NCERT Solutions · Class 9 Mathematics Introduction to Linear Polynomials

A fixed starting amount plus the same addition every month.She starts with ₹$\displaystyle 500$ already in the account, and ₹$\displaystyle 150$ is added each month. (We assume she spends none of it, and we ignore any interest.)Working month by month, each figure is ₹$\displaystyle 150$ more than the one before it:

End of month	Amount in the account (₹)
$\displaystyle 1$	\(\displaystyle 500+150 = 650\)
$\displaystyle 2$	\(\displaystyle 650+150 = 800\)
$\displaystyle 3$	\(\displaystyle 800+150 = 950\)
$\displaystyle 4$	\(\displaystyle 950+150 = 1100\)
$\displaystyle 5$	\(\displaystyle 1100+150 = 1250\)

So from the second month onwards she has ₹$\displaystyle 800$, ₹$\displaystyle 950$, ₹$\displaystyle 1100$, ₹$\displaystyle 1250$, … — the amount goes up by exactly ₹$\displaystyle 150$ each time.To get the expression, notice that adding ₹$\displaystyle 150$ once a month for \(\displaystyle n\) months adds \(\displaystyle 150 \times n\) rupees in all, on top of the ₹$\displaystyle 500$ she began with: \[A_n = 500+150n \]Check. For \(\displaystyle n=2\), \(\displaystyle A_2 = 500+150(2) = 500+300 = 800\), which matches the table; for \(\displaystyle n=4\), \(\displaystyle A_4 = 500+600 = 1100\), which also matches.The highest power of \(\displaystyle n\) here is \(\displaystyle 1\), so \(\displaystyle A_n\) is a linear expression in \(\displaystyle n\). The constant \(\displaystyle 500\) is her starting balance and the coefficient \(\displaystyle 150\) is the fixed monthly increase.\(\displaystyle A_n = 500+150n\) rupees at the end of the \(\displaystyle n^{\text{th}}\) month; from the second month onwards the balances are ₹$\displaystyle 800$, ₹$\displaystyle 950$, ₹$\displaystyle 1100$, ₹$\displaystyle 1250$, …

A fixed starting number, with the same amount subtracted every hour.The rally begins with $\displaystyle 120$ members, and $\displaystyle 9$ leave in each hour that passes. Because members are lost rather than gained, the fixed number is subtracted each time:

Hours elapsed	Members remaining
$\displaystyle 1$	\(\displaystyle 120-9 = 111\)
$\displaystyle 2$	\(\displaystyle 111-9 = 102\)
$\displaystyle 3$	\(\displaystyle 102-9 = 93\)
$\displaystyle 4$	\(\displaystyle 93-9 = 84\)

After \(\displaystyle n\) hours, \(\displaystyle 9\) members have dropped out \(\displaystyle n\) times over, which is \(\displaystyle 9n\) members in total. So the number remaining is \[M_n = 120-9n \]Check. \(\displaystyle M_1 = 120-9 = 111\) and \(\displaystyle M_3 = 120-27 = 93\), both matching the table.This is a linear expression in \(\displaystyle n\) (the highest power of \(\displaystyle n\) is \(\displaystyle 1\)). The negative coefficient \(\displaystyle -9\) is what makes the count fall instead of rise.One sensible caution: the expression only describes the real situation while there are still members left. \(\displaystyle M_{13} = 120-117 = 3\), but \(\displaystyle M_{14} = 120-126 = -6\), and a rally cannot have \(\displaystyle -6\) members. So the pattern holds for \(\displaystyle n \le 13\) hours, after which everyone has gone.\(\displaystyle M_n = 120-9n\) members remain at the end of the \(\displaystyle n^{\text{th}}\) hour: $\displaystyle 111$ after $\displaystyle 1$ hour, $\displaystyle 102$ after $\displaystyle 2$ hours, $\displaystyle 93$ after $\displaystyle 3$ hours, and so on (valid up to \(\displaystyle n=13\)).

Start from the total and subtract a fixed number of pages each day.The book has $\displaystyle 500$ pages, and $\displaystyle 20$ pages disappear from the "still to read" pile every day. After \(\displaystyle d\) days she has read \(\displaystyle 20 \times d = 20d\) pages, so the number of pages still left is \[P = 500-20d \]After $\displaystyle 15$ days: \[P = 500-20(15) = 500-300 = 200 \]Check the other way round. In $\displaystyle 15$ days she reads \(\displaystyle 20 \times 15 = 300\) pages out of $\displaystyle 500$, and \(\displaystyle 500-300 = 200\) pages remain — the same answer reached by a different route.\(\displaystyle P = 500-20d\) is a linear pattern in \(\displaystyle d\): the highest power of \(\displaystyle d\) is \(\displaystyle 1\), the constant \(\displaystyle 500\) is the length of the book, and the coefficient \(\displaystyle -20\) is the fixed daily reading that makes the number fall.This pattern makes sense only until the book is finished. Setting \(\displaystyle 500-20d=0\) gives \(\displaystyle d=25\), so she finishes on the 25th day; beyond that the expression would give a negative number of pages, which is impossible.$\displaystyle 200$ pages will be left after $\displaystyle 15$ days; the linear pattern is \(\displaystyle P = 500-20d\) (valid for \(\displaystyle 0 \le d \le 25\)).