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

 It is currently 23 Mar 2019, 02:05

### 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

# Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 53795
Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?  [#permalink]

### Show Tags

20 Dec 2017, 06:18
00:00

Difficulty:

45% (medium)

Question Stats:

68% (01:35) correct 32% (02:03) wrong based on 38 sessions

### HideShow timer Statistics

Is $$\frac{x}{p}(p^2 + q^2 + r^2) = xp + yq + zr$$ ?

(1) $$yq = r^2$$ and $$zr = q^2$$

(2) $$x = p$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 53795
Re: Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?  [#permalink]

### Show Tags

20 Dec 2017, 06:20
Bunuel wrote:
Is $$\frac{x}{p}(p^2 + q^2 + r^2) = xp + yq + zr$$ ?

(1) $$yq = r^2$$ and $$zr = q^2$$

(2) $$x = p$$

Similar question from OG: https://gmatclub.com/forum/is-x-m-m-2-n ... 99423.html
_________________
examPAL Representative
Joined: 07 Dec 2017
Posts: 906
Re: Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?  [#permalink]

### Show Tags

21 Dec 2017, 11:34
1
1
Bunuel wrote:
Is $$\frac{x}{p}(p^2 + q^2 + r^2) = xp + yq + zr$$ ?

(1) $$yq = r^2$$ and $$zr = q^2$$

(2) $$x = p$$

We'll go for a Precise approach of simplification since this looks like a very technical question.

Multiplying by p (p cannot be zero as it is in denominator) gives:
$$xp^2 + xq^2 + xr^2 = xp^2 + yqp + zrp$$. Cancelling common factors and changing sides gives:
$$xq^2 - yqp + xr^2 - rzp = 0$$
Substituting the equations in (1) gives $$xzr - yqp + xyq - zrp = zr(x - p) + yq(x - p) = (zr + yq)(x-p)$$
Without knowing if $$zr + yq = 0$$ or if $$p = x$$ we cannot know if this is 0.
Insufficient!

Substituting the equations in (2) gives
$$xq^2 - yqx + xr^2 - rzx = x(q^2 - yq + r^2 - rz)$$
Since we don't know if either side is 0, this is also insufficient.

But, if we substitute $$x = p$$ into our result from (1) we get a result of 0!
This means that the original equation is true.
_________________
Re: Is x/p(p^2 + q^2 + r^2) = xp + yq + zr ?   [#permalink] 21 Dec 2017, 11:34
Display posts from previous: Sort by