Let G = Girls
B=Boys

1) This can be written as G<3B. Using real numbers, let the number of boys equal 10. Hence, G<3(10)= G<30.

To satisfy G<30:

G could be 5 and B is 10, in which case G<B
G could also be 15 and B is 10, in which case G>B.

This statement is thus insufficient.

2)This can be written as B<(3/4)G. As B and G are positive integers (you cannot have a negative or a fraction amount of boys and girls), B<G.

Hope this helps!

I've found that some of these inequality problems often introduce a level of ambiguity making them insufficient. Anytime you can refute the established case, or establish a case that contradicts the earlier case, the answer is going to be insufficient.

b = Boys
g = Girls

b > g?

(1) g < 3b

b= 5
3b = 15

The number of girls can be 4 can be less than the number of 5 boys or the number of girls can 10 which is greater than 5 boys but still supports the statement that g < 3b or 10 < 15. Since there can be multiple cases to either support contradict the answer, (1) is insufficient.

(2) b < 3/4g or 4b < 3g. Since 4 times the number of boys is still less than 3 times the number of girls, this establishes that there are more girls than boys and the answer is sufficient, hence B.

heshamelaziry wrote:In a certain daycare, are there more boys than girls?

(1) The number of girls is less than 3 times the number of boys.

(2) The number of boys is less than 3/4 the number of girls.


EXPLANATION FOR STATEMENT 1 PLEASE


OAB

From statement 1 :

G < 3(B)

Case 1 : B < G

B= 1, G = 2 -> 2 < 3(1) ..i,e 2 < 3

case 2 : B > G

B = 4, G =1 -> 1 < 3(4) ..i.e 1< 12 ..

hence, we are not able to determine from statement 1...

from statement 2, anyhow boys is less than 3/4 of girls .. it states the answer..