Question

Question 5(b) :

Complete the addition – subtraction box:

Answer 1

In the first row, we have to add $\frac{1}{2}$ and $\frac{1}{3}$.

$\frac{1}{2}+\frac{1}{3}=\frac{1\times 3+1\times 2}{6}=\frac{5}{6}$

Similarly for the second row

$\frac{1}{3}+\frac{1}{4}=\frac{1\times 4+1\times 3}{12}=\frac{7}{12}$.

In the first column, we have to subtract $\frac{1}{2}$ and $\frac{1}{3}$

$\frac{1}{2}-\frac{1}{3}=\frac{1\times 3-1\times 2}{6}=\frac{1}{6}$

Similarly for the second column

$\frac{1}{3}-\frac{1}{4}=\frac{1\times 4-1\times 3}{12}=\frac{1}{12}$

Now, when we add the fractions in the third row we get:

$\frac{1}{6}+\frac{1}{12}=\frac{1\times 2+1\times 1}{12}=\frac{3}{12}$

On subtracting the fractions in the third coulmn we get:

$\frac{5}{6}-\frac{7}{12}=\frac{5\times 2-7\times 1}{12}=\frac{3}{12}$

$\therefore$ On substituting the values in the box we get: