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# Fill in the boxes:i) $\dfrac{{\boxed{}}}{{\boxed{}}} - \dfrac{5}{8} = \dfrac{1}{4}$ ii) $\dfrac{{45}}{{60}} = \dfrac{{15}}{{\boxed{}}}$iii) $\dfrac{1}{6} = \dfrac{{\boxed{}}}{{30}}$  Verified
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Hint: To find the missing fractions of numbers, we will use the simple mathematical operations and simplification. For part (i) we will use the sign change while taking a fraction from one side to another in an equation. For (ii) and (iii) we will simply express a bigger fraction into its smaller equivalent to find the missing numbers.

Now, we have to solve three parts, so starting with :
(i) We have to find the missing fraction in $\dfrac{{\boxed{}}}{{\boxed{}}} - \dfrac{5}{8} = \dfrac{1}{4}$, Let us assume the unknown fraction to be ‘x’, then the equation will now change to $x - \dfrac{5}{8} = \dfrac{1}{4}$.
Now solving the equation for x by taking $\dfrac{5}{8}$ from theLHS to the RHS, such that the minus sign of $\dfrac{5}{8}$changes to a plus sign and the operation also changes from subtraction to that of addition on the RHS .
So, we will get :
$x = \dfrac{1}{4} + \dfrac{5}{8}$
Now, for solving this equation, we have to find the LCM of 4 and 8 and for that we have to find the prime factors of 4 and 8.
Now prime factorization of 4 and 8 will give u:
$2\left| \!{\underline {\, 4 \,}} \right. \\ 2\left| \!{\underline {\, 2 \,}} \right. \\ 1\left| \!{\underline {\, 1 \,}} \right. \\$ $2\left| \!{\underline {\, 8 \,}} \right. \\ 2\left| \!{\underline {\, 4 \,}} \right. \\ 2\left| \!{\underline {\, 2 \,}} \right. \\ 1\left| \!{\underline {\, 1 \,}} \right. \\$
So, clearly
$4 = 2 \times 2 \times 1 \\ 8 = 2 \times 2 \times 2 \times 1 \\$
Now, we know that LCM of two numbers is the product of their prime factors raised to the highest power.
So accordingly the LCM of 4 and 8 will be ${2^3}$ or 8.
So, Now, with the LCM of 4 and 8 as 8, we will solve the equation for x and get:
$x = \dfrac{{1 \times 2 + 1 \times 5}}{8} \\ \Rightarrow x = \dfrac{{2 + 5}}{8} \\ \Rightarrow x = \dfrac{7}{8} \\$
So, the value of x comes to be equal to $\dfrac{7}{8}$.
Which means that the value of the missing fraction is $\dfrac{7}{8}$.
(ii) In this we need to find the value of the fraction equivalent to $\dfrac{{45}}{{60}}$. Let the missing number be ‘y’.
So, the equation will now become:
$\dfrac{{45}}{{60}} = \dfrac{{15}}{y}$
Now, we will solve for ‘y’ so the equation will become:
$y = \dfrac{{15 \times 60}}{{45}} \\ \Rightarrow y = \dfrac{{60}}{3} \\ \Rightarrow y = 20 \\$
So, the missing number is ‘y’ whose value i s 20.
(iii) In this we need to find the value of the fraction equivalent to $\dfrac{1}{6}$. Let the missing number be ‘z’.
So, the equation will now become:
$\dfrac{1}{6} = \dfrac{z}{{30}}$
Now, we will solve for ‘x’ so the equation will become:
$z = \dfrac{{1 \times 30}}{6} \\ \Rightarrow z = \dfrac{{30}}{6} \\ \Rightarrow z = 5 \\$
So, the missing number is ‘z’ whose value is 5.

Note: We need to take care while changing the sides of the fraction, since by changing the sides, not only does the sign change but the operation also changes together. So, pluson LHS will become minus sign on RHS, and addition will become subtraction. Similarly, division on the LHS, will become multiplication of its reciprocal on the RHS.