
\[\text{If }\dfrac{1}{5}:\dfrac{1}{x}=\dfrac{1}{x}:\dfrac{1}{1.25}\], then the value of x is:
(a) 1.5
(b) 2
(c) 2.5
(d) 3.5
Answer
585.9k+ views
Hint: First convert the given ratio into a fraction. Then solve both the ratios separately for clarity and then equate the results of both the cases to get the relation of x. Then apply basic numerical operations to get the value of x. The value of x is the required result in this case.
Complete step-by-step answer:
We are given the ratio in the question and can be written in the form of:
\[\dfrac{1}{5}:\dfrac{1}{x}=\dfrac{1}{x}:\dfrac{1}{1.25}....\left( i \right)\]
Case 1: Solving the ratio on the left-hand side, we get,
\[LHS=\dfrac{1}{5}:\dfrac{1}{x}\]
We can convert the ratio into a fraction by the relation:
\[a:b=\dfrac{a}{b}=a\times \dfrac{1}{b}\]
By using this, we can write our ratio as follows:
\[LHS=\dfrac{\dfrac{1}{5}}{\dfrac{1}{x}}=\dfrac{1}{5}\times \dfrac{1}{\dfrac{1}{x}}\]
We know the relation of ‘a’ in the fraction which is \[\dfrac{1}{\dfrac{1}{a}}=a\]. By using this, we can write our equation in the form
\[LHS=\dfrac{1}{5}\times x\]
By simplifying, we get the value of the left-hand side as:
\[LHS=\dfrac{x}{5}.....\left( ii \right)\]
Case 2: Solving the ratio on the right-hand side, we get,
\[RHS=\dfrac{1}{x}:\dfrac{1}{1.25}\]
We can convert the ratio into a fraction by the relation:
\[a:b=a\times \dfrac{1}{b}\]
By using this, we get the value of our expression as:
\[RHS=\dfrac{1}{x}\times \dfrac{1}{\dfrac{1}{1.25}}\]
By basic knowledge of the fractions, we have a relation given as \[\dfrac{1}{\dfrac{1}{a}}=a\]
By substituting this, we can write our expression as:
\[RHS=\dfrac{1}{x}\times 1.25\]
By simplifying the above value, we get the value of the RHS as
\[RHS=\dfrac{1.25}{x}....\left( iii \right)\]
By substituting equation (ii) and (iii) in equation (i), we get it as:
\[\dfrac{x}{5}=\dfrac{1.25}{x}\]
By multiplying with x on both the sides, we get the equation as:
\[\dfrac{x}{5}\times x=\dfrac{1.25}{x}\times x\]
By multiplying with 5 on both the sides, we get the equation as
\[\dfrac{x}{5}\times x\times 5=\dfrac{1.25}{x}\times x\times 5\]
By canceling the like terms, we get the equation as
\[{{x}^{2}}=1.25\times 5\]
By simplifying the above equation, we get its value as
\[{{x}^{2}}=6.25\]
By applying the square root on both the sides, we get it as
\[x=\sqrt{6.25}\]
By simplifying the root value, we get the value of x as 2.5. So, the value of x is 2.5 to satisfy the given relation.
Therefore, option (c) is the right answer.
Note: Be careful while taking the ratio and don’t forget the \[\dfrac{1}{\text{term}}\] condition \[\dfrac{1}{\dfrac{1}{\text{term}}}=\text{term}\]. After the ratio, the terms change. Students confuse and write \[\dfrac{1}{x}\] in place of x. It will change the whole result. So, solve each and every step carefully while cross multiplying. Also, don’t forget to take the square root on the right-hand side as well.
Complete step-by-step answer:
We are given the ratio in the question and can be written in the form of:
\[\dfrac{1}{5}:\dfrac{1}{x}=\dfrac{1}{x}:\dfrac{1}{1.25}....\left( i \right)\]
Case 1: Solving the ratio on the left-hand side, we get,
\[LHS=\dfrac{1}{5}:\dfrac{1}{x}\]
We can convert the ratio into a fraction by the relation:
\[a:b=\dfrac{a}{b}=a\times \dfrac{1}{b}\]
By using this, we can write our ratio as follows:
\[LHS=\dfrac{\dfrac{1}{5}}{\dfrac{1}{x}}=\dfrac{1}{5}\times \dfrac{1}{\dfrac{1}{x}}\]
We know the relation of ‘a’ in the fraction which is \[\dfrac{1}{\dfrac{1}{a}}=a\]. By using this, we can write our equation in the form
\[LHS=\dfrac{1}{5}\times x\]
By simplifying, we get the value of the left-hand side as:
\[LHS=\dfrac{x}{5}.....\left( ii \right)\]
Case 2: Solving the ratio on the right-hand side, we get,
\[RHS=\dfrac{1}{x}:\dfrac{1}{1.25}\]
We can convert the ratio into a fraction by the relation:
\[a:b=a\times \dfrac{1}{b}\]
By using this, we get the value of our expression as:
\[RHS=\dfrac{1}{x}\times \dfrac{1}{\dfrac{1}{1.25}}\]
By basic knowledge of the fractions, we have a relation given as \[\dfrac{1}{\dfrac{1}{a}}=a\]
By substituting this, we can write our expression as:
\[RHS=\dfrac{1}{x}\times 1.25\]
By simplifying the above value, we get the value of the RHS as
\[RHS=\dfrac{1.25}{x}....\left( iii \right)\]
By substituting equation (ii) and (iii) in equation (i), we get it as:
\[\dfrac{x}{5}=\dfrac{1.25}{x}\]
By multiplying with x on both the sides, we get the equation as:
\[\dfrac{x}{5}\times x=\dfrac{1.25}{x}\times x\]
By multiplying with 5 on both the sides, we get the equation as
\[\dfrac{x}{5}\times x\times 5=\dfrac{1.25}{x}\times x\times 5\]
By canceling the like terms, we get the equation as
\[{{x}^{2}}=1.25\times 5\]
By simplifying the above equation, we get its value as
\[{{x}^{2}}=6.25\]
By applying the square root on both the sides, we get it as
\[x=\sqrt{6.25}\]
By simplifying the root value, we get the value of x as 2.5. So, the value of x is 2.5 to satisfy the given relation.
Therefore, option (c) is the right answer.
Note: Be careful while taking the ratio and don’t forget the \[\dfrac{1}{\text{term}}\] condition \[\dfrac{1}{\dfrac{1}{\text{term}}}=\text{term}\]. After the ratio, the terms change. Students confuse and write \[\dfrac{1}{x}\] in place of x. It will change the whole result. So, solve each and every step carefully while cross multiplying. Also, don’t forget to take the square root on the right-hand side as well.
Recently Updated Pages
Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Which is the Longest Railway Platform in the world?

India Manned Space Mission Launch Target Month and Year 2025 Update

Trending doubts
Which among the following has the highest peak in Eastern class 9 social science CBSE

Explain the importance of pH in everyday life class 9 chemistry CBSE

Which type of resource is iron ore ARenewable BBiotic class 9 social science CBSE

What is the Bhabar class 9 social science CBSE

Distinguish between Khadar and Bhangar class 9 social science CBSE

How do you find the valency of chlorine sulphur and class 9 chemistry CBSE


