April 10, 2026
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CHAPTER 2
Introduction to Linear Polynomials
Questions & Solutions | GANITA MANJARI | Class 9 Maths | Part I
EXERCISE SET 2.1 (Page 18)
1.
Find the degrees of the following polynomials:
(i) \(2x^2 - 5x + 3\)
(ii) \(y^3 + 2y - 1\)
(iii) \(-9\)
(iv) \(4z - 3\)
(i) \(2x^2 - 5x + 3\)
(ii) \(y^3 + 2y - 1\)
(iii) \(-9\)
(iv) \(4z - 3\)
Solution
(i) 2 (ii) 3 (iii) 0 (iv) 1
2.
Write polynomials of degrees 1, 2 and 3.
Solution
Degree 1: \(2x + 5\)
Degree 2: \(x^2 + 3x + 1\)
Degree 3: \(x^3 - 2x^2 + x + 4\)
Degree 2: \(x^2 + 3x + 1\)
Degree 3: \(x^3 - 2x^2 + x + 4\)
3.
What are the coefficients of \(x^2\) and \(x^3\) in the polynomial \(x^4 - 3x^3 + 6x^2 - 2x + 7\) ?
Solution
Coefficient of \(x^3\) is \(-3\), coefficient of \(x^2\) is \(6\).
4.
What is the coefficient of \(z\) in the polynomial \(4z^3 + 5z^2 - 11\) ?
Solution
0
5.
What is the constant term of the polynomial \(9x^3 + 5x^2 - 8x - 10\) ?
Solution
\(-10\)
Think and Reflect
Page 17
1. Can you identify the terms, variables and coefficients of this algebraic expression?
2. How is it different from the algebraic expression in Example 1?
2. How is it different from the algebraic expression in Example 1?
Answer
1. Terms: \(200l\), \(160w\), \(50lw\). Variables: \(l\) and \(w\). Coefficients: 200, 160, 50.
2. Example 1 has two variables but no product term; Example 2 has a product term \(lw\).
2. Example 1 has two variables but no product term; Example 2 has a product term \(lw\).
Page 17
1. Can you identify the terms, variables and coefficients of this algebraic expression?
2. Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?
2. Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?
Answer
1. Terms: \(10x\) and \(-x^2\). Variable: \(x\). Coefficients: 10 and \(-1\).
2. Example 1 has two variables; Example 3 has one variable. Both have constant difference pattern.
2. Example 1 has two variables; Example 3 has one variable. Both have constant difference pattern.
Page 19
Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?
Answer
Perimeters: 4 cm, 6 cm, 8 cm, 10 cm, 12 cm. Perimeter increases by 2 cm each time.
Page 19
If a player paid ₹750, how many matches did he play?
Answer
200 + 50m = 750 → 50m = 550 → m = 11 matches.
Page 20
We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, \(10x - x^2\), is a function of \(x\). Can you interpret this as an input-output process? What value does the expression take when \(x = 6\) cm?
Answer
When x = 6, area = 10(6) - 6^2 = 60 - 36 = 24 square cm.
Page 22
Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern.
Answer
Stage 5: 9, Stage 6: 11, Stage 7: 13. Sequence: 1, 3, 5, 7, 9, 11, 13.
Page 22
Using the expression \(2n - 1\), can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?
Answer
15th stage: 2(15)-1 = 29; 26th stage: 2(26)-1 = 51. 21 tiles → n = 11 (11th stage). 47 tiles → n = 24 (24th stage).
Page 23
What amount will be left on the 15th day? How many days will it take for the entire amount to be spent?
Answer
Amount left on 15th day = 100 - 15×5 = 25 rupees. Entire amount spent when 100 - 5n = 0 → n = 20 days.
Page 23
For how many km will the fare be ₹130?
Answer
For n ≥ 2: Fare = 15n - 5. 15n - 5 = 130 → 15n = 135 → n = 9 km.
Page 24
What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be ₹700?
Answer
For 15 km: 100 + 60×15 = ₹1000. For ₹700: 100 + 60d = 700 → d = 10 km.
Page 25
What will be the height of the water at the end of 5 months?
Answer
h(5) = 3 - 0.5×5 = 3 - 2.5 = 0.5 metres.
Page 27
Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent?
Answer
20 = cost per GB of data, 150 = fixed monthly fee.
Page 28
Identify other points on the line by completing the following table.
x: 1, 2, 5, 7, 9, 12, 20
y: 3, 15
x: 1, 2, 5, 7, 9, 12, 20
y: 3, 15
Answer
For y = 2x + 1: x=1→3, x=2→5, x=5→11, x=7→15, x=9→19, x=12→25, x=20→41.
Page 33
Differentiate between the graphs of the equations y = 3x + 1, and y = -3x + 1.
Answer
y = 3x + 1 has positive slope (increasing line); y = -3x + 1 has negative slope (decreasing line). Both have same y-intercept (0,1).
Page 35
Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?
