Let us assume that there are k steps on the escalator. In the first run, she took 25 steps in 60 seconds. In the same time, the escalator also moved k-25 steps .
=> Speed of madhu=25/60=5/12 steps/sec and speed of escalator= (k-25)/60 steps/sec
When she doubles her speed, it takes her 45 seconds to reach the top.
=> Total number of steps= $$=k=\frac{10}{12}\times 45 +\frac{k-25}{60}\times 45$$
=> $$\frac{150+3k-75}{4}=k$$ => k=75.
Time taken by Madhu to reach the top on a stationary escalator = $$\frac{75}{\frac{5}{12}}= 15 \times 12 $$= 180 seconds= 3 minutes

Let the time taken by A = t
Speed of A = 1000/t
Speed of B = 900/t
Speed of C = 900*9/10/t = 810/t
Time taken by C in the second race = 100t/81.
Time taken by B in the second race = 11t/9.
Time taken by A in the second race = 12t/10.
Out of these 12t/10 is the lowest. Hence A wins the race.
B is the second in race.
Distance covered by B in 12t/10 = 900/t*12t/10 = 1080
Distance = 1000+100-1080 = 20 m

Let the distance between the cities be 2x ( to simplify the calculations)
Hence after some time, Jessie and Jackie are at $$A_1$$ and $$B_1$$ respectively.
$$AA_1$$ = 60. Hence $$A_1C$$ = $$2x - 60$$.
Also Triangle $$A_1B_1C$$ is right angled with angle C = 60. Hence the side $$CB_1$$ will be half of the hypotenuse.
=> $$CB_1$$ = $$\frac{A_1C}{2}$$ = $$ x - 30$$ => $$BB_1$$ = $$2x - (x - 30)$$ = $$ x + 30$$
Also we know that Jessie is 90 KMs from the destination. Hence $$AA_2$$ = $$2x - 90$$

The ratio of distances traveled by Jessie and Jackie in the same time intervals should be same.
Hence, $$\frac{60}{x+30}$$ = $$\frac{2x-90}{2x}$$
Solving for $$x$$ we get x = 90. Hence we have $$AA_2$$ = 90.
=> Jessie traveled 90 Kms in 3hours. Hence time that Jessie will take to cover the remaining 90 Kms is 3 hours.

Time at which Ajay and Vijay meet is 40*(11-5)/(80-40)+11 = 17:00 pm. Assuming that Alay started at 't', (17-t)*10 = (t-5)*40 or t =7.4 which is 7:24 am
Alternatively,
Vijay travelled for 6 hours before Ajay started, so he would have covered 6*40 = 240 km. Now Relative speed between them is 40, so they will meet after 240/40 hours. i.e. at 5 pm. and by then the total distance traveled by them would have been 80*6 = 480 km. Alay travels at 50 km/hr, so to travel 480 he would require 480/50 = 9 hours 36 minutes. Hence, he would start at 7:24 am

Let the speeds of Kiran, Kumar, and Kishore be a, b and c respectively. Let the length of the track = L
The time at which Kiran and Kishore meet for the first time = $$\frac{L}{a + c}$$
When Kiran and Kishore meet, Kumar would be at a distance of L/2 from them.
The net relative speed of Kiran and Kumar = a - b and the time for them to be at a distance of L/2 is equal to $$\frac{L}{a + c}$$.
Thus, $$\frac{L}{a + c} \times (a - b) = L/2$$
On solving we get, a = 2b + c.
From the options only 8 : 3 : 2 satisfy the above equation

Let L and S be the length and speed of the train. L/(S-15) = 72/3600 and L/(S-20) = 81/3600. So, L = 0.9 km

Let the work done together by 10 men and 20 women in a day be W. Total work = 48W. After the first day, work remaining = 47W. Because of the replacement, let the additional work that can be done be D. So, 47W = 45(W+D). D=2/45 W. Similarly, no. of extra days needed in the second case = 46W/(W+2*2/45W) = 42.2

Since the round-up of a number is the nearest integer more than the given number. Hence A is the answer. (Note that there is a difference between round up, round off and round down )

Hence 43 is the answer.

Suppose by the time Ajay reaches C, Vijay reaches E.

Let the length of AB be 100 metres and let AC=X , thus BE also is equal to x as speed of Ajay and Vijay is same during that interval.

Let, CD=Y metres. Thus, ED = 2y as the speed of Vijay from E to D is twice that of Ajay's from C to D.

Hence, Total distance = AC + BE + CD + ED = 2X+3Y=100 and AD = AC + CD = X+Y=40

So, X=20 and Y=20. Hence the ratio is 1:2

Speed of Garibrath express = 72 km/hr = 20m/s
Speed of Durunto express = 54 km/hr = 15m/s
When the trains start to cross each other, Ram is at front end of Garibrath express and Shyam is at rear end of Durunto express. So the initial distance between them is equal to the length of Durunto express.
Hence initial distance between them = 180m
Shyam is moving in the same direction as the train so his effective speed is 15+4 = 19 m/s
Ram is moving in the direction opposite to the train, so his effective speed is 20-3 =17 m/s.
Hence with reference to the train Ram and Shyam are moving in the same direction but with reference to ground they are moving in the opposite direction with relative speed of 19+17 = 36 m/s
Total distance to be covered = 180 m
Hence required time = 180/36 = 5 seconds

B reaches 30 seconds after A and C reaches 20 seconds after B. So, C reaches 50 seconds after A

