# Find the total number of red and blue beads

• MHB
• anemone
In summary: Container B] (B) at (0,5.-1);\filldraw [fill=cyan, thin, draw=black] (0.3,3.6) rectangle (0.6,4.6);\filldraw [fill=cyan, thin, draw=black] (0.6,3.6) rectangle (0.9,4.6);\
anemone
Gold Member
MHB
POTW Director
There are some red and blue beads.
The beads were packed into 2 containers.
At first, Container A contained 500 beads and $\dfrac{2}{5}$ of them were blue beads. Container B contained 300 beads and $\dfrac{1}{5}$ of them were blue.

Find the total number of red and blue beads that must be moved from Container A to Container B such that $\dfrac{3}{5}$ of the beads in Container A are red and $\dfrac{1}{4}$ of the beads in Container B are blue.

I am wondering if this problem can be solved without using any algebra method...

Hi anemone,

Are you sure the problem is correctly stated ?

As I read it, you should still end up with $\dfrac25$ blue beads in container A. As the proportion does not change, you must move $3$ red beads for every $2$ blue.

However, this can only decrease the proportion of blue beads in container B, and it cannot increase from $\dfrac15$ to $\dfrac14$.

Did I miss something ?

Initially, 60 of the 300 beads in container B are blue. After adding $5n$ beads, $2n$ of which are blue, the proportion will be $\dfrac{60+2n}{300+5n}$, which could increase towards a limit of $\dfrac25$ for large enough $n$.

castor28 said:
Hi anemone,

Are you sure the problem is correctly stated ?

Thanks for the reply, castor28! Yes, I checked again with the source from where I got the problem, there isn't any typo.

castor28 said:
As I read it, you should still end up with $\dfrac25$ blue beads in container A. As the proportion does not change, you must move $3$ red beads for every $2$ blue.

However, this can only decrease the proportion of blue beads in container B, and it cannot increase from $\dfrac15$ to $\dfrac14$.

Did I miss something ?
(Mmm) I actually don't see why not...(Crying)

The following method (without using algebra but the Singapore model method) works, by separating one part of fraction in container A into 5 equal subparts and one part of fraction in container B into 3 equal subparts, where each subpart represents 20 beads as shown below:

Before:
[TIKZ]
\coordinate[label=left: Container A] (A) at (0,5.3);
\filldraw [fill=cyan, thin, draw=black] (0.3,5) rectangle (0.6,6);
\filldraw [fill=cyan, thin, draw=black] (0.6,5) rectangle (0.9,6);
\filldraw [fill=cyan, thin, draw=black] (0.9,5) rectangle (1.2,6);
\filldraw [fill=cyan, thin, draw=black] (1.2,5) rectangle (1.5,6);
\filldraw [fill=cyan, thin, draw=black] (1.5,5) rectangle (1.8,6);
\filldraw [fill=cyan, thin, draw=black] (1.8,5) rectangle (2.1,6);
\filldraw [fill=cyan, thin, draw=black] (2.1,5) rectangle (2.4,6);
\filldraw [fill=cyan, thin, draw=black] (2.4,5) rectangle (2.7,6);
\filldraw [fill=cyan, thin, draw=black] (2.7,5) rectangle (3,6);
\filldraw [fill=cyan, thin, draw=black] (3,5) rectangle (3.3,6);
\filldraw [fill=magenta, thin, draw=black] (3.3,5) rectangle (3.6,6);
\filldraw [fill=magenta, thin, draw=black] (3.6,5) rectangle (3.9,6);
\filldraw [fill=magenta, thin, draw=black] (3.9,5) rectangle (4.2,6);
\filldraw [fill=magenta, thin, draw=black] (4.2,5) rectangle (4.5,6);
\filldraw [fill=magenta, thin, draw=black] (4.5,5) rectangle (4.8,6);
\filldraw [fill=magenta, thin, draw=black] (4.8,5) rectangle (5.1,6);
\filldraw [fill=magenta, thin, draw=black] (5.1,5) rectangle (5.4,6);
\filldraw [fill=magenta, thin, draw=black] (5.4,5) rectangle (5.7,6);
\filldraw [fill=magenta, thin, draw=black] (5.7,5) rectangle (6,6);
\filldraw [fill=magenta, thin, draw=black] (6,5) rectangle (6.3,6);
\filldraw [fill=magenta, thin, draw=black] (6.3,5) rectangle (6.6,6);
\filldraw [fill=magenta, thin, draw=black] (6.6,5) rectangle (6.9,6);
\filldraw [fill=magenta, thin, draw=black] (6.9,5) rectangle (7.2,6);
\filldraw [fill=magenta, thin, draw=black] (7.2,5) rectangle (7.5,6);
\filldraw [fill=magenta, thin, draw=black] (7.5,5) rectangle (7.8,6);
\draw [very thick] (0.3,5) rectangle (1.8,6);
\draw [very thick] (1.8,5) rectangle (3.3,6);
\draw [very thick] (3.3,5) rectangle (4.8,6);
\draw [very thick] (4.8,5) rectangle (6.3,6);
\draw [very thick] (6.3,5) rectangle (7.8,6);
\coordinate[label=left: Container B] (B) at (0,5.-1);
\filldraw [fill=cyan, thin, draw=black] (0.3,3.6) rectangle (0.6,4.6);
\filldraw [fill=cyan, thin, draw=black] (0.6,3.6) rectangle (0.9,4.6);
\filldraw [fill=cyan, thin, draw=black] (0.9,3.6) rectangle (1.2,4.6);
\filldraw [fill=magenta, thin, draw=black] (1.2,3.6) rectangle (1.5,4.6);
\filldraw [fill=magenta, thin, draw=black] (1.5,3.6) rectangle (1.8,4.6);
\filldraw [fill=magenta, thin, draw=black] (1.8,3.6) rectangle (2.1,4.6);
\filldraw [fill=magenta, thin, draw=black] (2.1,3.6) rectangle (2.4,4.6);
\filldraw [fill=magenta, thin, draw=black] (2.4,3.6) rectangle (2.7,4.6);
\filldraw [fill=magenta, thin, draw=black] (2.7,3.6) rectangle (3,4.6);
\filldraw [fill=magenta, thin, draw=black] (3,3.6) rectangle (3.3,4.6);
\filldraw [fill=magenta, thin, draw=black] (3.3,3.6) rectangle (3.6,4.6);
\filldraw [fill=magenta, thin, draw=black] (3.6,3.6) rectangle (3.9,4.6);
\filldraw [fill=magenta, thin, draw=black] (3.9,3.6) rectangle (4.2,4.6);
\filldraw [fill=magenta, thin, draw=black] (4.2,3.6) rectangle (4.5,4.6);
\filldraw [fill=magenta, thin, draw=black] (4.5,3.6) rectangle (4.8,4.6);
\draw [very thick] (0.3,3.6) rectangle (1.2,4.6);
\draw [very thick] (1.2,3.6) rectangle (2.1,4.6);
\draw [very thick] (2.1,3.6) rectangle (3,4.6);
\draw [very thick] (3,3.6) rectangle (3.9,4.6);
\draw [very thick] (3.9,3.6) rectangle (4.8,4.6);
\draw [<->] (0.3,6.2) -- (0.6,6.2);
[/TIKZ]

