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PS (potatoes) - m04q05

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Manager
Manager
User avatar
Joined: 15 Aug 2012
Posts: 96
Location: India
Concentration: Technology, Strategy
Schools: Merage '15 (A)
GPA: 3.6
WE: Consulting (Computer Software)
Followers: 0

Kudos [?]: 25 [0], given: 13

Re: PS (potatoes) - m04q05 [#permalink] New post 18 Apr 2013, 00:30
Robert can eat a 20 kg. bag of potatoes in 10 days. He and his wife Jane can eat the same bag in 7 days. How many days does it take Jane to eat a 20 kg. bag of potatoes by herself?

Let Jane eat 20kg bag of potatoes in x days.

The eqn=>1/x + 1/10 = 1/7

solving for x we get 70/3.

Hence B
Intern
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Joined: 24 Feb 2014
Posts: 2
Schools: ISB
Followers: 0

Kudos [?]: 0 [0], given: 7

Re: PS (potatoes) - m04q05 [#permalink] New post 19 Mar 2014, 19:13
Vercules wrote:
Policarpa wrote:
Can someone type out the Algebra from:
(1/10) + (1/x) = 1/7

Thanks


\frac{1}{10}+\frac{1}{x}=\frac{1}{7}

\frac{1}{x} = \frac{1}{7}-\frac{1}{10}

\frac{1}{x}=\frac{10-7}{70}

\frac{1}{x}=\frac{3}{70}

x=\frac{70}{3}

Hope this helps,

Vercules


dis is the right approach for such a problem..easy n crisp! cheers!
Intern
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Joined: 02 Dec 2013
Posts: 6
Location: India
Concentration: International Business, Technology
GMAT Date: 02-28-2014
GPA: 3.5
WE: Engineering (Other)
Followers: 0

Kudos [?]: 5 [0], given: 7

Re: PS (potatoes) - m04q05 [#permalink] New post 19 Mar 2014, 23:20
If robert eats 20kg of potatoes in 10 days, then per day he ll eat 2kg of potatoes.
Let "x" be the amount of potatoes, which will be eaten by his wife., then we can write a simple equation as follows

2+x=(20/7) since its given that if both eat together then 20kgs of potatoes will be finished in 7 days


therefore,
we get x=6/7

Let n be the number of days rob's wife needs to eat 20 kg of potatoes,

(X)x(n)=20
n=20x7/6
n=70/3...... Voilaaaaa
Re: PS (potatoes) - m04q05   [#permalink] 19 Mar 2014, 23:20
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