A car departs at 11:10am. It travels at 110km/hr until 2:30 then it reduces its speed by 5%for the remainder of trip. It reaches destination at 7:30. How much kilometer did he travel?
A cycled from P to Q at 10 kmph and returned at the rate of 9 kmph. B cycled both ways at 12 kmph. In the whole journey B took 10 minutes less than A. Find the distance between P and Q.
Lets assume distance between P and Q = d km.
Time taken by A in both side = d/10 + d/9 = 19d/90 hrs.
Time taken by B in both side = 2d/12 = d/6 hrs.
B tool 10 min. or 1/6 hrs. less than A.
So, 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
=> d = 15/4 km = 3.75 km.
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A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.
Average speed = (2*a*b)/(a + b) here a = 25 b = 4
Average speed = 2*25 * 4/(25 + 4) = 200/29 kmph.
Distance covered in 5 hours 48 minutes = Speed * time
Distance = (200/29)*(29/5) = 40 kms
Distance covered in 5 hours 48 minutes = 40 kms
Distance of the post office from the village = (40/2) =20 km.
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A car starts from rest and accelerates uniform upto speed 90 km/hrs in 5 sec. Total distance covered by car in this time will be
For 45kmph, distance to cover in one hour = 45 * 1000 = 45000 m.
Distance to be covered per minute (d) = 45000 / 60 = 750 m/min
Distance covered in 1 rotation of wheel =Pi*diameter=Pi*1.5 m.
So rotation /minute Required =750/(Pi*1.5)=750/[(22/7)*1.5]=159.09 ~ 159
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Juan is a gold medalist in athletics. In the month of May, if Juan takes 11 seconds to run "y" yards. How many seconds will it take him to run "x" yards at the same rate?
Juan takes 11 sec. to run "y" yards.
So, to cover 1 yard he need 11/y sec.
So to cover x yard he need 11x/y sec.
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Two vehicles A and B leaves from city Y to X. A overtakes B at 10:30 am and reaches city X at 12:00 pm. It waits for 2 hrs. and return to city Y. On its way it meets B at 3:00 pm and reaches city Y at 5:00 pm. B reaches city X, waits for 1hr and returns to city Y. After how many hours will B reach city Y from the time A overtook him for the first time?
Vehicle A overtaken B at 10.30 am and reached X at 12 pm.Â It started at 2 pm and met B at 3 pm at Q. It means, Vehicle A took one hour to cover distance 'n', So it should be at Q at 11 pm.Â It is clear that Vehicle A takes 0.5 hour to cover distance 'm'.Â
Now vehicle B travelled from 10.30 am to 3 pm to meet A. So it took 4.5 hours to cover m.Â So Speeds ratio = 4.5 : 0.5 = 9 : 1.
Now Vehicle A took a total of 1.5 + 3 = 4.5 hours to travel fro P to Y. So It must take 4.5 * 9 + 1 = 41.5 hours
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Two cars starts from A and B and travel towards each other at a speed of 50 kmph and 60 kmph respectively. At the time of their meeting the second car has travelled 120 kmph more than the first. The distance between A and B is:
Total kilometers travelled by 4 tyre = 40000 * 4 = 1,60,000.
This has to be share by 5 tyres.
So each tyre capacity = 1,60,000 / 5 = 32, 000.
You have a doubt, after we travel 32,000 km, we are left with 4 worn tyes and one new tyre.
But If the tyres are rotated properly after each 8000 km,
All the tyres are equally used.
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Jake and paul start together to walk 10kms jake walks 1.5kmph faster than paul and arrives at his destination 1.5 hours earlier than paul jakes speed walking is
Let speed of jake = x kmph.
So, speed of paul = ( x - 1.5 ) kmph.
Time difference =1.5hours
Total Distance = 10km
10/(x-1.5) -10/x = 1.5
By solving the above eq.
x = 4 kmph
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