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Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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28 Nov 2019, 00:05
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Competition Mode Question Let a, b, c, d, and e represent positive integers. Is \(ab + c = cd  e\)? (1) \(\sqrt{(cde)^2}\neq cde\) (2) \(e>cd\) Are You Up For the Challenge: 700 Level Questions
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Re: Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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28 Nov 2019, 01:08
(1) \(\sqrt{(cd−e)^2} ≠ cd−e\) > cd e is negative > cd  e < 0 We know that, Modulus can never be negative, So, ab+c can never be negative > We can DEFINITELY say ab+c ≠ cd−e > Sufficient
(2) e>cd > e < cd or e > cd > e > cd ONLY (Since c, d, e are positive integers) > 0 > cd  e > cd  e < 0
Is ab+c=cd−e ? We know that, Modulus can never be negative, So, ab+c can never be negative > We can DEFINITELY say ab+c ≠ cd−e > Sufficient
IMO Option D



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Re: Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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28 Nov 2019, 02:38
ab+c=cd−e.............THAT MEANS cd−e must be >0
(1) means that cd−e<0.............so sufficient
(2) e>cd...... since they are positive we can write it as e>cd......cde<0......sufficient
OA:D
then cde<0



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Re: Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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28 Nov 2019, 03:09
#1 √≠cd−e(cd−e)2≠cd−e l(cde)l≠ cde or say lab+cl≠lcd+el sufficient #2 e>cd in that case ab+c=cd−e LHS will be ve always so ab+c=cd−e not true IMO D
Let a, b, c, d, and e represent positive integers. Is ab+c=cd−e?
(1) (cd−e)2‾‾‾‾‾‾‾‾‾√≠cd−e(cd−e)2≠cd−e
(2) e>cd



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Re: Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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28 Nov 2019, 04:35
We know that a,b,c,d, and e are positive integers. We are to determine if ab+c=cde. What is worth noting is that, ab+c is a positive number. All that we need in order to make a decision is to determine that cde is negative. Once we are able to establish that, we can conclude that ab+c≠cde.
Statement 1: √{(cde)^2}≠cde. we don't know if cd>e or e>cd, so we cannot be able to determine if cde is negative or positive. If it is negative, we know ab+c cannot equal cde, but if cde is positive, there is a possibility that ab+c is equal to cde. Statement 1 is therefore insufficient.
Statement 2: e>cd This means that ecd is positive, and by extension, cde is negative. This is sufficient since we know that definitely ab+c can never equal cde. Since we know a,b, and c are all positive numbers, so ab+c will result in a positive number and an absolute value of a positive number cannot be negative.
Statement 2 alone is sufficient.
The answer is B.



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Re: Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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28 Nov 2019, 07:24
Quote: Let a, b, c, d, and e represent positive integers. Is ab+c=cd−e?
(1) \(\sqrt{(cd−e)^2}≠cd−e\)
(2) e>cd (a,b,c,d,e)>0 ab+c≥0…cd−e≥0…cd≥e? (1) \(\sqrt{(cd−e)^2}≠cd−e\) suficif (cde)≥0, cd≥e; if (cde)<0, cd<e; cde=cde when cd≥e; cde≠cde when cd<e; (2) e>cd sufic\(e>cd…e>cd\) Ans (D)



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Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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Updated on: 29 Nov 2019, 08:38
Q. Is \(ab+c=cd−e\)? Since the output of absolute value is always zero or positive, this question actually asks whether \(cde \geq{0}\)
(1) \(\sqrt{(cd−e)^2}≠cd−e\) This statement can be rewritten as \(cd−e≠cd−e\), meaning that \(cd−e <0\) as the output of absolute value is always zero or positive. This exactly answers the question above. SUFFICIENT
(2) e>cd If \(cd\) is positive and \(e\) is positive, then \(cd−e<0\) and this answers the question above. SUFFICIENT
Answer is (D)
Originally posted by chondro48 on 28 Nov 2019, 08:08.
Last edited by chondro48 on 29 Nov 2019, 08:38, edited 1 time in total.



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Re: Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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28 Nov 2019, 13:08
For cde to be equal to an absolute value, cd must be greater than or equal to e.
Both statements states that cd is smaller than e. Hence, ans to the question is no and each statement is sufficient.
IMO, Ans D
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Re: Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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28 Nov 2019, 17:48
Let a, b, c, d, and e represent positive integers. Is ab+c=cd−e ? > cde ≥0 > \(cd ≥e ?\)
(Statement1): \(\sqrt{(cde)}^{2} ≠ cde\) >\( \sqrt{(cde)}^{2} = cd e\)
if \(\sqrt{(cde)}^{2} ≠ cde\), then >\( \sqrt{(cde)}^{2} = ( cde)= e cd\)
that means that e is greater than cd \((e > cd)\) > Always NO sufficient
(Statement2): \(e > cd\) (Squaring both sides )> \(e^{2} > cd^{2}\) \(e^{2} cd^{2} >0\) \((e cd)(e+ cd) >0\)
NO info about whether e is greater than cd. Insufficient
The answer is A.



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Re: Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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29 Nov 2019, 01:47
I feel the answer is D As e, c & d are positive integers, If e > cd It will only imply e > cd, the other solution e<cd is not possible as all of them are given as positive Bunuel chetan2uPls clarify. Posted from my mobile device



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Re: Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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29 Nov 2019, 02:19
Dillesh4096 wrote: I feel the answer is D As e, c & d are positive integers, If e > cd It will only imply e > cd, the other solution e<cd is not possible as all of them are given as positive Bunuel chetan2uPls clarify. Posted from my mobile deviceYes you are correct. Since e and both c and d are positive. e=e and cd=cd This e>cd means e>cd And this too is sufficient.
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Re: Let a, b, c, d, and e represent positive integers. Is ab + c = cd 
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29 Nov 2019, 02:19






