News2 mins ago
Sunday Express Mensa Teaser
4 Answers
Can somebody please explain to me the easiest way of working out one of these, where the columns equal certain numbers?
Answers
Best Answer
No best answer has yet been selected by vivfrost. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.Hi, The answer assuming I'm looking at the right mensa question is to put the information into equations like this:-
1. 2x+2y=58
2.c+2z+y=69
3.3y+z=71
then using simutaneous equations you substitute
from 1. above y=29-x making y the subject
now using 3.
3(29-x)+z=71 giving z=-16+3x
now using 2. x+2(-16+3x)+(29-x)=69
you get x=12
now you can get z and y using the same process
1. 2x+2y=58
2.c+2z+y=69
3.3y+z=71
then using simutaneous equations you substitute
from 1. above y=29-x making y the subject
now using 3.
3(29-x)+z=71 giving z=-16+3x
now using 2. x+2(-16+3x)+(29-x)=69
you get x=12
now you can get z and y using the same process
-- answer removed --
Hi Viv
I do so hope this is the right puzzle.
An easier way to understand this is to again put the equations as :-
1. 2x+2y=58
2.x+2z+y=69
3.3y+z=71
Multiply equation 2 by 2:-
2. 2x+4z+2y=138
now deduct equation 1 from it which eliminates2x and 2y and leaves 4z=80 or z=20
just swap 20 for z in 3 giving y=17 etc.
I do so hope this is the right puzzle.
An easier way to understand this is to again put the equations as :-
1. 2x+2y=58
2.x+2z+y=69
3.3y+z=71
Multiply equation 2 by 2:-
2. 2x+4z+2y=138
now deduct equation 1 from it which eliminates2x and 2y and leaves 4z=80 or z=20
just swap 20 for z in 3 giving y=17 etc.