In a certain class consisting of 36 students, some boys and some girls, exactly 1/3 of the boys and exactly 1/4 of the girls walk to school. What is the greatest possible number of students in this class who walk to school?

A) 9
B) 10
C) 11
D) 12
E) 13

The important thing here to recognize here is that the number of girls and the number of boys must be positive INTEGERS. For example, we can't have 5 1/3 boys.

Also recognize that we're told that we have some boys and some girls
Since "some" means 1 OR MORE, we cannot have zero boys or zero girls.

Okay, now onto the question...

We want to MAXIMIZE the number of students who walk to school. Since a greater proportion of boys walk to school, we want to MAXIMIZE the number of boys in the class.
The greatest number of boys is 35 (since 36 boys would mean 0 girls, and we must have at least 1 girl)

35 boys
This is no good, because 1/3 of the boys walk to school, and 35 is not divisible by 3.

So, let's try ...
34 boys
This is no good, because 1/3 of the boys walk to school, and 34 is not divisible by 3.

As you can see, we need only consider values where the number of boys is divisible by 3. So, that's what we'll do from here on...

33 boys
If 1/3 of the boys walk to school, then 11 boys walk. Fine.
HOWEVER, if there are 33 boys, then there must be 3 girls.
If 1/4 of the girls walk to school, then there can't be 3 girls, since 3 is not divisible by 4.

30 boys
This means there are 6 girls
If 1/4 of the girls walk to school, then there can't be 6 girls, since 6 is not divisible by 4.

27 boys
This means there are 9 girls
If 1/4 of the girls walk to school, then there can't be 9 girls, since 9 is not divisible by 4.

24 boys and 12 girls
1/3 of the boys walk to school, so 8 boys walk
1/4 of the girls walk to school, so 3 girls walk
PERFECT - it works!!
So, a total of 11 children walk

Answer: C

Cheers,
Brent

jayhawk2001 wrote: ↑

Fri Mar 23, 2007 6:12 pm

We essentially need to find

max (1/3 * boys + 1/4 * girls)

We also know that boys + girls = 36. For max value, the boy count needs
to be higher (as girl count is divided by 4). We also know that girl count
is not zero. So...

1/3 * 33 + 1/4 * 3 --> number of girls is not divisible by 4
1/3 * 30 + 1/4 * 6 --> same...
...
1/3 * 24 + 1/4 * 12 --> both are divisible

Hence total count = 8+3 = 11

Solution:

We can let n = the number of boys; thus 36 - n = the number of girls. We are given that (1/3)n boys walk to school and (1/4)(36 - n) = 9 - (1/4)n girls walk to school.

Since (1/3)n and 9 - (1/4)n must be an integer, we see that n must be divisible by 3 and 4. In other words, n must be divisible by 12. Thus n can be either 12 or 24 (we exclude 0 and 36 since if n = 0, there will be no boys in the class, and if n = 36, there will be no girls in the class).

If n = 12, then (1/3)(12) = 4 boys and 9 - (1/4)(12) = 9 - 3 = 6 girls walk to school. That is, a total of 4 + 6 = 10 students walk to school.

If n = 24, then (1/3)(24) = 8 boys and 9 - (1/4)(24) = 9 - 6 = 3 girls walk to school. That is, a total of 8 + 3 = 11 students walk to school.

Thus the greatest possible number of students in this class who walk to school is 11.

Answer: C