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If x, y, and z are nonzero integers and x > yz, which of the
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10 Apr 2020, 13:15
10 Apr 2020, 13:15
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Expert Reply
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Question Stats:
63% (01:04) correct
36% (00:57) wrong based on 87 sessions
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If x, y, and z are nonzero integers and \(x > yz\), which of the following must be true ?
I. \(\frac{x}{y }> z\)
II. \(\frac{x}{z}> y\)
III. \(\frac{x}{(yz) }> 1\)
A. None of the above
B. I only
C. III only
D. I and II only
E. I, II, III
I. \(\frac{x}{y }> z\)
II. \(\frac{x}{z}> y\)
III. \(\frac{x}{(yz) }> 1\)
A. None of the above
B. I only
C. III only
D. I and II only
E. I, II, III
Re: If x, y, and z are nonzero integers and x > yz, which of the
[#permalink]
10 Apr 2020, 13:18
10 Apr 2020, 13:18
2
Expert Reply
Given: \(x > yz\).
If \(y<0\), then when we divide both parts by \(y\) we'll get \(\frac{x}{y}<z\). Thus option I is not necessarily true;
If \(z<0\), then when we divide both parts by \(z\) we'll get \(\frac{x}{z}<y\). Thus option II is not necessarily true;
If \(yz<0\), then when we divide both parts by \(yz\) we'll get \(\frac{x}{yz}<1\). Thus option III is not necessarily true.
So, none of the options must be true.
Answer: A.
If \(y<0\), then when we divide both parts by \(y\) we'll get \(\frac{x}{y}<z\). Thus option I is not necessarily true;
If \(z<0\), then when we divide both parts by \(z\) we'll get \(\frac{x}{z}<y\). Thus option II is not necessarily true;
If \(yz<0\), then when we divide both parts by \(yz\) we'll get \(\frac{x}{yz}<1\). Thus option III is not necessarily true.
So, none of the options must be true.
Answer: A.
Re: If x, y, and z are nonzero integers and x > yz, which of the
[#permalink]
13 Sep 2020, 10:06
13 Sep 2020, 10:06
1
Carcass wrote:
Given: \(x > yz\).
If \(y<0\), then when we divide both parts by \(y\) we'll get \(\frac{x}{y}<z\). Thus option I is not necessarily true;
If \(z<0\), then when we divide both parts by \(z\) we'll get \(\frac{x}{z}<y\). Thus option II is not necessarily true;
If \(yz<0\), then when we divide both parts by \(yz\) we'll get \(\frac{x}{yz}<1\). Thus option III is not necessarily true.
So, none of the options must be true.
Answer: A.
If \(y<0\), then when we divide both parts by \(y\) we'll get \(\frac{x}{y}<z\). Thus option I is not necessarily true;
If \(z<0\), then when we divide both parts by \(z\) we'll get \(\frac{x}{z}<y\). Thus option II is not necessarily true;
If \(yz<0\), then when we divide both parts by \(yz\) we'll get \(\frac{x}{yz}<1\). Thus option III is not necessarily true.
So, none of the options must be true.
Answer: A.
Hi Carcass,
For the 3rd option,how can we take the case of yz<0 ? Its mentioned in the question that x > yz so whatever cases we assume must first meet this condition otherwise it would violate the first condition given in the question.
case 1:
all are +ve i.e,
+x,+y,+z
in which case as per condition x>yz, x/yz would be > 1
case 2:
x is less -ve
yz together more -ve than x
in which case -ve/-ve is +ve,therefore x>yz is met but could be less than 1 for eg,
x=-5,y=2,z=-3
-5/2*-3
=0.8333 which is <1
I am not sure if your explanation for 3rd option is correct
