Substitute sensible numbers and work out the problem. Then change the numbers back to letters. For example if the machine puts 6 caps on bottles in 2 minutes, it will put 6 /2 caps on per minute, or (6 /2) x 60 caps per hour. Putting letters back this is 60c/m.If you divide the required number of caps (b) by the caps per hour you will get time taken in hours. This gives bm/60c
In this type of questions we need to get the relative speed between them,
The relative speed of the boys = 5.5kmph ? 5kmph
= 0.5 kmph
Distance between them is 8.5 km
Time = Distance/Speed
Time= 8.5km / 0.5 kmph = 17 hrs
Let, Distance between A and B = d
Distance travelled by P while it meets Q = d + 11
Distance travelled by Q while it meets P = d ? 11
Distance travelled by Q while it meets R = d + 9
Distance travelled by R while it meets Q = d ? 9
Here the ratio of speeds of P & Q => SP : SQ = d + 11 : d ? 11
The ratio of speeds of Q & R => SQ : SR = d + 9 : d ? 9
But given Ratio of speeds of P & R => P : R = 3 : 2
=> = 3/2
=> d = 1, 99
=> d = 99 satisfies.
Therefore, Distance between A and B = 99
Time taken to meet for the first time anywhere on the track
= length of the track / relative speed
= 300 / (15 + 25)5/18 = 300x 18 / 40 x 5 = 27 seconds.
Let t be the time after Kajol starts, when she meets Ajay, then
t = 300/(x+y)
This should be less than 2.5 or (x+y)>120
Since y= 3x/2 => y > 72
This (y>72) is greater than 67.5 km/h and hence Shahrukh will always overtake Ajay before he meets Kajol.
Let the initial speed of Akhil be '4p' kmph
Then speed after decrease in speed = '3p' kmph
We know that,
Change in speed == Change in time
According to the given data,
Hence, the initial speed of Akhil = 4p = 4 x 12 = 48 kmph.
Let the speed of river and boat be 'r' m/min and 'b' m/min.
so relative speed in upstream (b-r)m/min and in downstream (b+r)m/min.
Now in upstream distnce covered in 5 min is 5(b-r)miles
so total distnce covered => 1 + 5(b-r)miles in upstream
In downstream distance covered in 5min is 5(b+r)miles
Now 1 + 5(b-r) = 5(b+r)
1+5b-5r = 5b+5r
1 = 10r
r = 1/10 mile/min => 6 mile/hr.
Second gun shot take 30 sec to reach rahul imples distance between two.
given speed of sound = 330 m/s
Now, distance = 330 m/s x 30 sec
Hence, speed of the bus = d/t = 330x30/(14x60 + 30) = 330/29 m/s.
The distance is constant in this case.
Let the time taken for travel with a speed of 10 kmph be 't'.
Now the speed of 15 kmph is 3/2 times the speed of 10 kmph.
Therefore, time taken with the speed of 15 kmph will be 2t/3 (speed is inversely proportional to time)
Extra time taken = t - 2t/3 = t/3
=> 1pm - 11am = 2hrs
=> t/3 = 2h
=> t = 6 hrs.
Now, Distance = speed x time = 10 x 6 = 60 kms
Time he takes to reach at noon = 6 - 1 = 5 hrs
Now, Speed = 60/5 = 12 kmph.
Distance d = 1200km
let S be the speed
he walks 15 hours a day(i.e 24 - 9)
so totally he walks for 70 x 15 = 1050hrs.
S = 1200/1050 => 120/105 = 24/21 => 8/7kmph
given 1 1/2 of previous speed
so 3/2 * 8/7= 24/14 = 12/7
New speed = 12/7kmph
Now he rests 10 hrs a day that means he walks 14 hrs a day.
time = 840 x 7 /12 => 490 hrs
=> 490/14 = 35 days
So he will take 35 days to cover 840 km.
Now, the distance covered by Chennai express in 2 hrs = 60 x 2 = 120 kms
Let the Charminar Express takes 't' hrs to catch Chennai express
=> 80 x t = 60 x (2 + t)
=> 80 t = 120 + 60t
=> t = 6 hrs
Therefore, the distance away from Hyderabad the two trains meet = 80 x 6 = 480 kms.
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