NCERT Solutions Class 9 Maths Exercise 2.2 — Introduction to Linear Polynomials

NCERT Class 9 Mathematics — Ganita Manjari Part I, Introduction to Linear Polynomials, Exercise Set 2.2. All 7 questions solved.

This exercise has two halves that look unrelated and are not. Questions 1 and 2 ask you to evaluate a polynomial — put a number in place of the variable and work out what comes out. Questions 3 to 7 ask you to build a linear polynomial from a sentence and then find the value of the variable that makes it come out right.

The method for the word problems is the same every time:

1. Name the unknown you know least about with a letter.
2. Write every other quantity in the problem in terms of that letter.
3. Turn the sentence that contains a number into an equation.
4. Solve, then read the answer back into the words of the question — the letter is rarely the thing being asked for.

Key insight. Choose the unknown so that the other quantities come out without fractions. In Question 5, calling the number of five-rupee coins $x$ makes the two-rupee coins $3x$; calling the two-rupee coins $x$ would make the five-rupee coins $x/3$ and the arithmetic messier for no gain. The equation is always easier to solve if you let the smallest or most-referred-to quantity be the letter.

Question 1

Find the value of the linear polynomial $5x – 3$ if:

Solution. Substituting means replacing every $x$ by the given number and then evaluating. Brackets are worth writing when the number is negative.

(i) $x = 0$:

$$5(0) – 3 = 0 – 3 = -3$$

(ii) $x = -1$:

$$5(-1) – 3 = -5 – 3 = -8$$

(iii) $x = 2$:

$$5(2) – 3 = 10 – 3 = 7$$

Notice the pattern: as $x$ goes up by $1$, the value goes up by $5$ — from $-8$ at $x = -1$, to $-3$ at $x = 0$, to $2$ at $x = 1$, to $7$ at $x = 2$. That constant step is the defining feature of a linear polynomial, and the rest of the chapter is built on it.

Question 2

Find the value of the quadratic polynomial $7s^2 – 4s + 6$ if:

Solution. Same method, with one extra care: square before multiplying, and remember that the square of a negative number is positive.

(i) $s = 0$:

$$7(0)^2 – 4(0) + 6 = 0 – 0 + 6 = 6$$

(ii) $s = -3$:

$$7(-3)^2 – 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81$$

Both of the first two terms turn out positive here: $(-3)^2 = 9$, and subtracting $4 \times (-3)$ means adding $12$.

(iii) $s = 4$:

$$7(4)^2 – 4(4) + 6 = 7(16) – 16 + 6 = 112 – 16 + 6 = 102$$

Unlike Question 1, the steps between values are not constant here — from $6$ to $81$ to $102$ — because the polynomial is quadratic, not linear.

Question 3

The present age of Salil’s mother is three times Salil’s present age. After $5$ years, their ages will add up to $70$ years. Find their present ages.

Solution. Let Salil’s present age be $x$ years. Then his mother’s present age is $3x$ years.

In $5$ years both of them are $5$ years older — this is the step that is easy to get wrong, because it applies to each person separately:

$$\text{Salil} = x + 5, \qquad \text{mother} = 3x + 5$$

Their sum is $70$:

$$(x + 5) + (3x + 5) = 70$$ $$4x + 10 = 70$$ $$4x = 60$$ $$x = 15$$

So Salil is $15$ and his mother is $3 \times 15 = 45$.

Check against the words: in five years they will be $20$ and $50$, and $20 + 50 = 70$. ✓

Question 4

The difference between two positive integers is $63$. The ratio of the two integers is $2 : 5$. Find the two integers.

Solution. A ratio of $2 : 5$ means the numbers are $2k$ and $5k$ for some positive number $k$ — writing them this way builds the ratio in, so only the difference is left to use.

$$5k – 2k = 63$$ $$3k = 63$$ $$k = 21$$

So the integers are

$$2k = 42 \qquad \text{and} \qquad 5k = 105$$

Check: $105 – 42 = 63$, and $42 : 105 = 2 : 5$ after dividing both by $21$. ✓

Question 5

Ruby has $3$ times as many two-rupee coins as she has five-rupee coins. If she has a total of ₹$88$, how many coins does she have of each type?

