sanju09 wrote:Six countries in a certain region sent 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.

OA E

What should a be a good approach to solve this kind of problems quickly and efficiently?

Here's how I tried to solve this problem: A >= 10? (Y/N)

Conditions: A is Second Greatest to send reps.
Stmt-I: (except A)- B or C or D or E = 41 reps

Disproving: A<10
Suppose B=41
=>A=9, C=8, D=7, E=6
=> Total = 71 reps.
Cannot disprove

Suppose B=41
=> A=12, C=11, D=10, E=1
=> Total = 75 reps
OR
=> A=10, C=9, D=8, E=7
=> Total = 75 reps

Suff?

Stmt-II: A has fewer than 12 reps at congress (Does fewer mean <= 12)
Disproving: A<10
A=9, B=41, C=8, D=7, E=6
Total reps = 71 reps
Cannot disprove

Suppose A = 11
=> B=41, C=9, D=8, E=6
=> Total = 75 reps.

Suff?

[D]

But OA [spoiler][E][/spoiler]

Please provide alternate ways to resolve such problems!!! Much thanks.

Six countries in a certain region sent 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.

Let the countries be A,B,C,D,E,F
St 1
B has 41 meaning the remaning have 33.

If C D E and F have 1,2,3,4, = 10 A will have 23. suff
BUT
CDEG could be 5,6,7,8 = 26 meaning A would have 7.

So a number of combinations possible

St 2
A could have 11 or any number between 1 -10. so not sufficient

Together

With B =41 and A less than 12, we still have same scenarios as before

so ans is E

The way I tackled this problem is I viewed it as an ascending (or descending) number problem, since we know that no two countries can have the same number of representatives. In addition, we are dealing with positive integers here since ethically speaking, you can't cut a representative in half. After reading statement one, we know that the remaining number of representatives must be 75-41= 34. Thus, we can view the remaining number of representatives as n-1 + n-2 + n-3 + n-4 + n-5 = 34. Or 5n = 49. 49 divided by 5 is 9 with a remainder of 4. Thus, the number of representatives will be 41, 12 (9+4-1), 7 (9-2), 6 (9-3), 5 (9-4), 4 (9-5). Thus, we can cancel A, because although 12 is the second highest number, from the sequence we derived we could also have 9 as the second highest number and spread out the remaining three among the rest of representatives. IE: 41, 9, 8,7,6. Now we can move on to statement 2, and using the sequence we derived for statement one, we can see that Country A can send 11 representatives, 9 representatives, or a number less than 9.

Hope this helps.