cleetus wrote:
Towns X and Y are in two different countries. X is 3600 km to the west of Y. An airline operates non-stop flights between X and Y. Its schedule is tabulated below. The given time is the local time on the same day.
Departure Arrival
Town time Town Time
Y 4:00PM X 7.30PM
X 3:00AM Y 1:30PM
The planes cruise at the same speed in both directions. however, a steady wind blows from Y to X at 75 Km/hr. The effective speed of each plane is affected by this.
Question 1
Find the time difference between X and Y
A) 2hrs
B)2.5 hrs
C)3 hrs
D)4.5 hrs
E) 1.5 hrs
Question 2
Find the speed at which each plane cruises in Km/hr
A) 425
B) 525
C) 475
D) 550
E) 450
In both the questions official answer is choice B.
Plz explain how to get the answers in both the cases. Thanks in advance.
First thing to note in this question is let's say a plane starts from X at 3:00 AM reaches Y at 1:30 pm (according to the given data), stays there for 2.5 hrs (because 1:30 to 4:00 is 2.5 hrs), starts from Y at 4:00 pm and comes back to X at 7:30 pm (according to the given data). Therefore, out of the total 16.5 hrs (from 3:00 am to 7:30 pm) that it was away from X, 2.5 hrs were spent resting in Y. Rest of the time, it was flying in the air.
Total journey time = 14 hrs. (this is the actual journey time)
Now, this question is like boats/streams question (If you are comfortable with it, it might help you visualize better). When the plane is going from Y to X, the speed of the plane increases by 75 km/hr. When it is going from X to Y, its speed decreases by 75 km/hr. Let the plane's original speed be p.
Make a quick equation:
Time taken to travel from Y to X + Time taken to travel from X to Y = 14
3600/(p + 75) + 3600(p - 75) = 14
Put in the values of p from 2nd question to get your answer. Try 475 first; it is in the middle and will make p - 75 = 400 so 3600/400 will be 9. You have to check if the first term is 5 or not... You find that it is greater than 5 so you need to go higher on plane speed. Try 525. You get the answer.
So the plane takes 6 hrs from Y to X (3600/(525 + 75) = 6)
But according to given data, it left Y at 4:00 and arrived at x at 7:30 i.e. 3.5 hrs. The 6- 3.5 = 2.5 hrs must be the time difference between the two cities. X must be behind Y by 2.5 hrs.
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