Hi Ender, just that certain memory didn't fail me. I recall reading your solution and you nailed it without the assumption made.Ender wrote:Things to note,Sruthi wrote: please help me with this question.

X,Z, Y, are divided equally into 3 groups. Hence if they say the number of male in X equals to the number of female in Y, to me, it says the male in X plus the male in Y is 1/3 of the school teachers, like wise the same for female teachers in X and Y together is 1/3 of the school population.

a) Let S be total school teachers

Male teachers in group X and Y is S/3

Since male in group Z is 1/4 (1u)of the school male teacher, there's 3/4 (3u)of the school males teacher in group X and Y.

3u->S/3, u-> S/9

Total Male in school is 4u, -> S/3+S/9 = 4/9(S),

Hence the female would be 5/9(S).

Ratio of M:F -> 4:5

b)45 females in X and Y constitute to 1/3 the of school population, therefore school population is 45X3 = 135

the number of males teacher is 4/9 X 135 = 60

My method was similar to what I did in my previous post:

X:Y:Z = 1:1:1

(I let Xmales = Yfemales = u)

Since Zmales:Tmales = 1p:4p => deduce Xmales+Ymales = 3p

Since Xmales=Yfemales => deduce that Xmales + Xfemales = 3p = Xtotal

Go back to X:Y:Z=1:1:1 = 3p:3p:3p => deduce in School Z, since Zmales=1p then Zfemales=2p

So Tfemales = 3p+2p=5p and all the rest follows.

There is no assumption made as well. It was then I knew value of "u" does not matter.

(Sorry for earlier typo)

Hope the above is clear. Cheers!

(NB: It is as if the setter of "SCHOOL ABC" question did not think "SCHOOL XYZ" question of ratio 1:1:1 is challenging enough that he/she has to set with ratio 1:2:3.)