An old man and a young man are working together in an office and staying together in a near by apartment. The old man takes 30 minutes and the young 20 minutes to walk from appartment to office. If one day the old man started at 10.00 AM and the young man at 10:05AM from the apartment to office, when will they meet?
Ratio of old man speed to young man speed = 2:3
The distance covered by old man in 5 min = 10
The 10 unit is covered with relative speed=10/(3-2)=10 min
so, they will meet at 10:15 am.
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In a potato race, 20 potatoes are placed in a line of intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
Given, total number of potatos = 20.
First potato 24 metres from the starting point. There are 4 meters in the intervals. A contestant is required to bring the potatoes back to the starting place one at a time. So for the first potato he has to travel 48 meters, for second 56 meters ...
48,56,64............20 terms.
a = 48, d= 8, n = 20.
Important Point:
Sum of n terms in A.P = Sn=n2[2a+(n?1)d]
Ferrari is leading car manufacturer. Ferrari S.P.A. is an Italian sports car. It has enjoyed great success. If Mohan's Ferrari is 3 times faster than his old MERCEDES which gave him 35 km/ph if Mohan travelled 490 km in his Ferrari the how much time in hours he took?
Time = (Distance/Speed) = (490/35)
Therefore the total time = time/3 = (490/35)/3 = 4.6666 ~ 4.7
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My flight takes off at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N 70W.What is the local time when my plane landed?
Latitudianal difference=(10+70)=80
So, time difference=80X4=320mins=5 hrs 20 mins
Since the place is located towards west
Therefore local time of that place during take off=8:40 p.m.
So, local time during landing=8:40p.m.+10hrs=6:40a.m
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Two people A and B starts running at 10:00 AM towards north at a speed of 40Kmph and 60Kmph respectively. After 2 hours(12:00 noon ), A turned left and B turned right. They stopped at 2:00 PM.Find the distance between A and B at 2:00 PM.
We should only focus on the distances travelled by them when A & B start running in opposite direction i.e after 12 noon. Both run for two hours till 2pm
So distance travelled by A= 40*2=80km And distance travelled by B = 60*2=120km
So separation between them is 120+80=200km
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The time taken to travel from city X to city Y is 12 hrs. If a person starts at 6 A.M from city X and another person leaves the city Y at 8 A.M. When they will meet. (Speed of Both person is same.)
Let's assume Person "A" travel from City X to City Y at 6:00 AM.
and Person "B" travel from City Y to City X at 08:00 AM.
If the distance between two cities (City X and City Y) = 12 KM
Person A travel distance of 2 hrs = 2 KM.
Distance remaining = 10 Km to be traveled by both person
The time required to travel 10 Km by both person = 5 hrs.
So, They will meet after 5 hrs. when Y start. i.e 1:00 PM
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A man leaves office daily at 7:00 PM. A driver with car comes from his home to pick him from office and bring back home. One day he gets free at 5:30 PM and instead of waiting for driver he starts walking towards home. In the way he meets the car and returns home on car. He reaches home 20 minutes earlier than usual. In how much time does the man reach home usually?
Let the distance be d. Then time taken at x speed is d/x ( This d/x is the value which we have to find.
Now the new speed is 4/5x. So, time taken will be= 5d/4x
Acc. to the question= 5d/4x-d/x=1/4
Solving we get d/x = 1hr
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A boy standing idle sounds a whistle to his friend at a distance of 1200 m moving away from him in a speeding car at 108 kmph. Find the duration (seconds) after which his friend is going to hear him. (Speedof sound = 330m/sec).
The logic behind this question is to get that extra distance which the car will cover while the sound approaches him while traveling 1200m.
Time taken by sound to cover 1200m =1200/330= 40/11 sec= 3.63sec (i)
In this time of 40/11sec the car will cover an extra distance of 30m/sec*40/11sec= 109.09m
So extra time taken by sound to cover this distance will be= 109.09/330= 0.33sec (ii)
Adding both the times we get 3.96 approx = 4 sec.
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A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is
Let his walking speed be x km/hr and riding speed be ykm/hr
First case: d/x +d/y = 23/4
Second case: 2d/y= 15/4 (ii)
We have to calculate 2d/x
Putting the value of d/y from eq.(ii) in eq (i) we get d/x as 31/8
So, 2d/x will be equal to 62/8= 7hrs 45 minutes.
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