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ab = 0?

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ab = 0? [#permalink]

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New post 07 Nov 2010, 12:49
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ab = 0?

(1) a + b + ab = 0
(2) a - b = 0
[Reveal] Spoiler: OA

Kudos [?]: 29 [2], given: 0

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Joined: 02 Sep 2009
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Re: ab=0? [#permalink]

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New post 07 Nov 2010, 13:17
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Creeper300 wrote:
ab=0?

1. a+b+ab=0
2. a-b=0


(1) a+b+ab=0 --> if both unknowns equal to zero then the answer will be YES but if a=1 and b=-1/2 then the answer will be NO. Not sufficient.
(2) a=b. Clearly insufficient.

(1)+(2) a+b+ab=0 --> as a=b then a+b+ab=b+b+b^2=2b+b^2=0 --> b(2+b)=0 --> either b=a=0 and the answer will be YES or b=a=-2 and the answer will be NO. Not sufficient.

Answer: E.
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Re: ab = 0? [#permalink]

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New post 22 Jan 2015, 21:45
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Creeper300 wrote:
ab = 0?

(1) a + b + ab = 0
(2) a - b = 0


Is ab = 0?
implies Is either a or b or both = 0? i.e. is it necessary that at least one of them has to be 0?

(1) a + b + ab = 0

a(1+b) = -b
a = -b/(1+b)
If b = 0, a = 0
If b = 1, a = -1/2
So ab may or may not be 0. Not sufficient.

(2) a - b = 0
a = b
If a = 0, b = 0
If a = 5, b = 5
So ab may or may not be 0. Not sufficient.

Using both, a + a + a^2 = 0
2a + a^2 = 0
a(2+a) = 0
So either a = 0 (so b = 0) or a = -2 (so b = -2)
In first case, ab = 0. In second case ab = 4.
So ab may or may not be 0. Not sufficient.

Answer (E)
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Re: ab = 0? [#permalink]

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New post 24 Nov 2017, 15:25
Creeper300 wrote:
ab = 0?

(1) a + b + ab = 0
(2) a - b = 0



Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables and 0 equation, C is most likely to be the answer and so we need to consider both conditions 1) and 2) together.

Condition 1) & 2)

a = b
a + b + ab = a + a + a^2 = a^2 + 2a = a(a+2) = 0
a = 0 or a = -2
Since a = b, ab = 0 or ab = (-2)(-2) = 4.

Therefore, the answer is E.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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