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.
NO. PROPERTIES
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 87
- Joined: Sun Jul 24, 2011 10:36 am
- cans
- Legendary Member
- Posts: 1309
- Joined: Mon Apr 04, 2011 5:34 am
- Location: India
- Thanked: 310 times
- Followed by:123 members
- GMAT Score:750
B) x > 3^20 or x>9^10. Insufficient. x can be 11^10 or (9.5)^10.Is x < 10^10
1) x = 3^22
2) x > 3^20
A) x= 9^11. Sufficient. (its either less or not less but we have only one answer)
IMO A
If my post helped you- let me know by pushing the thanks button
Contact me about long distance tutoring!
[email protected]
Cans!!
Contact me about long distance tutoring!
[email protected]
Cans!!
- cans
- Legendary Member
- Posts: 1309
- Joined: Mon Apr 04, 2011 5:34 am
- Location: India
- Thanked: 310 times
- Followed by:123 members
- GMAT Score:750
median = (l+m)/2 and mean = (j+k+l+m+n+o)/6Q 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.
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
If my post helped you- let me know by pushing the thanks button
Contact me about long distance tutoring!
[email protected]
Cans!!
Contact me about long distance tutoring!
[email protected]
Cans!!
-
- Newbie | Next Rank: 10 Posts
- Posts: 5
- Joined: Tue Jul 26, 2011 9:22 am
- Followed by:1 members
[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.