IMO A

limit is 2<q<7<p

s1) we need to prove if 7p/q is an int , then p/q is also an int
if p=8
7p/q=56/q=int for q=2,4 only ,and for all these value p/q=int

if p=9
7p/q=63/q=int only for q=3, and p/q=int for q=3

again if p=11

7p/q=77/q ; for the given value of q 77/p can not be int

ifp=12
7p/q=84/q=int for q=2,3,4 , and for all these values p/q is an int

hence it is equally distributable
hence suff...

s2) insuff...

if p=8; p^2=64...equally ditributable for q=2,4 , and p/q is also an int and hence equal distribution is possible

but if p=10;p^2=100
for q=2and 5 p^2 is divisble and so is p
but for q=4 p is not div by q

hence bth yes for 1 case and no for another
hence insuff

bhumika.k.shah wrote:A soccer captain needs to distribute p soccer balls among q practice groups. If 2<q<7<p , is it possible to give each of the p balls to one of the q practice groups such that each group has the same number of soccer balls?

It is possible to distribute 7p soccer balls among the q practice groups such that each group has the same number of soccer balls.
It is possible to distribute p^2 soccer balls among the q practice groups such that each group has the same number of soccer balls.


how to begin solving this sum

The question is whether p/q is an integer (if it is, it is possible to distribute p/q balls to each of the groups)

A gives 7p/q is an integer - q is less than 7 and cannot be a multiple of 7 so, p must be a multiple of q and hence p/q is indeed an integer - Sufficient

B p^2/q is an integer - If p^2/q is an integer there is no guarantee that p/q will be an integer. Take, p =10 and q =4
P^2/q is an integer and p/q is not - NOT SUFFICIENT

Hence A