Answer
Lines shift vertically but remain parallel (same slope).
EXERCISE SET 2.2 (Page 21)
1.
Find the value of the linear polynomial \(5x - 3\) if:
(i) \(x = 0\)
(ii) \(x = -1\)
(iii) \(x = 2\)
(i) \(x = 0\)
(ii) \(x = -1\)
(iii) \(x = 2\)
Solution
(i) -3 (ii) -8 (iii) 7
2.
Find the value of the quadratic polynomial \(7s^2 - 4s + 6\) if:
(i) \(s = 0\)
(ii) \(s = -3\)
(iii) \(s = 4\)
(i) \(s = 0\)
(ii) \(s = -3\)
(iii) \(s = 4\)
Solution
(i) 6 (ii) 81 (iii) 102
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 age = x. Mother's age = 3x. (x+5)+(3x+5)=70 → 4x+10=70 → x=15. Salil is 15 years, mother is 45 years.
4.
The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Solution
Let integers be 2k and 5k. 5k - 2k = 63 → k = 21. Integers are 42 and 105.
5.
Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total ₹88, how many coins does she have of each type?
Solution
Let ₹5 coins = x. Then ₹2 coins = 3x. 5x + 2(3x) = 88 → 11x = 88 → x = 8. ₹5 coins = 8, ₹2 coins = 24.
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
Shorter = x, longer = 4x. x + 4x = 300 → x = 60. Pieces: 60 feet and 240 feet.
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
Width = x, Length = 2x + 3. 2(2x+3 + x) = 24 → 2(3x+3) = 24 → 6x+6=24 → x=3. Width = 3 cm, Length = 9 cm.
EXERCISE SET 2.3 (Pages 23-24)
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 nth month.
Solution
Amount = 500 + 150n
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, ... hours? Find a linear expression to represent the number of members at the end of the nth hour.
Solution
Members = 120 - 9n
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
(i) 156 sq cm, (ii) 130 sq cm, (iii) 104 sq cm. Area = 13 × breadth
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
(i) 385 cubic cm, (ii) 693 cubic cm, (iii) 1001 cubic cm. Volume = 77 × height
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
Pages read in 15 days = 300. Pages left = 200. Pages left = 500 - 20d
EXERCISE SET 2.4 (Pages 25-26)
1.
Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.
(i) Find the height after 7 months.
(ii) Make a table of values for t varying from 0 to 10 months.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
(i) Find the height after 7 months.
(ii) Make a table of values for t varying from 0 to 10 months.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
Solution
(i) 1.75 + 7×0.5 = 5.25 feet
(ii) t=0→1.75, t=1→2.25, t=2→2.75, t=3→3.25, t=4→3.75, t=5→4.25, t=6→4.75, t=7→5.25, t=8→5.75, t=9→6.25, t=10→6.75
(iii) h = 1.75 + 0.5t. Constant increase → linear growth.
(ii) t=0→1.75, t=1→2.25, t=2→2.75, t=3→3.25, t=4→3.75, t=5→4.25, t=6→4.75, t=7→5.25, t=8→5.75, t=9→6.25, t=10→6.75
(iii) h = 1.75 + 0.5t. Constant increase → linear growth.
2.
A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.
(i) Find the value after 3 years.
(ii) Make a table for t from 0 to 8 years.
(iii) Find an expression for v and t, and explain why it represents linear decay.
(i) Find the value after 3 years.
(ii) Make a table for t from 0 to 8 years.
(iii) Find an expression for v and t, and explain why it represents linear decay.
Solution
(i) 10000 - 3×800 = ₹7600
(ii) t=0→10000, t=1→9200, t=2→8400, t=3→7600, t=4→6800, t=5→6000, t=6→5200, t=7→4400, t=8→3600
(iii) v = 10000 - 800t. Constant decrease → linear decay.
(ii) t=0→10000, t=1→9200, t=2→8400, t=3→7600, t=4→6800, t=5→6000, t=6→5200, t=7→4400, t=8→3600
(iii) v = 10000 - 800t. Constant decrease → linear decay.
3.
The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
(i) Find population after 6 years.
(ii) Make a table for t from 0 to 10 years.
(iii) Find expression for P and t, and explain why it represents linear growth.
(i) Find population after 6 years.
(ii) Make a table for t from 0 to 10 years.
(iii) Find expression for P and t, and explain why it represents linear growth.
Solution
(i) 750 + 6×50 = 1050
(ii) t=0→750, t=1→800, t=2→850, t=3→900, t=4→950, t=5→1000, t=6→1050, t=7→1100, t=8→1150, t=9→1200, t=10→1250
(iii) P = 750 + 50t. Constant increase → linear growth.
(ii) t=0→750, t=1→800, t=2→850, t=3→900, t=4→950, t=5→1000, t=6→1050, t=7→1100, t=8→1150, t=9→1200, t=10→1250
(iii) P = 750 + 50t. Constant increase → linear growth.
4.
A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.
