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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, 09:11
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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, 09:33
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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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 > \(x^3 + y^2 + z=1+162=13\), 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.
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Re: Divisibility by 7 [#permalink]
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13 Jun 2010, 10:04
excellent explanation..thanks. kudos



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Re: Divisibility by 7 [#permalink]
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21 Jul 2010, 09:21
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, 09:33



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Re: Arithmetic [#permalink]
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07 Jan 2012, 03:37
X^2=Y+5 Y=Z2=2X2 Z=2X X^2=2X=3 X^22X3=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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Re: Arithmetic [#permalink]
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07 Jan 2012, 15:39
Rephrase the stem: x^2 = y + 5 x^2  y + 5 = 0 x^2 (z2) + 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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If x^2 = y + 5, y = z  2 and z = 2x, is x^3 + y^2 + z [#permalink]
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Updated on: 10 Jul 2013, 02:29
If x^2 = y + 5, y = z  2 and z = 2x, is x^3 + y^2 + z divisible by 7?
(1) x > 0 (2) y = 4
Originally posted by keiraria on 29 Apr 2012, 02:29.
Last edited by Bunuel on 10 Jul 2013, 02:29, edited 2 times in total.
Edited the question



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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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29 Apr 2012, 05:24
keiraria wrote: If x^2 = y + 5, y = z  2 and z = 2x, is x^3 + y^2 + z divisible by 7?
(1) x>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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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 > \(x^3 + y^2 + z=1+162=13\), 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.
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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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22 Sep 2012, 10:28
Bunuel wrote: keiraria wrote: If x^2 = y + 5, y = z  2 and z = 2x, is x^3 + y^2 + z divisible by 7?
(1) x>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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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 > \(x^3 + y^2 + z=1+162=13\), 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. Your explanation is good but i want to know where am i making a mistake. z=2x y= 2x2 x^2 = 2x+3 \(x^3 + y^2 +z = x(2x+3) + (2x2)^2 + 2x = 6x^2 3x +4\) Why cant we use this equation to solve the question.
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Re: Divisibility by 7 [#permalink]
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27 Apr 2014, 06:17
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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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 > \(x^3 + y^2 + z=1+162=13\), 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.



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Re: Divisibility by 7 [#permalink]
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28 Apr 2014, 02:41
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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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 > \(x^3 + y^2 + z=1+162=13\), 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 \(z = 2x\). (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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21 May 2015, 10:16
If x^2 = y + 5, y = z  2 and z = 2x, is x^3 + y^2 + z divisible by 7?
What is provided to us: z = 2x and y = z  2 => y = 2x 2. and x^2 = y + 5 => x^2 = 2x 2 + 5 => x^2  2x  3 = 0. Solving this Quad. Eqn. we get x = 1 or x = 3. So, following sets for possible values of x,y and z :: (x=1,y=4,z=2) OR (x=3,y=4,z=6)
Now, (1) x > 0 if x > 0, then we will have x =3, y = 4 and z = 6. And we can easily evaluate the expression: x^3 + y^2 + z = 49. So Yes divisible by 7. So, Sufficient. (2) y = 4 Again we have, x =3, y = 4 and z = 6. And x^3 + y^2 + z = 49. So Yes divisible by 7. So, Sufficient. Hence, Ans:: 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, 07:08
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
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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, 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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