Questions & Answers

Question

Answers

i) $\dfrac{{\boxed{}}}{{\boxed{}}} - \dfrac{5}{8} = \dfrac{1}{4}$

ii) $\dfrac{{45}}{{60}} = \dfrac{{15}}{{\boxed{}}}$

iii) $\dfrac{1}{6} = \dfrac{{\boxed{}}}{{30}}$

Answer

Verified

128.4k+ views

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.