question_answer 41)

Complete the addition-subtraction box.

Answer:

(a) Here \[\frac{7}{8}.\] and \[\frac{1}{4},\frac{2}{4},\frac{3}{4}\] Now, \[\frac{4}{4}\] and \[\frac{1}{8},\frac{2}{8},\frac{3}{8}\] Also, we have 2 ? 1 = 1 and \[\frac{7}{8}.\] Hence, the required complete box is (b) Here, \[\frac{1}{2}+\frac{1}{3}=\frac{1\times 3}{2\times 3}+\frac{1\times 2}{3\times 2}\] [\[\frac{3}{5}\] LCM of 2 and 3 = 6] \[\frac{4}{5}.\] and \[\frac{1}{3}+\frac{1}{4}=\frac{1\times 4}{3\times 4}+\frac{1\times 3}{4\times 3}\] [\[\frac{2}{8},3\]LCM of 3 and 4 = 12] \[\frac{3}{8}\] Now, \[\frac{1}{2}-\frac{1}{3}=\frac{1\times 3}{2\times 3}-\frac{1\times 2}{1\times 2}\] [\[\because \]LCM of 2 and 3 = 6]] \[\frac{1}{8},\frac{2}{8},\frac{3}{8}\] and \[\frac{2}{5},\frac{3}{5},\frac{8}{5}\] [\[\frac{7}{8}\]LCM of 2 and 4 = 12] \[\frac{1}{5}.\] Also, we have \[\frac{2}{5},3\] [\[\frac{3}{5}\] LCM of 6 and 12 = 12] \[\frac{4}{5}.\] [dividing numerator and denominator by 3] and \[\frac{2}{5},\frac{3}{5}\] [\[\frac{4}{5}\] LCM of 6 and 12= 12] \[\frac{8}{5},\] [dividing numerator and denominator by 3] Hence, the required complete box is