We were busy setting up our fledgling organisation Takshzila. Now that she has learnt how to walk, that gives us some free time to post blogs. So here we go...
My idea is that in each post i take an innovative question or a question which has an innovative solution (read oral solution which does not need the use of any pencil-work). So i am not going to post a lot of theory notes here (for that you can visit our site www.takshzila.com *), just interesting solutions to questions....
Starting with a caselet of CAT 2007 question...
Cities A and B are in different time zones. A is located 3000 km east of B. The table below describes the schedule of an airline operating non-stop flights between A and B. All times indicated are local and are of the same day.
Departure Arrival
City Time City Time
B 8:00 am A 3:00 pm
A 4:00 pm B 8:00 pm
Assume that planes cruise at the same speed in both directions. However the effective speed is influenced by a steady wind blowing from east to west at 50 kmph.
Q. 1: What is the time difference between A and B?
1. 2 h 2. 2 h 30 m 3. 1 h 4. 1 h 30 m 5. Cannot be determined
Q. 2: What is the plane's cruising speed in kmph?
1. 550 2. 600 3. 500 4. 700 5. Cannot be determined
SOLUTIONS:
Rather than worry about the time difference in the two cities, one should have just understood that the plane leaves city B at 8:00 am and arrives back at city B at 8:00 pm. Thus it is away for 12 hours, 1 hour of which it is not travelling (between 3:00 pm and 4:00 pm at city A)
Thus, in a total of 11 hours, the plane travels 3000 km at speed (x – 50) and 3000 km at speed of (x + 50). Thus,
Just plugging the given options of the speeds, we see that x = 550 satisfies the equation. In-fact the choice should be obvious as 3000 is not completely divisible by any of the other options when they are either reduced or increased by 50.
Now, time taken from B to A is 3000/(550 – 50) = 6 hours, and hence the plane reaches A at 8:00 am + 6 hrs i.e. 2 pm. But the time then at A is 3 pm. Thus the time difference between the two cities is 1 hour.
Some Learning Opportunities...
Once one realises that in a total of 11 hours, the plane travels 3000 km at speed (x – 50) and 3000 km at speed of (x + 50), one could have proceeded even by the following two alternate ways, each of which has its own learning...
Alternate method 1: To refresh funda of average speed...
If same distances are same are travelled at speeds of a and b, then the average speed is given by 
Since both ways the distances are same, the average speed in this case i.e. 6000 kms/ 11 hrs will be given by 
Looking at the denominator, it should be obvious that x has to be a multiple of 11 and the only option satisfying this is 550. Plugging x = 550, one sees that it satisfies the equation.
Alternate method 2: To refresh funda of use of ratios...
Since the distances covered both ways is the same, the time taken has to be in inverse ratio of the speed i.e. in the ratio of (x + 50) : (x – 50). Further the addition of the time should be 11 hours. Again plugging the given options of speed, and seeing for which option would the reduced ratio add up to 11, we find that 600 : 500 is 6 : 5 and the addition is 11 hours. Thus speed of plane is 550 and while going from B to A the plane takes 6 hours, from which the difference in time between the cities can be found.
Cheers
* The site is just been created and the notes will be available at the Student Resource Center in a couple of weeks.
6 comments:
Amazing post. [But detailed explanation may help beginners too]
thanx for the solution..:)
thankx for the solution
but can u also guide for questions that do not give total time = 11 or smthn and yet we have to find the time difference
woah.. thanks man !
Superb explanation.... You make concepts crystal clear
Let the speed of aeroplane is x and the time difference between two time zones is t. Now when an aeroplane departs from B to A it moves against the direction of moving air in that case resultant speed of the aeroplane will be " x-50" Km/hr.
Now time required to travel from B to A
3000/(x-50)
Now the equation will be
8+(3000/(x-50))+t= 12+3 ... 1
t is added because A is situated in the East of B so A will be t unit ahead of B. 12 has been added to 3 because 8 is in am and 3 is in pm.
But when aeroplane departs from A and moves towards B, in that case it moves along the direction of air. So resultant speed of aeroplane will be
x+50 Km/hr
Now time required to reach from A to B
3000/(x+50)
And the equation will be
4+(3000/(x+50))-t = 8 ..... 2
The concept of subtracting t is reverse of the above explanation given in equation 1 i.e. Zone B will lag by t, and since both the arrival and departure time is given in pm so there is no need to add or subtract 12.
Now the equation 1 can be rewritten in simplified was as
(3000/(x-50))+t = 7 .... 3
And eqn 2 can be rewritten as
(3000/(x+50))-t = 4 ..... 4
Adding eqn 3 and 4
(3000/(x+50))+(3000/(x-50)) = 11
Or, 3000((1/(x+50))+((1/(x-50)))=11
Or, 3000((x-50+x+50)/(x^2-2500))=11
Or, 3000×2x/(x^2-2500)=11
Or, 6000x=11(x^2-2500)
Or, 6000x=11x^2-27500
Or, 11x^2-6000x-27500=0
Or, 11x^2-6050x+50x-27500=9
Or, 11x(x-550)+50(x-550)=0
Or, (11x+50)(x-550)=0
Or, 11x+50=0
Or, x=-50/11 we ignore this value because speed should be in positive
Or, x-550=0
Or, x=550 it is acceptable
So, speed of aeroplane will be 550Km/hr Ans.
Next time difference can be calculated from any of the eqn 3 or 4
Here we choose 3
Putting value of x in eqn 3
(3000/500)+t=7
Or, 6+t=7
Or, t=1
So time difference is of 1hr Ans.
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