Question

If $xy = p$, $xz = p_{}^2$ and $yz = p_{}^3$, also $x + y + z = 13$ and $x_{}^2 + y_{}^2 + z_{}^2 = 91$ then $\dfrac{z}{y}$ is equal to :
A. $\left( {\text{1}} \right){\text{3}}$
B. $\left( {\text{2}} \right)\dfrac{{\text{7}}}{{\text{3}}}$
C. $\left( 3 \right)13$
D. $\left( 4 \right)\dfrac{{13}}{3}$

Answer 1

Hint: Here we expand the given terms by using the formula and doing some substitution in the given data and find the value.
Finally we get the answer.

Formula used: $(a + b + c)_{}^2 = a_{}^2 + b_{}^2 + c_{}^2 + 2ab + 2bc + 2ca$

Complete step-by-step answer:
It is given that $x + y + z = 13$
Now, by squaring both the sides of the above equation we get-
$(x + y + z)_{}^2 = 13_{}^2$
Here we have to apply the formula $(a + b + c)_{}^2 = a_{}^2 + b_{}^2 + c_{}^2 + 2ab + 2bc + 2ca$ in the above equation and we get-
$x_{}^2 + y_{}^2 + z_{}^2 + 2xy + 2yz + 2zx = 169$
Taking $2$ as common we get-
$x_{}^2 + y_{}^2 + z_{}^{} + 2(xy + yz + zx) = 169$
Since it is mentioned in the question that $xy = p$, $xz = p_{}^2$, $yz = p_{}^3$, we can put those values in the above equation and we get
$x_{}^2 + y_{}^2 + z_{}^2 + 2(p + p_{}^2 + p_{}^3)$ $ = 169$
Now it is also given in the question that $x_{}^2 + y_{}^2 + z_{}^2 = 91$, therefore we can write –
$91 + 2(p + p_{}^2 + p_{}^3) = 169$
Subtracting $91$ from $169$ after taking it on the right side we get-
$2(p + p_{}^2 + p_{}^3) = 169 - 91$
$2(p + p_{}^2 + p_{}^3) = 78$
Now we have to divide $78$ by $2$ and we get-
$(p + p_{}^2 + p_{}^3) = 39$
Now we have to break $39$ in such a way that we get one number, the next one is the square of the first number and the third one is the cube of the first number.
So we can write-
$(p + p_{}^2 + p_{}^3) = 3 + 9 + 27$
Therefore, $(p + p_{}^2 + p_{}^3) = 3 + 3_{}^2 + 3_{}^3$
So value of $p = 3$
Now by dividing $xz = p_{}^2$ by $xy = p$ we get-
We can write $\dfrac{z}{y} = $ $\dfrac{{xz}}{{xy}}$ $ = \dfrac{{p_{}^2}}{{p_{}^{}}} = p$ $ = 3$
Hence value of $\dfrac{z}{y} = 3$

So, the correct answer is “Option A”.

Note: While solving this question the main critical point is to break the number $39$ in such a way that the third number is the cube of first number and second number is the square of first number and here is comes the chance of mistake so try to focus on the concept.