Ratio Question (Primary 6 Singapore): Solution found but require explanation

[khairyfarhan]

Hi everyone,

First off, I'd like to say thank you for helping out. I'm a home tutor who specializes in teaching English, but sometimes help my students for Mathematics. I'm new here so I look forward to learn more from other experienced Maths teachers.

So this is the question:

Megan had some cupcakes. She decided to pack the cupcakes into boxes.
1/4 of the boxes had only 1 cupcake and 5/9 of the remaining boxes had 2 cupcakes each. The rest of the boxes had 3 cupcakes.
Given that there were 3525 cupcakes, how many boxes did Megan have?

The following is the solution that I came up with: (Another person actually provided the solution using algebra, but I was sure this question had something to do with ratio and my student hadn't learned any algebra manipulation yet since he's only Primary 6 - Grade 6 equivalent I think)

Type A Boxes: Boxes which had 1 cupcake
Type B Boxes: Boxes which had 2 cupcakes
Type C Boxes: Boxes which had 3 cupcakes

A : B : C
3 : 5 : 4
This is the ratio between the types of boxes.

I had to find the common link between the ratio of boxes and the ratio of cupcakes per Type. So I multiplied the ratio of boxes with the number of cupcakes they had.

I'm thinking of explaining to my student that we multiply this so we can find out the ratio of cupcakes according to their Types. Does this make sense? This is the first confused part. Or does it tell us the number of parts of cupcakes in the parts of the boxes? (I don't think that makes any sense.)

A : B : C
3 : 5 : 4
x1 : x2 : x3
3 : 10 : 12

So the total parts for the cupcakes should add up to the total number of cupcakes.

3 + 10 + 12 = 25
3525 / 25 = 141

This is the second confused part. I don't know what 141 really represents. I think it represents the number of boxes per part of cupcakes, but I'm sure at this point, I'm confusing myself further.

So since in the beginning we have 12 total parts ( 3 + 5 + 4 ),

141 x 16 = 1692 (answer)

Thank you for the help. I really appreciate it.

God bless.

 

[khairyfarhan]

Oops. Yes you're right, Denis. It should be 141 * 12 = 1692.

Well, yes, that's using algebra to solve it, with the unknown n.

Thing is the 12-year-olds here aren't able to comprehend how to work out this: n(3/12 * 1) + n(5/12 * 2) + n(4/12 * 3) = 3525

Would you advise another way of solving it?

I know this can be frustrating, and I've spent my time on it today for several hours already.

Appreciate the help nonetheless.

 

[stapel]

Megan had some cupcakes. She decided to pack the cupcakes into boxes. One-fourth of the boxes had only 1 cupcake and 5/9 of the remaining boxes had 2 cupcakes each. The rest of the boxes had 3 cupcakes. Given that there were 3525 cupcakes, how many boxes did Megan have?

Use the "bars" they've been learning for so many years.

Draw a bar, and split it into four parts:

*-----*-----*-----*-----*
|     |     |     |     |
|     |     |     |     |
*-----*-----*-----*-----*

The first portion is the "one-fourth" of the boxes, each of which contains one cupcake.

Split the other three portions into thirds:

*-----*-----*-----*-----*
|     | | | | | | | | | |
|     | | | | | | | | | |
*-----*-----*-----*-----*

This splits "the remaining boxes" into nine portions, of which five contain two cupcakes each.

By dividing each of the original four boxes into thirds, the bar is split into twelve portions. Four of the portions each contain three items each; three contain one item each; five contain two items each.

If you combine the four three-items-each portions with four of the two-items-each portions, you'll have four five-items-each portions. This will leave you with one two-items-each portion and three one-item-each portions. Since each of these portions contains the same number of boxes, then the items can be re-sorted to form one five-items-each portion.

This means that the 3,525 cupcakes can be split into five five-items-each portions. Each portion must then contain 3,525 / 5 = 705 cupcakes. Then each five-items-each portion must contain 705 / 5 = 141 sets of five.

Working backwords, what can you obtain for the numbers of each type of box?