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[email protected]: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Sun Jul 24, 2011 10:36 am

by [email protected] » Sat Sep 17, 2011 12:38 pm

Is x < 10^10

1) x = 3^22
2) x > 3^20

WHY ANS D IS WRONG

Q 18 In a set of positive integers {j, k, l, m, n, o}, in which j < k < l < m < n < o, is the mean larger than the median?

1) The sum of n and o is more than twice the sum of j and k.

2) The sum of k and o is the sum of l and m.

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Source: Beat The GMAT — Data Sufficiency | Sat Sep 17, 2011 12:38 pm

cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

by cans » Sat Sep 17, 2011 9:10 pm

Is x < 10^10

1) x = 3^22
2) x > 3^20

B) x > 3^20 or x>9^10. Insufficient. x can be 11^10 or (9.5)^10.
A) x= 9^11. Sufficient. (its either less or not less but we have only one answer)
IMO A

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cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

by cans » Sat Sep 17, 2011 9:28 pm

Q 18 In a set of positive integers {j, k, l, m, n, o}, in which j < k < l < m < n < o, is the mean larger than the median?

1) The sum of n and o is more than twice the sum of j and k.

2) The sum of k and o is the sum of l and m.

median = (l+m)/2 and mean = (j+k+l+m+n+o)/6
mean>median???

B) k+o = l+m. also j<k and n<o, then j+n < l+m
thus mean < 3(l+m)/6 < median.
Sufficient

A) n+o > 2(j+k)
thus (j+k+n+o+l+m ) /6 > (3(j+k) + l+ m ) /6
mean > (3(j+k) + l+ m ) /6
(3(j+k) + l+ m ) /6 > 2/3 (j+K)
mean > 2(j+K)/3 and median = (l+M)/2 > (j+k)/2
insufficient
IMO B

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simonsays452: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Tue Jul 26, 2011 9:22 amFollowed by: 1 member

by simonsays452 » Sun Sep 18, 2011 6:03 am

[email protected] wrote: Is x < 10^10

1) x = 3^22
2) x > 3^20

WHY ANS D IS WRONG

IMO the way to work this problem is to recognize that the stimulus asks if X is larger than a known single number (10^10).
Statement 1 says x equals one identified single number. One number, regardless of what it equals will ALWAYS be larger or smaller than another known number. Therefore 1 is SUFF.

For statement 2: x>3^20=9^10; 9^10<10^10 but because X is larger than 9^10 it could also be 11^10 which is larger than 10^10. Because we get yes and no from statement 2: insufficient.

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