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12 Oct 2021, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
40% (02:07) correct
60%
(01:42)
wrong
based on 113
sessions
History
Date
Time
Result
Not Attempted Yet
If a is a positive integer, is \(t^a > w^a \)?
(1) \(t > w\)
(2) \(t = 2w\)
(1) \(t > w\)
(2) \(t = 2w\)
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
Posts: 3,087
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
12 Oct 2021, 06:25
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1) t>w
Let T = -1 and W = -2
Case 1:
A = 2
1<2
T^a < W^a
Case 2:
A = 3
-1 > -8
T^a > W^a
Let T = 2 and W = 1
Case 3:
A = 2
4>1
T^a > W^a
Not sufficient
2) T = 2W
Case 1: Let T = W = 0
A = 2
T^a = W^a
Case 2:
Let T = 2 and W = 1
A = 2
4>1
T^a > W^a
Not sufficient
Combining both.
T and W are positive and T > W
Then,
T^a will always be greater than W^a
Irrespective of any value of T, W, and A
Answer is C
Posted from my mobile device
Let T = -1 and W = -2
Case 1:
A = 2
1<2
T^a < W^a
Case 2:
A = 3
-1 > -8
T^a > W^a
Let T = 2 and W = 1
Case 3:
A = 2
4>1
T^a > W^a
Not sufficient
2) T = 2W
Case 1: Let T = W = 0
A = 2
T^a = W^a
Case 2:
Let T = 2 and W = 1
A = 2
4>1
T^a > W^a
Not sufficient
Combining both.
T and W are positive and T > W
Then,
T^a will always be greater than W^a
Irrespective of any value of T, W, and A
Answer is C
Posted from my mobile device
Archit3110
Major Poster
12 Oct 2021, 08:20
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target
is \(t^a > w^a \)
where a is a +ve integer
#1
\(t > w\)
t=-1 and w = -2
if a is odd we get yes and if a even we get no
insufficient
#2
\(t = 2w\)
if w=-1 ; t=-2
if a is even we get yes if a is odd we get no
from 1 &2
possible only when both t& w are+ve
therefore for all +ve values of a target is always yes
sufficient option C
is \(t^a > w^a \)
where a is a +ve integer
#1
\(t > w\)
t=-1 and w = -2
if a is odd we get yes and if a even we get no
insufficient
#2
\(t = 2w\)
if w=-1 ; t=-2
if a is even we get yes if a is odd we get no
from 1 &2
possible only when both t& w are+ve
therefore for all +ve values of a target is always yes
sufficient option C
Bunuel
