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Re M0319
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15 Sep 2014, 23:20



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Bunuel wrote: Official Solution:
In order \(abc = 0\) to be true at least one of the unknowns must be zero. (1) \(a^2 = 2a\). Rearrange: \(a^22a=0\). Factor out \(a\): \(a(a2)=0\). Either \(a=0\) or \(a=2\). If \(a=0\) then the answer is YES but if \(a=2\) then \(abc\) may not be equal to zero (for example consider: \(a=2\), \(b=3\) and \(c=4\)). Not sufficient. (2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\) \(b= \frac{c*(a+b)^2}{(a+b)^2}  c\); \(b=cc\); \(b=0\). Sufficient.
Answer: B I DISAGREE WITH B........2 actually reduces to b*(a+b)^2=0 which gives us b=0 or a=b...(we cannot simply strike of (a+b)^2 in numerator and denominator since we do not know that a+b is not equal to 0.). so it should be E.



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27 May 2015, 06:33
suhasreddy wrote: Bunuel wrote: Official Solution:
In order \(abc = 0\) to be true at least one of the unknowns must be zero. (1) \(a^2 = 2a\). Rearrange: \(a^22a=0\). Factor out \(a\): \(a(a2)=0\). Either \(a=0\) or \(a=2\). If \(a=0\) then the answer is YES but if \(a=2\) then \(abc\) may not be equal to zero (for example consider: \(a=2\), \(b=3\) and \(c=4\)). Not sufficient. (2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\) \(b= \frac{c*(a+b)^2}{(a+b)^2}  c\); \(b=cc\); \(b=0\). Sufficient.
Answer: B I DISAGREE WITH B........2 actually reduces to b*(a+b)^2=0 which gives us b=0 or a=b...(we cannot simply strike of (a+b)^2 in numerator and denominator since we do not know that a+b is not equal to 0.). so it should be E. That's not true. If a+b were 0, then \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\) would not be correct for any real a, b, and c. When having something like x^2/x=y you can ALWAYS reduce by x and write x = y. The answer is B.
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Bunuel wrote: Official Solution: (2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\) \(b= \frac{c*(a+b)^2}{(a+b)^2}  c\); \(b=cc\); \(b=0\). Sufficient.
Answer: B Can someone explain the second step in this simplification process. It looks like factoring out the \sqrt{c}, but somehow \sqrt{c} became c. Is there some rule that allows this?



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Re: M0319
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27 May 2015, 16:24
The square root of c needs to be squared too



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I guess what I'm saying is that I've never seen this operation before. I believe it, because I basically get the same answer the long way:
(a\(\sqrt{c}\) + b\(\sqrt{c}\))^2 = (a\(\sqrt{c}\) + b\(\sqrt{c}\))(a\(\sqrt{c}\) + b\(\sqrt{c}\)) ...and then the foil method
If it's a valid shortcut then I wonder if there is a proof or something for it so I know when I can use it in other situations.



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M0319
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Updated on: 10 Jun 2015, 18:02
meshackb wrote: I guess what I'm saying is that I've never seen this operation before. I believe it, because I basically get the same answer the long way:
(a\(\sqrt{c}\) + b\(\sqrt{c}\))^2 = (a\(\sqrt{c}\) + b\(\sqrt{c}\))(a\(\sqrt{c}\) + b\(\sqrt{c}\)) ...and then the foil method
If it's a valid shortcut then I wonder if there is a proof or something for it so I know when I can use it in other situations. This is like the example below: (2*5+3*5)^2 = 5^2 * (2+3) = 25*5 = 125 The way you are mentioning the equation is also correct but instead of the traditional foil method , try to see it this way: (a\(\sqrt{c}\) + b\(\sqrt{c}\))^2 = (a\(\sqrt{c}\) + b\(\sqrt{c}\))(a\(\sqrt{c}\) + b\(\sqrt{c}\)) Take sqroot(c) common from both the terms on RHS of the equation above. sqroot (c)*sqroot (c) = c as sqroot (any number)= number ^0.5 (and x^p * x^q = x^ (p+q)). So after doing this you get, sqroot (c) * sqroot (c) * (a+b)^2 = c * (a+b)^2. This is what Bunuel has done.
Originally posted by ENGRTOMBA2018 on 27 May 2015, 17:18.
Last edited by ENGRTOMBA2018 on 10 Jun 2015, 18:02, edited 1 time in total.



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Re: M0319
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27 May 2015, 20:11
Thanks, that clears it up!