When the two meet for the first time, the total distance travelled by them is l, where l is the distance between P and Q.
When they meet for the second time the total distance travelled by them is 3l
The first meeting point was 720 m from P, so this means that Ashu runs 720 m before their first meeting. Since the speed remains constant, the distance travelled by Ashu in the time that they cover 3d distance = 720*3 = 2160m
Since the second meeting point is 800 m from Q,
2160 = l+800
=> l = 2160-800 = 1360
So length of track is 1360.
By the time of first meeting, Ashu had travelled 720, so distance travelled by Nitesh is 1360-720 = 640
So ratio of speeds of Ashu and Nitesh is 9:8
Total distance travelled by two when they meet for the 3rd time = 5l = 1360*5 = 6800m
=> Distance travelled by Ashu = 6800*9/17 = 3600 m
=> Distance of third meeting point from Q = 4080 - 3600 = 480m

The frisbee travels in the air until Ram and Shyam meet. Assume that the speeds of Ram and Shyam are a m/s and b m/s respectively. So, they will meet after 100/(a+b) seconds. The frisbee travels throughout this time.

=> Distance travelled by frisbee= $$\frac{100}{a+b}\times 20= 12000$$ => $$a+b=1/6 $$m/s or 10 m/min

Ratio of a:b is 2:3 => a= 4 m/min and b = 6 m/min

Let the total volume of the tank be 120 litres. Therefore, the inlet pipes fill 30, 20 and 15 litres/ hour respectively. The outlet pipes empty 60 and 12 litres/hour respectively. Therefore, the rate of outflow is more than the rate of inflow. Hence the tank can never be full.

The time taken by Y to reach Town B is 50/10 = 5 hours.
X covers 6*5 = 30 Km by the time Y reaches Town B.

At that time, the distance between X and Y is 50-30 = 20 km and their relative speed is 6+10 = 16 Km/hr
So, the time taken by them to meet is 20/16 = 1.25 hours
Distance covered by X in this time is 1.25*6 = 7.5 Km

So, distance between Town A and Town C is 30+7.5 = 37.5 Km

Speed of Ajay, A = 150%V. So, relative speed is V/2. Time taken to run 400m at V/2 = 4 minutes. So, V=200m/min. Ajay's speed is = 150%*200m/min = 300m/min. Time taken by Ajay to run 2400m is 8 minutes.

Let the efficiency of each person be 1 unit of work per day. So, number of units of work to be done in total = 100 units
Work done on the first day = 1 unit
Work done on the second day = 2 units
Work done on the third day = 3 units and so on
Let the work be completed in ‘n’ days
Work completed in ‘n’ days = 1 + 2 + 3 +…+ n = n(n+1)/2 units
So, n(n+1)/2 = 100 => n(n+1) = 200
Solving this, we get n = 13.65 days
So, the work gets completed on the 14th day

The hands of a correct clock coincide after every 720/11 minutes.
So, 720/11 minutes of the first clock, equal 60 minutes on a correct clock and 720/11 minutes of the second clock equal 72 minutes on a correct clock.
So, in 1 minute on a correct clock, the difference in times of the two clocks is 720/11*[1/60 - 1/72] = 2/11 minutes.
So, time difference in a day is 2880/11 = 262 minutes

Suppose by the time Ajay reaches C, Vijay reaches E.

Let the length of AB be 100 metres and let AC=X , thus BE also is equal to x as speed of Ajay and Vijay is same during that interval.

Let, CD=Y metres. Thus, ED = 2y as the speed of Vijay from E to D is twice that of Ajay's from C to D.

Hence, Total distance = AC + BE + CD + ED = 2X+3Y=100 and AD = AC + CD = X+Y=40

So, X=20 and Y=20. Hence the ratio is 1:2

Relative Speed of the two guys is 40+60=100 km/hr. So, in 3 minutes, they cover 3/60*100 = 5 km

Speed of Ajay, A = 150%V. So, relative speed is V/2. Time taken to run 400m at V/2 = 4 minutes. So, V=200m/min. Ajay's speed is = 150%*200m/min = 300m/min. Time taken by Ajay to run 2400m is 8 minutes.

Let the speed of A be $$S_A$$ and the speed of B be $$S_B$$

Everyday, A and B travel 15 mins each. But on this specific day, B traveled 12 mins and A traveled 17 mins.
So, $$15*(S_A+S_B) = 12*S_B + 17*S_A$$
Or, $$3*S_B = 2*S_A$$

Therefore, the walking speeds of A and B are in the ratio 3:2.

So, B covers 2/5 th of the distance between their houses in 15 min.

Therefore, to cover the entire distance, he takes 37.5 min.

Let the speed of the water be W and the speed of rowing in still water be R. Let the distance between Haridwar and Rishikesh be S.
So, $$\frac{S}{R-W} = 5$$
And, $$\frac{S}{R} - \frac{S}{R+W} = \frac{5}{6}$$
Therefore, $$\frac{SW}{R(R+W)} = \frac{5}{6}$$
Dividing first equation by this, we get, $$\frac{R(R+W)}{W(R-W)} = 6$$
 =$$R^2 + WR = 6WR - 6W^2$$ 

=$$R^2 - 5WR + 6W^2 = 0$$
Therefore, $$R = 2W$$ or $$R=3W$$
If $$R=2W$$, $$S = 5W$$ and time taken to row downstream is $$\frac{5}{3} = 100$$ minutes.
If $$R=3W$$, $$S = 10W$$ and time taken to row downstream is $$\frac{10}{4} = 2.5$$ hours.

As time taken to row downstream is given to be greater than 2 hours, the correct answer is 2.5 hours.

B reaches 30 seconds after A and C reaches 20 seconds after B. So, C reaches 50 seconds after A