Please help: number of students; maximum number of cubes

nubianj

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May 7, 2006
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i have a couple of questions so pls bear with me

1st is... in a history class that consisted of 30students,the number of seniors was 3 more than twice the number of juniors,and 3/10 of the students wer neither junirs nor seniors, which is greater : the number of juniors in the class OR 6
so far i did 3/10 of 30 which is =9 are not seniors or juniors,so that means there are 21juniors and senior ad this is where i get stuck

Question # 2 if u have a rectangular box 15cm ength,6cm and 12, what is the maximum numbeer of cubes with an edge of 3centimeters that can be packed in the rectangular box with the above dimensions.
 
nubianj said:
1st is... in a history class that consisted of 30students,the number of seniors was 3 more than twice the number of juniors,and 3/10 of the students wer neither junirs nor seniors, which is greater : the number of juniors in the class OR 6

so far i did 3/10 of 30 which is =9 are not seniors or juniors,so that means there are 21juniors and senior ad this is where i get stuck
Are you trying to solve it in your head?

Name stuff! WRITE DOWN clear, concise, accurate, and helpful definitions.

S = # of Seniors in the Class
J = # of Juniors in the Class
F = # of "Other" students. I'll call them "F"reshmen, just so they can have a name.

Now, decode one piece at a time.

"30 students"

S + J + F = 30

"number of seniors was 3 more than twice the number of juniors"

S = (2*J) + 3

"3/10 of the students were neither juniors nor seniors"

F = (3/10)*30

That's all there is. Now look around for simplifications. You got one already.

F = (3/10)*30
F = 9

This makes
S + J + F = 30
S + J + 9 = 30
S + J = 21
You got this, too!

We still have another piece you didn't use.
S = (2*J) + 3

Substitution!!!!! With these two,
S + J = 21
S = (2*J) + 3

We have
[(2*J) + 3] + J = 21

Can you now solve for J?

Note: Be careful. Make sure you reread the problem statement and answer the right question.
 
nubianj said:
if u have a rectangular box 15cm ength,6cm and 12, what is the maximum numbeer of cubes with an edge of 3centimeters that can be packed in the rectangular box with the above dimensions.

a = volume of box : 15 * 6 * 12
b = volume of cubes: 3 * 3 * 3

number of cubes = a / b
 
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