Question : $x$ varies inversely as the square of $y$. Given that $y=2$ for $x=1$, the value of $x$ for $y=6$ will be equal to:
Option 1: 3
Option 2: 9
Option 3: $\frac{1}{3}$
Option 4: $\frac{1}{9}$
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Correct Answer: $\frac{1}{9}$
Solution : $x$ varies inversely as a square of $y$. $⇒x=\frac{a}{y^2}$, where $a$ is constant At $y=2, x=1$ $⇒1=\frac{a}{2^2}$ $⇒a=4$ If $y=6$, then, $\therefore x=\frac{4}{6^2}=\frac{4}{36}=\frac{1}{9}$ Hence, the correct answer is $\frac{1}{9}$.
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Question : If $\mathrm{X: Y=3: 2}$, then what is the value of $\mathrm{(X-Y):(X+Y) }$?
Option 1: 4 : 5
Option 2: 1 : 5
Option 3: 8 : 9
Option 4: 3 : 7
Question : If $(7 x+y):(7 x–y)=7: 2$, then what is the value of $x: y$?
Option 1: 9 : 35
Option 2: 11 : 42
Option 3: 6 : 41
Option 4: 17 : 52
Question : If $x=-1$, then the value of $\frac{1}{x^{99}}+\frac{1}{x^{98}}+\frac{1}{x^{97}}+\frac{1}{x^{96}}+\frac{1}{x^{95}}+\frac{1}{x^{94}}+\frac{1}{x}-1$ is:
Option 1: 1
Option 2: 0
Option 3: –2
Option 4: –1
Question : If $A=\frac{2}{3} \div \frac{4}{9}, B=\frac{6}{11} \times\frac{33}{12}$ and $C=\frac{1}{3} \times \frac{9}{2}$, then what is the value of $A \times B-C$?
Option 1: $\frac{3}{8}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{3}{4}$
Option 4: $\frac{3}{16}$
Question : If (7x + y) : (7x – y) = 7 : 3, then what is the value of x : y?
Option 1: 4 : 3
Option 2: 5 : 14
Option 3: 15 : 6
Option 4: 6 : 15
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