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

Sent from my Lenovo K53a48 using GMAT Club Forum mobile app


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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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]
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]
Wrong OA.

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

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

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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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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