v1rok wrote:

ps_dahiya wrote:

I know B gives the answer but not sure about A.

Somehow I feel A will give the answer like 30 individual, 3 cartons of 6 and 9 cartons of 12.

So I go for D.

Exactly my feelings too! Somehow after solving numerous problems of various difficulty I am starting to get a sixth sense on these traps. I hope that my gut feeling would not let me down on the exam date...

So, anybody knows how to show that (A) is sufficient too?

A mathematical approach to solve this would be:

x = No of dozens

y = No of half-dozens

z = No of individual eggs

From the stem,

x + 0.6y+0.16z = 15 --(A)

From 2:

x = 9

z = 30

We can find y (as everyone here has found) and only one such y exists. And hence SUFF.

Condition #1.

Given

x + 0.6y+0.16z = 15 --(A)

12x+6y+z = 150 --(B)

(From #2) x = 9, y = 2 and z = 30 is defenitely one of the solutions. We are looking for alternate solution

(A)*10 - (B) gives

10x = 3z

We ALREADY KNOW that

x = 9, z = 30 --> y = 2 is a valid solution
Try to find other integer solutions for ths:

x = 3, z = 10 --> y = 104/6 Not an integer. Invalid solution

x = 6, z = 20 --> y = 58/6 Not an integer. Invalid solution

x = 12, z = 40 --> y becomes -ve. No neeto to check further

#1. is also sufficient. Hence D.

But as everyone here mentioned, all of this analysis takes a hell lot of time. It is not worth spending more than 3 minutes.

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