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Intern  B
Joined: 07 May 2015
Posts: 35
Re: Four extra-large sandwiches of exactly the same size were ordered for  [#permalink]

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Hi all,

I don't mean to bump an old post but this leads me to think I am really over-thinking problems. I took the same approach as many of you, picking numbers, but I got lost in the question stem. I kept thinking "is Carol included in M? is she an outsider that eats them after they've been divided evenly among the M students?" This seems to be a common error I make. Any tips for getting over this?
Director  P
Joined: 14 Dec 2017
Posts: 522
Location: India
Re: Four extra-large sandwiches of exactly the same size were ordered for  [#permalink]

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Bunuel wrote:
Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 4. Three of the sandwiches were evenly divided among the students. Since 4 students did not want any of the fourth sandwich, it was evenly divided among the remaining students. If Carol ate one piece from each of the four sandwiches, the amount of sandwich that she ate would be what fraction of a whole extra-large sandwich?

A) (m+4)/[m(m-4)]

B) (2m-4)/[m(m-4)]

C) (4m-4)/[m(m-4)]

D) (4m-8)/[m(m-4)]

E) (4m-12)/[m(m-4)]

Kudos for a correct solution.

Let X be the area of each sandwich.

Given 3 sandwiches are evenly divided among m students. Hence each student received one piece of area 3X/m

Now the 4th sandwich is refused by 4 students & is evenly divided among the rest of the students. Hence each of these students received a X/(m-4) area a piece of this sandwich.

Carol ate one piece from each sandwich, hence total area of sandwich consumed by carol = 3X/m + X/(m-4) = X(4m-12)/[m(m-4)]

Hence Carol ate (4m-12)/[m(m-4)] fraction of the extra large sandwich.

Answer E.

Thanks,
GyM
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Senior Manager  G
Joined: 04 Aug 2010
Posts: 415
Schools: Dartmouth College
Re: Four extra-large sandwiches of exactly the same size were ordered for  [#permalink]

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Bunuel wrote:
Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 4. Three of the sandwiches were evenly divided among the students. Since 4 students did not want any of the fourth sandwich, it was evenly divided among the remaining students. If Carol ate one piece from each of the four sandwiches, the amount of sandwich that she ate would be what fraction of a whole extra-large sandwich?

A) (m+4)/[m(m-4)]

B) (2m-4)/[m(m-4)]

C) (4m-4)/[m(m-4)]

D) (4m-8)/[m(m-4)]

E) (4m-12)/[m(m-4)]

Let m=5, implying that there are 5 students.
Let each sandwich = 5 units, implying that 3 sandwiches = 3*5 = 15 units.

Since 3 sandwiches are distributed among all 5 students -- including Carol -- the number of units received by Carol from these 3 sandwiches = 15/5 = 3 units.

Since 4 of the 5 students do not share in the last sandwich, and Carol eats a portion of EVERY sandwich, all 5 units of the last sandwich must be given to Carol.
Thus, total units for Carol = 3+5 = 8 units.

Resulting fraction:
(Carol's units)/(units per sandwich) = 8/5. This is our target.
Now plug m=5 into the answer choices to see which yields our target of 8/5.

Every answer choice has the same denominator:
m(m-4) = 5(5-4) = 5.
To yield our target of 8/5, the correct answer choice must have a numerator of 8.
Only E works:
4m-12 = (4*5) - 12 = 8.

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Manager  S
Joined: 11 Apr 2018
Posts: 128
Location: India
GPA: 4
WE: Information Technology (Computer Software)
Re: Four extra-large sandwiches of exactly the same size were ordered for  [#permalink]

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I feel that this question could be best solved by substituting some value to m.
Let m = 12.

Scenario 1:
3 sandwiches --- 12 students
? --- 1 student
$$= \frac{3}{12} = \frac{1}{4}$$

Scenario 2:
1 sandwich --- 8 students
? --- 1 student
$$= \frac{1}{8}$$

$$Total : \frac{1}{4}+ \frac{1}{8} = \frac{3}{8}$$

When substituted m=12 in all of the options, only E gives the correct value.
E is the correct answer.
Intern  B
Joined: 15 Sep 2018
Posts: 30
Re: Four extra-large sandwiches of exactly the same size were ordered for  [#permalink]

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If the $$4$$ sandwiches are to be divided among $$m$$ students, this means that each of student will have $$\frac{4}{m}$$ of the entire portion.

Given that $$4$$ of the students didn’t want their share on the fourth sandwich, these were then divided among the total students (minus those $$4$$); that’s $$1$$ divided by $$(m-4)$$ students.

This means each of the students has received an additional $$\frac{1}{(m-4)}$$ of the sandwiches.

Carol ate $$\frac{3}{m}$$ and $$\frac{1}{(m-4)}$$ of the portions. That adds up to :

$$\frac{3}{m}+\frac{1}{(m-4)}$$
$$=\frac{3(m-4)}{m(m-4)} +\frac{m}{m(m-4)}$$
$$=\frac{(3m-12+m)}{(m(m-4))}$$
$$=\frac{(4m-12)}{(m(m-4))}$$

The final answer is .
Manager  B
Joined: 13 Aug 2018
Posts: 63
Re: Four extra-large sandwiches of exactly the same size were ordered for  [#permalink]

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Hi, I assumed m = 6 and what I got was 3/6 (3 sandwiches divided among 6) + 1/2 (6-4= 2 people equally sharing the one sandwich) and got 1 and selected E, which was correct.

However, I had a little problem with the language of the question when it said Carol had each of the four sandwiches. I was confused because I thought it said that Carol should have one piece of four sandwiches which would mean that she should have 1/2 (the share equally divided for all the sandwiches) * 4 which is 2.

Any help will be appreciated to how to translate this part properly  Re: Four extra-large sandwiches of exactly the same size were ordered for   [#permalink] 13 Oct 2018, 21:02

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# Four extra-large sandwiches of exactly the same size were ordered for

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