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

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

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.

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.

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.

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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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

Let \(m=5\), implying that there are 5 students.
Let each sandwich = 5 pounds, implying that 3 sandwiches = 3*5 = 15 pounds.

Since 3 sandwiches are distributed among all 5 students -- including Carol -- the weight received by Carol from these 3 sandwiches \(= \frac{15}{5 }= 3\) pounds.

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

Resulting fraction:
\(\frac{Carol's-weight}{sandwich-weight} = \frac{8}{5} \)
The correct answer must yield \(\frac{8}{5}\) when \(m=5\).

Since each answer choice has the same denominator, we can ignore the denominators when evaluating thw answer choices.
To yield \(\frac{8}{5}\), the correct answer choice must have a numerator of 8.
Only E works:
\(4m-12 = (4*5) - 12 = 8\)


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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 .

Hello Experts,

EMPOWERgmatRichC, VeritasKarishma, IanStewart, Bunuel, chetan2u, GMATGuruNY, AaronPond, ccooley, RonPurewal, MentorTutoring
--The wording does not make sense to me. What does it mean by ''fourth'' sandwich? Shouldn't it be ''four'' sandwiches?
''Any of the fourth sandwich'' indicates that there are so many ''fourth'' sandwiches. But, my thinking says that there are no ''fourth'' sandwiches; there is JUST ''four'' sandwiches.
Appreciate your help...
Thanks__
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Hi Asad

Yes, the wordings are a bit confusing, but the inference is clear.
Each sandwich is divided in m students, however the 4th sandwich is divided between m-4 as 4 students do not want it.

Solution..

Best is to take a number for m, and the smart number should be a multiple of 3, as we have to divide the 3 sandwiches in equal proportion, and could be a smaller number so as to make your calculations easy.

Let m=6,
(I) So if each of these 3 sandwiches are evenly distributed amongst 6 students, then each student gets \(\frac{1}{6}\) of each sandwich.
(II) Now the fourth sandwich is to be shared between 6-4 or 2 students, and each of these 2 students will get \(\frac{1}{2}\) of 4th sandwich.

So carol gets, 1/6 of 3 sandwiches and 1/2 of the fourth one => \(3*\frac{1}{6}+\frac{1}{2}=1\)

Now substitute m as 6, and look for the choice that gives you 1 as the answer
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Asad, "any" is used for countable as well as uncountable nouns.
Do we have any apples?
I don't want any bread.

"Any of the fourth sandwich" means that they did not want even a bite of the fourth sandwich.

"Any of the four sandwiches" has a very different meaning. It means they did not want even one of the four sandwiches.
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Hello, Asad, and thank you for tagging me. It might help you to think of these sandwiches as numbers instead and apply the information from the problem directly.

1: evenly divided among m students
2: evenly divided among m students
3: evenly divided among m students
4: evenly divided among m - 4 students (since 4 students did not want any of it)

You should definitely understand how to turn a cardinal number (think counting) into an ordinal number (think order) for the sake of ease of access in math topics. The list is pretty easy after the first few (cardinal/ordinal):

1/first
2/second
3/third
4/fourth
.
.
.

From fourth on, a -th is put on the end of the number stem until you get to twenty-first, whereupon the pattern repeats for the next ten numbers (until thirty-first), and so on.

I can see why the phrasing of this question confused you! Good luck with your studies.

- Andrew
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Hi

It is given that Carol has a piece from each sandwich, so I could infer that each sandwich is separately divided amongst 6, so 1/6 for each.
Although when you calculate for 3 sandwiches, it becomes 3*1:6=1/2.
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Hi All,

We’re told that 4 extra-large sandwiches of the same size were ordered by M students, where M > 4. The first three sandwiches were EVENLY DIVIDED among the students, but since 4 of the students did not want any of the 4th sandwich, that sandwich was evenly divided among the remaining students. We’re told that Carol ate a piece from EACH of the 4 sandwiches. We’re asked what fraction of a WHOLE sandwich Carol’s portion was. While this question is wordy, we can solve it rather easily by TESTing VALUES.

