barrel’s contents

[gmatcraze]

If 1/3 of a full barrel’s contents is emptied with each turn of the wheel, what fraction of the barrel remains after 2 turns of the wheel? (A) 1/9
(B) 1/3
(C) 4/9
(D) 2/3
(E) 8/9

OA:

SPOILER: C

[flex_manny]

After first turn, remaining = 2/3
Second term empties (1/3) * (2/3) = 2 /9
Total emptied = (1/3) + (2/9) = 5/9
Remaining 1 - (5/9)= 4/9

[ssky]

After 1 st turn barrel lost 1/3. remaning is 1=1/3=2/3
On second turn the barrel looses 1/3 of what was remaining.

1/3 of remaining is 1/3 * 2/3=2/9

total lost so far = 2/3 - 2/9=4/9 so answer is C

[NishantG]

C for me too

[bose]

you can use the formula x(1-1/3)^2=4/9x here x is the original contents of the barrel.

[sanjeev_sharma]

IMO nowhere it is written that 1/3 of the remaining in th barrel will get emptied after a turn of the wheel. Here it is only metioned that 1/3 of the full barrel is emptied with the turn of the wheel.

So after 2 turn of the wheel, barrel will be emptied 2/3 therefore the what is left in the barrel = 1-2/3 = 1/3.

[sanjeev_sharma]

IMO nowhere it is written that 1/3 of the remaining in th barrel will get emptied after a turn of the wheel. Here it is only metioned that 1/3 of the full barrel is emptied with the turn of the wheel.

So after 2 turn of the wheel, barrel will be emptied 2/3 therefore the what is left in the barrel = 1-2/3 = 1/3.

[GmatG]

Originally Posted by flex_manny

After first turn, remaining = 2/3
Second term empties (1/3) * (2/3) = 2 /9
Total emptied = (1/3) + (2/9) = 5/9
Remaining 1 - (5/9)= 4/9

Agree to flex_many's approach.. 4/9

[btit]

Although I got and=C, I think, Sanjiv_Sharma has a point. In the first reading, I did not pay much attention to word "Full"..
Isnt the stem indicates, each time wheel truns, X amt (=1/3 of FULL barrel) emptied.

Any thoughts? Or we should not read that deeeep?

[ssky]

First of all, the calcualtion to getting Answer C is correct, second remember that more than often the obvious easily seen answer is wrong (atleast that is the case with GMAT)