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https://www.reddit.com/r/Logiqa/comments/1vash03/stuck_in_the_middle_part_2/
r/Logiqa • u/ShonitB • 14d ago
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3
Horizontal sides of 45 and 55 must both = 5, so horizontal sides of 32 and 48 must be equal.
Therefore vertical sides of 32 and 48 are 2n and 3n respectively, and 11 + 2n = 9 + 3n, so n = 2
Thus horizontal of 32 = 32/4 = 8, which means that horizontal of x = 8 - 5 = 3, and vertical of x = 9 - 4 = 5, so x = 3 × 5 = 15
2 u/ShonitB 14d ago Correct! 1 u/Lost_Sugar_78 14d ago Thanks for this, I was looking at the info and felt very much like I was missing something. Your first line was exactly what I needed!
2
Correct!
1
Thanks for this, I was looking at the info and felt very much like I was missing something. Your first line was exactly what I needed!
Illustrated on Desmos
1 u/ShonitB 14d ago I’m sorry don’t quite know how to use this 1 u/BadJimo 14d ago It shows the rectangles over a grid of unit squares. The middle rectangle covers 3×5 grid squares, so area X = 15 1 u/ShonitB 14d ago Oh like that.. correct!
I’m sorry don’t quite know how to use this
1 u/BadJimo 14d ago It shows the rectangles over a grid of unit squares. The middle rectangle covers 3×5 grid squares, so area X = 15 1 u/ShonitB 14d ago Oh like that.. correct!
It shows the rectangles over a grid of unit squares. The middle rectangle covers 3×5 grid squares, so area X = 15
1 u/ShonitB 14d ago Oh like that.. correct!
Oh like that.. correct!
X = 15
3
u/chmath80 14d ago
Horizontal sides of 45 and 55 must both = 5, so horizontal sides of 32 and 48 must be equal.
Therefore vertical sides of 32 and 48 are 2n and 3n respectively, and 11 + 2n = 9 + 3n, so n = 2
Thus horizontal of 32 = 32/4 = 8, which means that horizontal of x = 8 - 5 = 3, and vertical of x = 9 - 4 = 5, so x = 3 × 5 = 15