NCERT Solutions Class 9 Maths Exercise 2.3 — Introduction to Linear Polynomials

NCERT Class 9 Mathematics — Ganita Manjari Part I, Introduction to Linear Polynomials, Exercise Set 2.3. All 5 questions solved.

Exercise Set 2.2 ended with equations to solve. This one stops one step earlier, and asks for the formula rather than a single number. Each question describes something that changes in equal steps, and wants the rule that gives its value at stage $n$.

Every answer here has the same shape:

$$\text{value after } n \text{ steps} \;=\; (\text{starting value}) \;+\; (\text{change each step}) \times n$$

with a plus when the quantity grows and a minus when it shrinks. That is exactly the form $an + b$ of a linear polynomial, with $b$ the starting value and $a$ the step.

Key insight. Test your formula at $n = 0$ before anything else. It should give the situation before any step has happened — ₹500 in the account, 120 members in the rally, 500 pages unread. A formula that fails this test has the starting value and the first step muddled, which is the commonest error in the whole exercise.

Solve the following:

Question 1

A student has ₹$500$ in her savings bank account. She gets ₹$150$ every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the $n^{th}$ month.

Solution. She starts with ₹$500$ and adds ₹$150$ at the end of each month. Tracking the first few months:

$$\text{end of month } 1: \; 500 + 150 = ₹650$$ $$\text{end of month } 2: \; 650 + 150 = ₹800$$ $$\text{end of month } 3: \; 800 + 150 = ₹950$$ $$\text{end of month } 4: \; 950 + 150 = ₹1100$$

From the second month onwards the amounts run ₹$800$, ₹$950$, ₹$1100$, ₹$1250$, … — going up by ₹$150$ each time.

After $n$ months she has received the ₹$150$ exactly $n$ times, so

$$A(n) = 500 + 150n$$

Check at $n = 0$: $A(0) = 500$, the amount before any pocket money arrives. ✓

Question 2

A rally starts with $120$ members. Each hour, $9$ members drop out of the group. How many members will remain after $1, 2, 3, \ldots$ hours? Find a linear expression to represent the number of members at the end of the $n^{th}$ hour.

Solution. This one shrinks, so the step is subtracted.

$$\text{after } 1 \text{ hour}: \; 120 – 9 = 111$$ $$\text{after } 2 \text{ hours}: \; 111 – 9 = 102$$ $$\text{after } 3 \text{ hours}: \; 102 – 9 = 93$$

After $n$ hours, $9$ members have left $n$ times over:

$$M(n) = 120 – 9n$$

Where the formula stops meaning anything. Unlike a bank balance, a rally cannot have a negative number of members. Setting $M(n) = 0$:

$$120 – 9n = 0 \quad\Longrightarrow\quad n = \frac{120}{9} = \frac{40}{3} = 13\tfrac{1}{3}$$

which is not a whole number of hours. So after $13$ hours $M(13) = 120 – 117 = 3$ members are left, and within the next hour the last three drop out. At $n = 14$ the formula would give $-6$, which describes nothing. The expression is a good model for $n = 0, 1, 2, \ldots, 13$ and no further.

Question 3

Suppose the length of a rectangle is $13$ cm. Find the area if the breadth is (i) $12$ cm, (ii) $10$ cm, (iii) $8$ cm. Find the linear pattern representing the area of the rectangle.

Solution. Area of a rectangle is length $\times$ breadth, and the length is fixed at $13$ cm, so only the breadth varies.

(i) $13 \times 12 = 156 \text{ cm}^2$

(ii) $13 \times 10 = 130 \text{ cm}^2$

(iii) $13 \times 8 = 104 \text{ cm}^2$

Writing $b$ for the breadth in cm:

$$A(b) = 13b$$

This is linear in $b$ — and it is worth being clear about why, because area is usually the standard example of something that is not linear. Area depends on two dimensions; here one of them has been nailed down at $13$, so only one thing is free to change, and the area moves in proportion to it. Each extra centimetre of breadth adds exactly $13 \text{ cm}^2$, as the table confirms: from $b = 10$ to $b = 12$ the area rises by $26$, which is $2 \times 13$.

