Wine Problem

[knight247]

No OA. Detailed explanations would be appreciated. Thanks

[GMATGuruNY]

Since 4/40 = 1/10, the volume of wine decreases by 1/10 each time 4 liters are drawn from the 40-liter cask.
To determine the resulting ratio of wine to water, plug in an easy value for the total volume.

Let the original volume of wine = 1000 liters.
Remaining amount of wine after the first drawing = 1000 - .1(1000) = 1000 - 100 = 900.
Remaining amount of wine after the second drawing = 900 - .1(900) = 900 - 90 = 810.
Remaining amount of wine after the third drawing = 810 - .1(810) = 810 - 81 = 729.
Resulting volume of water = 1000-729 = 271.
Wine : water = 729:271.

Wine/total = 729/1000.
Since the actual volume of the cask is 40 liters, the remaining amount of wine after the 3 drawings = (729/1000) * 40 = 29.6 liters.

[vishugogo]

Since 4/40 = 1/10, the volume of wine decreases by 1/10 each time 4 liters are drawn from the 40-liter cask.
To determine the resulting ratio of wine to water, plug in an easy value for the total volume.

Let the original volume of wine = 1000 liters.
Remaining amount of wine after the first drawing = 1000 - .1(1000) = 1000 - 100 = 900.
Remaining amount of wine after the second drawing = 900 - .1(900) = 900 - 90 = 810.
Remaining amount of wine after the third drawing = 810 - .1(810) = 810 - 81 = 729.
Resulting volume of water = 1000-729 = 271.
Wine : water = 729:271.

Wine/total = 729/1000.
Since the actual volume of the cask is 40 liters, the remaining amount of wine after the 3 drawings = (729/1000) * 40 = 29.6 liters.

[hemant_rajput]

Since 4/40 = 1/10, the volume of wine decreases by 1/10 each time 4 liters are drawn from the 40-liter cask.
To determine the resulting ratio of wine to water, plug in an easy value for the total volume.

Let the original volume of wine = 1000 liters.
Remaining amount of wine after the first drawing = 1000 - .1(1000) = 1000 - 100 = 900.
Remaining amount of wine after the second drawing = 900 - .1(900) = 900 - 90 = 810.
Remaining amount of wine after the third drawing = 810 - .1(810) = 810 - 81 = 729.
Resulting volume of water = 1000-729 = 271.
Wine : water = 729:271.

Wine/total = 729/1000.
Since the actual volume of the cask is 40 liters, the remaining amount of wine after the 3 drawings = (729/1000) * 40 = 29.6 liters.

Guru has already provide a detailed explanation for this kind of problem. So I'm going to show you a little trick to solve this quickly in exam.

just remember this formula:-

wine/(water + wine) =(1 - (b/a))^n

b = amount of wine replaced by water.
a = amount of wine initially.
n = no. of time the replacement process was repeated.


so now applying this method to the problem above.

[1 -(4/40)]^3 ==> [ 1 -(1/0)]^3 = [9/10]^3 = 729/1000.

a.
so final quantity of wine left (729/1000)*40.

b. 729/(1000-729) = 729/271