After:
[TIKZ]
\coordinate[label=left: Container A] (A) at (0,5.3);
\filldraw [fill=cyan, thin, draw=black] (0.3,5) rectangle (0.6,6);
\filldraw [fill=cyan, thin, draw=black] (0.6,5) rectangle (0.9,6);
\filldraw [fill=cyan, thin, draw=black] (0.9,5) rectangle (1.2,6);
\filldraw [fill=cyan, thin, draw=black] (1.2,5) rectangle (1.5,6);
\filldraw [fill=cyan, thin, draw=black] (1.5,5) rectangle (1.8,6);
\filldraw [fill=cyan, thin, draw=black] (1.8,5) rectangle (2.1,6);
\filldraw [fill=cyan, thin, draw=black] (2.1,5) rectangle (2.4,6);
\filldraw [fill=cyan, thin, draw=black] (2.4,5) rectangle (2.7,6);
\filldraw [fill=cyan, thin, draw=black] (2.7,5) rectangle (3,6);
\filldraw [fill=cyan, thin, draw=black] (3,5) rectangle (3.3,6);
\filldraw [fill=magenta, thin, draw=black] (3.3,5) rectangle (3.6,6);
\filldraw [fill=magenta, thin, draw=black] (3.6,5) rectangle (3.9,6);
\filldraw [fill=magenta, thin, draw=black] (3.9,5) rectangle (4.2,6);
\filldraw [fill=magenta, thin, draw=black] (4.2,5) rectangle (4.5,6);
\filldraw [fill=magenta, thin, draw=black] (4.5,5) rectangle (4.8,6);
\filldraw [fill=magenta, thin, draw=black] (4.8,5) rectangle (5.1,6);
\filldraw [fill=magenta, thin, draw=black] (5.1,5) rectangle (5.4,6);
\filldraw [fill=magenta, thin, draw=black] (5.4,5) rectangle (5.7,6);
\filldraw [fill=magenta, thin, draw=black] (5.7,5) rectangle (6,6);
\filldraw [fill=magenta, thin, draw=black] (6,5) rectangle (6.3,6);
\filldraw [fill=magenta, thin, draw=black] (6.3,5) rectangle (6.6,6);
\filldraw [fill=magenta, thin, draw=black] (6.6,5) rectangle (6.9,6);
\filldraw [fill=magenta, thin, draw=black] (6.9,5) rectangle (7.2,6);
\filldraw [fill=magenta, thin, draw=black] (7.2,5) rectangle (7.5,6);
\filldraw [fill=magenta, thin, draw=black] (7.5,5) rectangle (7.8,6);
\draw [very thick] (0.3,5) rectangle (1.8,6);
\draw [very thick] (1.8,5) rectangle (3.3,6);
\draw [very thick] (3.3,5) rectangle (4.8,6);
\draw [very thick] (4.8,5) rectangle (6.3,6);
\draw [very thick] (6.3,5) rectangle (7.8,6);
\coordinate[label=left: Container B] (B) at (0,5.-1);
\filldraw [fill=cyan, thin, draw=black] (0.3,3.6) rectangle (0.6,4.6);
\filldraw [fill=cyan, thin, draw=black] (0.6,3.6) rectangle (0.9,4.6);
\filldraw [fill=cyan, thin, draw=black] (0.9,3.6) rectangle (1.2,4.6);
\filldraw [fill=magenta, thin, draw=black] (1.2,3.6) rectangle (1.5,4.6);
\filldraw [fill=magenta, thin, draw=black] (1.5,3.6) rectangle (1.8,4.6);
\filldraw [fill=magenta, thin, draw=black] (1.8,3.6) rectangle (2.1,4.6);
\filldraw [fill=magenta, thin, draw=black] (2.1,3.6) rectangle (2.4,4.6);
\filldraw [fill=magenta, thin, draw=black] (2.4,3.6) rectangle (2.7,4.6);
\filldraw [fill=magenta, thin, draw=black] (2.7,3.6) rectangle (3,4.6);
\filldraw [fill=magenta, thin, draw=black] (3,3.6) rectangle (3.3,4.6);
\filldraw [fill=magenta, thin, draw=black] (3.3,3.6) rectangle (3.6,4.6);
\filldraw [fill=magenta, thin, draw=black] (3.6,3.6) rectangle (3.9,4.6);
\filldraw [fill=magenta, thin, draw=black] (3.9,3.6) rectangle (4.2,4.6);
\filldraw [fill=magenta, thin, draw=black] (4.2,3.6) rectangle (4.5,4.6);
\filldraw [fill=magenta, thin, draw=black] (4.5,3.6) rectangle (4.8,4.6);
\filldraw [fill=cyan, thin, draw=black] (4.8,3.6) rectangle (5.1,4.6);
\filldraw [fill=cyan, thin, draw=black] (5.1,3.6) rectangle (5.4,4.6);
\filldraw [fill=magenta, thin, draw=black] (5.4,3.6) rectangle (5.7,4.6);
\filldraw [fill=magenta, thin, draw=black] (5.7,3.6) rectangle (6,4.6);
\filldraw [fill=magenta, thin, draw=black] (6,3.6) rectangle (6.3,4.6);
\draw [very thick] (0.3,3.6) rectangle (1.2,4.6);
\draw [very thick] (1.2,3.6) rectangle (2.1,4.6);
\draw [very thick] (2.1,3.6) rectangle (3,4.6);
\draw [very thick] (3,3.6) rectangle (3.9,4.6);
\draw [very thick] (3.9,3.6) rectangle (4.8,4.6);
\draw [very thick, dotted] (4.8,3.6) rectangle (6.3,4.6);
\draw (2.7,5) --(3,6);
\draw (3,5) --(3.3,6);
\draw (6.9,5) --(7.2,6);
\draw (7.2,5) --(7.5,6);
\draw (7.5,5) --(7.8,6);
[/TIKZ]

