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

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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)]


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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)]

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Three of the sandwiches were evenly divided among the students
i.e. Amount of Sandwiches with each students = 3/m

Amount of 4th Sandwich with remaining (m-4) students = 1/(m-4)

The amount of Sandwich that carol ate = 3/m + 1/(m-4) = (3m-12+m)/[m(m-4)] = (4m-12)/[m(m-4)]

Answer: option E
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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 20 Oct 2015, 03:22
Since first three sandwiches were equally divided among m students , Carol ate = 3/m
Fourth sandwhich was divided equally among m-4 students , carol ate = 1/(m-4)
Amount of sandwhich Carol ate = 3/ m + 1/(m-4) = 4m-12/m(m-4)

Answer E
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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)]


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I feel most comfortable to plug in a number for m since the question does allow so.

We have 4 sandwiches to share among m students. m>4. Let m = 6.

3 of the 4 sandwiches are shared equally among 6 students. Therefore we have 3/4 of the 4 sandwiches to share among 6. \(\frac{3}{4}\)/6 = \(\frac{1}{8}\). So we know that one piece is 1/8 of the 4 sandwiches.

Finally, the fourth sandwich is divided among the remaining students. In our case this will be 2 students. We have \(\frac{1}{4}\)/2 = \(\frac{1}{8}\)

Now hungry Carol eats from each one piece. Hence she eats 1/4 of the 4 sandwiches which is a whole sandwich (1 = Target Value) in total.

Plug in 6 in the equations below. Answer D will provide you with a 1 as a result.
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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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Let's assume \(m=10\)

Then Carol ate

\(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{6}=\frac{28}{60}\) of the sandwhich.

The nominator needs to equal 28 when 10 is plugged in for m. Obviously, only E satisfies the condition.

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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 23 Oct 2015, 21:05
reto wrote:
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.



I feel most comfortable to plug in a number for m since the question does allow so.

We have 4 sandwiches to share among m students. m>4. Let m = 6.

3 of the 4 sandwiches are shared equally among 6 students. Therefore we have 3/4 of the 4 sandwiches to share among 6. \(\frac{3}{4}\)/6 = \(\frac{1}{8}\). So we know that one piece is 1/8 of the 4 sandwiches.

Finally, the fourth sandwich is divided among the remaining students. In our case this will be 2 students. We have \(\frac{1}{4}\)/2 = \(\frac{1}{8}\)

Now hungry Carol eats from each one piece. Hence she eats 1/4 of the 4 sandwiches which is a whole sandwich (1 = Target Value) in total.

Plug in 6 in the equations below. Answer D will provide you with a 1 as a result.

I think you are having some mistakes here.
First of all, the first 3 sandwiches are shared equally among 6 students. So each one will eat: 3/6 = 0.5 (sandwich)
Then 4 students do not eat the fourth one, so there are only two members eating including Carol so carol will eat: 1/2 = 0.5 (sandwich)
So totally, Carol will eat exactly 1 sandwich, using m = 6, the answer must be E where as your answer D will be that Carol eats 4/3 sandwich (not correct)
Using m as a variable or using a particular figure, the answer still must be E.

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

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New post 12 Dec 2015, 02:53
the moment I see the answer choices ‘all variables’ I know I can plug in numbers


Let m=6 so a pie eaten by Carol from 3 sandwiches is 3/6= ½
A pie eaten by Carol from the fourth sandwich = ½
So she had 1 whole sandwich
M(m-4)= 6*2=12 so we want 12 in the numerator only E gives the numerator of 12 as (4m-12)=(4*6-12)=12
Ans E
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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 10 Jan 2016, 12:16
[quote="NoHalfMeasures"]the moment I see the answer choices ‘all variables’ I know I can plug in numbers


Let m=6 so a pie eaten by Carol from 3 sandwiches is 3/6= ½
A pie eaten by Carol from the fourth sandwich = ½
So she had 1 whole sandwich
M(m-4)= 6*2=12 so we want 12 in the numerator only E gives the numerator of 12 as (4m-12)=(4*6-12)=12
Ans E[/quote

I have a question here... Here's my calculation. Please let me know what I'm doing wrong.

Let m=6. so each of the student gets 3/6 or 1/2 of the sandwich.
4th sandwich is shared by only 2 students equaling 1/2 sandwich each.
Carol's bite from sandwiches is (1/2)*3+1/4=2.

Where am i going wrong?

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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 11 Apr 2016, 21:08
Flexxice 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)]


I get what I have to do and more or less know how to do it, but in the solution, the OG makes a really weird calculation. Unfortunately, this weird step is needed and I hope someone can explain this step to me. :)


Carol ate [3][/m] of the first three sandwiches and [1][/m - 4] of the last sandwich. Thus, I have to add both fractions. The following is the OG's way:


[3][/m] + [1][/m - 4] = [3(m - 4) + m][/m(m - 4)] = [4m - 12][/m(m - 4)]


I have no idea what has been done in the second step. Could someone please explain this to me? The rest is clear.

Thanks a lot in advance!


it took me some time to understand the solution,but can help you little bit.

there are 4 sandwiches and no of students m , whatever the value may be.

3 sandwiches divided amount student each will get 3/m part.
now the 4th sandwich is divided amount m-4 student because 4 student dont want to eat now no of students pending is m-4

carol is among those student who was in both groups
so first group got 3/m + second grp got 1/m-4 , becauase only 1 sanwich is left to divide among m-4 students
add and calculate.

