Class 6 Operations on whole Numbers Exercise 4.2-1

\begin{array}{l} \text {Q1. A magic square is an array of numbers having the same number of rows and } \\ \text {columns and the sum of numbers in each row, column or diagonal being the same. } \\ \text {Fill in the blank cells of the following magic squares:} \\ (i) \end{array} \begin{array}{|c|c|c|} \hline & 8 & 13 \\ \hline & 12 & \\ \hline 11 & & \\ \hline \end{array}\begin{array}{l} \text {Sol. (i) Sum of diagonal values } =13+12+11=36 \\ \text {No. in the first cell of the first row }=36-(8+13)=15 \\ \text {No. in the first cell of the second row }=36-(15+11)=10 \\ \text {No. in the third cell of the second row }=36-(10+12)=14 \\ \text {No. in the second cell of the third row }=36-(8+12)=16 \\ \text {No. in the third cell of the third row} =36-(11+16)=9 \\ \end{array}\begin{array}{|c|c|c|} \hline \color{red}{\bf 15} & 8 & 13 \\ \hline \color{red}{\bf 10} & 12 & \color{red}{\bf 14} \\ \hline 11 & \color{red}{\bf 16} & \color{red}{\bf 9} \\ \hline \end{array}\begin{array}{l} (ii) \end{array} \begin{array}{|c|c|c|c|c|} \hline 22 & & 6 & 13 & 20 \\ \hline & 10 & 12 & 19 & \\ \hline 9 & 11 & 18 & 25 & \\ \hline 15 & 17 & 24 & 26 & \\ \hline 16 & & & 7 & 14 \\ \hline \end{array}\begin{array}{l} \text {Sol. (ii) Sum of diagonal values } =20+19+18+17+16=90 \\ \text {No. in the second cell of the first row }=90-(22+6+13+20)=29 \\ \text {No. in the first cell of the second row} =90-(22+9+15+16)=28 \\ \text {No. in the fifth cell of the second row }=90-(28+10+12+19)=21 \\ \text {No. in the fifth cell of the third row }=90-(9+11+18+25)=27 \\ \text {No. in the fifth cell of the fourth row }=90-(15+17+24+26)=8 \\ \text {No. in the second cell of the fifth row }=90-(29+10+11+17)=23 \\ \text {No. in the third cell of the fifth row }=90-(6+12+18+24)=30 \\ \end{array}\begin{array}{|c|c|c|c|c|} \hline 22 & \color{red}{\bf 29} & 6 & 13 & 20 \\ \hline \color{red}{\bf 28} & 10 & 12 & 19 & \color{red}{\bf 21} \\ \hline 9 & 11 & 18 & 25 & \color{red}{\bf 27} \\ \hline 15 & 17 & 24 & 26 & \color{red}{\bf 8} \\ \hline 16 & \color{red}{\bf 23} & \color{red}{\bf 30} & 7 & 14 \\ \hline \end{array}\begin{array}{l} \text {Q2. Perform the following subtractions and check your results by performing } \\ \text {corresponding additions: } \\ (i) \quad 57839-2983 \\ (ii) \quad 92507-10879 \\ (iii) \quad 400000-98798 \\ (iv) \quad 5050501-969696 \\ (v) \quad 200000-97531 \\ (vi) \quad 3030301-868686 \\ \\\text {Sol. } (i) \quad 57839-2983=54856 \\ \text {Verification : } 54856+2983=57839 \\ \\(ii) \quad 92507-10879=81628 \\ \text {Verification : } 81628+10879=92507 \\ \\(iii) \quad 400000-98798=301202 \\ \text {Verification : } 301202+98798=400000 \\ \\(iv) \quad 5050501-969696 =4080805 \\ \text {Verification : } 4080805+969696=5050501 \\ \\(v) \quad 200000-97531 =102469 \\ \text {Verification : } 102469+97531=200000 \\ \\(vi) \quad 3030301-868686=2161615 \\\text {Verification : } 2161615+868686=3030301 \\ \\\text {Q3. Replace each * by the correct digit in each of the following: } \\ (i) \end{array}\begin{array}{ccc} & 8 & 7 & 6 \\ – & * & 3 & * \\ \hline & 6 & * & 7 \\ \hline \end{array}\begin{array}{l} \text {At unit place 6 – * = 7 means we need to borrow 1 from tens place } \\ \text { because } 6 – * \neq 7 \text { as } 7 \gt 6. \\ \therefore \quad 16 – * = 7 \\ \Rightarrow \quad \text {* at unit place} = 16-7 = 9 \\ \text {When 7 is reduced by 1 it gives 6 so 6 – 3 = 3 } \\ \text {And finally, at hundreds place 8 – * = 6 so we get * value as 2.