Six countries in a certain region sent a total of 75 representatives to an international congress, and no two countries sent the same number of representatives. Of the six countries, if Country A sent the second greatest number of representatives, did Country A send at least 10 representatives?

(1) One of the six countries sent 41 representatives to the congress
(2) Country A sent fewer than 12 representatives to the congress

I selected B on this one when the answer is E. I can kind of see why I am not correct, but am not completely sure. Can someone prove this to me mathematically? Thanks!

so this problem is basically concerned with EXTREMES: you're trying to figure the least, or greatest, number of representatives (or both) that could be sent in certain situations, in order to determine the range of possibilities. remember, then, if you want to figure extreme values, you have to consider extreme situations. here's one way you can progress through the problem:

-- (1) alone --

clearly 41 representatives = greatest number sent by any one country.
therefore, the countries from second place on down sent a total of 75 - 41 = 34 representatives.

EXTREME CASE 1: smallest possible # for the second country
in this case, you want to spread the remaining 34 representatives out as evenly as possible, so that the 2nd, 3rd, 4th, 5th, and 6th place countries are as near each other as possible.
34/5 = 6.8, so try to cluster the numbers around this average: the distribution with the least possible amount of variation is 9, 8, 7, 6, 4 (you can't get consecutive integers - try it for yourself)
therefore, the second greatest number of representatives must be at least 9

EXTREME CASE 2: largest possible # for the second country
in this case, you want to make the 3rd, 4th, 5th, and 6th values as small as possible. this is straightforward: make them 4, 3, 2, and 1 respectively.
this means that the 2nd place country sent 34 - 4 - 3 - 2 - 1 = 24 representatives
therefore, the second greatest number of representatives must be 24 or less

9 < second highest number < 24

insufficient


-- (2) alone --

in this case, there are no further restrictions on the numbers of representatives.

the highest number of representatives that country a could send is clearly 11.
therefore, the second greatest number of representatives must be 11 or less

to make the number as small as possible, just let the 2nd, 3rd, 4th, 5th, 6th place numbers be 5, 4, 3, 2, 1 respectively, and give all the rest of the representatives to the first place country.
therefore, the second greatest number of representatives must be at least 5

5 < second highest number < 11

insufficient


-- together --

we have
9 < second highest number < 24
AND
5 < second highest number < 11

therefore
9 < second highest number < 11

still insufficient

answer = e

In order to solve this problem, I thought of one more way.
(1) The question asks if country A sent at least 10 representatives. If we will show that A can send 10 representatives or 9 representatives, we will rule out option 1.

I plugged in 10 as the number of representatives that country A sent, and saw that the remaining 4 countries should send 75-41-10=24 representatives. Possible.

Than I try to plug in 9 as the number of representatives that country A sent, and saw that the remaining 4 countries should send 75-41-9=25 representatives. That's also Possible.

Thus, Insufficient.

(2) Same example for (1) shows that it's insufficient.

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Moty Keret

this method works fine on this problem -- i assume you've chosen 9 and 10 because 9 is the largest value that's not "at least 10" and 10 is the smallest value that's "at least 10".

however, this won't work on every problem.
if you get a problem on which there are other restrictions besides inequalities - say, for instance, you have to have equal numbers of representatives from a few different countries, this bringing divisibility into the mix - it's possible that you might have to test other numbers, and that the "border" numbers (in this case, 9 and 10) won't be the examples you're looking for.

but, on this problem, well played.

I still have an issue with this problem. Problem says "no two countries sent the same number of representatives". This means to me that the other 4 countries sent the same amount of representatives, leaving choice 1 as sufficient.

Can someone clarify on this matter?

Highest: 41, A: 10, Rest: 8, 7, 5, 4
Highest: 41, A: 9, Rest: 8, 7, 6, 4
Which certifies both the statements. Hence E.

"No two countries sent the same number" means that, if you pick ANY two of the countries, the countries will NOT have sent the same number. In other words, every country sent its own unique number of representatives, and that number does not match the number from any of the other countries.

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

SIX Countries sent a total of 75 reps, no two countries sent the same number. Country A sent the second greatest number of representatives, did Country A send at least 10 representatives?

(1) One of the six countries sent 41 representatives to the congress


(2) Country A sent fewer than 12 representatives to the congress


1. 41 is the greatest number (as 75-41=34 < 41). Yet it does not
allow an exact minimum value of the second greatest. (Below)

2. Where does the second greatest "bottom out"? 11? 9?

Combined,

n2 + R(n) = D, where

n2 = second greatest
R(n) = an approximating function
where n is the integer value of the 3rd greatest
and the numbers are consecutively summed
from 3rd greatest to least.

D = the sum of the 2nd greatest, 3rd greatest....to the least.

Minimum value of n2
occurs at the maximum value of R(n)

At D = 34,
the greatest value for R(n) CANNOT be specified.

Where R(n) = 7654, and n2 = 12

Consider the following counter, where x is the approximate value of the second greatest and y is the approximate value of the third greatest.

Flight path from tuner
as x approaches lower values,
the next greatest, y, approaches b.

Paths must not cross.

At the tune, note differences between integers.

X Y Alternate Channel

12[22 7654] [22 8653]
11[23 8654] [23 8753]
10[24 9654][2 spaces - back-forward ok]

9 [25 8764][1 space - no back-forward]

ccamankulor, your contribution is appreciated. however, i believe that, by including such things as "approximating function", "flight path from tuner", and "r(n) values", you are unnecessarily complicating the issue.
after all, the solution to this problem doesn't require anything beyond simple arithmetic:
post9689.html#p9689

this is actually the whole point of problems like these -- they don't demand anything beyond simple addition and subtraction. that doesn't necessarily mean that the problem is easy -- you may still have to figure out a bunch of things in context, and it's not always straightforward to figure out what you have to add and subtract -- but we don't want people thinking that things are too much more complicated than they actually are.