Et Tu Mindsporte !

STOI, October 23, Endgame
The only participants in a fine arts workshop were handsome teenagers from institute A and B. In one assignment each boy had to draw sketches of all other participating boys, separate sheet for each sketch. Participating girls did likewise for other girls. Boys used 66 more sketches than girls. In photography session each participant had to take separate photographs of participants of the opposite sex from other institute. When all 150 photographs were displayed, B girls formed the happiest lot. How many A girls participated?

My email to Mindsport/Endgame

Dear Sir,

After spending quite some time on today's question I have come to the conclusion that it is ambiguous in nature. I might have to eat my words but I am prepared to do so. Endgame is one of the few things I like about STOI and if this tower also falls then I don't know what will happen. The ambiguity is because of the statement, " When all 150 photographs were displayed, B girls formed the happiest lot". Just using the fact that there are a total of 150 photographs does not lead to a unique solution and so it seems that the second part of the statement is necessary to make any progress. What am I supposed to infer from it? Am I supposed to be a psychological expert who knows what makes a girl happy? If you are going to use the phrase "common sense" then I suppose you start writing Shobha De's column instead of Mindsport / Endgame.

However, here is the solution to the problem. Consider the variables X, x1, x2, Y, y1, y2.
X is the total number of boys. x1 is number of boys from A and x2 is number of boys from B. Similarly for Y. So, we have X = x1 + x2 and Y = y1 + y2.
Now, by first condition X*(X-1) - Y*(Y-1) = 66 and considering the fact that X and Y are positive integers, we get the three possible solutions for (X,Y) as (9,3), (13,10), (18,16).
By the second condition we get the equation,
2 * (x1*y2 + x2*y1) = 150
Solving this equation with the values obtained above and keeping in mind that all x1, x2, y1, y2 are positive integers we get,
x1 = 4, x2 = 9, y1 = 7, y2 = 3
OR
x1 = 9, x2 = 4, y1 = 3, y2 =7

I am unable to arrive at a unique solution which will give me number of girls in B i.e. the variable y2. It can be both 3 or 7 unless you come up with an explanation to the ambiguity mentioned at the beginning of this mail.

I expect that you will at least reply.
Regards,
Ravi Handa

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TOI continues to amuse me with their editorial page. Gurcharan Das's "Men And Ideas" has an article with the title "Amoral Familism". The last line of the article is:
So, dear reader, it is time to shake off your complacency, and the next time around, don't vote for "hereditary asses, imbeciles, and this
Yes, thats it. The article ends exactly like this. If you don't believe me, check page number 14 of today's paper. It is on the right side of the cartoon "Passing Thought". Someone in the proof-reading section goofed up again. If TOI cannot even afford good proof-readers how do they manage to stay at the top?