Question : $\text { If } x^2+y^2+z^2=x y+y z+z x \text { and } x=1 \text {, then find the value of } \frac{10 x^4+5 y^4+7 z^4}{13 x^2 y^2+6 y^2 z^2+3 z^2 x^2}$.
Option 1: 2
Option 2: 0
Option 3: –1
Option 4: 1
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Correct Answer: 1
Solution : $x^2+y^2+z^2=x y+y z+z x$ $⇒\frac{1}{2}[(x-y)^2+(y-z)^2+(z-x)^2] = 0$ $⇒ x= y, y=z,$ and $z=x$ As $x =1$, so $y=z=1$ So, $\frac{10 x^4+5 y^4+7 z^4}{13 x^2 y^2+6 y^2 z^2+3 z^2 x^2}$ $= \frac{10(1)^4+5(1)^4+7(1)^4}{13(1)^2(1)^2+6(1)^2(1)^2+3(1)^2(1)^2}$ $=\frac{22}{22}$ $=1$ Hence, the correct answer is 1.
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Question : If $x+y+z=0$ and $x^2+y^2+z^2=40$, then what is the value of $x y+y z+z x?$
Option 1: –20
Option 2: 5
Option 3: –5
Option 4: –10
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Option 1: 1
Option 3: 2
Option 4: 3
Question : If $\frac{(x+y)}{z}=2$, then what is the value of $[\frac{y}{(y-z)}+\frac{x}{(x-z)}]?$
Option 1: $0$
Option 2: $1$
Option 3: $2$
Option 4: $–1$
Question : If $\frac{1}{x+\frac{1}{y+\frac{2}{z+\frac{1}{4}}}}=\frac{29}{79}$, where x, y, and z are natural numbers, then the value of $(2 x+3 y-z)$ is:
Option 1: 0
Option 2: 4
Option 3: 1
Option 4: 2
Question : Let $x, y, z$ be fractions such that $x<y<z$. If $z$ is divided by $x$, the result is $\frac{5}{2}$, which exceeds $y$ by $\frac{7}{4}$. If $x+y+z=1 \frac{11}{12}$, then the ratio of $(z-x):(y-x)$ is:
Option 1: 6 : 5
Option 2: 9 : 5
Option 3: 5 : 6
Option 4: 5 : 9
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