If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z : GMAT Data Sufficiency (DS)
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# If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z

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If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]

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13 Jun 2010, 08:11
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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) $$y = 4$$
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Re: Divisibility by 7 [#permalink]

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13 Jun 2010, 08:33
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study 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. $$y = 4$$

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:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$

$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$

$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.

Hope it helps.
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Re: Divisibility by 7 [#permalink]

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13 Jun 2010, 09:04
excellent explanation..thanks. kudos
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Re: Divisibility by 7 [#permalink]

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21 Jul 2010, 08:21
shouldn't the answer be B? when x=1 or x=2, the answer is not divisible by 7, while when x=3, it is. Hence A is insufficient.
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Re: Divisibility by 7 [#permalink]

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21 Jul 2010, 08:33
anandnat wrote:
shouldn't the answer be B? when x=1 or x=2, the answer is not divisible by 7, while when x=3, it is. Hence A is insufficient.

Pleas read the solution above.

What are the values of $$z$$ and $$y$$ when $$x=1$$ or $$x=2$$? What is the value of $$x^3+y^2+z$$ when $$x=1$$ or $$x=2$$?

$$x=1$$ and $$x=2$$ do not satisfy the system of equations given in the stem, hence are not the valid solutions for $$x$$.
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07 Jan 2012, 02:10
Can anyone help me in solving the below problem in 2 min
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07 Jan 2012, 02:37
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X^2=Y+5
Y=Z-2=2X-2
Z=2X

X^2=2X=3
X^2-2X-3=0
X=3, -1
Y=4,-4
Z=6,-2
X^3+Y^2+Z

1.
X>0
X=3, Y=4, Z=6
X^3+Y^2+Z= Some value we don't care about but can determine whether or not the value is divisible by 7.

Sufficient

2.
y=4, X=3, Z=6
X^3+Y^2+Z= Some value we don't care about but can determine whether or not the value is divisible by 7.

Sufficient

Hence D
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07 Jan 2012, 02:52
X^2=Y+5
Y=Z-2=2X-2
Z=2X

X^2=2X=3
X^2-2X-3=0
X=3, -1
Y=4,-4
Z=6,-2
X^3+Y^2+Z

1.
X>0
X=3, Y=4, Z=6
X^3+Y^2+Z= Some value we don't care about but can determine whether or not the value is divisible by 7.

Sufficient

2.
y=4, X=3, Z=6
X^3+Y^2+Z= Some value we don't care about but can determine whether or not the value is divisible by 7.

Sufficient

Hence D

Thanks i made things complex........
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07 Jan 2012, 14:39
Rephrase the stem:
x^2 = y + 5
x^2 - y + 5 = 0
x^2 -(z-2) + 5 = 0
x^2 - z - 3 = 0
x^2 - 2x - 3 = 0
x = 3, -1
Substituting gives z = 6, -2 and y = 4, -4

Notice that both statements ask for same information, that is 1) x>0 -> x=3 and corresponding y=4 and z=6. 2) y=4 means x=3 and z=6

Substituting these values in the stem gives us 51, which is not div by 7. So, both suff.

D
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Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]

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21 Sep 2013, 14:07
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Re: Divisibility by 7 [#permalink]

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27 Apr 2014, 05:17
Bunuel wrote:
study 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. $$y = 4$$

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:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$

$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$

$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.

Hope it helps.

Hi bunnel,

In second statement y=4 if we replace we found x^2 = 4+5 = 9

so can we write x= +3 and -3. if we replace x= -3 then sum will be -24+16+6 = -1. then how it can be divisible by 7

Thanks.
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Re: Divisibility by 7 [#permalink]

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28 Apr 2014, 01:41
PathFinder007 wrote:
Bunuel wrote:
study 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. $$y = 4$$

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:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$

$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$

$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.

Hope it helps.

Hi bunnel,

In second statement y=4 if we replace we found x^2 = 4+5 = 9

so can we write x= +3 and -3. if we replace x= -3 then sum will be -24+16+6 = -1. then how it can be divisible by 7

Thanks.

Given: $$x^2 = y + 5$$, $$y = z - 2$$ and $$z = 2x$$.

(2) $$y = 4$$:

$$y = z - 2$$ --> $$z=6$$;
$$z = 2x$$ --> $$x=3$$.

So, x cannot be -3.

Does this make sense?
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Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]

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23 Jul 2015, 02:47
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Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]

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30 Jul 2016, 08:44
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Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]

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21 Aug 2016, 06:08
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Remember => we dont need to solve this scary looking question.
we need the value of x,y,z
now solving the system of three equations its clear that xcan be 3 and -1
now statement 1=> x>0 => sufficient as x=3 hence we can get the value of y and z
statement 2 => again if y=4 => x=3 and z=6 so we have the values of x,y,z

Note => divisiblity by 7 can be replaced by 13 or 5 or 17 or 20 even 138483461349273794739939738972 that would still be immaterial here. we just need the number a.k.a the values of x,y,z

Smash that D

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Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]

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21 Aug 2016, 09:34
stonecold wrote:
Remember => we dont need to solve this scary looking question.
we need the value of x,y,z
now solving the system of three equations its clear that xcan be 3 and -1
now statement 1=> x>0 => sufficient as x=3 hence we can get the value of y and z
statement 2 => again if y=4 => x=3 and z=6 so we have the values of x,y,z

Note => divisiblity by 7 can be replaced by 13 or 5 or 17 or 20 even 138483461349273794739939738972 that would still be immaterial here. we just need the number a.k.a the values of x,y,z

Smash that D

Stone Cold

I also followed the same approach and got x =3 or -1. I never looked for the scary equation and both the options gave me x=3 and hence will always give a unique value of the scary equation.
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Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z   [#permalink] 21 Aug 2016, 09:34
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