## Four extra-large sandwiches

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### Four extra-large sandwiches

by ska7945 » Mon Sep 08, 2008 5:23 pm

00:00

A

B

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D

E

## Global Stats

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?

1. m+4/m(m-4)
2. 2m-4/m(m-4)
3. 4m-4/m(m-4)
4. 4m-8/m(m-4)
5. 4m-12/m(m-4)
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by just_do_it » Mon Sep 08, 2008 5:35 pm
IMO E,

each of the 3 sandwiches is divided among m students. Each one gets 1/m
the 4th sandwich is divided among (m-4) studetns . each one gets 1/(m-4)

since carol ate parts of each of the 4 sandwiches, she would have eaten

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

hope this helps.

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by vishubn » Fri Oct 24, 2008 8:35 pm
can any one elaborate on this more plzzzzzz

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by cramya » Fri Oct 24, 2008 10:31 pm
Let x be the size of each sandwich

Carol ate x/m+x/m+x/m + x/m-4 (first 3 divided equally among m students and the 4th among m-4 students so each part would be x/m in the first 3 and x/m-4 for the 4th)

=3x/m + x/m-4

Fraction Carol ate per whole large sandwich(size x)

(3x/m + x/m-4 ) / x

(3x(m-4)+ mx / m(m-4) ) *1/x

x(3m-12+m) / m(m-4) * 1/x

x cancels out

4m-12/m(m-4) is the fraction she ate of a whole extra-large sandwich

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by GMATGuruNY » Thu Sep 10, 2015 9:09 am
ska7945 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?

1. m+4/m(m-4)
2. 2m-4/m(m-4)
3. 4m-4/m(m-4)
4. 4m-8/m(m-4)
5. 4m-12/m(m-4)
I received a PM requesting that I solve this problem.

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.

Each 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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by krutik » Thu Sep 10, 2015 11:26 am
thanks a ton. You are the best

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### Re: Four extra-large sandwiches

by [email protected] » Fri Apr 23, 2021 2:54 pm
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.