Class 6 Operations on whole Numbers Exercise 4.4-1

\begin{array}{l} \text {Q1. Does there exist a whole number a such that } a \div a=a? \\ \\\text {Sol. Yes. There exists a whole number ‘a’ such that } a \div a = a. \\ \text {The whole number is 1 where } 1 \div 1=1 \\ \\\text {Q2. Find the value of:} \\ (i) \quad 23457 \div 1 \\ (ii) \quad 0 \div 97 \\ (iii) \quad 476+(840 \div 84) \\ (iv) \quad 964-(425 \div 425) \\ (v)\quad (2758 \div 2758)-(2758 \div 2758) \\ (vi) \quad 72450 \div(583-58) \\ \\\text {Sol. } (i) \quad 23457 \div 1 = {\large \frac {23457}{1} } = 23457 \\ \\(ii) \quad 0 \div 97 = {\large \frac {0}{97} } =0 \\ \\(iii) \quad 476+(840 \div 84) = 476+{\large \frac {840}{84} } \\ =476+10 =486 \\ \\(iv) \quad 964-(425 \div 425) =964- {\large \frac {425}{425} } \\ =964-1 =963 \\ \\(v)\quad (2758 \div 2758)-(2758 \div 2758) \\ ={\large \frac {2758}{2758} } – {\large \frac {2758}{2758} } =1-1 = 0 \\ \\(vi) \quad 72450 \div(583-58) = {\large \frac {72450}{(583-58)} } \\ ={\large \frac {72450}{525} } \\ =138 \\ \\\text {Q3. Which of the following statements are true: } \\ (i) \quad 10 \div(5 \times 2)=(10 \div 5) \times(10 \div 2) \\ (ii) \quad (35-14) \div 7=35 \div 7-14 \div 7 \\ (iii) \quad 35-14 \div 7=35 \div 7-14 \div 7 \\ (iv) \quad (20-5) \div 5=20 \div 5-5 \\ (v) \quad 12 \times(14 \div 7)=(12 \times 14) \div(12 \times 7) \\ (vi) \quad (20 \div 5) \div 2=(20 \div 2) \div 5 \\ \\\text {Sol. (i) False. } \\ LHS = {\large \frac{10}{(5 \times 2)}} \\ ={\large \frac{10}{10}} = 1 \\ \\RHS =({\large \frac{10}{5}}) \times ({\large \frac{10}{2}}) \\ =2 \times 5 = 10 \\ \text {Hence } LHS \neq RHS \\ \\\text {(ii) True } \\ LHS = {\large \frac{(35-14)}{7}} \\ ={\large \frac{21}{7}} = 3 \\RHS = {\large \frac{35}{7}} – {\large \frac{14}{7}} \\ =5-2 = 3 \\ \text {Hence } LHS = RHS \\ \\\text {(iii) False. } \\ LHS = 35- {\large \frac{14}{7}} \\ =35-2=33 \\RHS = 35 \div 7 – 14 \div 7 \\ ={\large \frac {35}{7}} – {\large \frac {14}{7}} \\ =5-2 = 3 \\ \text {Hence } LHS \neq RHS \\ \\\text {(iv) False. } \\ LHS ={\large \frac{(20-5)}{5}} \\ ={\large \frac{15}{5}} =3 \\RHS = {\large \frac{20}{5}} – 5 = 4-5 = -1 \\\text {Hence } LHS \neq RHS \\ \\\text {(v) False. } \\ LHS = 12 \times ({\large \frac{14}{7}}) \\ =12 \times 2 = 24 \\RHS = {\large \frac{(12 \times 14)}{(12 \times 7)} }\\ ={\large \frac{168}{84}} = 2 \\ \text {Hence } LHS \neq RHS \\ \\\text {(vi) True } \\ LHS = {\large \frac{(\frac {20}{5})}{2}} \\ ={\large \frac{4}{2}} = 2 \\RHS = {\large \frac{(\frac{20}{2})}{5} } \\ ={\large \frac {10}{5} } = 2 \\ \text {Hence } LHS = RHS \\ \\ \end{array}