NCERT Solutions · Class 9 Mathematics The World of Algorithms

Take \(\displaystyle 3847\) and \(\displaystyle 2795\), aligned from the right, and follow Steps $\displaystyle 2$–4.\[7+5=12 \;\Rightarrow\; \text{write } 2,\ \text{carry}=1 \] \[4+9+1=14 \;\Rightarrow\; \text{write } 4,\ \text{carry}=1 \] \[8+7+1=16 \;\Rightarrow\; \text{write } 6,\ \text{carry}=1 \] \[3+2+1=6 \;\Rightarrow\; \text{write } 6,\ \text{carry}=0 \] \[3847+2795=6642 \]Correct. But the steps say only "less than $\displaystyle 10$" and "more than $\displaystyle 10$". Take \(\displaystyle 2345+1255\):\[5+5=10 \quad \text{(neither rule applies)} \]Following the steps literally, we are stuck; they should say "$\displaystyle 10$ or more".Answer: \(\displaystyle 3847+2795=6642\), correct; the steps stall only when a column total is exactly \(\displaystyle 10\).

Not as written. Step $\displaystyle 3$ says "add the two digits", but \(\displaystyle 934\) has no digit in the thousands and ten-thousands columns. Reading a missing digit as \(\displaystyle 0\) repairs this. Take \(\displaystyle 45678+934\):\[8+4=12 \;\Rightarrow\; \text{write } 2,\ \text{carry}=1 \] \[7+3+1=11 \;\Rightarrow\; \text{write } 1,\ \text{carry}=1 \] \[6+9+1=16 \;\Rightarrow\; \text{write } 6,\ \text{carry}=1 \] \[5+0+1=6 \;\Rightarrow\; \text{write } 6,\ \text{carry}=0 \] \[4+0+0=4 \;\Rightarrow\; \text{write } 4 \] \[45678+934=46612 \]The carry from the hundreds column must reach a column that holds only one digit.Answer: No, not as written; treating the missing digits as \(\displaystyle 0\) gives \(\displaystyle 45678+934=46612\).

Each column is one place value, and only digits of the same place value may be added.\[345 = 3\times 100+4\times 10+5 \] \[27 = 2\times 10+7 \] \[345+27 = 3\times 100+(4+2)\times 10+(5+7) = 372 \]Right alignment pairs units with units and tens with tens, whatever the lengths. Aligning from the left would add \(\displaystyle 3\) hundreds to \(\displaystyle 2\) tens:\[345+270 = 615 \neq 372 \]A carry also moves one place to the left, so we start at the right.Answer: Aligning from the right puts equal place values in one column; otherwise the sum is wrong (\(\displaystyle 615\) instead of \(\displaystyle 372\)).

The carry entering any column is at most \(\displaystyle 1\): after Step $\displaystyle 2$ it is \(\displaystyle 0\) or \(\displaystyle 1\), and Step $\displaystyle 3$ sets it to \(\displaystyle 0\) or \(\displaystyle 1\) again. A digit is at most \(\displaystyle 9\), so a column total is at most\[9+9+1=19 \] \[10 \le \text{total} \le 19 \;\Rightarrow\; \text{total}=10\times 1+\text{units digit} \]A total below \(\displaystyle 20\) holds at most one ten, so the new carry is at most \(\displaystyle 1\).Answer: The largest column total is \(\displaystyle 9+9+1=19<20\), so at most one ten is carried.

Without Step $\displaystyle 5$, a carry produced by the leftmost column is never written, so the sum loses its leading \(\displaystyle 1\).Works when the leftmost column total is below \(\displaystyle 10\):\[3+6=9,\quad 2+5=7,\quad 1+4=5 \;\Rightarrow\; 123+456=579 \]Fails when it is \(\displaystyle 10\) or more:\[5+2=7,\quad 8+6=14 \Rightarrow 4,\quad 7+4+1=12 \Rightarrow 2,\ \text{carry}=1 \] \[\text{Steps 1–4 give } 247, \text{ but } 785+462=1247 \]Step $\displaystyle 5$ writes the last carry \(\displaystyle 1\) on the left, giving \(\displaystyle 1247\).Answer: It works when the final carry is \(\displaystyle 0\) (\(\displaystyle 123+456=579\)) and fails when it is \(\displaystyle 1\) (\(\displaystyle 785+462\) gives \(\displaystyle 247\), not \(\displaystyle 1247\)).