What rational number is halfway between $\dfrac{1}{5}$ and $\dfrac{1}{3}$ ?
Answer
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Hint: At first, we need to convert the fractions into their equivalent form. This is done by multiplying $6$ with both the numerator and the denominator for the first fraction and multiplying $10$ with both the numerator and the denominator for the second fraction. The denominator of the two becomes equal, but the numerator doesn’t. We find a numerator which lies midway between the two new numerators. The resultant fraction contains this numerator and the equal denominator.
Complete step-by-step solution:
The given fractions that we have in this problem are $\dfrac{1}{5}$ and $\dfrac{1}{3}$ . We can convert the fractions into some comparable form without actually changing them, by turning them into equivalent fractions by multiplying the numerator and denominator by the same numbers. For the first fraction, we multiply $6$ with both the numerator and the denominator and get,
$\dfrac{1}{5}\times \dfrac{6}{6}=\dfrac{6}{30}....\left( i \right)$
For the second fraction, we multiply $10$ with both the numerator and the denominator and get,
$\dfrac{1}{3}\times \dfrac{10}{10}=\dfrac{10}{30}....\left( ii \right)$
Now, we see that the fractions i and ii have the same denominator and the number between $6$ and $10$ is $8$ . The corresponding fraction is $\dfrac{8}{30}$ . We can notice that this fraction is not in its simplest form. To bring it into the simplest form, we divide the numerator and denominator by $2$ and get,
$\Rightarrow \dfrac{8\div 2}{30\div 2}=\dfrac{4}{15}$
Thus, we can conclude that the rational number that lies halfway between $\dfrac{1}{5}$ and $\dfrac{1}{3}$ is $\dfrac{4}{15}$.
Note: We should be careful while converting the fractions to their equivalent form and multiply both the numerator and the denominator. We can also solve the problem using another method. We know that the arithmetic mean of two numbers lies exactly midway between them. The formula for AM is $\dfrac{a+b}{2}$ . Here, $a=\dfrac{1}{5}$ and $b=\dfrac{1}{3}$ . So, the rational number becomes $\dfrac{\dfrac{1}{5}+\dfrac{1}{3}}{2}=\dfrac{4}{15}$ .
Complete step-by-step solution:
The given fractions that we have in this problem are $\dfrac{1}{5}$ and $\dfrac{1}{3}$ . We can convert the fractions into some comparable form without actually changing them, by turning them into equivalent fractions by multiplying the numerator and denominator by the same numbers. For the first fraction, we multiply $6$ with both the numerator and the denominator and get,
$\dfrac{1}{5}\times \dfrac{6}{6}=\dfrac{6}{30}....\left( i \right)$
For the second fraction, we multiply $10$ with both the numerator and the denominator and get,
$\dfrac{1}{3}\times \dfrac{10}{10}=\dfrac{10}{30}....\left( ii \right)$
Now, we see that the fractions i and ii have the same denominator and the number between $6$ and $10$ is $8$ . The corresponding fraction is $\dfrac{8}{30}$ . We can notice that this fraction is not in its simplest form. To bring it into the simplest form, we divide the numerator and denominator by $2$ and get,
$\Rightarrow \dfrac{8\div 2}{30\div 2}=\dfrac{4}{15}$
Thus, we can conclude that the rational number that lies halfway between $\dfrac{1}{5}$ and $\dfrac{1}{3}$ is $\dfrac{4}{15}$.
Note: We should be careful while converting the fractions to their equivalent form and multiply both the numerator and the denominator. We can also solve the problem using another method. We know that the arithmetic mean of two numbers lies exactly midway between them. The formula for AM is $\dfrac{a+b}{2}$ . Here, $a=\dfrac{1}{5}$ and $b=\dfrac{1}{3}$ . So, the rational number becomes $\dfrac{\dfrac{1}{5}+\dfrac{1}{3}}{2}=\dfrac{4}{15}$ .
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