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15 Jul 2019, 00:20
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Difficulty:
95%
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Question Stats:
36% (02:47) correct
64%
(03:02)
wrong
based on 58
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The operator a# is defined on a number a as equal to (–1)^m*a, where m is an integer. If a is not equal to zero, what is (a#)# – a ?
(1) a# + a = 0
(2) (a#)*a + a^2 = 0
(1) a# + a = 0
(2) (a#)*a + a^2 = 0
15 Jul 2019, 10:30
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Bunuel
The operator a# is defined on a number a as equal to (–1)^m*a, where m is an integer. If a is not equal to zero, what is (a#)# – a ?
(1) a# + a = 0
(2) (a#)*a + a^2 = 0
(1) a# + a = 0
(2) (a#)*a + a^2 = 0
giving a try though not sure
#1
a=1 and m=1
so a# + a = 0 ;
a#'(–1)^m*a ; -1 and a=1 so -1+1 ; 0
sufficient
(a#)# – a ; -1-1 ; -2
#2
(a#)*a + a^2 = 0
a=1 so a# ; -1 and a^2 = 1 ; -1+1 ; 0
sufficient
IMO D
15 Jul 2019, 13:56
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Archit3110
Bunuel
The operator a# is defined on a number a as equal to (–1)^m*a, where m is an integer. If a is not equal to zero, what is (a#)# – a ?
(1) a# + a = 0
(2) (a#)*a + a^2 = 0
(1) a# + a = 0
(2) (a#)*a + a^2 = 0
giving a try though not sure
#1
a=1 and m=1
so a# + a = 0 ;
a#'(–1)^m*a ; -1 and a=1 so -1+1 ; 0
sufficient
(a#)# – a ; -1-1 ; -2
#2
(a#)*a + a^2 = 0
a=1 so a# ; -1 and a^2 = 1 ; -1+1 ; 0
sufficient
IMO D
Why didn't you checked with a=-1, because that option will give you different results
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