Solution. Let the number of five-rupee coins be $x$. Then the number of two-rupee coins is $3x$.

The trap here is to add the coins when the question is about money. Each five-rupee coin is worth ₹$5$ and each two-rupee coin is worth ₹$2$, so the total value is

$$5 \times x \;+\; 2 \times 3x \;=\; 5x + 6x \;=\; 11x$$

Setting this equal to the total:

$$11x = 88 \quad\Longrightarrow\quad x = 8$$

So Ruby has $8$ five-rupee coins and $3 \times 8 = 24$ two-rupee coins.

Check: $8 \times 5 = ₹40$ and $24 \times 2 = ₹48$, and $40 + 48 = ₹88$. ✓ (She has $32$ coins in all, which is not what the question asked for.)

Question 6

A farmer cuts a $300$ feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?

Solution. Let the shorter piece be $x$ feet. Then the longer piece is $4x$ feet, and together they make up the whole fence:

$$x + 4x = 300$$ $$5x = 300$$ $$x = 60$$

So the pieces are $60$ feet and $4 \times 60 = 240$ feet.

Check: $60 + 240 = 300$ feet, and $240$ is indeed four times $60$. ✓

Question 7

If the length of a rectangle is three more than twice its width and its perimeter is $24$ cm, what are the dimensions of the rectangle?

Solution. The length is described in terms of the width, so let the width be $x$ cm. Then

$$\text{length} = 2x + 3$$

“Three more than twice its width” is $2x + 3$, not $2(x + 3)$ — the doubling happens first.

The perimeter of a rectangle is twice the sum of the length and the width:

$$2\bigl[(2x + 3) + x\bigr] = 24$$ $$2(3x + 3) = 24$$ $$3x + 3 = 12$$ $$3x = 9$$ $$x = 3$$

So the width is $3$ cm and the length is $2(3) + 3 = 9$ cm.

Check: the perimeter is $2(9 + 3) = 24$ cm, and $9$ is three more than twice $3$. ✓

Common mistakes

• Question 2(ii), writing $7(-3)^2 = -63$. Squaring comes before multiplying, and $(-3)^2 = +9$, so the term is $+63$. Writing the bracket in rather than the bare $-3$ is what prevents this.
• Question 2(ii), subtracting $12$ instead of adding it. The term is $-4s$ with $s = -3$, so it is $-4 \times (-3) = +12$. Two minus signs make a plus.
• Question 3, adding $5$ once. Writing $x + 3x + 5 = 70$ treats the five years as happening to only one of them. Both people age, so it is $(x+5) + (3x+5)$, which is $10$ more, not $5$.
• Question 4, taking the numbers as $2$ and $5$ themselves. Their difference is $3$, not $63$. A ratio fixes the proportion, not the size, which is exactly why the multiplier $k$ is needed.
• Question 5, adding the coins instead of their value. Writing $x + 3x = 88$ gives $x = 22$, so $22$ five-rupee and $66$ two-rupee coins — worth ₹$242$, nearly three times the total stated. Multiply each count by what that coin is worth.
• Question 5, answering “$32$ coins”. That is the total number of coins; the question asked how many she has of each type.
• Question 6, reading “four times as long” as a difference. Writing $x + (x + 4) = 300$ gives pieces of $148$ and $152$ feet, which are barely different sizes. “Four times as long” is multiplication.
• Question 7, writing the perimeter as $\text{length} + \text{width}$. A rectangle has two lengths and two widths, so the perimeter is $2(l + w)$. Forgetting the $2$ gives a width of $7$ cm and a length of $17$ cm, whose perimeter is $48$ cm.
• Question 7, reading “three more than twice its width” as $2(x+3)$. That is $2x + 6$, and it gives a width of $2$ cm. Translate the sentence in the order it is written: take the width, double it, then add three.

Practise next

• Exercise Set 2.3 — the same idea of building an expression from a situation, but now the expression is left as a formula in $n$ rather than solved for one value. It is the natural next step from Questions 3 to 7 here.
• The End-of-Chapter Exercises, Questions 3 and 4 — two more word problems of exactly this shape, one with fractions and one where a wrong choice of unknown leads to a negative answer that has to be rejected.