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PS (potatoes) - m04q05 [#permalink]
 29 Aug 2007, 05:29
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? (A) 46\frac{2}{3}(B) 23\frac{1}{3}(C) 17(D) 12(E) 11\frac{2}{3} Source: GMAT Club Tests - hardest GMAT questions Please, explanations ONLY.
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Robert must be extremely fat... 2kgs of potatoes in a day  
robert eats 20/10 in one day
jane eats 20/j in one day
robert + jane = 20/10 +20/j
given robert+jane eats in 7 days so perday they will eat 20/7
hence,
20/10+20/j=20/7
j = 23+ 1/3 or 70/3
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Re: PS (potatoes) - m04q05 [#permalink]
21 Sep 2009, 12:49
gmat, can you break down the equation 20/10 + 20/j = 20/7 -----> j= 23 1/3 or 70/3 for me? For some reason I just can't see it worked out right now. Thanks. Posted from my mobile device 
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Re: PS (potatoes) - m04q05 [#permalink]
24 Mar 2010, 08:19
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Answer (B)
Total potatoes = 20Kg (redundant data)
In 1 day Robert can eat (1/10) of total potatoes (A)
In 1 day Robert + Jane can eat (1/7) of total potatoes (B)
Hence, In 1 day Jane alone can eat (B - A) total potatoes
= (1/7) - (1/10) = 3/70
3/70 of total potato in 1 Day
total potatos in 1/(3/70) days
= 70/3 = 23*(1/3) days
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Re: PS (potatoes) - m04q05 [#permalink]
24 Mar 2010, 09:29
Vlad77 wrote: 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? (A) 46\frac{2}{3}(B) 23\frac{1}{3}(C) 17(D) 12(E) 11\frac{2}{3} Source: GMAT Club Tests - hardest GMAT questions Please, explanations ONLY. 1/10 + 1/x = 1/7 x = 70/3 = 23\frac{1}{3}hence B
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Re: PS (potatoes) - m04q05 [#permalink]
10 Apr 2010, 05:36
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yes, the fractional approach is simpler. 1/10 + 1/x = 1/7 x = 70/3(23 1/3)
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Re: PS (potatoes) - m04q05 [#permalink]
05 Jul 2010, 02:04
Not sure, if it is the right way to approach it.
Let's say Robert and Robert/Jane consume potato at a constant rate.
That is 2kg and 20/7kg a day respectively.
Then difference in the rate can be ascribed to Jane.
20/7-14/7=6/7
# of days Jane would need to consume 20kg of potato is 20/6/7=70/3
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Re: PS (potatoes) - m04q05 [#permalink]
28 Mar 2011, 06:07
think what type of problem is it. is it a work problem or rate problem? since the rate at which robert and jane consume potatoes are given, it's a rate problem. however i am writing down the formulae for work problem as well as rate problems work problem\ 1/X + 1/Y=1/Z WHERE X AND Y ARE THE RESPECTIVE AMOUNT OF WORK DONE BY X AND Y RATE PROBLEM rates should be added directly. hence, x + y=z where x and y are the respective rates of X and Y and Z is the combined rate. note that z>(x and y). applying in this question, 2+x=20/7, x=6/7. since this the rate at which jane consumes potatoes i.e. 6 poltatoes in 7 days, and we are supposed to find out the time, therefore 20=6/7 * days days = 70/3. ans B 
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Re: PS (potatoes) - m04q05 [#permalink]
28 Mar 2011, 10:36
Work problem
1/10 + 1/X=1/7
Solving for X
x=70/3 (23 1/3)
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Re: PS (potatoes) - m04q05 [#permalink]
28 Mar 2011, 18:07
The 20 kg information is not relevant as it's same for both. In 1 day Robert -> 1/10 part of the bag In 1 day Jane - > 1/x part of bag 1/10 + 1/x = 1/7 => 1/x = (10 - 7)/70 => x = 70/3 = 23 1/3 Answer - B
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Re: PS (potatoes) - m04q05 [#permalink]
28 Mar 2011, 21:04
Scithian wrote: Not sure, if it is the right way to approach it.
