Just look at the figures TOTAL students = 36
B / 3 and g / 4;
dividing the class with MAX multiple of 3 and 4,
36 / 4 = 9 ====> Not possible -- Zero boys....
32 / 4 = 8 ====> Not possible as 4 / 3 is not exactly divisible.
28 / 4 = 7 ====> Not possible as 8 / 3 is not exactly divisible.
Look at 24 / 4 = 6 ==> 12 / 3 = 4 Exactly divisible. Thus, total students walking = 6 + 3 = 9.
Now we need to find whether this is the MAX number of students who can possibly walk. Anyways, we know that this is the distribution of the class whatsoever. Just interchange the figures,
We get 24 / 3 = 8 ==> 12 / 4 = 3 => Exactly divisible. Thus, 8 + 3 = 11 is the maximum no of students who can walk. So, our earlier distribution is contraversial and thus this distribution is the correct one.
