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PS (potatoes) - m04q05 : Retired Discussions [Locked]

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PS (potatoes) - m04q05 [#permalink]
29 Aug 2007, 04:29

2

This post was BOOKMARKED

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?

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.

Re: PS (potatoes) - m04q05 [#permalink]
24 Mar 2010, 08:29

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}

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

Re: PS (potatoes) - m04q05 [#permalink]
28 Mar 2011, 05:07

Expert's post

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 _________________

Re: PS (potatoes) - m04q05 [#permalink]
28 Mar 2011, 20:04

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.

Re: PS (potatoes) - m04q05 [#permalink]
30 Mar 2012, 05:45

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?

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 _________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Re: PS (potatoes) - m04q05 [#permalink]
30 Mar 2012, 06:04

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]

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.

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 _________________

Please +1 KUDO if my post helps. Thank you.

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