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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. \(y = 4\)

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.

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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I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

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, 15:07

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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. \(y = 4\)

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.

Answer: D.

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

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.

Answer: D.

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

Please clarify

Thanks.

Given: \(x^2 = y + 5\), \(y = z - 2\) and \(z = 2x\).

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, 03:47

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).

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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, 09:44

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).

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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, 07:08

1

This post received KUDOS

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

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, 10:34

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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