As we can see, after moving 2 subparts of blue beads and 3 subparts of red beads from container A into container B, we get what we wanted in the end, $\dfrac{3}{5}$ of the beads in Container A are red and $\dfrac{1}{4}$ of the beads in Container B are blue.

Therefore, the total number of red and blue beads that must be moved from container A to container B is $5\times 20 =100$.

You're welcome, morgancol!

You're also encouraged to fire away with whatever math problems that you can't solve and wish to get some help at our forum! Just so you know, for problem involving fraction/ratio/percentage like this one, it can be solved using algebra or without algebra way. (Happy)

## 1. How do you find the total number of red and blue beads?

To find the total number of red and blue beads, you need to count the number of red beads and the number of blue beads separately, then add the two numbers together.

## 2. What if there are other colors of beads mixed in?

If there are other colors of beads mixed in, you still need to count the red and blue beads separately. The total number of beads will be the sum of the red and blue beads, regardless of any other colors present.

## 3. Is there a specific formula for finding the total number of beads?

No, there is no specific formula for finding the total number of beads. It simply involves counting the number of red and blue beads and adding them together.

## 4. Can you use a scientific method to determine the total number of beads?

No, determining the total number of beads is not a scientific experiment. It is a simple counting task that does not require the use of a scientific method.

## 5. Do you need to know the ratio of red to blue beads in order to find the total number?

No, you do not need to know the ratio of red to blue beads in order to find the total number. You just need to count the number of each color and add them together.

Replies
1
Views
1K
Replies
1
Views
635
Replies
2
Views
4K
Replies
2
Views
1K
Replies
1
Views
861
Replies
1
Views
1K
Replies
69
Views
4K
Replies
1
Views
3K
Replies
6
Views
948
Replies
2
Views
2K