This is a linear polynomial with constant term $0$ — at zero breadth there is no rectangle and no area.

Question 4

Suppose the length of a rectangular box is $7$ cm and breadth is $11$ cm. Find the volume if the height is (i) $5$ cm, (ii) $9$ cm, (iii) $13$ cm. Find the linear pattern representing the volume of the rectangular box.

Solution. Volume is length $\times$ breadth $\times$ height. Two of the three are fixed, and $7 \times 11 = 77$, so the volume is $77$ times the height.

(i) $77 \times 5 = 385 \text{ cm}^3$

(ii) $77 \times 9 = 693 \text{ cm}^3$

(iii) $77 \times 13 = 1001 \text{ cm}^3$

With $h$ for the height in cm:

$$V(h) = 77h$$

Again, volume is famously a three-dimensional quantity, yet the pattern here is linear — for the same reason as in Question 3. Fixing two of the three dimensions leaves a single free variable, and every extra centimetre of height adds one more $7 \times 11$ slab of $77 \text{ cm}^3$.

Question 5

Sarita is reading a book of $500$ pages. She reads $20$ pages every day. How many pages will be left after $15$ days? Express this as a linear pattern.

Solution. The number of pages left falls by $20$ each day, starting from $500$. After $d$ days she has read $20d$ pages, so

$$P(d) = 500 – 20d$$

After $15$ days:

$$P(15) = 500 – 20 \times 15 = 500 – 300 = 200 \text{ pages left}$$

Check at $d = 0$: $P(0) = 500$, the whole book still unread. ✓

She finishes when $P(d) = 0$, that is $20d = 500$, so $d = 25$ days — and as in Question 2, the formula only describes the reading for $d = 0$ to $25$.

Common mistakes

• Questions 1 and 2, muddling the starting value with the first step. Writing $A(n) = 650 + 150n$ for Question 1 counts the first month’s pocket money twice, and gives ₹$800$ at the end of month $1$ instead of ₹$650$. Testing the formula at $n = 0$ catches it at once: it should return the amount before anything happens.
• Question 2, writing $9n – 120$. This has the subtraction the wrong way round and gives $-111$ after one hour. The quantity that shrinks is the membership, so the starting number comes first: $120 – 9n$.
• Question 2, using the formula past its range. $M(20) = -60$ is arithmetically fine and physically meaningless. Any decreasing model has a last valid step, and it is worth finding it.
• Question 3, answering “area is not linear”. Area is not linear in general, but here the length is fixed, which leaves only one variable. The same reasoning makes the volume in Question 4 linear in the height.
• Question 3, writing $A = 13 + b$. The two dimensions multiply, they do not add. That formula would make a $13$ cm by $12$ cm rectangle have an area of $25 \text{ cm}^2$.
• Question 4, multiplying only two of the three dimensions. Using $V = 11h$ or $V = 7h$ forgets the other fixed side. Multiply both fixed dimensions together first — $7 \times 11 = 77$ — and the pattern becomes clear.
• Question 5, giving the number of pages read instead of left. $20 \times 15 = 300$ is how much she has read; the question asked how many are left, which is $500 – 300 = 200$.
• Units dropped from the formula. $A(b) = 13b$ is in square centimetres and $V(h) = 77h$ in cubic centimetres. The number alone does not say which, and in Question 4 the difference is between a box and a rectangle.

Practise next

• Exercise Set 2.4 — the same formulas, now named: growing quantities are linear growth and shrinking ones linear decay, and the exercise asks for tables of values alongside the expression.
• The End-of-Chapter Exercises, Question 5 and Question 12 — Question 5 is a savings problem of exactly this kind, and Question 12 asks for the $n^{th}$ term of a matchstick pattern, which is this same skill applied to a picture instead of a sentence.