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Difficulty:
5%
(low)
Question Stats:
80% (00:53) correct
20%
(00:49)
wrong
based on 581
sessions
History
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If x and y are negative integers, which of the following must be true?
I. \(x - y > 0\)
II. \(\frac{x}{y} > y\)
III. \(x^2 > y\)
A. I only
B. II only
C. III only
D. I and III
E. II and III
I. \(x - y > 0\)
II. \(\frac{x}{y} > y\)
III. \(x^2 > y\)
A. I only
B. II only
C. III only
D. I and III
E. II and III
ArjunJag1328
Joined: 24 Dec 2017
Last visit: 12 Jan 2025
Posts: 135
Given Kudos: 48
Location: India
Concentration: Strategy, Real Estate
Schools: Johnson '21
31 Oct 2018, 23:29
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X and Y are -ve integers
(I) x - y>0
Case 1: x=-2 and y =-3
-2+3 = 1>0 (YES)
Case 2: x=-3 and y=-2
-3+2 = -1>0 (NO)
Hence A & D can be ruled out
(II) x/y >y
Case 1: x= -2 and y=-3
-2/-3 > -3
0.66 > -3 (YES)
LHS will be always positive since numerator and denominator are -ve and RHS is always -ve
Hence it must be true
(III) x^2>y
Case 1: x = -2 and y=-3
=>(-2)^2>-3
=>4>-3
This case is always true because a -ve integer will turn +ve once squared and RHS will be always -ve
Hence it must be true
Answer IMO is E
(I) x - y>0
Case 1: x=-2 and y =-3
-2+3 = 1>0 (YES)
Case 2: x=-3 and y=-2
-3+2 = -1>0 (NO)
Hence A & D can be ruled out
(II) x/y >y
Case 1: x= -2 and y=-3
-2/-3 > -3
0.66 > -3 (YES)
LHS will be always positive since numerator and denominator are -ve and RHS is always -ve
Hence it must be true
(III) x^2>y
Case 1: x = -2 and y=-3
=>(-2)^2>-3
=>4>-3
This case is always true because a -ve integer will turn +ve once squared and RHS will be always -ve
Hence it must be true
Answer IMO is E
10 Nov 2018, 02:30
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pushpitkc
both x and y are negative.
I. x - y>0. could be when negative y is greater than negative x. Here we have to be careful about sign.
II. all time true. whenever we divide a negative number by another it yields positive one. thus the result will be greater than negative y.
III.\(x^2\)= positive. True.
