Latest cocubes Aptitude Question SOLUTION: If x1/p = y1/q = z1/r and y2 = xz, what is the value of p+r?

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If x1/p = y1/q = z1/r and y2 = xz, what is the value of p+r?

Asked In cocubes Shilpi Shandilya (9 years ago)

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Gaurav Priyadarshi  (9 years ago)

let x^(1/p) = y^(1/q) = z^(1/r) = k
therefore, x = k^p
y = k^q
z = k^r
now,
y^2 = xz
=> k^2q = k^p.k^r
=> 2q=p+r......ans

Is this solution Helpfull? Yes (43) | No (3)  

anonymous  (9 years ago)

according to question, x^(q/p)=y and also z^(q/r)=y..so put this in LHS
lets come to RHS ...we can say that x^(r/p) =z. put this in RHS....
now u will get x^(q/p).z^(q/r)=x.z ...in place of z put above expression containing z
so x^(q/p).x^(q/p)=x.x^(r/p)..( do it carefully)
equate powers:- 2q/p=1+r/p
u will get 2q=p+r

Is this solution Helpfull? Yes (4) | No   

Urumadla VaraPrasad Reddy  (9 years ago)

x1/p1=y1/q1==>> x1/y1=p/q==>>p=x1, q=y1
similarly r=z1
p+r=x1+z1..
this is the answer as there r no values given

Is this solution Helpfull? Yes (1) | No (3)  

gaurav  (6 years ago)

2q
by taking log of the equations given,it becomes
2q=p+r

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Mohan Avinash  (7 years ago)

let x^(1/p) = y^(1/q) = z^(1/r) = k
therefore, x = k^p;
y = k^q;
z = k^r;
now,
y^2 = xz;
=> k^2q = k^p.k^r;
=> 2q=p+r ans
Priya "Use ';' at the end of the statement "

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