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Is (a + b) < (c + d)? (1) c and d are negative integers such that

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Is (a + b) < (c + d)? (1) c and d are negative integers such that  [#permalink]

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New post Updated on: 09 Apr 2018, 04:40
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A
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D
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Question Stats:

37% (01:51) correct 63% (02:04) wrong based on 62 sessions

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Is (a + b) < (c + d)?


(1) c and d are negative integers such that \((a + b)^3-(c + d)^3 = 0\).

(2) a and b are positive integers such that \((a + b)^2-(c + d)^2 = 0\).


source: Time4education

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Originally posted by Nixondutta on 08 Apr 2018, 23:45.
Last edited by amanvermagmat on 09 Apr 2018, 04:40, edited 2 times in total.
Edited the question.
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Re: Is (a + b) < (c + d)? (1) c and d are negative integers such that  [#permalink]

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New post 09 Apr 2018, 00:59
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Nixondutta wrote:
Is (a + b) < (c + d)?
source: Time4education

(1) c and d are negative integers such that \((a + b)^3\) - \((c + d)^3\) = 0.
(2) a and b are positive integers such that \((a + b)^2\) - \((c + d)^2\) = 0.



From 1: a+b=c+d

sufficient

From 2: a+b=c+d

sufficient

hence D
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Re: Is (a + b) < (c + d)? (1) c and d are negative integers such that  [#permalink]

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New post 09 Apr 2018, 03:49
1
Statement (1) - (a+b)^3 = (c+d)^3
Thus, (a+b) = (c+d).

Sufficient.

Statement (2) - (a+b)^2 = (c+d)^2
+/- (a+b) = +/- (c+d)

Thus, one may be positive and other may be negative and vice versa, at the same time. Otherwise both may be positive and equal.
Thus, insufficient.

Hence option A.

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Re: Is (a + b) < (c + d)? (1) c and d are negative integers such that  [#permalink]

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New post 09 Apr 2018, 04:08
2
SonalSinha803 wrote:
Statement (1) - (a+b)^3 = (c+d)^3
Thus, (a+b) = (c+d).

Sufficient.

Statement (2) - (a+b)^2 = (c+d)^2
+/- (a+b) = +/- (c+d)

Thus, one may be positive and other may be negative and vice versa, at the same time. Otherwise both may be positive and equal.
Thus, insufficient.

Hence option A.

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From option b it is given that a+b>0. so,

either a+b>c+d or a+b=c+d

a+b<c+d is not possible. hence sufficient
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Re: Is (a + b) < (c + d)? (1) c and d are negative integers such that  [#permalink]

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New post 09 Apr 2018, 05:39
Nixondutta wrote:
Is (a + b) < (c + d)?


(1) c and d are negative integers such that \((a + b)^3-(c + d)^3 = 0\).

(2) a and b are positive integers such that \((a + b)^2-(c + d)^2 = 0\).


source: Time4education


we don't know the individual values of a,b,c,d. we are to find out the greater one between a+b and c+d

statement 1: we are given that (a+b)^3 - (c+d)^3=0

(a+b)^3=(c+d)^3
remove the exponents as both are same. we get a+c = c+d. sufficient as c+d is not greater than a+b.

statement 2 is the same as statement 1.

thus both statement are individually sufficient. Answer will be D.
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Re: Is (a + b) < (c + d)? (1) c and d are negative integers such that  [#permalink]

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New post 21 May 2019, 20:18
kunalcvrce wrote:
Nixondutta wrote:
Is (a + b) < (c + d)?
source: Time4education

(1) c and d are negative integers such that \((a + b)^3\) - \((c + d)^3\) = 0.
(2) a and b are positive integers such that \((a + b)^2\) - \((c + d)^2\) = 0.



From 1: a+b=c+d

sufficient

From 2: a+b=c+d

sufficient

hence D


Why are we not using the formula [a 3 − b 3 = (a − b) (a 2 + a b + b 2 )] here? Am I missing something? Also it does not say anything like a, b needs to be integers.
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Re: Is (a + b) < (c + d)? (1) c and d are negative integers such that  [#permalink]

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New post 21 May 2019, 21:16
Wrong OA.

In 2. a+b = +- (c+d).

Posted from my mobile device
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Is (a + b) < (c + d)? (1) c and d are negative integers such that  [#permalink]

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New post Updated on: 23 May 2019, 00:30
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Dushyant20 wrote:
Wrong OA.

In 2. a+b = +- (c+d).

Posted from my mobile device



Hi Dushyant20,

The OA is correct, both statements are sufficient.

The question is \((a + b) < (c + d)\) ?
Or in other words \((a + b) - (c + d) < 0\) ?

The answer is NO, if left side is equal to zero or greater than 0
The answer is YES, if left side is less than 0


Statement 2: \(a\) and \(b\) are positive integers such that \((a+b)^2−(c+d)^2=0\)

So \((a+b)^2−(c+d)^2=0\) and \((a+b) > 0\) while \((c+d)\) can be both positive and negative

First let \((c+d)\) be postive and , for example, equal to \(5\). For the equality in Statement 2 to be true \((a+b)\) should also be equal to \(5\)
Now answer the question whether \((a+b) - (c+d)<0\) ? Or is \(5 - 5\) less than \(0\)? The answer is NO because \(0\) is not less than \(0\).

Second let \((c+d)\) be negative and equal to \(-5\). Now \((a+b)\) is still positive according to Statement 2 and thus equal to \(5\).
Again answer the question whether \((a+b) - (c+d)<0\) ? Or is \(5 - (-5)\) less than \(0\)? The answer is again NO because 10 is not less than 0.

Thus, Statement 2 is Sufficient.
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Originally posted by ShukhratJon on 22 May 2019, 02:32.
Last edited by ShukhratJon on 23 May 2019, 00:30, edited 2 times in total.
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Re: Is (a + b) < (c + d)? (1) c and d are negative integers such that  [#permalink]

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New post 22 May 2019, 02:51
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Dushyant20 wrote:
Wrong OA.

In 2. a+b = +- (c+d).

Posted from my mobile device


However, pay attention to an important fact that the question itself is wrong because Statement 1 and Statement 2 contradict each other.

Statement 1: \(c\) and \(d\) are negative integers such that \((a+b)^3−(c+d)^3=0\)

If \(c\) and \(d\) are negative integers, then \((c+d)\) is also negative. In this case \((a+b)\) also has to be negative so that \((a+b)^3−(c+d)^3\) be equal to \(0\). For example, \(-5 - (-5) = 0\)

But Statement 2 says that \(a\) and \(b\) are positive integers. Hence \((a+b)\) also is positive. How can \((a+b)\) be simultaneously positive and negative? That's how Statement 1 and Statement 2 contradict each other.

The question is wrong.
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Re: Is (a + b) < (c + d)? (1) c and d are negative integers such that   [#permalink] 22 May 2019, 02:51
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