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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 27 Apr 2016, 19:33
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)]



i put m=6.
3 sandwiches were divided among 6 people. thus, each ate half.
last one - 4 refused, and 2 ate half of the fourth.
so last 2 ate in total 1 "whole" (by size) sandwich

since we have same denominator, we must find a value for numerator to be equal with it.
m(m-4)=6*2=12. so numerator must be 12.
only E works.

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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 28 Apr 2016, 05:57
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.



We are originally given that 4 sandwiches were ordered and that 3 of those sandwiches were divided by m students. Thus, each student had 3/m sandwiches.

However, since 4 students did not want any of the 4th sandwich, the 4th sandwich was actually divided amongst the remaining students, or m – 4 students. Thus, each of the remaining students received 1/(m-4) sandwiches.

We are given that Carol actually ate one piece from each of the four sandwiches. Thus, Carol ate:

3/m + 1/(m- 4)

Getting a common denominator of m(m-4) we have:

(3m-12)/[m(m-4)] + m/[m(m-4)] = (4m-12)/[m(m-4)]

Since a whole sandwich is equal to 1, Carol had (4m-12)/[m(m-4)] of a full sandwich.

Answer: E
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Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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Yep, picking numbers works nicely here. Here's a video that goes over a value substitution solution for this question:



Algebra also works well:




Hope the videos are helpful.

Last edited by GMATAcademy on 06 Jul 2016, 06:13, edited 1 time in total.

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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 06 Jul 2016, 07:02
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.


3 sandwiches divided among m students.
Hence each student gets 3/m sandwiches

For the last one, 4 students do not want to eat, hence remaining students = m - 4
Each student gets 1/ (m-4) sandwiches

Total sandwiches eaten by Carol = 3/m + 1/(m-4) = 3m - 12 + m / m(m-4) = 4m-12/m(m-4)

Correct Option: E

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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 25 Mar 2017, 05:46
took me a little longer because of tuff wording:
smart numbers technique
let's 6 be the total number of students
students ordered 4 sandwiches, 3 ones were divideded between 6 students, each got 3/6
then the left sandwich were divided between 2 (6-4) students. hence 1/2 got each student

that chick Carol ate one piece in each party: 3/6+1/2=1
question asks what was her fraction of a sandwich: 1/1=1

Let's put 6 in each answer choice and test

E is one that works

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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 27 Mar 2017, 23:29
JeffTargetTestPrep wrote:
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.



We are originally given that 4 sandwiches were ordered and that 3 of those sandwiches were divided by m students. Thus, each student had 3/m sandwiches.

However, since 4 students did not want any of the 4th sandwich, the 4th sandwich was actually divided amongst the remaining students, or m – 4 students. Thus, each of the remaining students received 1/(m-4) sandwiches.

We are given that Carol actually ate one piece from each of the four sandwiches. Thus, Carol ate:

3/m + 1/(m- 4)

Getting a common denominator of m(m-4) we have:

(3m-12)/[m(m-4)] + m/[m(m-4)] = (4m-12)/[m(m-4)]

Since a whole sandwich is equal to 1, Carol had (4m-12)/[m(m-4)] of a full sandwich.

Answer: E


can you please explain, all the students would have had 4m-12/(m(m-4)) pieces except 4 students who refused?
because if each student gets 3/m portion - it is similar to 1/m + 1/m + 1/m
am i right?

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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 28 Mar 2017, 00:02
Avinash_R1 wrote:

can you please explain, all the students would have had 4m-12/(m(m-4)) pieces except 4 students who refused?
because if each student gets 3/m portion - it is similar to 1/m + 1/m + 1/m
am i right?


Hi,

I think you haven't understood the question correctly. Since Carol ate a piece from all the 4 cakes she belongs to the category of people who did not refuse to eat the 4th cake.

In the first 3 cakes she could eat 3/m
In the 4th cake she ate 1/(m - 4)

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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 18 Apr 2017, 00:16
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)]

This is my 2 cents.
I plugged m = 12
Then, 3 breads / 12 equally gives = 1/4
As 4 didn't want anymore, 1 bread / 8 gives = 1/8
So, Carol ate (1/4) + (1/8) = 3/8.
(3/8) of bread divide by 4 breads gives = 3/24 <--this is the fraction of the bread Carol ate.

What this means is that the numerator needs to be a factor of 3.
Only E is --> (4*12)-12 = 36

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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]

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New post 16 Nov 2017, 15:48
Method - 1 (number picking).

Let us pick a number which is more than 4 and gets divided by 4. The first such number is 8, so let m = 8


so Carol will have \(\frac{1}{8}\) from 1st sandwich, \(\frac{1}{8}\)from 2nd sandwich, and \(\frac{1}{8}\) from 3rd sandwich - total = \(\frac{3}{8}\)

The 4th sandwich will be divided into 4 less part; so it will be divided into 4 parts, of which Carol will have 1/4

So Carol will have a total of \(\frac{3}{8}\) + \(\frac{1}{4}\)= \(\frac{3}{8}\)+ \(\frac{2}{8}\) = \(\frac{5}{8}\)


Now let us apply m = 8 in each of the 5 options to see, which one gives us \(\frac{5}{8}\), the answer will be option E


Method - 2 (algebra).

Carol will have \(\frac{1}{m}\) of a sandwich from 1st, 2nd, and 3rd sandwich, and \(\frac{1}{(m-4)}\) from the 4th sandwich


So Carol will have a total of 3 * \(\frac{1}{m}\) + \(\frac{1}{(m-4)}\), which will lead to option E

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Re: Four extra-large sandwiches of exactly the same size were ordered for   [#permalink] 16 Nov 2017, 15:48
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