25=(x+y+z)^2=(x+y)+z)^2=x^2+2xy+y^2+z^2+2z(x+y)
less xy+yz+zx=3 multiplied by 2 [spoiler]i.e. 2xy+2z(x+y)=6[/spoiler]

x^2+y^2=19, the greatest value of x is found when y=0 as y^2 returns +ve

hence x^2+0=19 and x=sqroot(19)
c

quantskillsgmat wrote:x,y and z are real numbers such that x+y+z=5 and xy+yz+zx is 3. what can be greatest value of x.
a)5/3
b)13/3
c)root19
d)none

Couple of ways to do this..

https://www.artofproblemsolving.com/Reso ... .proofreed

Most definitely nowhere close to a GMAT problem..

pemdas wrote:25=(x+y+z)^2=(x+y)+z)^2=x^2+2xy+y^2+z^2+2z(x+y)
less xy+yz+zx=3 multiplied by 2 [spoiler]i.e. 2xy+2z(x+y)=6[/spoiler]
x^2+y^2+z^2=19, the greatest value of x is found when y=0 as y^2 returns +ve
hence x^2+0=19 and x=sqroot(19)
c

[/quote]

You have missed the term z^2 (in red above.) The value of (y,z) cannot (0,0) because we already know that the value of xy+yz+zx is 3(which will be 0 if the value of (y,z) cannot (0,0)).

pemdas wrote:25=(x+y+z)^2=(x+y)+z)^2=x^2+2xy+y^2+z^2+2z(x+y)
less xy+yz+zx=3 multiplied by 2 [spoiler]i.e. 2xy+2z(x+y)=6[/spoiler]

x^2+y^2=19, the greatest value of x is found when y=0 as y^2 returns +ve

hence x^2+0=19 and x=sqroot(19)
c

I don't follow you...

(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 25

subtract 2xy + 2yz + 2xz = 6

and get

x^2 + y^2 + z^2 = 19

We can not just set y=z=0 here.

shankar.ashwin wrote:Couple of ways to do this..

https://www.artofproblemsolving.com/Reso ... .proofreed

Most definitely nowhere close to a GMAT problem..

Ah.... interesting site. And, ya, I would have to agree with your assessment there too.

I agree with chieftang. if we set y=z=0 then initial equations will fail. especial the one with xy+yz+zx

user123321

I wont expect this problem to come in GMAT.
in these kind of problems especially if equations are homogeneous like these and asked us to find the max of one variable. then we can assume rest of variables same and proceed. I dont know whether it is true or not but it works most of the times.

x+2y = 5; y(x+2y) = 3
solving both you get y=3 or 1/3
since y can't be 3, we have y=z=1/3
=>x=13/3

comment on this approach are highly welcome
Thanks,
user123321

Although it works in this problem, you wouldn't want to follow it as a generic approach.

If the same question were to ask integer solutions of 'x', 'y' would not be equal to 'z'

Integer solutions would be x = 3, then y = -1 and z = 3 (or) y = 3 and z = -1, which is contrary to your assumption y=z. Infact y(or)z would be equal to x in that case. Actually I think we are wasting time discussing this problem here A legitimate solution would cover topics beyond scope of the GMAT.

Another approach here https://www.algebra.com/algebra/homework ... 49251.html

user123321 wrote:I wont expect this problem to come in GMAT.
in these kind of problems especially if equations are homogeneous like these and asked us to find the max of one variable. then we can assume rest of variables same and proceed. I dont know whether it is true or not but it works most of the times.

x+2y = 5; y(x+2y) = 3
solving both you get y=3 or 1/3
since y can't be 3, we have y=z=1/3
=>x=13/3

comment on this approach are highly welcome
Thanks,
user123321

shankar.ashwin wrote:Actually I think we are wasting time discussing this problem here A legitimate solution would cover topics beyond scope of the GMAT.

+1

Thanks

by attending math forums, i'm used to keep a couple of posts in short-term memory and then link these posts together to derive a use from them

it's amazing how this question right here https://www.beatthegmat.com/integers-t101375.html#440837 is similar to the original question (quoted below). In both cases we select the largest no as possible answer. Actually, yes I missed some z^2 in previous sol. and later reviewed the others sol-s, but all in all it comes nowhere close to what GMAT would test. By applying the methods and approaches discussed for integer question, I am able now to understand that (x+y+z) has arithmetic mean 5/3 and the largest no must be >5/3. This cold be any no, I would select the largest possible, which is 13/3 for the next intake of such tricky question in exam. I don't know why, I am pretty sure such kind of question may show in GMAT but in different suit

BTW, just checked 13/3 is very close to root(19) -> root(169/9) vs. root(19)

quantskillsgmat wrote:x,y and z are real numbers such that x+y+z=5 and xy+yz+zx is 3. what can be greatest value of x.
a)5/3
b)13/3
c)root19
d)none