Wine Problem

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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Wed Jan 23, 2013 9:56 am

No OA. Detailed explanations would be appreciated. Thanks

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Source: Beat The GMAT — Problem Solving | Wed Jan 23, 2013 9:56 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Jan 23, 2013 1:19 pm

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.

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vishugogo: Master | Next Rank: 500 Posts; Posts: 131; Joined: Tue Aug 30, 2011 4:50 amFollowed by: 6 members; Location: India; Thanked: 28 times

by vishugogo » Fri Jan 25, 2013 5:12 am

GMATGuruNY wrote: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.

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hemant_rajput: Master | Next Rank: 500 Posts; Posts: 447; Joined: Sun Apr 22, 2012 7:13 amFollowed by: 13 members; Thanked: 46 times; GMAT Score:700

by hemant_rajput » Sat Jan 26, 2013 10:34 am

GMATGuruNY wrote: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

I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.

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