Question

If \[x\% \] of \[y\] is equal to \[1\% \] of \[z\], \[y\% \] of \[z\] is equal to \[1\% \] of \[x\], and \[z\% \] of \[x\] is equal to \[1\% \] of \[y\], then the value of \[xy + yz + zx\] is
(a) 1 (b) 2 (c) 3 (d) 4

Answer 1

Hint:
Here, we need to find the value of the expression \[xy + yz + zx\]. First, we will use the given information to find the squares of the three numbers. Then, we will use the equations formed and the algebraic identity for the sum of three numbers to form a quadratic equation in terms of \[xy + yz + zx\]. Finally, we will use the quadratic formula to find the roots of the quadratic equation, and hence, the value of the expression \[xy + yz + zx\].

Complete step by step solution:
First, we will find \[x\% \] of \[y\].
Thus, we get
\[x\% {\rm{ of }}y = \dfrac{x}{{100}} \times y = \dfrac{{xy}}{{100}}\]
Now, we will find \[1\% \] of \[z\].
Thus, we get
\[1\% {\rm{ of }}z = \dfrac{1}{{100}} \times z = \dfrac{z}{{100}}\]
It is given that \[x\% \] of \[y\] is equal to \[1\% \] of \[z\].
Thus, we get
\[ \Rightarrow \dfrac{{xy}}{{100}} = \dfrac{z}{{100}}\]
Simplifying the equation, we get
\[ \Rightarrow xy = z\]……….\[\left( 1 \right)\]
Now, we will find \[y\% \] of \[z\].
Thus, we get
\[y\% {\rm{ of }}z = \dfrac{y}{{100}} \times z = \dfrac{{yz}}{{100}}\]
We will find \[1\% \] of \[x\].
Thus, we get
\[1\% {\rm{ of }}x = \dfrac{1}{{100}} \times x = \dfrac{x}{{100}}\]
It is given that \[y\% \] of \[z\] is equal to \[1\% \] of \[x\].
Thus, we get
\[ \Rightarrow \dfrac{{yz}}{{100}} = \dfrac{x}{{100}}\]
Simplifying the equation, we get
\[ \Rightarrow yz = x\]…………..\[\left( 2 \right)\]
Now, we will find \[z\% \] of \[x\].
Thus, we get
\[z\% {\rm{ of }}x = \dfrac{z}{{100}} \times x = \dfrac{{zx}}{{100}}\]
We will find \[1\% \] of \[y\].
Thus, we get
\[1\% {\rm{ of }}y = \dfrac{1}{{100}} \times y = \dfrac{y}{{100}}\]
It is given that \[z\% \] of \[x\] is equal to \[1\% \] of \[y\].
Thus, we get
\[ \Rightarrow \dfrac{{zx}}{{100}} = \dfrac{y}{{100}}\]
Simplifying the equation, we get
\[ \Rightarrow zx = y\]…………\[\left( 3 \right)\]
Now, we will use the three equations to simplify the expressions.
Substituting \[y = zx\] in equation \[\left( 2 \right)\], we get
\[\begin{array}{l} \Rightarrow \left( {zx} \right)z = x\\ \Rightarrow {z^2}x = x\end{array}\]
Dividing both sides by \[x\], we get
\[ \Rightarrow {z^2} = 1\]
Substituting \[y = zx\] in equation \[\left( 1 \right)\], we get
\[\begin{array}{l} \Rightarrow x\left( {zx} \right) = z\\ \Rightarrow {x^2}z = z\end{array}\]
Dividing both sides by \[z\], we get
\[ \Rightarrow {x^2} = 1\]
Substituting \[z = xy\] in equation \[\left( 2 \right)\], we get
\[\begin{array}{l} \Rightarrow y\left( {xy} \right) = x\\ \Rightarrow x{y^2} = x\end{array}\]
Dividing both sides by \[x\], we get
\[ \Rightarrow {y^2} = 1\]
Now, we will use algebraic identity to find the value of the expression \[xy + yz + zx\].
Using the algebraic identity \[{\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right)\], we get
\[ \Rightarrow {\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right)\]
Substituting \[{x^2} = 1\], \[{y^2} = 1\], and \[{z^2} = 1\] in the equation, we get
\[ \Rightarrow {\left( {x + y + z} \right)^2} = 1 + 1 + 1 + 2\left( {xy + yz + zx} \right)\]
Adding the terms in the expression, we get
\[ \Rightarrow {\left( {x + y + z} \right)^2} = 3 + 2\left( {xy + yz + zx} \right)\]
Substituting \[z = xy\], \[x = yz\], and \[y = zx\] from equations \[\left( 1 \right)\], \[\left( 2 \right)\], and \[\left( 3 \right)\] in the equation, we get
\[ \Rightarrow {\left( {xy + yz + zx} \right)^2} = 3 + 2\left( {xy + yz + zx} \right)\]
Let \[\left( {xy + yz + zx} \right) = t\].
Rewriting the equation, we get
\[\begin{array}{l} \Rightarrow {t^2} = 3 + 2t\\ \Rightarrow {t^2} - 2t - 3 = 0\end{array}\]
We can observe that this is a quadratic equation.
We will now use the quadratic formula to find the roots of the above quadratic equation.
First, we will find the value of the discriminant of the equation.
Comparing the equation \[{t^2} - 2t - 3 = 0\] with the standard form of a quadratic equation \[a{x^2} + bx + c = 0\], we get
\[a = 1\], \[b = - 2\], and \[c = - 3\]
Substituting \[a = 1\], \[b = - 2\], and \[c = - 3\] in the formula for discriminant \[D = {b^2} - 4ac\], we get
\[ \Rightarrow D = {\left( { - 2} \right)^2} - 4\left( 1 \right)\left( { - 3} \right)\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow D = 4 + 12\\ \Rightarrow D = 16\end{array}\]
Now, substituting \[a = 1\], \[b = - 2\], and \[D = 16\] in the quadratic formula \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\], we get
\[ \Rightarrow t = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {16} }}{{2 \times 1}}\]
Simplifying the expression, we get
\[ \Rightarrow t = \dfrac{{2 \pm 4}}{2}\]
Factoring out 2 from the numerator, we get
\[\begin{array}{l} \Rightarrow t = \dfrac{{2\left( {1 \pm 2} \right)}}{2}\\ \Rightarrow t = 1 \pm 2\end{array}\]
Therefore, either \[t = 1 + 2\] or \[t = 1 - 2\].
Thus, we get \[t = 3, - 1\].
Substituting \[t = \left( {xy + yz + zx} \right)\], we get
\[xy + yz + zx = 3\]

Thus, the correct option is option (c).

Note:
We have used the algebraic identity \[{\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right)\] in the solution. The square of the sum of three numbers is given by the algebraic identity \[{\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right)\].
We used the quadratic formula to find the roots of the quadratic equation. The quadratic formula states that the roots of a quadratic equation \[a{x^2} + bx + c = 0\] are given by \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\], where \[D\] is the discriminant given by the formula \[D = {b^2} - 4ac\].