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

 It is currently 16 Oct 2019, 07:49 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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Let a,b,c, and d be nonzero real numbers. If the quadratic equation

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Intern  Joined: 15 Dec 2015
Posts: 5
Concentration: Sustainability, Strategy
Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

10
85 00:00

Difficulty:   65% (hard)

Question Stats: 63% (02:14) correct 37% (02:32) wrong based on 660 sessions

HideShow timer Statistics

Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$

B. $$-\frac{ac}{bd}$$

C. $$-\frac{ad}{bc}$$

D. $$\frac{ab}{cd}$$

E. $$\frac{ad}{bc}$$
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

105
36
amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$
B. $$-\frac{ac}{bd}$$
C. $$-\frac{ad}{bc}$$
D. $$\frac{ab}{cd}$$
E. $$\frac{ad}{bc}$$

Having trouble with this problem and would appreciate if someone could walk me through a solution. Thanks!

$$ax(cx+d)=-b(cx+d)$$

You solve a quadratic by splitting it into factors.

$$ax(cx+d) + b(cx+d) = 0$$

But recognise that the factors are already there in this equation. Take (cx + d) common.
$$(ax + b)*(cx + d) = 0$$

Roots are -b/a and -d/c

Their ratio could be bc/ad or ad/bc.

_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Manager  Status: 2 months to go
Joined: 11 Oct 2015
Posts: 105
GMAT 1: 730 Q49 V40 GPA: 3.8
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

8
1
Let a, b, c and d be nonzero real numbers. If the quadratic equation ax(cx+d)=-b(cx+d) is solved for x, which of the following is a possible ration of the 2 solutions?

$$Let\mbox{'}s\; say\; a=1,\; b=2,\; c=3\; and\; d=4$$

the solutions would be:
a) -$$\frac{1}{6}$$
b) -$$\frac{3}{8}$$
c) -$$\frac{2}{3}$$
d) $$\frac{1}{6}$$
e) $$\frac{2}{3}$$

$$Now\; let\mbox{'}s\; solve\; the\; equation,\; after\; substituting\; a=1,\; b=2,\; c=3\; and\; d=4\; into\; the\; eq.\; we\; find: 3x^{2}\; +\; 10x\; +\; 8\; =\; 0$$
$$which\; solved\; gives\; us\; \left( x+\frac{4}{3} \right)\left( x+2 \right),\; where\; the\; roots\; are\; x_{a}=\; -\frac{4}{3}\; and\; x_{b}=\; -2,\; if\; we\; divide\; them:\; \frac{-\frac{4}{3}}{-2}\; we\; find\; \frac{xa}{xb}\; =\; \frac{2}{3}$$

$$Answer\; E$$
General Discussion
Math Expert V
Joined: 02 Aug 2009
Posts: 7958
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

19
12
amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$
B. $$-\frac{ac}{bd}$$
C. $$-\frac{ad}{bc}$$
D. $$\frac{ab}{cd}$$
E. $$\frac{ad}{bc}$$

Having trouble with this problem and would appreciate if someone could walk me through a solution. Thanks!

Hi,
one method would be take advantage of SUM and PRODUCT of roots of a quadratic equation..
$$ax(cx+d)=-b(cx+d)$$ ..
$$acx^2+(ad+bc)x+bd = 0$$..
Let the roots be S and T
Sum of roots = $$S+T = -(\frac{ad+bc}{ac})$$..
Product of roots = $$S*T = \frac{bd}{ac}$$..

two methods..

1) equate from SUM..
we know $$S+T = -(\frac{ad+bc}{ac})$$..
$$S+T = -(\frac{ad}{ac}+\frac{bc}{ac})$$
let $$S = -\frac{ad}{ac}$$and $$T = -\frac{bc}{ac}$$
so$$\frac{S}{T} = \frac{ad}{bc}$$
E

2)we are to find$$\frac{S}{T} or \frac{T}{S}$$..
divide SUM by PRODUCT--

$$\frac{S+T}{ST} = -(\frac{ad+bc}{ac})/\frac{bd}{ac}$$..
$$\frac{S}{ST}+\frac{T}{ST} = -( \frac{ad}{bd} + \frac{bc}{bd})$$ ..
$$\frac{1}{T} + \frac{1}{S} = -\frac{a}{b} - \frac{c}{d}$$..

so JUST for finding the ratio, we will equate two sides--
Here it doesn't matter what do you equate with, because we are just finding any of the ratio

$$\frac{1}{T}= -\frac{a}{b}$$and $$\frac{1}{S}= -\frac{c}{d}$$..

$$\frac{\frac{1}{T}}{\frac{1}{S}}$$ = $$\frac{-\frac{a}{b}}{-\frac{c}{d}}$$..

$$\frac{S}{T} = \frac{ad}{bc}$$

E
_________________
Current Student D
Joined: 12 Aug 2015
Posts: 2569
Schools: Boston U '20 (M)
GRE 1: Q169 V154 Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

2
1
Don't Even Think about calculating the roots by opening the brackets
It becomes a mess..!
Here take cx+d common and solve for x
we can see that the ratio may be ad/bc or bc/ad
Hence E
_________________
SVP  B
Joined: 06 Nov 2014
Posts: 1873
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

4
1
2
dgeorgie wrote:
Let a, b, c and d be nonzero real numbers. If the quadratic equation ax(cx+d)=-b(cx+d) is solved for x, which of the following is a possible ration of the 2 solutions?

A. -ab/(cd)
B. -ac/(bd)
D. ab/(cd)

Actually, I dont understand why do we have a quadratic equation. Isn't it the same as ax=-b ? Then we would have only 1 solution
I appreciate your suggestions.
Thank you in advance!

No, you cannot cancel (cx+d) from both the sides. Because (cx+d) can be 0 too and you cannot cancel 0 from both sides.
You can easily cancel out any numbers, but not variables.

For Example: x(x+2) = 5(x+2). In this case, you cannot simply cancel out (x+2) from both the sides.

Therefore ax(cx+d)=-b(cx+d) should be written as (ax+b)(cx+d) = 0
Now in the quadratic equation, any of the terms can be equal to zero.

Therefore, the two solutions will be x = -b/a and x = -d/c

Ratio of two roots = (-d/c) / (-b/a) = ad/bc
Correct option: E

Does this help?
Target Test Prep Representative D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8069
Location: United States (CA)
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

8
3
amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$
B. $$-\frac{ac}{bd}$$
C. $$-\frac{ad}{bc}$$
D. $$\frac{ab}{cd}$$
E. $$\frac{ad}{bc}$$

ax(cx + d) = -b(cx + d)

ax(cx + d) + b(cx + d) = 0

Next we can factor out (cx + d):

(cx + d)(ax + b) = 0

Using the zero product property, we can set the expression in each set of parentheses to zero:

cx + d = 0

cx = -d

x = -d/c

Or

ax + b = 0

ax = -b

x = -b/a

So a possible ratio is:

(-d/c)/(-b/a)

_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern  Joined: 20 Oct 2014
Posts: 3
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

Hello, can someone please explain why this works:
ax(cx + d) = -b(cx + d)
axc^2 + axd = -bcx - bd
ax^2 + axd + bcx + bd = 0
ax^2 + (ad + bc)x + bd = 0

Then take the ratio of the ad/bc (in the brackets) - is it just luck that I got it right this way? Thanks for the help!

Posted from my mobile device
Board of Directors V
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3582
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

1
AfricanPrincess wrote:
Hello, can someone please explain why this works:
ax(cx + d) = -b(cx + d)
axc^2 + axd = -bcx - bd
ax^2 + axd + bcx + bd = 0
ax^2 + (ad + bc)x + bd = 0

Then take the ratio of the ad/bc (in the brackets) - is it just luck that I got it right this way? Thanks for the help!

Posted from my mobile device

In the highlighted portion above, you have missed c in acx^2.

Also, in the bracket you have the sum of ad and bc not the product.

Now, if I consider your statement

acx^2 + (ad + bc)x + bd = 0

There is a very good solution provided by chetan2u in the forum. You can refer to that solution.

Else, you can simply do it in this way :

ax(cx + d) = -b(cx + d)

=> either cx+d=0 => x=-d/c ---(1)
or
ax=-b => x=-b/a ---(2)

Divide (1) by (2) or vice versa, you will get ad/bc or bc/ad.
_________________
My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place

GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.
New! Best Reply Functionality on GMAT Club!
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free
Check our new About Us Page here.
Intern  B
Joined: 06 Apr 2017
Posts: 29
Location: United States (OR)
Schools: Haas EWMBA '21
GMAT 1: 730 Q48 V44 GMAT 2: 730 Q49 V40 GPA: 3.98
WE: Corporate Finance (Health Care)
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

This is a 'could be' question - don't overly complicate it.

$$ax(cx+d)=-b(cx+d) \rightarrow acx^2 +adx=-bcx-bd \rightarrow acx^2+x(ad+bc)+bd$$

Quadratics are easier to work with when the leading coefficient is $$1$$

Since this is a COULD BE question, then let's say $$a=c=1$$

Now we have

$$x^2+x(ad+bc)+bd$$

Since this is a COULD BE question with a well-formed quadratic, the roots could definitely be $$ad/bc$$, since $$ad$$ and $$bc$$ sum to the coefficient of the second term and multiply to the third term

Intern  B
Joined: 18 Jul 2017
Posts: 5
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

one method would be take advantage of SUM and PRODUCT of roots of a quadratic equation..
can anyone plz explain how to solve this equation
thank you
Intern  B
Joined: 16 Oct 2017
Posts: 1
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

1
ax(cx+d)=-b(cx+d)

either cx+d = 0:
then, x = -d/c ..... (solution 1)

or cx + d not equal to 0:
then cancel it from both sides and;
ax = -b
i.e. x = -b/a ........ (solution 2)

therefore; ratio of 2 solutions = (-d/c) / (-b/a)
Intern  B
Joined: 10 Jun 2017
Posts: 23
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

VeritasPrepKarishma wrote:
amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$
B. $$-\frac{ac}{bd}$$
C. $$-\frac{ad}{bc}$$
D. $$\frac{ab}{cd}$$
E. $$\frac{ad}{bc}$$

Having trouble with this problem and would appreciate if someone could walk me through a solution. Thanks!

$$ax(cx+d)=-b(cx+d)$$

You solve a quadratic by splitting it into factors.

$$ax(cx+d) + b(cx+d) = 0$$

But recognise that the factors are already there in this equation. Take (cx + d) common.
$$(ax + b)*(cx + d) = 0$$

Roots are -b/a and -d/c

Their ratio could be bc/ad or ad/bc.

