Critical Path
Logical Reasoning

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Common Information

The flow of energy from Generator (G) to Motor (M) in a power distribution network is as shown. Points P, Q, R and S are capacitor banks (denoted by CB) in the network. The arrows mark the direction of the energy flow. The energy loss (in Rs. ’00) in the power lines during the flow is indicated by the numbers adjacent to the arrows.

Common information image for Critical Path, Logical Reasoning:2296-1

An energy analyst has to make sure that when the energy flows from G to M, it flows through a path where the total energy loss (in Rs. ’00) is minimal. There is an additional energy loss (in Rs. ’00) at the capacitor banks. This loss can be regulated. For example, if the energy analyst selects a path G – P – M (using CB P) then the total energy loss would be Rs. 1,000 + Rs. 600 + the regulated energy loss at the CB P.

Q.

Common Information Question: 4/4

If the power line between R and M is not working, then which capacitor bank has the highest regulated energy loss such that the energy loss from G to M across all the paths is the same?

 A.

Q only

 B.

P only

 C.

S

 D.

Q or P

 E.

Insufficient information

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Solution:
Option(E) is correct

Let us analyze the given information.

Let us assume that the energy losses (in Rs. ’00) at the capacitor banks P, Q, R and S are p, q, r and s respectively.

Draw a table showing the energy losses (in Rs. ’00) for all possible routes between G and M in terms of the variables above.

Table below can be scrolled horizontally

Route Energy loss(in Rs. ’00)
G – P – M 10 + p + 6 = 16 + p
G – Q – P – M 3 + q + 3 + p + 6 = 12 + q + p
G – Q – R – M 3 + q + 4 + r + 3 = 10 + q + r
G – S – M 8 + s + 7 = 15 + s
G – S – R – M 8 + s + 2 + r + 3 = 13 + s + r

Two of our paths are eliminated because the path R – M is not working. Our redrawn table looks like this:

Table below can be scrolled horizontally

Route Energy loss(in Rs. ’00)
G – P – M 10 + p + 6 = 16 + p
G – Q – P – M 3 + q + 3 + p + 6 = 12 + q + p
G – S – M 8 + s + 7 = 15 + s

We now equate the energy losses on the three paths that we have.

Equating the first two expressions, we get,

16 + p = 12 + q + p

q = 4

Equating the first and third expressions, we get,

16 + p = 15 + s

s - p = 1

From the equations that we have, we cannot determine the exact values of p and s. Thus, we cannot compare p, q and s. So the information is insufficient to get a unique answer.

Hence, option E is the correct choice.


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