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PS (potatoes) - m04q05

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PS (potatoes) - m04q05 [#permalink]

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29 Aug 2007, 05:29
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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) $$11\frac{2}{3}$$

[Reveal] Spoiler: OA
B

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29 Aug 2007, 06:34
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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]

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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.

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Re: PS (potatoes) - m04q05 [#permalink]

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24 Mar 2010, 08:19
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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]

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24 Mar 2010, 09:29
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) $$11\frac{2}{3}$$
[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

1/10 + 1/x = 1/7
x = 70/3 = $$23\frac{1}{3}$$
hence B

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Re: PS (potatoes) - m04q05 [#permalink]

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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]

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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]

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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]

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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]

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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

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Re: PS (potatoes) - m04q05 [#permalink]

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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]

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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]

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30 Mar 2012, 06:45
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) $$11\frac{2}{3}$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

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]

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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]

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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]

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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]

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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]

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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]

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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]

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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,

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Re: PS (potatoes) - m04q05   [#permalink] 04 Apr 2013, 12:53

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