The Kingdom X

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Thu May 20, 2010 4:11 am

Six kingdoms in a certain county send a total of 75 legislature to the global senate and no two kingdoms send the same number of legislature. Of the 6 kingdoms, if Kingdom X sent the second highest number of legislature, did it send atleast 10 legislature?

(1) One kingdom sent 41 legislature.

(2) The Kingdom X sent less than 12 legislature.

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Source: Beat The GMAT — Data Sufficiency | Thu May 20, 2010 4:11 am

liferocks: Legendary Member; Posts: 576; Joined: Sat Mar 13, 2010 8:31 pmFollowed by: 1 member; Thanked: 97 times

by liferocks » Thu May 20, 2010 4:40 am

From 1
one kingdom has sent 41..so other 5 together has sent 75-41=34

Now second highest will be minimum when the range of the number of legislature will be minimum for these 5 kingdoms.

34/5..nearest integer is 6..so minimum range will be for 4,6,7,8,9..when all values are different..so minimum number of legislature sent by X is 9
maximum number of legislature sent by X is {34-(1+2+3+4)}=24...cannot conclude anything

From 2
The Kingdom X sent less than 12 legislature...as shown for condition 1...range of legislature from X will be 5 to 12 ..cannot conclude any thing

combining
9<=number of legislature sent by X<12 ...so cannot conclude that X has definitely sent at least 10 legislature...insufficient

Ans option E

Last edited by liferocks on Thu May 20, 2010 5:47 pm, edited 2 times in total.

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sk818020: Master | Next Rank: 500 Posts; Posts: 214; Joined: Mon Mar 29, 2010 1:46 pm; Location: Houston, TX; Thanked: 37 times; GMAT Score:700

by sk818020 » Thu May 20, 2010 9:22 am

(1) tells that one kingdome sent 41. This also tells us that the country that sent the most sent 41 reps because,

75-41=34

We also know that country X was the second highest, so we can determine a range for the number of reps X sent (x).

To maximize x we can assume every the other four countries sent the miminum possible,

1+2+3+4=10 rep. So,

x=34-10=24; 24 is x's maximum value.

To minimize x; we know the other four countries must be less than x, so;

34/5=6+4/5

When we distribute the remainder we find the minimum x can be is 9; because

6 6 6 6 6
+ + + + -
3 2 1 0 2 (3+2+1-2=4=the remainder)
= = = = =
9 8 7 6 4 = 34

So;

9<x<24

Because x can be less than or greater than 10, (1) is not sufficient.

(2) Simply tells us that x<12. This leaves the possibility that x is 10 or 11, so (1) is insufficient.

If you combine (1) and (2) you get;

9<x<12

Again, this includes the possibility that x is less than or greater than ten. So,

IMO E is the answer.

What is the OA?

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sk818020: Master | Next Rank: 500 Posts; Posts: 214; Joined: Mon Mar 29, 2010 1:46 pm; Location: Houston, TX; Thanked: 37 times; GMAT Score:700

by sk818020 » Thu May 20, 2010 9:28 am

liferocks wrote: so minimum number of legislature sent by X is 9

This tells you that x<9. But in your conclusion you say;

9<number of legislature sent by X

I beleive you forgot that 9 could equal x, thus, this quote should read,

9<number of legislature sent by x;

This would lead you to the same conclusion I got, E.

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liferocks: Legendary Member; Posts: 576; Joined: Sat Mar 13, 2010 8:31 pmFollowed by: 1 member; Thanked: 97 times

by liferocks » Thu May 20, 2010 5:46 pm

sk818020 wrote:
liferocks wrote: so minimum number of legislature sent by X is 9
This tells you that x<9. But in your conclusion you say;

9<number of legislature sent by X
I beleive you forgot that 9 could equal x, thus, this quote should read,

9<number of legislature sent by x;

This would lead you to the same conclusion I got, E.

agreed...edited first post.Thnx!

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