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

 It is currently 21 Nov 2018, 00:04

### 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

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

November 22, 2018

November 22, 2018

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
• ### Free lesson on number properties

November 23, 2018

November 23, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.

# M05-28

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50712

### Show Tags

17 Aug 2018, 00:17
bonion wrote:
Bunuel wrote:
Official Solution:

We have system of equations with three distinct equations and three unknowns, so we can solve it. As $$x$$ is squared we'll get two values for it and also two values for $$y$$ and $$z$$: two triplets. Hence we'll get two values for $$x^3 + y^2 + z$$, one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:

$$x^2 = y + 5$$;

$$z = 2x$$;

$$y = z - 2$$, or which is the same $$y=(2x)-2$$

Solve for $$x$$: $$x^2=(2x-2) + 5$$. Rearrange: $$x^2-2x-3=0$$. Either $$x=3$$ or $$x=-1$$

So, the first triplet would be: $$x=3$$, $$y=4$$, $$z=6$$. Which gives: $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;

The second triplet would be: $$x=-1$$, $$y=-4$$, $$z=-2$$. Which gives: $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$. We deal with the first triplet: $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. Sufficient.

(2) $$y = 4$$. We deal with the first triplet: $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. Sufficient.

Hello Bunuel I am not sure what i missed, but in statement 2, y = 4, then x^2 = 9, then x = 3 or -3,
If x = 3 the equation results 49 which is divisible by 7
if x = -3 the equation results -5 which is not divisible by 7
So, i think statement 2 is not sufficient.

x = -3 is not a solution.

y = 4, then z = 6 (from y = z - 2) and x = 3 (from z = 2x).
_________________
Intern
Joined: 19 Jun 2017
Posts: 9

### Show Tags

24 Oct 2018, 22:23
HI @Bunuel/anyone.

I interpreted the given equation x^3+y^2+z divisible by 7 as x^3+(x^2-5)^2+2x divisible by 7
x^2=y+5 so y=x^2-5 ,z=2x. i have substituted these values in the above equation.
so now according to stmnt 1 x>0,
if x=1 value =19 not divisible by 7
if x=2 value =13 not divisible by 7
but if x=3 value =49 divisible by 7

so i have ignored a and marked b as answer
can anyone explain where am i wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 50712

### Show Tags

24 Oct 2018, 22:44
MaheshDuggirala wrote:
HI @Bunuel/anyone.

I interpreted the given equation x^3+y^2+z divisible by 7 as x^3+(x^2-5)^2+2x divisible by 7
x^2=y+5 so y=x^2-5 ,z=2x. i have substituted these values in the above equation.
so now according to stmnt 1 x>0,
if x=1 value =19 not divisible by 7
if x=2 value =13 not divisible by 7
but if x=3 value =49 divisible by 7

so i have ignored a and marked b as answer
can anyone explain where am i wrong.

If you read the solution carefully, you'll see that the system of equation we have ($$x^2 = y + 5$$, $$y = z - 2$$ and $$z = 2x$$) has the following roots: $$x=3$$, $$y=4$$, $$z=6$$ OR $$x=-1$$, $$y=-4$$, $$z=-2$$. No other values of x, y and z satisfy these three equations (so x can be neither 1 nor 2). If x = 3, then $$x^3+y^2+z$$ IS divisible by 7 but if x = -1 then it's NOT.

(1) says that x > 0, so x = 3 and $$x^3+y^2+z$$ IS divisible by 7.
(2) says that y = 4, so x = 3 and $$x^3+y^2+z$$ IS divisible by 7.

Hope it's clear.
_________________
Intern
Joined: 19 Jun 2017
Posts: 9

### Show Tags

25 Oct 2018, 20:22
Thanks Bunuel for the great explanation.

So u meant to say that for any given equation we should try to find the variable values and then interpret the equation.
If we don't do so we gonna miss the limitations on those variable values.
is my understanding correct?.
Manager
Joined: 28 Jun 2018
Posts: 57
GMAT 1: 490 Q39 V18
GMAT 2: 640 Q47 V30
GMAT 3: 670 Q50 V31
GMAT 4: 700 Q49 V36
GPA: 4

### Show Tags

16 Nov 2018, 02:05
bonion wrote:
Bunuel wrote:
Official Solution:

We have system of equations with three distinct equations and three unknowns, so we can solve it. As $$x$$ is squared we'll get two values for it and also two values for $$y$$ and $$z$$: two triplets. Hence we'll get two values for $$x^3 + y^2 + z$$, one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:

$$x^2 = y + 5$$;

$$z = 2x$$;

$$y = z - 2$$, or which is the same $$y=(2x)-2$$

Solve for $$x$$: $$x^2=(2x-2) + 5$$. Rearrange: $$x^2-2x-3=0$$. Either $$x=3$$ or $$x=-1$$

So, the first triplet would be: $$x=3$$, $$y=4$$, $$z=6$$. Which gives: $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;

The second triplet would be: $$x=-1$$, $$y=-4$$, $$z=-2$$. Which gives: $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$. We deal with the first triplet: $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. Sufficient.

(2) $$y = 4$$. We deal with the first triplet: $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. Sufficient.

Hello Bunuel I am not sure what i missed, but in statement 2, y = 4, then x^2 = 9, then x = 3 or -3,
If x = 3 the equation results 49 which is divisible by 7
if x = -3 the equation results -5 which is not divisible by 7
So, i think statement 2 is not sufficient.

Hey bonion

I am not sure if u still have the same doubt. But here is the simple logic.

If we take x = -3 then does it satisfy ALL the other equations?
That is IF $$x = -3$$ ,
Plug it in here $$z = 2x$$ ----- We get $$z = -6$$
Next plug it here $$y = z - 2$$ ----- We get $$y = -8$$

Now most importantly check $$y = -8$$ in the first equation
Plug it here $$x^2 = y + 5$$
$$x^2 = -8 + 5$$
$$x^2 = 3$$

Do u see now that we cannot take x = -3? Because it does not satisfy all the equations.

Hope it helps!
M05-28 &nbs [#permalink] 16 Nov 2018, 02:05

Go to page   Previous    1   2   [ 25 posts ]

Display posts from previous: Sort by

# M05-28

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.