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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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Updated on: 09 Apr 2018, 04:40
1
6
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Difficulty:

95% (hard)

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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09 Apr 2018, 00:59
1
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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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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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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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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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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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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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Updated on: 23 May 2019, 00:30
1
1
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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22 May 2019, 02:51
2
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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