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Divisible by 7

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Source: — Data Sufficiency |
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by rijul007 » Sat Oct 22, 2011 11:44 am
x^2=y+5
y= z-2
z=2x


x^2 = 2x-2+5 = 2x+3
x^2-2x-3=0
(x+1)(x-3) = 0

x= -1 or 3

Statement 1
x>0

therefore the value of x = 3

x^3+y^2+z
=>x^3+(2x-2)^2 +2x
=>x^3+4x^2-8x+4+2x
=>x^3+4x^2-6x+2x
putting value of x
=>27+36-18+6
=>51

Statement 1 alone is sufficient


Statement 2
y=4

y=z+4
z=6
z=2x=6
x=3

Even Statement 2 alone is sufficient


Both the statements alone are sufficient.
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by vaibhavgupta » Sat Oct 22, 2011 12:29 pm
GmatKiss 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
IMO D
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