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If x, y, and z are positive numbers, is x - y < z ?

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If x, y, and z are positive numbers, is x - y < z ?  [#permalink]

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New post 30 Sep 2019, 22:14
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A
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D
E

Difficulty:

  55% (hard)

Question Stats:

56% (02:00) correct 44% (02:19) wrong based on 52 sessions

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Re: If x, y, and z are positive numbers, is x - y < z ?  [#permalink]

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New post 30 Sep 2019, 22:38
1
We are given that x,y, and z are positive numbers. We are to determine whether x-y<z

Statement 1: x+y<z.
Sufficient. This is because if z is greater than the sum of x and y, then z>x-y and z>y-x.

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 x-y(8)>z(4)
and another situation where x=3, y=1, and z=4, xy(3)<z^2 (16) and x-y(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 ?  [#permalink]

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New post 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 x-y< z .....suff

(2) xy<z^2 .... either of x,y must be less than z ...and the diff will not be > z ..from which x-y< z....suff


OA:D
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Re: If x, y, and z are positive numbers, is x - y < z ?  [#permalink]

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New post 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 x-y is less than z,
If x= 2, y=1/2 and Z = 2 (which satisfies the condition of statement 2), then x-y is less than z,
If x= 1/3, y=1/4 and Z = 1/3 (which satisfies the condition of statement 2), then x-y is less than z.

Hence, Ans. D.
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Re: If x, y, and z are positive numbers, is x - y < z ?  [#permalink]

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New post 30 Sep 2019, 23:53
1
Is \(x−y < z\) ?
or Is \(x < y+z\) ?

(1) \(x+y < z\)
--> \(x < z-y\)
If x is less than 'z-y', 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 ?  [#permalink]

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New post 01 Oct 2019, 00:43
1
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 ?  [#permalink]

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New post 01 Oct 2019, 05:15
1
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>x-y; ie. 3>2+0.5, then 3>2-0.5…; sufic.

(2) \(xy<z^2\): insufic.
if (2)(0.25)<1, then 2-0.25>1, condition is false;
if (2)(0.25)<4, then 2-0.25<2, condition is true;

Answer (A)
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Re: If x, y, and z are positive numbers, is x - y < z ?  [#permalink]

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New post 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 ?  [#permalink]

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New post 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.
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Re: If x, y, and z are positive numbers, is x - y < z ?   [#permalink] 22 Oct 2019, 17:48
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