# Annoying Puzzle

• Dec 8th 2009, 11:43 AM
Mukilab
Annoying Puzzle
What should be the value of the fourth column?

• Dec 8th 2009, 12:03 PM
TheEmptySet
Quote:

Originally Posted by Mukilab
What should be the value of the fourth column?

Write a system of equations using each Blue and Red as a variables and the 2nd and 3rd columns.

Then use the first column to find yellow.

Then the final answer is $\displaystyle R+2Y+B=41$
• Dec 8th 2009, 12:10 PM
Mukilab
Quote:

Originally Posted by TheEmptySet
Write a system of equations using each Blue and Red as a variables and the 2nd and 3rd columns.

Then use the first column to find yellow.

Then the final answer is $\displaystyle R+2Y+B=41$

That's exactly what I did but I ended up scrawling all over my notepad. I'll try again and see if I get your answer.
• Dec 8th 2009, 01:12 PM
manhmanh17vn
Quote:

Originally Posted by Mukilab
What should be the value of the fourth column?

call red is R, yellow is Y and blue is B
so on the first column we have $\displaystyle 2R+Y+B=43$
second column we have $\displaystyle Y+3B=55$
last column we have $\displaystyle 3R+B=45$
so you have
$\displaystyle 2R+Y+B=43$
$\displaystyle Y+3B=55$
$\displaystyle 3R+B=45$
then you can find R, B, and Y by yourself. try this and talk your answer.
• Dec 9th 2009, 10:40 AM
Mukilab
Sorry, I don't know how I got muddled up. I think it's because i named them a, b and c then got a mixed with c and so on...

Anyways I got

Y+2=R
39-3Y=B

SUB INTO: R+2Y+B=?

Y+2+2Y+39-3Y

41=?

Thanks
• Dec 9th 2009, 11:32 AM
Soroban
Hello, Mukilab!

Quote:

What should be the value of the fourth column?

We have a system of equations:

. . $\displaystyle \begin{array}{ccccc}\text{Column 1:} & 2R + Y + B &=& 43 \\ \text{Column 2:} & R \qquad+ 3B &=& 55 \\ \text{Colukn 3:} & 3R \qquad + B &=& 45 \end{array}$

Solution: .$\displaystyle R = 10,\;Y = 8,\;B = 15$

Columkn 4: .$\displaystyle R + Y + B + Y \;=\;10+8+15+8\;=\;41$

• Dec 10th 2009, 09:53 AM
Mukilab
Quote:

Originally Posted by Mukilab
Sorry, I don't know how I got muddled up. I think it's because i named them a, b and c then got a mixed with c and so on...

Anyways I got

Y+2=R
39-3Y=B

SUB INTO: R+2Y+B=?

Y+2+2Y+39-3Y

41=?

Thanks

Quote:

Originally Posted by Soroban
Hello, Mukilab!

We have a system of equations:

. . $\displaystyle \begin{array}{ccccc}\text{Column 1:} & 2R + Y + B &=& 43 \\ \text{Column 2:} & R \qquad+ 3B &=& 55 \\ \text{Colukn 3:} & 3R \qquad + B &=& 45 \end{array}$

Solution: .$\displaystyle R = 10,\;Y = 8,\;B = 15$

Columkn 4: .$\displaystyle R + Y + B + Y \;=\;10+8+15+8\;=\;41$

Sorry for the misunderstanding, I thought my previous post showed I got it. Thanks for the contribution though!