Question: The speed of a railway engine is 42 Km per hour when no compartment is attached, and the reduction in speed is directly proportional to the square root of the number of compartments attached. If the speed of the train carried by this engine is 24 Km per hour when 9 compartments are attached, the maximum number of compartments that can be carried by the engine is?
- 49
- 48
- 46
- 47
- 44
“The speed of a railway engine is 42 Km per hour when no compartment is attached”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.
Solution and Explanation:
Approach Solution 1:
There is only one approach to this question.
Given us that the speed of a railway engine is 42km per hour when no compartment is attached. The reduction in speed is directly proportional to the square root of the number of compartments attached. The speed of the train carried by the engine is 24km per hour when 9 compartments are attached. It is asked what is the maximum number of compartments that can be carried by the engine.
So it can be deduced that when there is no compartment attached to the engine the engine will run at its original speed. Adding compartments to the engine will increase resistance and hence the speed will decrease. So if we keep adding compartments to the engine then after adding some amount of compartments the train engine will not be able to pull all the compartments. Hence will stop moving.
Now we are given that the reduction in speed is directly proportional to the square root of the number of compartments attached to it.
Mathematically it can be written as –
Let v be the speed of the train and n be the number of compartments attached.
\(Δvα\sqrt{n}\)
\(Δv=k\sqrt{n}\), k is proportionality constant
Let Vi be the original speed of the railway engine and Vf be the speed of the engine after adding n compartments.
We have,
Vi - Vf =\(k*\sqrt{n}\) where k is the proportionality constant
Vi has a fixed value which is 42 km per hour
We get
42 - Vf =\(k*\sqrt{n}\)— (1)
Now we are given a case where the railway engine is attached with 9 compartments.
So the speed of the railway engine should be reduced, which is now 24 km per hour
In this case n = 9 and the final speed is 24km per hour, so Vf = 24
Putting the value of n and Vf in the equation (1)
We get,
42 - 24 = \(k*(\sqrt{9})\)
18 = k * 3
K = 18 /3
K = 6
The value of proportionality constant is 6. So now the formula becomes
42 - Vf = 6 \(\sqrt{n}\)
We have to find the number of compartments to be attached so that the train will stop moving
The final velocity will be 0.
So,
42 - 0 = 6 \(\sqrt{n}\)
\(\sqrt{n}\)= 42 / 6
\(\sqrt{n}\) = 7
n = 49
So when the number of compartments is 49 or more then the velocity of the engine will be 0.
Therefore the maximum number of compartments that can be carried by the engine will be 48.
Correct Answer: B
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