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If x^2 + y^2 + z^2 < 4, is x^2 + z^2 < 1 ? [#permalink]
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14 Aug 2017, 01:59
Question Stats:
51% (01:21) correct 49% (01:28) wrong based on 106 sessions
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Re: If x^2 + y^2 + z^2 < 4, is x^2 + z^2 < 1 ? [#permalink]
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14 Aug 2017, 02:09
Bunuel wrote: If \(x^2 + y^2 + z^2 < 4\), is \(x^2 + z^2 < 1\) ?
(1) \(x^2 + y^2 < 1\)
(2) \(y^2 + z^2  < 1\) \(x^2 + y^2 + z^2 < 4 \iff x^2 + y^2 + z^2 < 4\) \(x^2 + z^2 < 1 \iff x^2+z^2<1\) (1) \(x^2+y^2 < 1 \implies x^2 < 1\) If \(z^2 = 3\) then \(x^2+z^2>1\). If \(z^2 = 0\) then \(x^2+z^2<1\). Insufficient. (2) \(y^2+z^2 < 1 \implies z^2 < 1\) If \(x^2 = 3\) then \(x^2+z^2>1\). If \(x^2 = 0\) then \(x^2+z^2<1\). Insufficient. Combine (1) and (2) \(x^2+y^2+y^2+z^2<2 \implies x^2+2y^2+z^2<2 \implies x^2+z^2<2\) We could have \(x^2+z^2=1\) or \(x^2+z^2<1\) or \(x^2+z^2>1\). Insufficient. The answer is E.
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If x^2 + y^2 + z^2 < 4, is x^2 + z^2 < 1 ? [#permalink]
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14 Aug 2017, 23:40
If \(x^2 + y^2 + z^2 < 4\), is \(x^2 + z^2 < 1\) ?
(1) \(x^2 + y^2 < 1\)
Let x^2 = y^2 =0 & z^2 =1/2........Answer to question is Yes
Let x^2 = 1/2 & y^2 =0 & z^2 =1/2........Answer to question is No
Insufficient
(2) \(y^2 + z^2  < 1\)
Let x^2 = y^2 =0 & z^2 =1/2........Answer to question is Yes
Let x^2 = 1/2 & y^2 =0 & z^2 =1/2........Answer to question is No
Insufficient
Combining 1 & 2
Same examples above for Statementa 1 & 2 prove Insufficient
Answer: E



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If x^2 + y^2 + z^2 < 4, is x^2 + z^2 < 1 ? [#permalink]
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15 Aug 2017, 05:19
1. x^2 +y^2 <1, this implies x^2 + y^2 lies in [0,1), and x^2 lies in [0,1) hence, z^2 < 4  (x^2 +y^2), implies z^2 lies between [0,4) also, since x^2 is always positive so x^2+z^2 lies between [0+x^2, 4+x^2) thus min is 0, and max is 5, insufficient.
2. y^2 +z^2 <1, this implies, y^2 +z^2 lies in [0,1) also, z^2 lies in [0,1) given x^2 < 4 (y^2 +z^2), implies x^2 lies in [0,4)
for the same reason as 1, insufficient.
answer E.



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Re: If x^2 + y^2 + z^2 < 4, is x^2 + z^2 < 1 ? [#permalink]
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15 Aug 2017, 07:19
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Waoooo, Some really really byzantine solutions above.
Let me try to put this in a laconic way => in such questions > ALWAYS (and I mean ALWAYS) try and use test cases or else the question can be pretty painful.
Statement 1 => Test case 1 => 0,0,0 => x^2+z^2<1 is true Test case 2 => 0.9,0,0.9 => x^2+z^2<1 is false
Statment 2 => Test case 1 => 0,0,0 => x^2+z^2<1 is true Test case 2 => 0.9,0,0.9 => x^2+z^2<1 is false
Combining them =>
Test case 1 => 0,0,0 => x^2+z^2<1 is true Test case 2 => 0.9,0,0.9 => x^2+z^2<1 is false
Hence E.
Notice how the test case is the same which makes the job 10x times easier
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Re: If x^2 + y^2 + z^2 < 4, is x^2 + z^2 < 1 ? [#permalink]
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19 Sep 2017, 00:08
St 1: x sq + y sq <1 +ve < 1 > either it adds to 0; or tend to 0.9 but less than 1
> but we don't know exact value of x or y
St 2: y sq + z sq < 1 +ve < 1 > either it adds to 0; or tend to 0.9 but less than 1 > but we don't know exact value of x or y
1) + 2) if we take max value 0.49+0.64+0.81< 4 > we still don't know the exact values of x,y,z square
so Ans E




Re: If x^2 + y^2 + z^2 < 4, is x^2 + z^2 < 1 ?
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19 Sep 2017, 00:08






