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gmatprep 5

This topic has 2 expert replies and 3 member replies
resilient Legendary Member Default Avatar
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Post Tue Apr 29, 2008 12:14 am
In a certain class consisting of 36 students, some boys and some girls, exactly 1/3 boys and of the girls walk to school. What is the greatest possible number of students in this class who walk to school?

9
10
11
12
13


Qa is 11

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amitansu Master | Next Rank: 500 Posts Default Avatar
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Post Tue Apr 29, 2008 1:03 am
We have to find a combination of no.s ,those sum up to give 36 and at the same time those no. should be divisible by 3 and 4 .

Say for e.g 30 Boys and 6 Gals
here 1/3 B is 10 but 6 is not divisible by 6.

So let's try for lower numbers.. like 29 and 7 not possible..
28 and 8 not possible; 27 and 9 nope; 26 and 10 nope ;
finally 24 and 12 yes.. this is the right combi and max . no. when you add their 1/3 and 1/4

so 1/3 of 24=8 and 1/4 of 12 =3 sum 'em up makes 11.

Any other quicker method...though i could be able to make it in about 80 seconds

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AleksandrM Legendary Member
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Post Tue Apr 29, 2008 1:52 pm
All you really need to do here is try the lowest whole number of boys and then the lowest whole number of girls. So... ladies first:

4 girls total in this class will give you 32 boys. 10 boys + 1 girl = 11 students walk to school

Now, let's try the boys:

3 boys total in this class will give you 33 girls. 8 girls + 1boy = 9 students walk to school.

Looks like 11 is the right answer.

The 12 is a trap answer. GMAC knows there will be people who'll remember something about solving problems involving fractions, and you multiply the 3 and 4 in the denominator. You could also get 12 by getting 1/3rd of 36, but that is a STUPID stupid mistake since you would also have to make the second stupid mistake of getting 9 as 1/4th of 36 and add them together, which would quickly show you that you are not on the wrong path. However, some might take 1/3rd of 9 to get 3 and add that to 9 to get 12, but I doubt very many people would do that.



Last edited by AleksandrM on Wed Apr 30, 2008 5:19 pm; edited 1 time in total

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zacharyz Senior | Next Rank: 100 Posts Default Avatar
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Post Wed Apr 30, 2008 11:36 am
12 is a trap, but for another reason.

The maximum number of walkers will occur by using the greater of the fractions (1/3). I want as many boys as possible in the class to get the maximum number of walkers.

This leads me to saying 1/3 of the 36 is 12. This IS the maximum number of walkers... if you didn't have any girls in the class. Since you have to have boys AND girls, you start taking out boys until you find a number that combines girls and boys to 36 and qualify as even multiples of 4 and 3, respectively.

Since I know I can't have a class of all boys, the lowest common multiple of 3 and 4 is 12. This means I should take out 12 boys, leaving 24 boys and 4 girls. (or you could have done this by jumping down multiples of 3... 33, 30, 27, 24... until you find a number that evens the class out with girls).

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Post Wed Nov 29, 2017 10:05 am
resilient wrote:
In a certain class consisting of 36 students, some boys and some girls, exactly 1/3 boys and 1/4 of the girls walk to school. What is the greatest possible number of students in this class who walk to school?

9
10
11
12
13
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

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Post Tue Dec 05, 2017 9:29 am
Quote:
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

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