M05- Question 28

Question

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

Please provide explanation as well

BarneyStinson · Answer

Answer must be B.

Stmt 1) simplifying the given equation in terms of x, 
x^3 + 4*x^2 - 14*x + 16.

for any value of x > 0, say x = 1, 2, it may or may not be divisible by 7. Insufficient.

Stmt 2) simplifying the above equation in terms of y,
(sqrt(y + 5))  +  .

Substituting y = 4, (+/-3)(-5) + 25, no matter 3 is positive or negative, it is still not divisible by 7. Standard answer, confirmed not divisible - sufficient.

ichha148 · Answer

OA is mentioned as D

Bunuel · Answer

We have system of equations with three 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.

jatt86 · Answer

d

x=-1 or x=3
x>0
x=3

ichha148 · Answer

Thanks a lot Bunuel , as always your explanation is great and lucid

+1 from my side

M05- Question 28

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