IF…. M = 5, then we have 5 people sharing the first 3 sandwiches (and Carol would be the 5th student who ultimately gets the ENTIRE 4th sandwich to herself – since the other 4 students don’t want any of it).

Thus, Carol received 1/5 of each of the first 3 sandwiches and ALL of the 4th sandwich. This is the equivalent of eating (3)(1/5) + 1 = 1 3/5 sandwiches. The answers are all written as mixed fractions though, so the correct answer will equal 8/5 when M = 5.

While we would normally have to do all of the math to check all 5 answers, several of the them are written in such as way that you can eliminate them by only checking the numerator….

Answer A: The numerator here would be (5+4) = 9… which is NOT 8 (nor a multiple of 8), so this cannot be the answer.

Answer B: The numerator here would be (10 – 4) = 6… which is also NOT 8 or a multiple of 8, so this cannot be the answer.

Answer C: The numerator here is (20 – 4) = 16… which is fine. The overall fraction equals 16/5 though, which is NOT a match for what we’re looking for.

Answer D: The numerator here is (20 – 8) = 12, which does not fit what we’re looking for.

Answer E: Here, the numerator is (20 – 12) = 8, which is a match. The overall fraction is 8/5, which is a MATCH for what we are looking for – and since it’s the only one that matches, it MUST be the correct answer.

Final Answer:

GMAT Assassins aren’t born, they’re made,
Rich

The question is asking for fraction of the whole that Carol ate.

So, what I did was to divide the fraction that Carol ate by 4 sandwiches to get proper fraction.

Where did I go wrong ?

Hi BhaveshGMAT,

This question did NOT ask "what fraction of the 4 sandwiches did Carol eat?" - it asked "what fraction of a WHOLE sandwich did Carol eat?"

As an example, if Carol at 1 1/2 sandwiches, then the answer would be 1.5 = 3/2... (NOT 1.5/4 = 3/8).

GMAT assassins aren't born, they're made,
Rich

Hi AndrewN - Please explain if you can -

I understood the calculation part that how much Carol ate and that's how I came to the right answer but what I do not understand is that the question is asking for - "the amount of sandwich that she ate would be what fraction of a whole extra-large sandwich?"

Why don't we do an extra step after we know the total of her portion to find what fraction of a WHOLE extra-large sandwich she's eating?

Hello, IN2MBB2PE. The answer to your question is that the two amounts are one and the same. As long as you keep your fractions straightened out, the partial amounts of each sandwich that she did eat will sum to an amount relative to one whole extra-large sandwich, whether there were 5 people, in which case your fraction will be slightly greater than 1—8/5 or 1 and 3/5 of one sandwich, to be exact—or, say, 24 people. Consider the latter case:

Sandwich #1 (evenly divided)—Carol eats 1/24 of the sandwich

Sandwich #2 (evenly divided)—Carol eats 1/24 of the sandwich

Sandwich #3 (evenly divided)—Carol eats 1/24 of the sandwich

Sandwich #4 (evenly divided among all but 4 students)—Carol eats 1/20 of the sandwich

\(\frac{1}{24} + \frac{1}{24} + \frac{1}{24} + \frac{1}{20}\)

\(3(\frac{1}{24}) + \frac{1}{20}\)

\(\frac{3}{24} + \frac{1}{20}\)

\(\frac{1}{8} + \frac{1}{20}\)

\(\frac{5}{40} + \frac{2}{40}\)

\(\frac{7}{40}\)

Of course, this 7/40 also uses one sandwich as the benchmark, so no extra step, such as dividing the sum by 4, is needed. (We have already performed the necessary division on each sandwich. The key here is that all the sandwiches are of exactly the same size.)

Thank you for following up. Perhaps that makes more sense now. Good luck with your studies.

- Andrew
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­Pick numbers or not- just keep it organized: