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

 It is currently 05 Dec 2019, 12:38 ### 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.  # If ab = x and a/b = y, and ab does not equal zero, then

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59561
If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

### Show Tags 00:00

Difficulty:   55% (hard)

Question Stats: 66% (02:31) correct 34% (02:33) wrong based on 171 sessions

### HideShow timer Statistics

Tough and Tricky questions: Algebra.

If $$ab = x$$ and $$\frac{a}{b} = y$$, and $$ab$$ does not equal zero, then $$a^2 - b^2 =$$

A. $$\frac{x(y - 1)(y + 1)}{y}$$
B. $$xy - \frac{y}{x}$$
C. $$x^2 - y^2$$
D. $$y(x^2 - \frac{1}{x^2})$$
E. $$\frac{(y + x)(y - x)}{x}$$

Kudos for a correct solution.
Manager  Joined: 23 Oct 2014
Posts: 84
Concentration: Marketing
Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

### Show Tags

1

ab = x and a/b = y, and ab does not equal zero, then a^2 - b^2 =

a=x/b and a=by
x/b=by
x/y=b^2

since a = by
a^2=(b^2)(y^2)

So a^2-b^2 = (b^2)(y^2)-(x/y)
Since b^2=x/y

=(xy^2)/y-(x/y)
=x(y^2-1)/y
=x(y-1)(y+1)/y
Manager  Joined: 21 Jan 2014
Posts: 61
WE: General Management (Non-Profit and Government)
Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

### Show Tags

1
given that a=by &ab=x
so dividing 1 with 2 results in 1/b=by/x
b^2=x/y

so a^2-b^2=(by)^2-(x/a)^2
=((aby)^2-x^2)/a^2
=((xy)^2-x^2)/a^2 ;from equation 2
=(x^2)(y^2-1)/a^2
=(ab^2)(y-1)(y+1)/a^2 ;from equation 2
=b^2(y-1)(y+1)
=x(y-1)(y+1)/y ; since b^2=x/y

Current Student B
Joined: 02 Sep 2014
Posts: 86
Location: United States
GMAT 1: 770 Q50 V44 GPA: 3.97
Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

### Show Tags

Bunuel wrote:

Tough and Tricky questions: Algebra.

If $$ab = x$$ and $$\frac{a}{b} = y$$, and $$ab$$ does not equal zero, then $$a^2 - b^2 =$$

A. $$\frac{x(y - 1)(y + 1)}{y}$$
B. $$xy - \frac{y}{x}$$
C. $$x^2 - y^2$$
D. $$y(x^2 - \frac{1}{x^2})$$
E. $$\frac{(y + x)(y - x)}{x}$$

Kudos for a correct solution.

ab=x a/b=y.
Multiply the two and a^2 = xy.
Divide the two and b^2 = x/y.
a^2-b^2 = xy-x/y = x*(y^2-1)*(1/y) = x*(y+1)*(y-1)/y = Choice A.
SVP  Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1727
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

### Show Tags

4
ab = x ............. (1)

$$\frac{a}{b} = y$$ ............. (2)

(1) * (2)

$$a^2 = xy$$ .............. (3)

a = by ............ From (2)

by*b = x ........... from (1)

$$b^2 = \frac{x}{y}$$ ................... (4)

(3) - (4)

$$a^2 - b^2 = xy - \frac{x}{y}$$

$$= x(y - \frac{1}{y})$$

$$= x(\frac{y^2 - 1}{y})$$

$$= \frac{x(y+1)(y-1)}{y}$$

Senior Manager  P
Joined: 17 Mar 2014
Posts: 426
Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

### Show Tags

1
Question is a^2-b^2 = ?

Solution:

As per question stem,
ab= x -- equation (1)
a= by --equation (2)

Let us substitute eq (2) int0 eq(1), we will get

by*b= x

(b^2)*y=x

hence, b^2 = x/y ---eq (3)

Now solve equation 3 & 2.
we will get a^2 = xy -- eq 4

using eq 3 & 4, we can solve for a^2 - b^2

a^2 - b^2 = xy- x/y = (xy^2-x)/y = x(y^2-1)/y = x(y-1)(y+1)/y

Regards,
Ammu
Intern  Joined: 20 Jan 2013
Posts: 33
Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

### Show Tags

1
$$a^{2}$$ - $$b^{2}$$ = $$b^{2}$$[ $$(a/b)^{2}$$ -1] ....1)

As given --> $$\frac{a}{b}$$ = y, ab = x

a = by => by*b = x =>$$b^{2}$$ = x/y

Let's put these values in equation 1

$$\frac{x[(y^{2} - 1)]}{y}$$ => $$\frac{x(y - 1)(y + 1)}{y}$$

Math Expert V
Joined: 02 Sep 2009
Posts: 59561
Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

### Show Tags

1
Official Solution:

If $$ab = x$$ and $$\frac{a}{b} = y$$, and $$ab$$ does not equal zero, then $$a^2 - b^2 =$$

A. $$\frac{x(y - 1)(y + 1)}{y}$$
B. $$xy - \frac{y}{x}$$
C. $$x^2 - y^2$$
D. $$y(x^2 - \frac{1}{x^2})$$
E. $$\frac{(y + x)(y - x)}{x}$$

There are two paths to the solution: using algebra and picking numbers.

Algebra

Many algebraic manipulations are possible. However, we should be guided by the form of the answer choices, which contain only $$x$$'s and $$y$$'s. Thus, we should look for ways to express $$a$$ in terms of only $$x$$ and $$y$$ (not $$b$$); likewise, we want $$b$$ in the same terms, without $$a$$.

Again, we have $$ab = x$$ and $$\frac{a}{b} = y$$.

If we solve the first equation for b and plug into the second equation, here is what we get:
$$b = \frac{x}{a}$$

Solve first eq. for $$b$$
$$\frac{a}{(\frac{x}{a})} = y$$

Substitute $$\frac{x}{a}$$ for $$b$$ in second eq.
$$\frac{a^2}{x} = y$$

Bring the $$a$$ up to the numerator (dividing by $$\frac{x}{a}$$ is the same as multiplying by $$\frac{a}{x}$$)
$$a^2 = xy$$

Multiply both sides by $$x$$

In fact, we could have gotten this result directly if we had multiplied the two original equations together:
$$ab = x$$
$$\frac{a}{b} = y$$

Multiply each side (canceling the $$b$$'s):
$$a^2 = xy$$

Similarly, we can get $$b^2 = \frac{x}{y}$$ either by substitution (solving one equation for $$a$$ and plugging into the other equation) or by dividing the first equation by the second, canceling the $$a$$'s:
$$ab = x$$
$$\frac{a}{b} = y$$

Divide each side:
$$b^2 = \frac{x}{y}$$

(Note that $$ab$$ divided by $$\frac{a}{b}$$ is the same as $$ab$$ times $$\frac{b}{a}$$. The $$a$$'s cancel, leaving $$b^2$$.)

Now that we have $$a^2 = xy$$ and $$b^2 = \frac{x}{y}$$, we can substitute into the expression we are looking for:
$$a^2 - b^2 = xy - \frac{x}{y}$$

Substitute the expressions above.

This does not match any answer choice, so we should manipulate further:

$$= x(y - \frac{1}{y})$$

Pull out the common factor $$x$$
$$= x(\frac{y^2}{y} - \frac{1}{y})$$

Make a common denominator of $$y$$
$$= \frac{x(y^2 - 1)}{y}$$

Pull out the common divisor $$y$$ (that is, $$y$$ in the denominator)
$$= \frac{x(y - 1)(y + 1)}{y}$$

Factor the special product $$y^2 - 1$$ into $$(y - 1)(y + 1)$$

This expression matches the first answer choice.

Plugging Numbers

Choose suitable numbers for $$a$$ and $$b$$. Here, we should use small numbers and make $$a$$ divisible by $$b$$. For instance, let $$a = 6$$ and $$b = 2$$.

Then, from $$ab = x$$, we have $$x = (6)(2) = 12$$.

Likewise, from $$\frac{a}{b} = y$$, we have $$y = \frac{6}{2} = 3$$.

Finally, we compute our target number:
$$a^2 - b^2 = 62 - 22 = 36 - 4 = 32.$$

Now, compute each answer choice and compare the result to the target of 32. Note that you should stop computing as soon as you see that you are not going to get 32, but you should check all the answer choices.

(A) $$\frac{x(y - 1)(y + 1)}{y} = \frac{(12)(3 - 1)(3 + 1)}{3} = \frac{(12)(2)(4)}{3} = (4)(2)(4) = 32$$ CORRECT

(B)$$xy - \frac{y}{x} = (12)(3) - \frac{3}{12} = 36 - \frac{1}{4} =$$ stop here (result is too big) INCORRECT

(C)$$x^2 - y^2 = 122 - 3^2 = 144 - 9 =$$ stop here (result is too big) INCORRECT

(D)$$y( x^2 - \frac{1}{x^2} ) = 3 ( 122 - \frac{1}{122}) = 3 (144 - \frac{1}{144} ) =$$ stop here (result is too big) INCORRECT

(E)$$\frac{(y + x)(y - x)}{x} = \frac{(3 + 12)(3 - 12)}{12} =$$ stop here (result is negative) INCORRECT

Non-Human User Joined: 09 Sep 2013
Posts: 13709
Re: If ab = x and a/b = y, and ab does not equal zero, then  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: If ab = x and a/b = y, and ab does not equal zero, then   [#permalink] 25 Apr 2019, 16:24
Display posts from previous: Sort by

# If ab = x and a/b = y, and ab does not equal zero, then  