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A train of 300 m is travelling with the speed of 45 km/h when it passes point A completely. At the same time, a motorbike starts from point A with the speed of 70 km/h. When it exactly reaches the middle point of the train, the train increases its speed to 60 km/h and motorbike reduces its speed to 65 km/h. How much distance will the motorbike travel while passing the train completely?
Speed of train while passing point
A = 70 x (5/18) m/s = VI
Speed of bike initially = 70 x (5/18) m/s = V2
Time taken by the bike to reach at the mid-point of the train = 1 5 0 /(V 2 - V I)
Again find out the new speeds of train and bike, and calculate the time taken by the bike to cover the rest 150 m distance relative to the train.
Two riders on the horseback with a gun and a bullet proof shield were moving towards each other at a constant speed of 20 km/h and 5 km/h respectively. When they were 100 km apart, they started firing bullets at each other at the speed of 10 km/h. When a bullet of rider 1 hits the shield of rider 2, rider 2 fires a bullet and the process continues vice versa. Neglecting the time lag at the instant when the bullet hits the shield and the rider fires the shot, find the total distance covered by all the bullets shot by both the riders.
This question has actually got nothing to do with bullets initially. We can see that it takes them 4 hours to reach each other. And this is the same time for which bullets will cover some distance.
So, the total distance covered by the bullet = 4 x 1 0 = 40 km
A train approaches a tunnel PQ which is 16 m long. Two rabbits A and B are standing at points which are 12 m and 8 m inside the tunnel with respect to the entrance P. When the train is x m away from P, A starts running towards P and B towards Q. Difference between the ratios of the speed of A to that of the train and the ratio of the speed of B to that of train is 1/8. How much can the distance x be, if both of them get caught at the ends of the tunnel?
Correct Answer :- c
Explanation : As D = S x T
D is directly proportional to S.
SA/ST = 12/x............(1)
SB/ST = 8/(x+16)........(2)
Subtract (1) from (2)
12/x - 8/(x+16) = 1/8
x = 48
Refer to the data below and answer the questions that follow.
There is a race between Sagar and Sapna. Both of them Delhi. Sagar started on a bike with a speed of 40 km/h and Sapna has started in a car with a speed of 60 km/h from Mumbai to Delhi After 31/2 h of the journey there was a snag in the car She tried to repair the car but in vain After half an hour she got a lift for another 500 km in a truck, which was travelling with a speed of 45 km/h. From there Delhi was at a distance of 200 km on road, instead, Sapna took a shorter route which was only 100 km away from Delhi She started running at the speed of 30 km/h on the shorter route to reach Delhi.
Q. If there was no snag in the car, by how much distance Sapna would have defeated Sagar?
Total distance from Mumbai to Delhi = 910 km Time taken by Sapna to cover this distance, had there been no snag = 910/60
Since their speeds are in the ratio 3:2, Manoj would have covered [910 - (910 x 40)/60] km less than Sapna.
A train’s journey is disrupted due to an accident on its track after it has travelled 30 km. Its speed then comes down to 4/5th of its original and consequently it runs 45 min late. Had the accident taken place 18 km farther away, it would have been 36 min late. Find the original speed of the train.
Let the original speed be X km/h According to the question, 18/(4/5x) - 18/x = 9/60 hr
x = 30 km/h
There are two swimmers A and B who start swimming towards each other from opposite banks of the lake. They meet at a point 900 ft from one shore for the first time. They cross each other, touch the opposite bank and return. They meet each other again at 300 ft from the other shore. What is the width of the lake?
Let us assume that the width of the lake = x. So, when one of the runners A covers 900 m, the other one B is covering (x - 900) m. To meet next time, A will be covering (x - 900 + 300) m whereas B will be covering (900 + X-300) m.
Now, 900/(x - 900) = (x - 900 + 300)/(x + 900 - 300)
Now use options to find the answer.
Ramesh and Somesh are competing in a 100 m race. Initially, Ramesh runs at twice the speed of Somesh for the first fifty m. After the 50 m mark, Ramesh runs at l/4th his initial speed while Somesh continues to run at his original speed. If Somesh catches up with Ramesh at a distance of ‘N ’ m from the finish line, then N is equal to
This question gives us the freedom to assume any value of speeds of Ramesh and Somesh. Let us assume the initial speed of Somesh = 20 m/s, then the initial speed of Ramesh = 40 m/s. Till 50 m they are running with this speed only. Time taken by Ramesh in covering 50 m = 1.25 sec. In the same time Somesh is covering 25 m.
After this stage, speed of Somesh is 20 m/s, whereas speed of Ramesh = 10 m/s. Now relative speed = 10 m/s and distance = 25 m. At 75 m from the starting, both of them will be meeting.
