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This is one of our many quizzes on Direct and inverse variation. Click on this to access more such competitive quizzes on Ratio and Proportions.

Direct and inverse variation Quiz 2

mathematics

 If $x = 40$ and $y = 80$, then $x$ and $y$ are:

 Use proportionality constant to solve this question.

 Given, $x = 40$ and $y = 80$ $ \\ $ We can see that, $80 = 2 \times 40 \\ \Rightarrow y = 2x \\ $ Hence, $y$ is directly proportional to $x$ i.e., $ \\ y \propto x \\ $ In which 2 is proportionality constant of the given terms.

 Solve using Direct variation $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/cf8087b5-032b-405d-812d-69e36d77d20b8351748736891657762.png"/>

 Apply direct variation i.e.,$y = kx$ and solve to find the variation.

 For Direct variation we have, $y = kx$, hence applying this let us find the variation i.e., just take any pair from the given table and substitute in the equation as: $ y = kx$ $ \\ $ Let us consider the first term substitute in the equation: $4 = k2$ $ \\ $ Now to solve for $k$ as: $\dfrac{4}{2} = k \\ \Rightarrow 2 = k \\ $ Therefore, we also consider the other terms of $x$ and $y$ and solve for direct variation we get the value of $k$ as $2$.

 An industry uses $8$ gallons of gasoline to run $290$ gallons of machines. How much gasoline will the industry require to run $400$ machines? To solve, use the direct variation method.

 Use direct variation with respect to the quantity of gasoline given.

 Given, $8$ gallons of gasoline $ \to $ $290$ machines. $ \\ $ To find the constant variation of the equation, $k = \dfrac{y}{x}$ $ \\ \Rightarrow k = \dfrac{{290}}{8} \\ \Rightarrow k = 36.25 \\ $ As, it is direct variation, we know that $y = kx$, $y = 36.25x$ $ \\ $ Let, $x$ be the amount of gasoline for $400$ machines, $400 = 36.25x$ $ \\ \Rightarrow x = \dfrac{{400}}{{36.25}} \\ \Rightarrow 11.03 = x \\ $ Therefore, $11.03$ gallons of gasoline is required to on $400$ machines.

 In Direct variation, $y$ is directly proportional to $\sqrt x $, if $y = 70$ and $x = 49$, then find the value of $y$ when the value of $x = 4$.

 Apply Direct variation i.e., $y = kx$, and find the equation to get the value of  $y$.

 W.K.T for Direct variation we have $y = kx$, $ \\ $ Then $ y = k\sqrt x $ $ \\ $ Now, substitute the value of $x$ and $y$ we get: $ \\ 70 = k\sqrt {49} \\ \Rightarrow 70 = k \times 7 \\ \Rightarrow \dfrac{{70}}{7} = k \\ \Rightarrow 10 = k \\ $ Now, substitute the value of $k$ in the obtained equation i.e., $ \\ y = k\sqrt x \\ \Rightarrow y = 10\sqrt 4 \\ \Rightarrow y = 10 \times 2 \\ \Rightarrow y = 20$

 If there are $15$ sheets of thick paper, each weighing $50$ grams, how many thick sheets would there be if the paper weighed $2500$ grammes?

 Use the direct proportion equation concept to solve this question.

 Given, $15$ sheets of thick paper$ \to $ $50$ grams $ \\ $ So, let us consider the number of sheets as $x$, hence now we need to use direct variation concept to find the value of $x$: $ \\ \dfrac{{15}}{{50}} = \dfrac{x}{{2500}} \\ \Rightarrow x = \dfrac{{\left( {15 \times 2500} \right)}}{{50}} \\ \Rightarrow x = \dfrac{{37500}}{{50}} \\ \Rightarrow x = 750 \\ $ Therefore, $750$ sheets of the thick paper would weigh if the paper is $2500$ grams.