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The given pair of values forms an inverse variation. Find the missing value:
\[\left( {2.6,4.5} \right)\], \[\left( {x,6.3} \right)\]

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Answer
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Hint: It is given in the question that the given pair of value forms an inverse variation. We will form an equation using the first variable, second variable and constant using the concept of inverse variation. We will use the first pair of values to get the value of the constant. This will give us a general equation including the variables. Then we will substitute the value of the other variable from the second pair to get the value of the missing number.

Complete step-by-step answer:
The given pair forms an inverse variation. So, the first variable i.e. \[x\] varies inversely as the second variable i.e. \[y\] and also when the value of \[x\] increases, then the value of \[y\] decreases.
Also, the product of \[x\] and \[y\] is constant.
\[xy = k\] ……….. \[\left( 1 \right)\]
Now, we will consider the value \[\left( {2.6,4.5} \right)\].
We will substitute the value of \[x\] with 2.6 and \[y\] with 4.5 in equation \[\left( 1 \right)\].
\[ \Rightarrow 2.6 \times 4.5 = k\]
On multiplying the numbers, we get
\[ \Rightarrow 11.7 = k\]
Thus, the general equation becomes;
\[ \Rightarrow xy = 11.7\] ……… \[\left( 2 \right)\]
Now, we will consider the values \[\left( {x,6.3} \right)\].
We will substitute the value of \[y\] with 6.3 in equation \[\left( 2 \right)\].
\[x \times 6.3 = 11.7\]
Dividing both sides by 6.3, we get,
\[ \Rightarrow \dfrac{{x \times 6.3}}{{6.3}} = \dfrac{{11.7}}{{6.3}}\]
On further simplification, we get
\[ \Rightarrow x = 1.857\]
Hence, the value of \[x\] is equal to 1.857.
Hence, the required missing number is equal to 1.857.

Note: Inverse variation means the first variable varies inversely as the second variable. Also, when the value of the first variable increases, then the value of the second variable decreases and vice versa. We will use the fact that the product of both the variables will be constant. We have used this property to get the value of the missing variable.