The ratio between the speeds of two trains is 7:8. If the second train runs 400 kms in 4 hours, then the speed of the first train is?
speed of second train = 400km in 4 hr
ratio of speeds is 7:8
7:8 = v:100
v = 87.5km/hr
Two persons Prabhat and Vinay are walking around a circular park of the length 960 m. Prabhat walks at the rate of 80 m/min while Vinay walks at the rate of 60 m/min. If both of them start from the same starting point at the same time in the same direction, when will they be together?
Relative speed of Vinay and Prabhat will be 20 m/minute to cover the track of 960 m. It will take 48 minutes.
Two cyclists start from the same place to ride in the same direction. Aflatoon starts at noon with a speed of 8 km/h and Bablajoon starts at 2 pm with a speed of 10 km/h. At what times Aflatoon and Bablajoon will be 5 km apart?
At 2 O’ clock Aflatoon has already covered 16 km @ 8 km/h, Bablajoon starts running in the same direction @ 10 km/h. The relative speed is 2 km/h.
Therefore to be 5km apart bablajoon has to cover 11 km. relative speed=2km/hr, therefore for 11km it will take 5hr 30 min more. Therefore they will meet at 7:30 the same day. And similarly 12.30 a.m. on the next day.
Read the passage given below and solve the questions based on it There are 8 days in a week from Sunday io Saturday and another day called Fund ay on planet North. There are 36 h in a day and each hour has 90 min while each minute has 60 s.
Find the approximate angle between the hands of a clock on North when the time is 16.50 am?
Since there are 3 6 hours in a day, 18 hours dial will be there in the clock.
At 16:50, the hour hand will be in between 10 and 11 of the actual clock and the minute hand will be in between 6 and 7 of the actual clock. Now we can eliminate the options.
Read the passage below and solve the questions based on it* A number o f runners, numbered 1, 2. 3 , , N and so on, start simultaneously at the same point on a circular track and keep on running continuously in the same direction, around the track. They run in such a way that the speed of the runner numbered N (N>1) isN times that of the runner numbered 1 .
If there are exactly six runners, then at how many distinct points on the track is the runner numbered 1 overtaken by any of the other five runners?
Assume that the track length is 1000 m.
Now, runner 1 and runner 2 will meet at one point, i.e., the starting point.
Runner 1 and runner 3 will meet at two points, at 500 m and at the starting point.
Runner 1 and runner 4 will meet at three points, at 333.33 m, at 666.66 m and at the starting point. Runner 1 and runner 5 will meet at four points, at 250 m, 500 m, 750 m and at starting point.
Runner 1 and runner 6 will meet at five points, at 200 m, at 400 m, at 600 m, at 800 m and at the starting point.
These are 10 distinct points.
A merchant can buy goods at the rate of Rs. 20 per good. The particular good is part of an overall collection and the value is linked to the number of items that are already on the market. So, the merchant sells the first good for Rs. 2, second one for Rs. 4, third for Rs. 6…and so on. If he wants to make an overall profit of at least 40%, what is the minimum number of goods he should sell?
Let us assume he buys n goods.
Total CP = 20n
Total SP = 2 + 4 + 6 + 8 ….n terms
Total SP should be at least 40% more than total CP
2 + 4 + 6 + 8 ….n terms ≥ 1.4 * 20 n
2 (1 + 2 + 3 + ….n terms) ≥ 28n
n(n + 1) ≥ 28n
n2 + n ≥ 28n
n2 - 27n ≥ 0
n ≥ 27
The question is " If he wants to make an overall profit of at least 40%, what is the minimum number of goods he should sell?"
He should sell a minimum of 27 goods.
Hence, the answer is 27.
Choice C is the correct answer.
Read the passage given below and solve the questions based on it.
Amit intended to travel a certain distance at a certain uniform speed. But after one hour, he increased his speed by 25%. As a result in the remaining part of the lime that he originally planned for the journey, he could now cover as much distance as he initially thought he would be able to cover
What is the total time taken for the journey?
If Amit would have increased his speed by 25% in the beginning, he would have saved one hour in covering the actual planned distance. So, 1/5 T = 1 hr. (Where T is the actual planned time). Hence, T = 5 hours.
Distance between Lucknow and Patna is 300 km. Mayank leaves at a speed of x km/h from Lucknow towards Patna. After three hours Sharat leaves at the speed of (x + 10) km/h from Lucknow towards Patna. If x and the number of hours taken to meet after Sharat starts are integers, how much distance can Mayank cover before they meet?
One of the ways of solving this question is going through equations. But after a certain stages we will be required to start assuming the values because all the data are not given.
Another way of doing this problem is: Start working by assuming some values. Let us assume the speed of Mayank =10 km/h. In three hours he has covered 30 km. Now Sharat starts with a speed of 20 km/h. He will take 3 hours to meet Mayank. Till that time, the total distance covered by Mayank = 60 km.