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Re: M0319
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31 May 2015, 00:12
B (1) Gives a= 2,0 Insuff (2) Gives b = 0 or a=b but a cannot be = b as then equation's value will become undefined
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Bunuel wrote: Is \(abc = 0\)?
(1) \(a^2 = 2a\)
(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\) Why in (1) \(a^2 = 2a\), I cannot divide both sides by 'a' and get \(a = 2\)? \(a^2 = 2a\) > \((1/a) * a^2 = 2a * (1/a)\) > \(a=2\) Is it wrong to do this way? Edit: Answering my own question with a passage from Manhattan's Guide Book #2: "The GMAT will often attempt to disguise quadratic equations by putting them in forms that do not quite look like the traditional form o f \(ax^2+ bx  c = 0\)."



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09 Nov 2017, 20:12
guireif wrote: Bunuel wrote: Is \(abc = 0\)?
(1) \(a^2 = 2a\)
(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\) Why in (1) \(a^2 = 2a\), I cannot divide both sides by 'a' and get \(a = 2\)? \(a^2 = 2a\) > \((1/a) * a^2 = 2a * (1/a)\) > \(a=2\) Is it wrong to do this way? Edit: Answering my own question with a passage from Manhattan's Guide Book #2: "The GMAT will often attempt to disguise quadratic equations by putting them in forms that do not quite look like the traditional form o f \(ax^2+ bx  c = 0\)." Yes, it's wrong. You cannot reduce this by a, because you'll loose a possible root a = 0. Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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Re: M0319
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13 Mar 2018, 07:05
Bunuel wrote: Official Solution:
In order \(abc = 0\) to be true at least one of the unknowns must be zero. (1) \(a^2 = 2a\). Rearrange: \(a^22a=0\). Factor out \(a\): \(a(a2)=0\). Either \(a=0\) or \(a=2\). If \(a=0\) then the answer is YES but if \(a=2\) then \(abc\) may not be equal to zero (for example consider: \(a=2\), \(b=3\) and \(c=4\)). Not sufficient. (2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\) \(b= \frac{c*(a+b)^2}{(a+b)^2}  c\); \(b=cc\); \(b=0\). Sufficient.
Answer: B Bunuel, In the way that you reduced the equation given in statement to. Should C also be squared. If I were to plug in 2 as A, 1 as B and 9 as C into that equation. I can’t get the equation to equal 81 unless the square root of C is ultimately squared. Posted from my mobile device



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wildhorn wrote: Bunuel wrote: Official Solution:
In order \(abc = 0\) to be true at least one of the unknowns must be zero. (1) \(a^2 = 2a\). Rearrange: \(a^22a=0\). Factor out \(a\): \(a(a2)=0\). Either \(a=0\) or \(a=2\). If \(a=0\) then the answer is YES but if \(a=2\) then \(abc\) may not be equal to zero (for example consider: \(a=2\), \(b=3\) and \(c=4\)). Not sufficient. (2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\) \(b= \frac{c*(a+b)^2}{(a+b)^2}  c\); \(b=cc\); \(b=0\). Sufficient.
Answer: B Bunuel, In the way that you reduced the equation given in statement to. Should C also be squared. If I were to plug in 2 as A, 1 as B and 9 as C into that equation. I can’t get the equation to equal 81 unless the square root of C is ultimately squared. Posted from my mobile deviceNot sure I can follow you... Anyway, we are factoring \(\sqrt{c}\) from the numerator in \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\): \(b= \frac{(\sqrt{c})^2*(a*+b)^2}{a^2+2ab+b^2}  c\); \(b= \frac{c*(a+b)^2}{(a+b)^2}  c\).
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M0319
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13 Mar 2018, 07:57
Bunuel wrote: wildhorn wrote: Bunuel wrote: Official Solution:
In order \(abc = 0\) to be true at least one of the unknowns must be zero. (1) \(a^2 = 2a\). Rearrange: \(a^22a=0\). Factor out \(a\): \(a(a2)=0\). Either \(a=0\) or \(a=2\). If \(a=0\) then the answer is YES but if \(a=2\) then \(abc\) may not be equal to zero (for example consider: \(a=2\), \(b=3\) and \(c=4\)). Not sufficient. (2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\) \(b= \frac{c*(a+b)^2}{(a+b)^2}  c\); \(b=cc\); \(b=0\). Sufficient.
Answer: B Bunuel, In the way that you reduced the equation given in statement to. Should C also be squared. If I were to plug in 2 as A, 1 as B and 9 as C into that equation. I can’t get the equation to equal 81 unless the square root of C is ultimately squared. Posted from my mobile deviceNot sure I can follow you... Anyway, we are factoring \(\sqrt{c}\) from the numerator in \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2}  c\): \(b= \frac{(\sqrt{c})^2*(a*+b)^2}{a^2+2ab+b^2}  c\); \(b= \frac{c*(a+b)^2}{(a+b)^2}  c\). Bunuel, thank you for the explanation. I see my mistake now. I have a follow up question. Would there be a simple way to reduce the numerator if there wasn't a constant multiple being used? For example, if the numerator was (A*C^1/2+B*D^1/2)^2 rather than (A*C^1/2+B*C^1/2)^2