Let's say Robert and Robert/Jane consume potato at a constant rate.
That is 2kg and 20/7kg a day respectively.
Then difference in the rate can be ascribed to Jane.
20/7-14/7=6/7
# of days Jane would need to consume 20kg of potato is 20/6/7=70/3 I think your approach is good. It's much easier to solve like this.
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Re: PS (potatoes) - m04q05 [#permalink]
30 Mar 2012, 06:27
Here is my approach:
Robert's rate is 2kg/day (He eats 20kg in 10 days.....sure he must be BIG hehe).
Robert and Jane, together, eat 20kg in 7 days, making their combined rate 20/7 kg per day.
So, 2 kg/day + Jane's rate = Their combined rate.
Jane's rate = 20/7 - 2 = (20 - 14)/7 = 6/7 kg per day.
Therefore, Jane can eat 20 kg potatoes in 20/(6/7) or 20 x (7/6) or 140/6 which comes out to be 23.33 or 23 1/3. Hence B.
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Re: PS (potatoes) - m04q05 [#permalink]
30 Mar 2012, 06:45
Vlad77 wrote: 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? (A) 46\frac{2}{3}(B) 23\frac{1}{3}(C) 17(D) 12(E) 11\frac{2}{3} Source: GMAT Club Tests - hardest GMAT questions Please, explanations ONLY. rate of husband & wife - husband...take this rate and 20/rate = gives the wife's num of days to complete the task.... so the answer is B
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Re: PS (potatoes) - m04q05 [#permalink]
30 Mar 2012, 07:04
A simpler formula exists when work related problem related to two people is presented:-
Suppose John takes "A" ( hours/day) to complete some work and Mary takes "B" (hours /days to complete the same work, then the time taken together can be calculated by using the formula [A X B]/[A + B]
In this case
If rate of work of Jane is J then:-
[10 X J]/[10+ J] = 7
Thus J = 23 1/3
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Re: PS (potatoes) - m04q05 [#permalink]
01 Apr 2012, 21:27
1/x+1/y=1/7 1/x=1/7-1/10=3/70 i.e. B
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Re: PS (potatoes) - m04q05 [#permalink]
12 Apr 2012, 04:15
Robert can eat 20kg in 10 days then rate is 2kg/day. R + J can eat 20 in 7 days then J rate is 20/7-2 = 6/7 KG/Day J can eat 20 kg in 20*7/6 days = 23 1/3 days. IMO B
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Re: PS (potatoes) - m04q05 [#permalink]
02 Apr 2013, 18:19
Good question. +1 for B. Rob finishes the bag alone in 10 day --> 1 day he eats 1/10 --> 7 days he eats 7/10 Rob + Jane finish the bag together in 7 day --> Rob eats 7/10, Jane eats 3/10. --> 7 days Jane eats 3/10 --> x days Jane eats 10/10 So x = 7*(10/10) / (3/10) = 70/3 = 23 1/3
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Re: PS (potatoes) - m04q05 [#permalink]
02 Apr 2013, 23:09
From the question,
R * T = W
Robert 2 10= 20KG
R & Wife 20/7 7= 20KG
----------------------------
Wife (20/7-2) * x= 20KG
Solving the equation
Rate=> (20/7-2)* x = 20
==> 20-14 * x = 140
==> 6 * x = 140
==> x = 140/6 = 70/3
==> x = 23(1/3)
Robert's wife can eat 20 kgs of potato in 23 (1/3) days
Wondering whether this falls under hard question.
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Re: PS (potatoes) - m04q05 [#permalink]
04 Apr 2013, 12:43
Can someone type out the Algebra from:
(1/10) + (1/x) = 1/7
Thanks
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Re: PS (potatoes) - m04q05 [#permalink]
04 Apr 2013, 12:53
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
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Re: PS (potatoes) - m04q05
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04 Apr 2013, 12:53
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