What should be the value of the fourth column?

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- Dec 8th 2009, 11:43 AMMukilabAnnoying Puzzle
What should be the value of the fourth column?

http://www.mensa.org.uk/download/3077/week-25.JPG - Dec 8th 2009, 12:03 PMTheEmptySet
- Dec 8th 2009, 12:10 PMMukilab
- Dec 8th 2009, 01:12 PMmanhmanh17vn
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 AMMukilab
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 AMSoroban
Hello, Mukilab!

Quote:

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

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 AMMukilab