
Find three rational numbers lying between \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\] (by average method)
Answer
496.5k+ views
Hint: To solve this problem, we use the concept of averages. And also, we use some methods of adding fractions (like fractions or dislike fractions). Average of two numbers \[a\] and \[b\] is evaluated as \[Average = \dfrac{{a + b}}{2}\] . And, we can insert infinitely many fractions, between two fractions.
The average of two numbers is also called the mean of those numbers, which gives exactly the middle value of those two numbers.
Complete step-by-step solution:
We need to find three rational numbers between \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\]
So, consider three rational numbers \[a,b,c\] lying between \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\] .
\[ \Rightarrow \dfrac{1}{5},a,b,c,\dfrac{1}{4}\]
If we take the average of two numbers, then we will get a number lying exactly in between those two numbers.
So, if we take average of \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\] , we get the value of \[b\] .
\[ \Rightarrow b = \dfrac{{\dfrac{1}{5} + \dfrac{1}{4}}}{2}\] (LCM of 5 and 4 is 20)
\[ \Rightarrow b = \dfrac{{\dfrac{4}{{20}} + \dfrac{5}{{20}}}}{2}\]
\[ \Rightarrow b = \dfrac{{\dfrac{9}{{20}}}}{2} = \dfrac{9}{{40}}\] (here \[b\] is also equal to \[\dfrac{{18}}{{80}}\] )
And similarly, to find \[c\] , we have to find the average of \[b\] and \[\dfrac{1}{4}\] .
\[ \Rightarrow c = \dfrac{{b + \dfrac{1}{{40}}}}{2} = \dfrac{{\dfrac{9}{{40}} + \dfrac{1}{4}}}{2}\]
LCM of 40 and 4 is 40 itself.
\[ \Rightarrow c = \dfrac{{\dfrac{9}{{40}} + \dfrac{{10}}{{40}}}}{2}\]
\[ \Rightarrow c = \dfrac{{\dfrac{{19}}{{40}}}}{2} = \dfrac{{19}}{{80}}\]
In the same way, to find \[a\] , we have to find the average between \[\dfrac{1}{5}\] and \[b\] .
\[ \Rightarrow a = \dfrac{{\dfrac{1}{5} + b}}{2} = \dfrac{{\dfrac{1}{5} + \dfrac{9}{{40}}}}{2}\]
LCM of 5 and 40 is 40 itself.
\[ \Rightarrow a = \dfrac{{\dfrac{8}{{40}} + \dfrac{9}{{40}}}}{2} = \dfrac{{\dfrac{{17}}{{40}}}}{2}\]
\[ \Rightarrow a = \dfrac{{17}}{{80}}\]
So, three rational number between \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\] are \[\dfrac{{17}}{{80}},\dfrac{{18}}{{80}},\dfrac{{19}}{{80}}\] .
Note: We can also find three rational numbers in another way also. In this method, we make the two given fractions into like fractions, and find the fractions between them.
The given fractions are, \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\] . So, LCM of 5 and 4 is 20. So, the like fractions will be \[\dfrac{4}{{20}}\] and \[\dfrac{5}{{20}}\].
So, we need to insert three rational numbers between \[\dfrac{4}{{20}}\] and \[\dfrac{5}{{20}}\]. So, here the denominators are the same, so we need to find numbers lying between numerators i.e. between 4 and 5.
But there are no numbers between 4 and 5, so multiply these fractions by 4 in both numerator and denominator.
So, the fractions become \[\dfrac{{16}}{{80}}\] and \[\dfrac{{20}}{{80}}\] .
Here, the numerators are 16 and 20, and there are numbers between these numbers.
So, numbers lying between the fractions given are, \[\dfrac{{17}}{{80}},\dfrac{{18}}{{80}},\dfrac{{19}}{{80}}\] .
The average of two numbers is also called the mean of those numbers, which gives exactly the middle value of those two numbers.
Complete step-by-step solution:
We need to find three rational numbers between \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\]
So, consider three rational numbers \[a,b,c\] lying between \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\] .
\[ \Rightarrow \dfrac{1}{5},a,b,c,\dfrac{1}{4}\]
If we take the average of two numbers, then we will get a number lying exactly in between those two numbers.
So, if we take average of \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\] , we get the value of \[b\] .
\[ \Rightarrow b = \dfrac{{\dfrac{1}{5} + \dfrac{1}{4}}}{2}\] (LCM of 5 and 4 is 20)
\[ \Rightarrow b = \dfrac{{\dfrac{4}{{20}} + \dfrac{5}{{20}}}}{2}\]
\[ \Rightarrow b = \dfrac{{\dfrac{9}{{20}}}}{2} = \dfrac{9}{{40}}\] (here \[b\] is also equal to \[\dfrac{{18}}{{80}}\] )
And similarly, to find \[c\] , we have to find the average of \[b\] and \[\dfrac{1}{4}\] .
\[ \Rightarrow c = \dfrac{{b + \dfrac{1}{{40}}}}{2} = \dfrac{{\dfrac{9}{{40}} + \dfrac{1}{4}}}{2}\]
LCM of 40 and 4 is 40 itself.
\[ \Rightarrow c = \dfrac{{\dfrac{9}{{40}} + \dfrac{{10}}{{40}}}}{2}\]
\[ \Rightarrow c = \dfrac{{\dfrac{{19}}{{40}}}}{2} = \dfrac{{19}}{{80}}\]
In the same way, to find \[a\] , we have to find the average between \[\dfrac{1}{5}\] and \[b\] .
\[ \Rightarrow a = \dfrac{{\dfrac{1}{5} + b}}{2} = \dfrac{{\dfrac{1}{5} + \dfrac{9}{{40}}}}{2}\]
LCM of 5 and 40 is 40 itself.
\[ \Rightarrow a = \dfrac{{\dfrac{8}{{40}} + \dfrac{9}{{40}}}}{2} = \dfrac{{\dfrac{{17}}{{40}}}}{2}\]
\[ \Rightarrow a = \dfrac{{17}}{{80}}\]
So, three rational number between \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\] are \[\dfrac{{17}}{{80}},\dfrac{{18}}{{80}},\dfrac{{19}}{{80}}\] .
Note: We can also find three rational numbers in another way also. In this method, we make the two given fractions into like fractions, and find the fractions between them.
The given fractions are, \[\dfrac{1}{5}\] and \[\dfrac{1}{4}\] . So, LCM of 5 and 4 is 20. So, the like fractions will be \[\dfrac{4}{{20}}\] and \[\dfrac{5}{{20}}\].
So, we need to insert three rational numbers between \[\dfrac{4}{{20}}\] and \[\dfrac{5}{{20}}\]. So, here the denominators are the same, so we need to find numbers lying between numerators i.e. between 4 and 5.
But there are no numbers between 4 and 5, so multiply these fractions by 4 in both numerator and denominator.
So, the fractions become \[\dfrac{{16}}{{80}}\] and \[\dfrac{{20}}{{80}}\] .
Here, the numerators are 16 and 20, and there are numbers between these numbers.
So, numbers lying between the fractions given are, \[\dfrac{{17}}{{80}},\dfrac{{18}}{{80}},\dfrac{{19}}{{80}}\] .
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