GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Oct 2019, 14:21

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# M03-19

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58453

### Show Tags

16 Sep 2014, 00:20
1
11
00:00

Difficulty:

25% (medium)

Question Stats:

73% (01:31) correct 27% (02:02) wrong based on 285 sessions

### HideShow timer Statistics

Is $$abc = 0$$?

(1) $$a^2 = 2a$$

(2) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 58453

### Show Tags

16 Sep 2014, 00:20
1
1
Official Solution:

In order $$abc = 0$$ to be true at least one of the unknowns must be zero.

(1) $$a^2 = 2a$$. Rearrange: $$a^2-2a=0$$. Factor out $$a$$: $$a(a-2)=0$$. 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) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$

$$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$;

$$b=c-c$$;

$$b=0$$. Sufficient.

_________________
Intern
Joined: 10 Jan 2015
Posts: 6

### Show Tags

27 May 2015, 07:23
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: $$a^2-2a=0$$. Factor out $$a$$: $$a(a-2)=0$$. 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) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$

$$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$;

$$b=c-c$$;

$$b=0$$. Sufficient.

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.
Math Expert
Joined: 02 Sep 2009
Posts: 58453

### Show Tags

27 May 2015, 07:33
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: $$a^2-2a=0$$. Factor out $$a$$: $$a(a-2)=0$$. 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) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$

$$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$;

$$b=c-c$$;

$$b=0$$. Sufficient.

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.

_________________
Intern
Joined: 14 Apr 2015
Posts: 29
Location: United States
Concentration: Nonprofit, Entrepreneurship
GMAT Date: 06-14-2015
GPA: 3.93
WE: Marketing (Non-Profit and Government)

### Show Tags

27 May 2015, 11:35
Bunuel wrote:
Official Solution:

(2) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$

$$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$;

$$b=c-c$$;

$$b=0$$. Sufficient.

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?
Intern
Joined: 24 Jan 2015
Posts: 2

### Show Tags

27 May 2015, 17:24
The square root of c needs to be squared too
Intern
Joined: 14 Apr 2015
Posts: 29
Location: United States
Concentration: Nonprofit, Entrepreneurship
GMAT Date: 06-14-2015
GPA: 3.93
WE: Marketing (Non-Profit and Government)

### Show Tags

27 May 2015, 17:48
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.
CEO
Joined: 20 Mar 2014
Posts: 2597
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)

### Show Tags

Updated on: 10 Jun 2015, 19:02
3
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, 18:18.
Last edited by ENGRTOMBA2018 on 10 Jun 2015, 19:02, edited 1 time in total.
Intern
Joined: 14 Apr 2015
Posts: 29
Location: United States
Concentration: Nonprofit, Entrepreneurship
GMAT Date: 06-14-2015
GPA: 3.93
WE: Marketing (Non-Profit and Government)

### Show Tags

27 May 2015, 21:11
Thanks, that clears it up!
Manager
Joined: 21 May 2015
Posts: 215
Concentration: Operations, Strategy
GMAT 1: 750 Q50 V41

### Show Tags

31 May 2015, 01:12
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
_________________
Apoorv

I realize that i cannot change the world....But i can play a part
Intern
Joined: 05 Jul 2016
Posts: 16
Location: Brazil
Concentration: Finance, Entrepreneurship
WE: Analyst (Investment Banking)

### Show Tags

09 Nov 2017, 15:04
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 $$ax^2+ bx - c = 0$$."
Math Expert
Joined: 02 Sep 2009
Posts: 58453

### Show Tags

09 Nov 2017, 21:12
1
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.
_________________
Manager
Joined: 06 Jan 2018
Posts: 51
Location: United States (TX)
GMAT 1: 730 Q49 V41
GPA: 3.31

### Show Tags

13 Mar 2018, 08:05
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: $$a^2-2a=0$$. Factor out $$a$$: $$a(a-2)=0$$. 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) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$

$$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$;

$$b=c-c$$;

$$b=0$$. Sufficient.

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
Math Expert
Joined: 02 Sep 2009
Posts: 58453

### Show Tags

13 Mar 2018, 08:24
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: $$a^2-2a=0$$. Factor out $$a$$: $$a(a-2)=0$$. 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) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$

$$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$;

$$b=c-c$$;

$$b=0$$. Sufficient.

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

Not sure I can follow you...

Anyway, we are factoring $$\sqrt{c}$$ from the numerator in $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$:

$$b= \frac{(\sqrt{c})^2*(a*+b)^2}{a^2+2ab+b^2} - c$$;

$$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$.
_________________
Manager
Joined: 06 Jan 2018
Posts: 51
Location: United States (TX)
GMAT 1: 730 Q49 V41
GPA: 3.31

### Show Tags

13 Mar 2018, 08:57
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: $$a^2-2a=0$$. Factor out $$a$$: $$a(a-2)=0$$. 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) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$

$$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$;

$$b=c-c$$;

$$b=0$$. Sufficient.

