Question Bank  100 CAT level questions on Time, Speed and Distance topic



12kmph
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Let say the bike ran @ 80 km/h for t hours
so 80 * t + 60 * (10  t) = 640
20t = 40
t = 2
Bike ran @ 60 km/h for 8 hours and consumed 8 * 2 = 16 L of fuel.
We know bike changed speed to 60 km/h when the fuel tank is half full, so the capacity of fuel tank should be 2 * 16 = 32 L

@mbatious ans kya h question ka

@mbatious if we consider the speeds as 5 ,1,1 respectively and the length if the circle as 60. Then i am getting 11 as the answer. Is this correct?

@mbatious 3.5 hours ? When you guys are planning to publish the answer keys?

Clocks can be considered as a circular race.
Here we can just apply some basic concepts
Minute and hour hands meet every 720/11 minutes
So in a usual day it meets 24 * 60 / (720/11) = 22 times
As the clock gains 10 minutes every hour, we have an extra 240 minutes so 240 / (720/11) = 3 extra meetings
so total 22 + 3 = 25 meetings.

credits : @shashank_prabhu
Let the circumference be 120 units and speed of A, B, C be 5 units, 1 unit and 1 unit respectively. So, A and B will meet once every 20 seconds and A and C will meet once every 30 seconds. So, in 1 minute, A gives B 3 flags and gives C 2 flags. So, we can safely say that in the first 30 distributions, B gets 18 flags and C gets 12 flags. Post that, after 20 seconds, B gets a flag, at the 30th second C gets a flag and at the 40th second, B gets another flag. So, in total, B gets 20 flags and C gets 13 flags. So, difference is 7.

Sir please provide solution

Let the distance PQ be d and speed of river be R and speed of boat be B
d/(B + R) = 4
d/(B  R) = 6
Solve for B.
4(B + R) = 6(B  R)
4B + 4R = 6B  6R
10R = 2B
R = B/5
d/(B + B/5) = d/(6B/5) = 5d/6B = 4
d = 24B/5If we put d = 24 units, then B = 5 and R = 1
So first hour boat will cover 6 units, second hour boat will cover 5 * 1.4 + 1 = 8 units and third hour 10 units.So 24 units will be covered in 3 hours

Funda : When running in the same direction : If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be a – b
A and B will meet at 1  2 = 1 point.
A and C will meet at 1  3 = 2 points
A and D will meet at 1  4 = 3 points
A and E will meet at 1  5 = 4 points
A and F will meet at 1  6 = 5 points
A and G will meet at 1  7 = 6 points
So A will put 1 + 2 + 3 + 4 + 5 + 6 = 21 flags.similarly B and C will meet at 2  3 = 1 point
B and D will meet at 2  4 = 2 points
B and E will meet at 2  5 = 3 points
B and F will meet at 2  6 = 4 points
B and G will meet at 2  7 = 5 points
So B will put 1 + 2 + 3 + 4 + 5 = 15 flagsSimilarly find for C, D, E and F.
We will get 21 + 15 + 10 + 6 + 3 + 1 = 56 flags

@mbatious 20 m? Sb:Sc =5:4 , so the difference would be 20?

@mbatious Let the starting point be P
let the distance travelled in 1min be X km.
Distance from P at the end of 4th minute = 4X
Distance from P at the end of 7th minute = 4X3X = X
Distance from P at the end of 9th minute = X+2X = 3X
Distance from P at the end of 10th minute = 3XX = 2X
2X = 2 x 4 x 1/60 km = 2/15km

@mbatious total time for second case is 13 hrs, Am i right ??

@mbatious time is 6:24 am

t + t/12 = 9 (time taken in the return journey is 1/12th the time taken for the onward journey)
t = 108/13 = 8.30 hours
t + t/6 = 9.68 hours?