with working pls
2006-06-07 06:59:29 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
a man burns a cigarrete to only 2/3rd part, if he uses the rest of the part of 3 cigarettes he makes 1 more cigrette.if he as 27 cigretes, how many did he burn in total?
2006-06-07 07:00:05 · update #1
He smokes the original 27 cigarettes and ends up with 27 butts. From these he can create 9 new cigarettes.
After he smokes these he has 9 butts from which he can create 3 new cigarettes.
Finally, he smokes these and has 3 butts to make 1 final cigarette.
When he smokes that he has 1 remaining butt.
27 + 9 + 3 + 1 = 40 cigarettes.
You are essentially asking how many 'smokable' portions of 27 cigarettes are there. And given that a smokable portion is 2/3 of a full cigarette, you could do this as:
27 divided by 2/3.
This is the same as:
27 times 3/2
Which comes out to:
81/2
40 1/2
So in the end he has 40 smokable portions (equal to 2/3) and 1/2 of a smokable portion (equal to 1/3).
The answer to your question is 40 'cigarettes' with one butt leftover.
2006-06-07 07:07:25 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
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2016-09-08 21:53:02 · answer #2 · answered by arruda 4 · 0⤊ 0⤋
26 2/3.
Only the last 1/3 is not burned.
2006-06-07 09:08:04 · answer #3 · answered by TfC_137 3 · 0⤊ 0⤋
of 3, you make 1 more
of 9, you make 3 more and 1 more
of 27, you make 9 more and 3 more and 1 more
27+9+3+1=40
2006-06-07 07:09:34 · answer #4 · answered by ringm 3 · 0⤊ 0⤋