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# Let a, b, c, d, and e represent positive integers. Is |ab + c| = cd -

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Joined: 02 Sep 2009
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Let a, b, c, d, and e represent positive integers. Is |ab + c| = cd -  [#permalink]

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28 Nov 2019, 00:05
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65% (hard)

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47% (01:35) correct 53% (01:52) wrong based on 62 sessions

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Let a, b, c, d, and e represent positive integers. Is $$|ab + c| = cd - e$$?

(1) $$\sqrt{(cd-e)^2}\neq cd-e$$

(2) $$|e|>|cd|$$

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Re: Let a, b, c, d, and e represent positive integers. Is |ab + c| = cd -  [#permalink]

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28 Nov 2019, 01:08
1
(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 -  [#permalink]

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28 Nov 2019, 02:38
1
|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......cd-e<0......sufficient

OA:D

then |cd-e|<0
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Re: Let a, b, c, d, and e represent positive integers. Is |ab + c| = cd -  [#permalink]

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28 Nov 2019, 03:09
1
#1
√≠cd−e(cd−e)2≠cd−e
l(cd-e)l≠ cd-e
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 -  [#permalink]

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28 Nov 2019, 07:24
1
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) $$\sqrt{(cd−e)^2}≠cd−e$$ sufic

if (cd-e)≥0, cd≥e;
if (cd-e)<0, cd<e;

|cd-e|=cd-e when cd≥e;
|cd-e|≠cd-e 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 -  [#permalink]

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Updated on: 29 Nov 2019, 08:38
1
Q. Is $$|ab+c|=cd−e$$?
Since the output of absolute value is always zero or positive, this question actually asks whether $$cd-e \geq{0}$$

(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

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 -  [#permalink]

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28 Nov 2019, 13:08
1
For cd-e 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 -  [#permalink]

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28 Nov 2019, 17:48
1
Let a, b, c, d, and e represent positive integers. Is |ab+c|=cd−e ?
--> cd-e ≥0 --> $$cd ≥e ?$$

(Statement1): $$\sqrt{(cd-e)}^{2} ≠ cd-e$$
-->$$\sqrt{(cd-e)}^{2} = |cd -e|$$

if $$\sqrt{(cd-e)}^{2} ≠ cd-e$$, then
-->$$\sqrt{(cd-e)}^{2} = -( cd-e)= e- cd$$

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

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Re: Let a, b, c, d, and e represent positive integers. Is |ab + c| = cd -  [#permalink]

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29 Nov 2019, 01:47
1
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 chetan2u
Pls clarify.

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Re: Let a, b, c, d, and e represent positive integers. Is |ab + c| = cd -  [#permalink]

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29 Nov 2019, 02:19
1
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 chetan2u
Pls clarify.

Posted from my mobile device

Yes 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 -  [#permalink]

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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|=cd-e.
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 cd-e is negative. Once we are able to establish that, we can conclude that |ab+c|≠cd-e.

Statement 1: √{(cd-e)^2}≠cd-e.
we don't know if cd>e or e>cd, so we cannot be able to determine if cd-e is negative or positive. If it is negative, we know |ab+c| cannot equal cd-e, but if cd-e is positive, there is a possibility that |ab+c| is equal to cd-e. Statement 1 is therefore insufficient.

Statement 2: |e|>|cd|
This means that e-cd is positive, and by extension, cd-e is negative. This is sufficient since we know that definitely |ab+c| can never equal cd-e. 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.

Re: Let a, b, c, d, and e represent positive integers. Is |ab + c| = cd -   [#permalink] 28 Nov 2019, 04:35
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