Test: Speed, Time & Distance - 3 - CAT MCQ

• Length of the train = 300 m
• Initial speeds: Train = 45 km/h (12.5 m/s), Motorbike = 70 km/h (19.44 m/s)
• At the middle of the train (after 150 m):
  • Train’s speed increases to 60 km/h (16.67 m/s)
  • Motorbike’s speed decreases to 65 km/h (18.06 m/s)

• Step 1: Time to reach the middle of the train

  Relative speed=19.44−12.5=6.94
  t 1 = 150/6.94 = 21.6 seconds.
• Step 2: Time to pass the remaining half of the train

  New relative speed=18.06−16.67=1.39 m/s
  t 2 ​= 150/ 1.39 ​≈107.9seconds

• Step 3: Total distance traveled by the motorbike

  d total=19.44×21.6+18.06×107.9≈2367

  The motorbike will travel 2367 meters while passing the train completely. = 2.37km (Aproxx)

• 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 10 = 40 km

Correct Answer :- c

Explanation :

• As D = S x T (Distance =  Speed x Time)
• 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
• On Solving We get, 
• x = 48

• 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.

• Let the original speed be X km/h According to the question, 18/(4/5x) - 18/x = 9/60 hr
• x = 30 km/h

• 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)
• The value that justifies the above equation is 2400
• Hence , Option A is correct.

• 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.

• speed of second train = 400km in 4 hr 
•  = 100km/hr
• ratio of speeds is 7:8
• 7:8 = v:100
• v = 87.5km/hr

• Relative speed of Vinay and Prabhat will be 20 m/minute to cover the track of 960 m.
• It will take 48 minutes.

• At 2 O’ clock Aflatoon has already covered 16 km at 8 km/h,
• Bablajoon starts running in the same direction at 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.

• 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.

• 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.

• 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.

• 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

• 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.