} \\ \end{array}\begin{array}{ccc} & 8 & 7 & 6 \\ – & \color{red}{\bf 2} & 3 & \color{red}{\bf 9} \\ \hline & 6 & \color{red}{\bf 3} & 7 \\ \hline \\ (ii) \end{array}\begin{array}{ccc} & 8 & 9 & 8 & 9 \\ – & * & * & 3 & 4 \\ \hline & 3 & 4 & * & * \\ \hline \end{array}\begin{array}{l} \text {Clearly in units place } 9-4 =5 \\ \text {And in tens place } 8-3=5 \\ \text {And in hundreds place } 9-*=3 \text { means * is 5} \\ \text {And in thousands place } 8-*=3 \text { means * is 5} \\ \end{array}\begin{array}{ccc} & 8 & 9 & 8 & 9 \\ – & \color{red}{\bf 5} & \color{red}{\bf 5} & 3 & 4 \\ \hline & 3 & 4 & \color{red}{\bf 5} & \color{red}{\bf 5} \\ \hline \\ (iii) \end{array}\begin{array}{ccc} & 6 & 0 & 0 & 0 & 1 & 0 & 7 \\ – & & * & * & 8 & 9 & 7 & 8 \\ \hline & 5 & 0 & 6 & * & * & * & * \\ \hline \end{array}\begin{array}{l} \text {In units place } 17-8=9 \text { because we can’t subract 8 from 7 so need to borrow 1} \\ \text {In tens place } 9-7=2 \\ \text {In hundreds place } 10-9=1 \\ \text {And, In thousands place } 9-8=1 \\ \therefore \text {The Difference } =5061129 \\ \text {So, in order to get the subtrhend, we will subtract 5061129 from 6000107} \\ =6000107 – 5061129 = 938978 \end{array}\begin{array}{ccc} & 6 & 0 & 0 & 0 & 1 & 0 & 7 \\ – & & \color{red}{\bf 9} & \color{red}{\bf 3} & 8 & 9 & 7 & 8 \\ \hline & 5 & 0 & 6 & \color{red}{\bf 1} & \color{red}{\bf 1} & \color{red}{\bf 2} & \color{red}{\bf 9} \\ \hline \\ (iv) \end{array}\begin{array}{ccc} & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ – & & & * & * & * & * & 1 \\ \hline & & * & 7 & 0 & 4 & 2 & * \\ \hline \end{array}\begin{array}{l} \text {In units place } 10-1=9 \\ \text {In tens place } 9-*=2 \text { means * = 7 }\\ \text {In hundreds place } 9-*=4 \text { means * = 5 }\\ \text {In thousands place } 9-*=0 \text { means * = 9 }\\ \text {In ten thousands place } 9-*=7 \text { means * = 2 }\\ \text {In lakhs place } 9-0=9 \\ \end{array}\begin{array}{ccc} & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ – & & & \color{red}{\bf 2} & \color{red}{\bf 9} & \color{red}{\bf 5} & \color{red}{\bf 7} & 1 \\ \hline & & \color{red}{\bf 9} & 7 & 0 & 4 & 2 & \color{red}{\bf 9} \\ \hline \\ (v) \end{array}\begin{array}{ccc} & 5 & 0 & 0 & 1 & 0 & 0 & 3 \\ – & & * & * & 6 & 9 & 8 & 7 \\ \hline & 4 & 8 & 4 & * & * & * & * \\ \hline \end{array}\begin{array}{l} \text {In units place } 13-7=6 \text { because we can’t subract 7 from 3 so need to borrow 1} \\ \text {In tens place } 9-8=1 \\ \text {In hundreds place } 9-9=0 \\ \text {In thousands place } 10-6=4 \\ \text {In ten thousands place } 9-*=4 \text { means * = 5 }\\ \text {In lakhs place } 9-*=8 \text { means * = 1 }\\ \end{array}\begin{array}{ccc} & 5 & 0 & 0 & 1 & 0 & 0 & 3 \\ – & & \color{red}{\bf 1} & \color{red}{\bf 5} & 6 & 9 & 8 & 7 \\ \hline & 4 & 8 & 4 & \color{red}{\bf 4} & \color{red}{\bf 0} & \color{red}{\bf 1} & \color{red}{\bf 6} \\ \hline \\ (vi) \end{array}\begin{array}{ccc} & 1 & 1 & 1 & 1 & 1 & 1 \\ – & & * & 6 & 7 & 8 & 9 \\ \hline & & 5 & 4 & 3 & 2 & * \\ \hline \end{array}\begin{array}{l} \text {In units place } 11-9=2 \text { because we can’t subract 9 from 1 so need to borrow 1} \\ \text {In ten thousands place } 10 – *=5 \text { means * = 5 }\\ \end{array}\begin{array}{ccc} & 1 & 1 & 1 & 1 & 1 & 1 \\ – & & \color{red}{\bf 5} & 6 & 7 & 8 & 9 \\ \hline & & 5 & 4 & 3 & 2 & \color{red}{\bf 2} \\ \hline \end{array}