Excellent explanation!!!! I got to the point of (cx+d)(ax + b) = 0 but then started to multiply everything and ended up in a huge mess... Obviously that either cx+d=0 or ax+b=0.... that's a basic quadratic expression rule..
Intern  B
Joined: 07 May 2015
Posts: 35
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

ScottTargetTestPrep wrote:
amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$
B. $$-\frac{ac}{bd}$$
C. $$-\frac{ad}{bc}$$
D. $$\frac{ab}{cd}$$
E. $$\frac{ad}{bc}$$

ax(cx + d) = -b(cx + d)

ax(cx + d) + b(cx + d) = 0

Next we can factor out (cx + d):

(cx + d)(ax + b) = 0

Using the zero product property, we can set the expression in each set of parentheses to zero:

cx + d = 0

cx = -d

x = -d/c

Or

ax + b = 0

ax = -b

x = -b/a

So a possible ratio is:

(-d/c)/(-b/a)

i don't mean to bump an old post, but i follow along your process all the way until the end where you get the ratio of ad/bc from (-d/c)/(-b/a). i'm not following. is there something basic i'm missing on why you're taking the denominator of one/numerator of the other, vis verse, and making them both positive?
Director  P
Joined: 02 Oct 2017
Posts: 720
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

1
Rephrasing statement of question

(ax+b)(cx+d)=0

Roots are -b/a or -d/c

So ratio would be bc/ad or ad/bc

Give kudos if it helps

Posted from my mobile device
_________________
Give kudos if you like the post
Director  V
Joined: 12 Feb 2015
Posts: 917
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$

B. $$-\frac{ac}{bd}$$

C. $$-\frac{ad}{bc}$$

D. $$\frac{ab}{cd}$$

E. $$\frac{ad}{bc}$$

The trick to answer this question correctly was to note that the equation is already written in the standard solution form. One can solve for the roots and take the fraction to get the correct answer.
_________________
________________
Manish "Only I can change my life. No one can do it for me"
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4005
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

Top Contributor
1
amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$

B. $$-\frac{ac}{bd}$$

C. $$-\frac{ad}{bc}$$

D. $$\frac{ab}{cd}$$

E. $$\frac{ad}{bc}$$

GIVEN: ax(cx + d) = -b(cx + d)
Add b(cx + d) to both sides to get: ax(cx + d) + b(cx + d) = 0
Rewrite as: (ax + b)(cx + d) = 0

This means either (ax + b) = 0 or (cx + d) = 0
Let's examine each case

Case a: ax + b = 0
So: ax = -b
Solve: x = -b/a

Case b: cx + d = 0
So: cx = -d
Solve: x = -d/c

Which of the following is a possible ratio of the 2 solutions?
ASIDE: Why does the question ask for a POSSIBLE ratio?
Well, the ratio of the solutions can be EITHER (-b/a)/(-d/c) OR (-d/c)/(-b/a)

Let's check the first ratio.
(-b/a)/(-d/c) = (-b/a)(c/-d) = bc/ad check the answer choices . . . not there.
However, answer choice E is the reciprocal of bc/ad, so it must be the correct answer.

Not convinced?
Let's confirm.
(-d/c)/(-b/a) = (-d/c)(a/-b) = ad/bc = answer choice E

Cheers,
Brent
_________________
SVP  P
Joined: 03 Jun 2019
Posts: 1689
Location: India
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

Show Tags

amitpaul527 wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

A. $$-\frac{ab}{cd}$$

B. $$-\frac{ac}{bd}$$

C. $$-\frac{ad}{bc}$$

D. $$\frac{ab}{cd}$$

E. $$\frac{ad}{bc}$$

Given: Let a,b,c, and d be nonzero real numbers.

Asked: If the quadratic equation $$ax(cx+d)=-b(cx+d)$$ is solved for x, which of the following is a possible ratio of the 2 solutions?

$$(ax+b)(cx+d)=0$$

x = - b/a or -d/c

Ratio of solutions = bc/ad or ad/bc

IMO E
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts."

Please provide kudos if you like my post. Kudos encourage active discussions.

My GMAT Resources: -

Efficient Learning
All you need to know about GMAT quant

Tele: +91-11-40396815
Mobile : +91-9910661622
E-mail : kinshook.chaturvedi@gmail.com Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation   [#permalink] 12 Sep 2019, 08:25
Display posts from previous: Sort by

Let a,b,c, and d be nonzero real numbers. If the quadratic equation

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  