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# M05-28

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Math Expert
Joined: 02 Sep 2009
Posts: 59587
Re: M05-28  [#permalink]

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17 Aug 2018, 01:17
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.

Answer: D

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.
Can you please clarify my answer.

x = -3 is not a solution.

y = 4, then z = 6 (from y = z - 2) and x = 3 (from z = 2x).
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M05-28  [#permalink]

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24 Oct 2018, 23:23
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: 59587
Re: M05-28  [#permalink]

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24 Oct 2018, 23:44
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.
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Re: M05-28  [#permalink]

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25 Oct 2018, 21:22
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?.
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M05-28  [#permalink]

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16 Nov 2018, 03:05
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.

Answer: D

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.
Can you please clarify my answer.

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!
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Re: M05-28  [#permalink]

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29 Jan 2019, 04:34
Bunuel

I tried to plug in some numbers as follows:
X=3, Y=4 and Z=6.....Solving the equation the sum equals 49 which is divisible by 7
X=5, Y=8 and Z=10.....Solving the equation the sum equals 199 which is not divisible by 7

How is the correct answer choice D in that case?
Math Expert
Joined: 02 Sep 2009
Posts: 59587
Re: M05-28  [#permalink]

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29 Jan 2019, 04:42
Megha1119 wrote:
Bunuel

I tried to plug in some numbers as follows:
X=3, Y=4 and Z=6.....Solving the equation the sum equals 49 which is divisible by 7
X=5, Y=8 and Z=10.....Solving the equation the sum equals 199 which is not divisible by 7

How is the correct answer choice D in that case?

First of all X=5, Y=8 and Z=10 does not satisfy $$x^2 = y + 5$$.

Also, (2) says that $$y = 4$$
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Re: M05-28  [#permalink]

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09 Feb 2019, 03:39
Hi,

I don't understand why statement A alone is sufficient. I simply tried different options for X>0 and found Y and Z by substitution. Why must I solve for X before and thus restrict the field to -1 and 3?

Many thanks in advance!
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Re: M05-28  [#permalink]

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09 Feb 2019, 04:02
Bunuel wrote:
If $$x^2 = y + 5$$, $$y = z - 2$$ and $$z = 2x$$, is $$x^3 + y^2 + z$$ divisible by 7?

(1) $$x \gt 0$$

(2) $$y = 4$$

The whole point was to utilize the given

$$x^2 = y + 5$$, $$y = z - 2$$ and $$z = 2x$$ when utilized, will give x^2 -2x-3=0

Which on solving will give 2 solutions for x i.e. -1 and 3

which when put back in y 2x -2 will give y = -4 or y = 4 and give z = -2 or z =6

from 1, x > 0

one case will work here
y = 4 & z = 6 , x = 3

$$x^3 + y^2 + z$$ divisible by 7
Sufficient

from 2, y =4
one case will work here
y = 4 & z = 6 , x = 3

$$x^3 + y^2 + z$$ divisible by 7
Sufficient

D
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Re: M05-28  [#permalink]

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09 Feb 2019, 04:07
pero88 wrote:
Hi,

I don't understand why statement A alone is sufficient. I simply tried different options for X>0 and found Y and Z by substitution. Why must I solve for X before and thus restrict the field to -1 and 3?

Many thanks in advance!

Hey pero88

If you were able to solve without looking for any values for x in A, that is really great

But since x is given as x^2, we can directly say that it will have 2 solutions, which will satisfy that particular expression.

Now when you know this, its difficult to guess and find the solutions rather than just utilize what is given, and try utilizing it in our benefit

x^2 in terms of y+5, wont be used in any way, but if we go ahead in the question, we see that we can substitute for y and find an expression in only one unknown value.
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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
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20 Feb 2019, 19:07
I think this is a high-quality question and I agree with explanation.
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Re: M05-28  [#permalink]

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02 Jul 2019, 16:26
This is such a sneaky, but good question!

You can solve for all variables based on the stem, there are going to be two conditions: x>0 or x<0 : that would alter the solutions as follows:

x=3 or x=-1
if x =3
y = 4
z = 6

if x=-1
y=-4
z=-2

(1) X> 0 therefore we can derive all other values to learn that the numerator = 49 which is divis by 7
sufficient

(2) y= 4
this can only happen when x>0. Thus x=3 and z = 6 -->sufficient
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M05-28  [#permalink]

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14 Sep 2019, 17:07
Bunuel wrote:
If $$x^2 = y + 5$$, $$y = z - 2$$ and $$z = 2x$$, is $$x^3 + y^2 + z$$ divisible by 7?

(1) $$x \gt 0$$

(2) $$y = 4$$

Another approach - It is a good idea in general to assess what you may be able to piece together from the given information before you jump in further by taking on either statement. Here, the three given equations each contain just two unknowns, so substitution is a straightforward process. As soon as I saw a quadratic ($$x^2$$), I thought that x would likely have two solutions. Thus, I looked to get the first equation in terms of x only. This led me from the second to the third equation and back again pretty quickly: if $$z = 2x$$, then, backtracking a bit, $$y = (2x) - 2$$, and, reaching back a little further, $$x^2 = (2x - 2) + 5$$, or $$x^2 - 2x -3 = 0$$. This is a factorable quadratic, with binomial roots of (x - 3) and (x + 1). The solutions will be 3 or -1. With all the hard work out of the way, I took on Statement (1) first, rather than what I consider to be the easier statement in (2).

Why would it be important that x would be greater than 0? Because that would lead only to the conclusion that x = 3. It is thus not even necessary to test what could happen if x were to equal -1. In fact, I did not even bother to see what would happen when I substituted 3 for x in the modified quadratic, since I knew that regardless of the outcome, it would be possible to tell whether the eventual answer would be divisible by 7. Statement (1) had to be SUFFICIENT, narrowing the answer choices down to (A) or (D).

Statement (2) requires little effort to justify, since all three given equations must be considered. The middle equation would tell us that z must equal 6, and then the third equation that x must equal 3. At that point, the original quadratic would no longer have two justifiable solutions for x, and we would also be able to answer the question. Thus, (D) had to be the correct response, and both the effort and time it took to arrive at such a conclusion were minimal.

In Yes/No questions, all you need to be able to determine is whether there is enough evidence to reach a certain conclusion. It may not be necessary to actually work the numbers out.

I sometimes call myself "the lazy mathematician" in my tutoring sessions. I love math, but I would rather work less to get a correct solution than put in more effort and get... the same thing. I have adopted this approach not only because of some innate tendency, but also no doubt due to knowing a brilliant mathematician in college, a guy who tied for 24th in the Putnam collegiate math competition as a sophomore, and who later finished first in the nation in another year. I discovered (via a friendship that extended into chess club) that his extraordinary abilities stemmed not from seeing something in the aether that no one else could see, but from finding ways to break down complex matters into simpler pieces that he could readily put together to lead to a solution.

Good luck with your studies.

- Andrew
M05-28   [#permalink] 14 Sep 2019, 17:07

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