The first thief takes (1/2 x + 1) . What remains ? x - (1/2x + 1)
So the 2nd thief takes 2/3 of [ x - (1/2x + 1) ]
What remains ? x - 2/3 [ x - (1/2x + 1) ]
So the 3rd thief takes 2/3 of { x - 2/3 [ x - (1/2x + 1) ] } and he takes 1 more .
What remains ? x - ( 2/3 { x - 2/3 [ x - (1/2x + 1) ] } + 1 )
And that whole ugly thing is equal to ' 1 ', so you can solve it for 'x'..
The whole problem from here on is an exercise in simplifying
an expression with a bunch of 'nested' parentheses in it.
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This is a lot harder than just solving the problem with logic and
waving your hands in the air. Here's how you would do that:
Start from the end and work backwards:
-- One diamond is left.
-- Before the 3rd thief took 1 more, there were 2.
-- That was 1/3 of what was there before the 3rd man took 2/3.
So he found 6 when he arrived.
-- 6 was 1/3 of what was there before the second thief helped himself.
So there were 18 when the 2nd man arrived.
-- 18 was 1 less than what was there before the first thief took 1 extra.
So he took his 1 extra from 19.
-- 19 was the remaining after the first man took 1/2 of all on the table.
So there were 38 on the table when he arrived.
Thank you for your generous 5 points.
For a regular tessellation, the shapes can be duplicated infinitely to fill a plane such that there is no gap. The only shapes that can form regular tessellations are equilateral traingle(all sides are equal. This means that it can be turned to any side and it would remain the same), square and regular hexagon. Looking at the given options, we have
Shape Tessellate?
Octagon No
Hexagon Yes
Pentagon No
Square Yes
Triangle No(unless it is specified that it is an equilateral triangle)
Make two shapes out of it.
The bottom is a rectangle 14 x 5 = 70 square cm
The top is a triangle 1/2 x 12 x 5 = 30 square cm
Total area = 70 + 30 = 100 square cm
Your answer would be 2.93.
