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If x, y, and z are positive numbers, is x  y < z ?
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30 Sep 2019, 22:14
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56% (02:00) correct 44% (02:19) wrong based on 52 sessions
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Competition Mode Question If x, y, and z are positive numbers, is \(x  y < z\)? (1) \(x + y < z\) (2) \(xy < z^2\)
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Re: If x, y, and z are positive numbers, is x  y < z ?
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30 Sep 2019, 22:38
We are given that x,y, and z are positive numbers. We are to determine whether xy<z
Statement 1: x+y<z. Sufficient. This is because if z is greater than the sum of x and y, then z>xy and z>yx.
Statement 2: xy<z^2. Not sufficient. This is because we can have a situation where x=9, y=1, and z=4. xy=9<z^2=16, and xy(8)>z(4) and another situation where x=3, y=1, and z=4, xy(3)<z^2 (16) and xy(2)<z(4).
Hence the answer is A.



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Re: If x, y, and z are positive numbers, is x  y < z ?
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30 Sep 2019, 23:21
If x, y, and z are positive numbers, is x−y<z?
(1) x+y<z .......this means both x and y are < z .....from which we can conclude that xy< z .....suff
(2) xy<z^2 .... either of x,y must be less than z ...and the diff will not be > z ..from which xy< z....suff
OA:D



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Re: If x, y, and z are positive numbers, is x  y < z ?
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30 Sep 2019, 23:45
If x, y, and z are positive numbers, is x−y<z?
(1) x+y<z Sufficient. If sum of two positive numbers (integral or fraction) is less than third positive number, then difference of the same two positive numbers will be less than third one.
(2) xy<z2 sufficient. If x= 1/2, y=1/4 and Z = 1 (which satisfies the condition of statement 2), then xy is less than z, If x= 2, y=1/2 and Z = 2 (which satisfies the condition of statement 2), then xy is less than z, If x= 1/3, y=1/4 and Z = 1/3 (which satisfies the condition of statement 2), then xy is less than z.
Hence, Ans. D.



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Re: If x, y, and z are positive numbers, is x  y < z ?
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30 Sep 2019, 23:53
Is \(x−y < z\) ? or Is \(x < y+z\) ?
(1) \(x+y < z\) > \(x < zy\) If x is less than 'zy', X will DEFINITELY be less than 'z+y' > Sufficient
(2) \(xy<z^2\) No relation between x & 'y+z' can be established E.g: If x = 1000, y = 1 & z = 32 > xy = 1000 is less than \(32^2 = 1024\) > Is x < y + z ?  NO If x = 0.5, y = 0.5 & z = 1 > xy = 0.25 is less than \(1^2 = 1\) > Is x < y + z ?  YES > Insufficient
IMO Option A



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Re: If x, y, and z are positive numbers, is x  y < z ?
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01 Oct 2019, 00:43
x, y and z are positive numbers Is x—y< z ???
Statement1: x+y< z We can add 2y to both sides of inequality—> x+y —2y< z—2y x—y< z—2y y is positive number —> z—2y’ is reduced, but still greater than ‘x—y’ —> x—y< z is correct Sufficient
Statement2: xy< z^{2} If x=1, y=1 and z=2 —> 1*1<4 1–1<2( Yes) If x=1,y=1/9 and z=1/2—> 1*1/9<1/4 1–1/9< 1/2 (No) Insufficient
The answer is A.
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Re: If x, y, and z are positive numbers, is x  y < z ?
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01 Oct 2019, 05:15
Quote: If x, y, and z are positive numbers, is x−y<z?
(1) \(x+y<z\) (2) \(xy<z^2\) \(x,y,z>0\) (1) \(x+y<z\): if z>x+y, then z>xy; ie. 3>2+0.5, then 3>20.5…; sufic. (2) \(xy<z^2\): insufic. if (2)(0.25)<1, then 20.25>1, condition is false; if (2)(0.25)<4, then 20.25<2, condition is true; Answer (A)



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Re: If x, y, and z are positive numbers, is x  y < z ?
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01 Oct 2019, 08:40
#1 (1) x+y<z sufficient to say that x−y<z #2 xy<z2 x=2,y=5 x*y=10 and z=4 ; 16 so x−y<z ; NO and yes when x=7,y=2 x*y=14 and z=4 ; 16 insufficient IMO A
If x, y, and z are positive numbers, is x−y<z?
(1) x+y<z
(2) xy<z2



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Re: If x, y, and z are positive numbers, is x  y < z ?
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22 Oct 2019, 17:48
Is there an algebraic way to show Statement 2 as insufficient? All the solutions I have come across so far (here and at other places including The Economist tutor that has this question on a sim test) involve plugging in values to demonstrate undesired results. Wondering whether a more intuitive solution is available.




Re: If x, y, and z are positive numbers, is x  y < z ?
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22 Oct 2019, 17:48






