Given that x and y are in inverse proportion, and $y_{1}$ , $y_{2}$ are values of y corresponding to the values $x_{1}, x_{2}$ of x respectively. Which of the following is correct?

$x_{1}$ $y_{2}$ = $x_{2}$ $y_{1}$

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$x_{1}$ $y_{1}$ = $x_{2}$ $y_{2}$

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$x_{1}$ $x_{2}$ = $y_{1}$ $y_{2}$

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$\frac{x_{1}}{y_{1}}$ = $\frac{x_{2}}{y_{2}}$

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Solution

The correct option is B
$x_{1}$ $y_{1}$ = $x_{2}$ $y_{2}$

If x and y are in inverse proportion, and $y_{1}$ , $y_{2}$ are values of y corresponding to the values $x_{1}$ , $x_{2}$ of x respectively, then

$x_{1}$ $y_{1}$ = $x_{2}$ $y_{2}$

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Similar questions

Q. Given that x and y are in inverse proportion, and

y_{1}, y_{2}

are values of y corresponding to the values

x_{1}, x_{2}

of x, respectively. Which of the following is correct?

Given that x and y are in direct proportion, and $y_{1}, y_{2}$ are values of $y$ corresponding to the values $x_{1}, x_{2}$ of $x$ respectively. Which of the following is correct?

Given that x and y are in inverse proportion, and $y_{1}$ $y_{2}$ are values of $y$ corresponding to the values $x_{1}$ $x_{2}$ of $x$ respectively. Which of the following is correct?

Q. If

x

and

y

are inversely proportional, find the values of

X_{1}, X_{2}

and

Y_{1}, Y_{2}

in the table given below:

$x$	$8$	$X_{1}$	$16$	$X_{2}$	$80$
$y$	$Y_{1}$	$4$	$5$	$2$	$Y_{2}$

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Given that x and y are in inverse proportion, and y1, y2 are values of y corresponding to the values x1,x2 of x respectively. Which of the following is correct?

The correct option is B x1y1=x2y2 If x and y are in inverse proportion, and y1, y2are values of y corresponding to the values x1, x2 of x respectively, then x1y1=x2y2

Given that x and y are in inverse proportion, and $y_{1}$ , $y_{2}$ are values of y corresponding to the values $x_{1}, x_{2}$ of x respectively. Which of the following is correct?

The correct option is B
$x_{1}$ $y_{1}$ = $x_{2}$ $y_{2}$

If x and y are in inverse proportion, and $y_{1}$ , $y_{2}$ are values of y corresponding to the values $x_{1}$ , $x_{2}$ of x respectively, then

$x_{1}$ $y_{1}$ = $x_{2}$ $y_{2}$