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If y=−m^2, which of the following must be true? [#permalink]
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If y=−m^2, which of the following must be true? I. y is negative. II. m is non-negative. III. If m is negative then y is negative. A. I only B. II only C. III only D. I and II only E. II and III only
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Re: If y=−m^2, which of the following must be true? [#permalink]
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Joy111 wrote: If y=−m^2, which of the following must be true?
I. y is negative.
II. m is non-negative.
III. If m is negative then y is negative.
A. I only
B. II only
C. III only
D. I and II only
E. II and III only Tricky question. +1. If y=−m^2, which of the following must be true?First of all notice m^2 is always non-negative, so -m^2 is non-positive (zero or negative), which means that y is zero when m=0 and y is negative for ANY other value of m. I. y is negative --> not necessarily true, if m=0 then y=-m^2=0; II. m is non-negative. m can take ANY value: positive, negative, zero. We don't have any restrictions on its value; III. If m is negative then y is negative. m is negative means that m is not zero. As discussed above if m is other than zero (positive or negative) then y is negative: y=-(negative^2)=-positive=negative (y=-(positive^2)=-positive=negative). So, this option is always true. Answer: C (III only). Hope it's clear.
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Re: If y=−m^2, which of the following must be true? [#permalink]
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01 Jun 2012, 03:14
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Re: If y=−m^2, which of the following must be true? [#permalink]
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Re: If y=−m^2, which of the following must be true? [#permalink]
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01 Jun 2012, 03:28
Joy111 wrote: If y=−m^2, which of the following must be true?
I. y is negative.
II. m is non-negative.
III. If m is negative then y is negative.
A. I only
B. II only
C. III only
D. I and II only
E. II and III only Condition I If m=0 then y=0, hence not always true COndition II m can be anything, +ve, 0 or -ve. m^2 will be either 0 (for m=0) or +ve Condition III if m is -ve (or to say m is less than 0) than m^2 is +ve. Hence -m^2 is always -ve. if m is -ve than y IS -ve Always True
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Re: If y=−m^2, which of the following must be true? [#permalink]
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01 Jun 2012, 04:03
Bunuel wrote: Joy111 wrote: If y=−m^2, which of the following must be true?
I. y is negative.
II. m is non-negative.
III. If m is negative then y is negative.
A. I only
B. II only
C. III only
D. I and II only
E. II and III only Tricky question. +1. If y=−m^2, which of the following must be true?First of all notice m^2 is always non-negative, so -m^2 is non-positive (zero or negative), which means that y is zero when m=0 and y is negative for ANY other value of m. I. y is negative --> not necessarily true, if m=0 then y=-m^2=0; II. m is non-negative. m can take ANY value: positive, negative, zero. We don't have any restrictions on its value; III. If m is negative then y is negative. m is negative means that m is not zero. As discussed above if m is other than zero (positive or negative) then y is negative: y=-(negative^2)=-positive=negative (y=-(positive^2)=-positive=negative). So, this option is always true. Answer: C (III only). Hope it's clear. If m is positive then too y is negative so statement iii is contradicted ?
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Re: If y=−m^2, which of the following must be true? [#permalink]
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01 Jun 2012, 04:09
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Re: If y=−m^2, which of the following must be true? [#permalink]
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01 Jun 2012, 04:42
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Joy111 wrote: If m is positive then too y is negative so statement iii is contradicted ? In response to the doubt (also on PM) Joy the question is asking which must always be true I will go thru the conditions 1 and 2 once again to help you understand the condition 3 Condition 1 says "y is -ve" That means it is saying that "y=-m^2" is always -ve But if m=0 then y becomes 0. So condition 1 is not true We rule out choices A and DCondition 2 says "m is non-negative" That means it is saying that in "y=-m^2" m is either 0 or +ve But it is not true, m can be -ve also From the question stem, there is no restriction on value of m, and hence m can be any number We rule out choices B and Esolved
Answer is CPursuing further Codition 3: "If m is negative then y is negative." This condition limits the values of m It is saying that IF m<0 then y is always -ve The only time when y is not -ve is when m=0 as the condition says that m is less than 0 (that is to say that m is not 0) Hence when m is -ve (less than 0) y is -veAlways True
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Re: If y=−m^2, which of the following must be true? [#permalink]
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01 Jun 2012, 14:33
manulath wrote: Joy111 wrote: If m is positive then too y is negative so statement iii is contradicted ? In response to the doubt (also on PM) Joy the question is asking which must always be true I will go thru the conditions 1 and 2 once again to help you understand the condition 3 Condition 1 says "y is -ve" That means it is saying that "y=-m^2" is always -ve But if m=0 then y becomes 0. So condition 1 is not true We rule out choices A and DCondition 2 says "m is non-negative" That means it is saying that in "y=-m^2" m is either 0 or +ve But it is not true, m can be -ve also From the question stem, there is no restriction on value of m, and hence m can be any number We rule out choices B and Esolved
Answer is CPursuing further Codition 3: "If m is negative then y is negative." This condition limits the values of m It is saying that IF m<0 then y is always -ve The only time when y is not -ve is when m=0 as the condition says that m is less than 0 (that is to say that m is not 0) Hence when m is -ve (less than 0) y is -veAlways TrueI think both of you are correct, I was thinking the question to be " Which of the following is ALWAYS true " Please correct me if I am wrong . I) which of the following MUST be true II) which of the following is ALWAYS true , I think for both of these the answer to this question will be different ?
