How do you write the direct variation equation if y varies directly with x and y=10 when x=2?
Answer
579.9k+ views
Hint: This type of problem is based on the concept of proportionality. First, we have to consider y varies directly with x, which means \[y\text{ }\propto \text{ }x\]. Here, \[\propto \] is proportional. Then, substitute a proportionality constant k, that is y=kx. And we have been given that y=10 when x=2. On substituting this value, we get the proportionality constant k and thus find the variation equation.
Complete step by step answer:
According to the question, we are asked to find the direct variation equation if y varies directly with x and y=10 when x=2.
We have been given that y varies directly with x.
We first have to consider that y varies directly with x.
This means that y is directly proportional to x.
\[y\text{ }\propto \text{ }x\] ------(1)
To remove the proportionality, we can substitute a proportionality constant to the equation (1).
We get,
\[y=kx\] ------(2)
Here, k is the proportionality constant.
And we have been given that y=10 when x=2.
Let us now substitute these values in the equation (2).
We get,
10=k(2)
But we know that 10=2x5.
Therefore, we get
2x5=k(2)
Let us now divide 2 on both the sides of the obtained equation.
We get,
\[\dfrac{2\times 5}{2}=k\times \dfrac{2}{2}\]
On further simplification, we get,
\[5=k\times 1\]
\[\Rightarrow 5=k\]
\[\therefore k=5\]
We have now found the value of the proportionality constant k.
On substituting the value of k in equation (2), we get
\[y=5x\]
Therefore, the direct variation equation is \[y=5x\].
Hence, the direct variation equation if y varies directly with x and y=10 when x=2 is \[y=5x\].
Note: Whenever you get this type of problem, we should always try to make the necessary calculations in the given equation to get the final answer. We should avoid calculation mistakes based on sign conventions. We should not forget to put proportionality constant. We may even be asked to find the value of x when y=5 using the direct variation equation.
Complete step by step answer:
According to the question, we are asked to find the direct variation equation if y varies directly with x and y=10 when x=2.
We have been given that y varies directly with x.
We first have to consider that y varies directly with x.
This means that y is directly proportional to x.
\[y\text{ }\propto \text{ }x\] ------(1)
To remove the proportionality, we can substitute a proportionality constant to the equation (1).
We get,
\[y=kx\] ------(2)
Here, k is the proportionality constant.
And we have been given that y=10 when x=2.
Let us now substitute these values in the equation (2).
We get,
10=k(2)
But we know that 10=2x5.
Therefore, we get
2x5=k(2)
Let us now divide 2 on both the sides of the obtained equation.
We get,
\[\dfrac{2\times 5}{2}=k\times \dfrac{2}{2}\]
On further simplification, we get,
\[5=k\times 1\]
\[\Rightarrow 5=k\]
\[\therefore k=5\]
We have now found the value of the proportionality constant k.
On substituting the value of k in equation (2), we get
\[y=5x\]
Therefore, the direct variation equation is \[y=5x\].
Hence, the direct variation equation if y varies directly with x and y=10 when x=2 is \[y=5x\].
Note: Whenever you get this type of problem, we should always try to make the necessary calculations in the given equation to get the final answer. We should avoid calculation mistakes based on sign conventions. We should not forget to put proportionality constant. We may even be asked to find the value of x when y=5 using the direct variation equation.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

When guru nanak ji was born in 1469 AD according to class 11 social studies CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Super glue is a polymer of ATerephthalic acid chloride class 11 chemistry CBSE

Cornea transplant in humans is almost never rejected class 11 biology NEET_UG