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

Not 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
Intern
Joined: 03 Jun 2019
Posts: 10
Location: India
Schools: YLP '20 (S)
GMAT 1: 630 Q41 V28

### Show Tags

24 Jun 2019, 05:39
1
Okay from my personal experience, whenever you get a DS question, solve the Second Statement first.

In the above question . solving the Second statment first -

Taking \sqrt{c} common from the RHS , we get
$$b =( \sqrt{c}^2 *(a+b)^2)/(a+b)^2 - c$$
=> b = $$c - c$$
=> b = 0.

Hence we see that b = 0. Therefore the product abc will be 0. We can conclude from this statement that we dont even require the 1st Statement anymore.

Therefore Option B (Statement 2 ALONE is sufficient to solve the question).

Kindly Give Kudos if you liked my explanation.
Intern
Joined: 15 Jan 2019
Posts: 3

### Show Tags

24 Jun 2019, 08:32
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: $$a^2-2a=0$$. Factor out $$a$$: $$a(a-2)=0$$. 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) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$

$$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$;

$$b=c-c$$;

$$b=0$$. Sufficient.

I'm getting confused on a particular step, when $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$ is being simplified to $$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$.
When solving it I'm getting $$b= \frac{c*(a^2+b^2)}{(a+b)^2} - c$$. And I'm not sure how $$(a^2+b^2)$$ is being written as $$(a+b)^2$$, as $$(a+b)^2 = {a^2+2ab+b^2}$$
Veritas Prep GMAT Instructor
Joined: 01 May 2019
Posts: 50

### Show Tags

24 Jun 2019, 09:04
3
banerjr1 wrote:

I'm getting confused on a particular step, when $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$ is being simplified to $$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$.
When solving it I'm getting $$b= \frac{c*(a^2+b^2)}{(a+b)^2} - c$$. And I'm not sure how $$(a^2+b^2)$$ is being written as $$(a+b)^2$$, as $$(a+b)^2 = {a^2+2ab+b^2}$$

Let's do this with some intermediate steps so the math is more clear:

$$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$

Factor out $$\sqrt{c}$$ from $$a*\sqrt{c}+b*\sqrt{c}$$

$$b= \frac{(\sqrt{c}*(a+b))^2}{a^2+2ab+b^2} - c$$

Distribute the square to $$\sqrt{c}$$ and $$a+b$$ (This is where you made your mistake — we can distribute a square to terms that are being multiplied together, but not terms that are being added together. This is why we get $$(a+b)^2$$ and not $$a^2 + b^2$$.)

$$b= \frac{\sqrt{c}^2*(a+b)^2}{a^2+2ab+b^2} - c$$

$$b= \frac{c*(a+b)^2}{a^2+2ab+b^2} - c$$

Either expand $$(a+b)^2$$ in numerator to $$a^2+2ab+b^2$$ OR factor $$a^2+2ab+b^2$$ in denominator to $$(a+b)^2$$

$$b= \frac{c*(a^2+2ab+b^2)}{a^2+2ab+b^2} - c$$ OR $$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$

Cancel like terms from numerator and denominator (either $$a^2+2ab+b^2$$ or $$(a+b)^2$$)

$$b= c - c$$

$$b= 0$$

So you're totally correct that $$a^2 + b^2$$ doesn't equal $$(a+b)^2$$, but we should never get $$a^2 + b^2$$ based on exponent rules for distributing exponents.
Manager
Joined: 15 Jun 2019
Posts: 205

### Show Tags

12 Jul 2019, 20:10
i am super confused,

Bunuel suppose a = (b* c)/b -c here as per this example we can write as a = c-c = 0

if a =(b* c)/b can we write as a = c by cancelling b rather if ab = bc can we cancel the b ???
_________________
please do correct my mistakes that itself a big kudo for me,

thanks
Senior Manager
Joined: 19 Nov 2017
Posts: 253
Location: India
Schools: ISB
GMAT 1: 670 Q49 V32
GPA: 4

### Show Tags

12 Jul 2019, 22:27
(1)
$$a^2 = 2a$$
$$a = 2$$

Don't know about any other value.
Not sufficient.

(2)
$$b = (a\sqrt{c} + b\sqrt{c})^2$$/$$(a + b)^2$$
$$b = {[a^2*c + b^2*c + 2abc]/[(a + b)^2]} - c$$
Taking C common from the numerator
$$b = {c*[a^2 + b^2 + 2ab]/[(a + b)^2]} - c$$
$$b = {c*[(a + b)^2]/[(a + b)^2]} - c$$
$$b = {c*[(a + b)^2]/[(a + b)^2]} - c$$
$$b = c - c$$
$$b = 0$$

if $$b = 0$$, then $$abc = 0$$. Hence, (2) is sufficient.

_________________

Vaibhav

Sky is the limit. 800 is the limit.

~GMAC
Re: M03-19   [#permalink] 12 Jul 2019, 22:27

Go to page    1   2    Next  [ 21 posts ]

Display posts from previous: Sort by

# M03-19

Moderators: chetan2u, Bunuel