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Re: If y=−m^2, which of the following must be true? [#permalink]
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02 Jun 2012, 07:26
Joy111 wrote: manulath wrote: Joy111 wrote: If m is positive then too y is negative so statement iii is contradicted ? In response to the doubt (also on PM) Joy the question is asking which must always be true I will go thru the conditions 1 and 2 once again to help you understand the condition 3 Condition 1 says "y is -ve" That means it is saying that "y=-m^2" is always -ve But if m=0 then y becomes 0. So condition 1 is not true We rule out choices A and DCondition 2 says "m is non-negative" That means it is saying that in "y=-m^2" m is either 0 or +ve But it is not true, m can be -ve also From the question stem, there is no restriction on value of m, and hence m can be any number We rule out choices B and Esolved
Answer is CPursuing further Codition 3: "If m is negative then y is negative." This condition limits the values of m It is saying that IF m<0 then y is always -ve The only time when y is not -ve is when m=0 as the condition says that m is less than 0 (that is to say that m is not 0) Hence when m is -ve (less than 0) y is -veAlways TrueI think both of you are correct, I was thinking the question to be " Which of the following is ALWAYS true " Please correct me if I am wrong . I) which of the following MUST be true II) which of the following is ALWAYS true , I think for both of these the answer to this question will be different ? No, they are the same. Option III must be true because it's always true.
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Re: If y=−m^2, which of the following must be true? [#permalink]
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02 Jun 2012, 07:55
hm, I ignored m=0, since 0 cant have a sign. is it ok to write y=-0!?
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Re: If y=−m^2, which of the following must be true? [#permalink]
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Re: If y=−m^2, which of the following must be true? [#permalink]
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02 Jun 2012, 08:13
Bunuel wrote: LalaB wrote: hm,
I ignored m=0, since 0 cant have a sign. is it ok to write y=-0!? y=-0 simply means that y=0, so there is nothing wrong with that. aha, I found in wiki - http://en.wikipedia.org/wiki/Sign_%28mathematics%29The number zero is neither positive nor negative, and therefore has no sign. In arithmetic, +0 and −0 both denote the same number 0, and the negation of zero is zero itself.I thought it is not allowed to write -0 ,since 0 is neither negative, nor positive. but wiki says -0 is ok and it means just 0. ok, Bunuel, now I agree with u 
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Re: If y=−m^2, which of the following must be true? [#permalink]
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Re: If y=−m^2, which of the following must be true? [#permalink]
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19 Jun 2013, 08:00
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Joy111 wrote: If y=−m^2, which of the following must be true?
I. y is negative.
II. m is non-negative.
III. If m is negative then y is negative.
A. I only
B. II only
C. III only
D. I and II only
E. II and III only Given that \(y = -m^2\) If m=0, then y =0 --> This eliminates option I. Again, m can be a non-negative value as there is no restriction on that, thus statement II gets eliminated. Multiply both sides by m assuming \(m\neq{0}\). Thus, \(my = -m^3\). Now, if m<0,then \(m^3<0\)and \(-m^3>0\). Hence, as my>0, m and y must have the same sign. C.
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Re: If y=−m^2, which of the following must be true? [#permalink]
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30 May 2014, 13:26
The biggest issue here is obviously noticing that m could be 0, thus making y=0 as well. The first option is a tricky one, but normally the other options will at least offer a hint and/or insight into the trick. In this case, option II mentions "non-negative," which suggests that the reader considers that m could be 0. Summary: Read all of the option choices! They're trying to give you hints
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If y=−m2, which of the following must be true? I) y is negative. [#permalink]
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14 Jun 2015, 12:07
If y=−m^2, which of the following must be true? I) y is negative. II) m is non-negative. III) If m is negative then y is negative. I only II only III only I and II only II and III only Please help with this question. Thanks 
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Re: If y=−m^2, which of the following must be true? [#permalink]
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Re: If y=−m^2, which of the following must be true? [#permalink]
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25 Nov 2016, 09:34
yep, that m=0 does the trick. tricky one!
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