen is 3 hours 18 minutes and 15 seconds.
Hence, the watch loses 1 minute and 54 seconds after every 3 hours 18 minutes and 15 seconds.
18 minutes and 15 seconds = 1095 seconds = 1095/3600 $$\approx$$ .304 hours.
=> 3 hours 18 minutes and 15 seconds = 3.304 hours
So, time lost in a day is $$1\frac{54}{60}*\frac{24}{3.304} = \frac{114}{60}*\frac{24}{3.304} \approx 13.8$$
So, the time lost by the watch in a day is approximately equal to 13 minutes and 48 seconds.

Since we know that Neera's husband drives at a uniform speed to and from his residence.
If he saved 10 mins overall travel time, he should have driven 5 mins less towards railway station and 5 mins less while driving towards residence.
If he saved 5 minutes in his return journey, he should have started to return 5 minutes before his actual return time.
When the husband met Neera, he should have met her 5 minutes before the actual meeting time i.e. at 5.55 PM. 
So, Neera must have walked for 55 minutes from 5PM.

Time difference, when second time car's engine failed at a distance of 30 km., is of 9 min.
Hence putting this in equation:
$$\frac{12}{\frac{4v}{5}} - \frac{12}{v} = \frac{9}{60}$$ hr. (Because difference of time is considered with extra travelling of 12 km. in second case)
We will get $$v (velocity) = 20$$ km/hr.
Now for distance $$ \frac{d-18}{16} + \frac{18}{20} - \frac{d}{20} = \frac{45}{60}$$ hr. (As car is 45 min. late after engine's faliure in first case)
So $$d$$ = 78 km. 
Hence none of these will be our answer.

let's $$l_s$$ is length of slower train and $$l_f$$ is length of faster train.
So according to second condition when two trains are moving in same direction
$$l_s = v_{fs} \times t$$  (where $$v_{fs}$$ is relative velocity of faster train w.r.t. slower train and t is time taken to cross it)
or $$l_s = \frac{(60-50) \times 5}{18} \times 18$$ = 50 meter
Only option which has length of slower train as 50 is C.