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.
Answer:
12cm^2
Step-by-step explanation:
1/2 x diagonal x diagonal
I would prefer to solve this in a decimal form
I will show this to u in 2 ways
1:.75::X:2.25
(1*2.25)/.75x
2.25/.75x
X=3
2.25/.75=3
when we divide the number of hours played by hors for one game we get the number of games that could be played
Answer:
28
Step-by-step explanation:
12+[(15-5)]+(9-3)]
PEMDAS
Parentheses first
12+[(10)]+(6)]
Then add
12+10+6