(i) Write an equation for the remaining balance b(x) after x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table for x from 1 to 10 days.
(i) Write an equation for the remaining balance b(x) after x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table for x from 1 to 10 days.
Solution
(i) b(x) = 600 - 15x. Constant decrease → linear decay.
(ii) 600 - 15x = 0 → x = 40 days.
(iii) x=1→585, x=2→570, x=3→555, x=4→540, x=5→525, x=6→510, x=7→495, x=8→480, x=9→465, x=10→450
(ii) 600 - 15x = 0 → x = 40 days.
(iii) x=1→585, x=2→570, x=3→555, x=4→540, x=5→525, x=6→510, x=7→495, x=8→480, x=9→465, x=10→450
EXERCISE SET 2.5 (Page 27)
1.
A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. When she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If y = ax + b, find a and b.
Solution
400 = 10a + b, 500 = 14a + b → 100 = 4a → a = 25, b = 150. y = 25x + 150.
2.
A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. For 10 hours, bill was ₹800. For 15 hours, bill was ₹1100. If y = ax + b, find a and b.
Solution
800 = 10a + b, 1100 = 15a + b → 300 = 5a → a = 60, b = 200. y = 60x + 200.
3.
Consider the relationship between temperature measured in degrees Celsius (C) and degrees Fahrenheit (F), which is given by C = aF + b. Find a and b, given that ice melts at 0°C = 32°F and water boils at 100°C = 212°F.
Solution
0 = 32a + b, 100 = 212a + b → 100 = 180a → a = 5/9, b = -160/9. C = (5/9)(F - 32).
EXERCISE SET 2.6 (Page 36)
1.
Draw the graphs of the following sets of lines. In each case, reflect on the role of 'a' and 'b'.
y = 4x, y = 2x, y = x
y = -6x, y = -3x, y = -x
y = 5x, y = -5x
y = 3x - 1, y = 3x, y = 3x + 1
y = -2x - 3, y = -2x, y = 2x + 3
y = 4x, y = 2x, y = x
y = -6x, y = -3x, y = -x
y = 5x, y = -5x
y = 3x - 1, y = 3x, y = 3x + 1
y = -2x - 3, y = -2x, y = 2x + 3
Solution
'a' (slope) controls steepness and direction. 'b' (y-intercept) shifts the line vertically. Lines with same slope are parallel.
END-OF-CHAPTER EXERCISES (Pages 36-40)
1.
Write a polynomial of degree 3 in the variable x, in which the coefficient of the x^2 term is -7.
Solution
x^3 - 7x^2 + 2x + 1 (any such polynomial)
2.
Find the values: 5x^2 - 3x + 7 if x = 1, and 4t^3 - t^2 + 6 if t = a.
Solution
5(1)^2 - 3(1) + 7 = 9. 4a^3 - a^2 + 6.
3.
If we multiply a number by 5/2 and add 2/3 to the product, we get -7/12. Find the number.
Solution
Let number = x. (5/2)x + 2/3 = -7/12 → (5x/2) = -15/12 = -5/4 → 5x = -10/4 = -5/2 → x = -1/2
4.
A positive number is 5 times another number. If 21 is added to both, then one becomes twice the other. Find the numbers.
Solution
Smaller = x, larger = 5x. 5x + 21 = 2(x + 21) → 3x = 21 → x = 7. Numbers are 7 and 35.
5.
If you have ₹800 and save ₹250 every month, find the amount after (i) 6 months (ii) 2 years. Express as a linear pattern.
Solution
Amount = 800 + 250n. (i) ₹2300 (ii) ₹6800.
*6.
The digits of a two-digit number differ by 3. If the digits are interchanged and added to the original number, we get 143. Find both numbers.
Solution
Let tens = x, units = y. 10x+y + 10y+x = 143 → 11(x+y)=143 → x+y=13. Also |x-y|=3. Solving: x=8,y=5 or x=5,y=8. Numbers are 85 and 58.
*7.
Draw the graphs, identify slopes and y-intercepts, find where they cut y-axis. Are any parallel?
(i) y = -3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x - 10
(iv) 3y = 6x - 11
(i) y = -3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x - 10
(iv) 3y = 6x - 11
Solution
(i) slope = -3, y-int = 4, cuts (0,4)
(ii) y = 2x + 7/2 → slope = 2, y-int = 7/2, cuts (0,3.5)
(iii) y = (6/5)x - 2 → slope = 6/5, y-int = -2, cuts (0,-2)
(iv) y = 2x - 11/3 → slope = 2, y-int = -11/3, cuts (0,-3.67)
(ii) and (iv) have slope 2 → parallel.
(ii) y = 2x + 7/2 → slope = 2, y-int = 7/2, cuts (0,3.5)
(iii) y = (6/5)x - 2 → slope = 6/5, y-int = -2, cuts (0,-2)
(iv) y = 2x - 11/3 → slope = 2, y-int = -11/3, cuts (0,-3.67)
(ii) and (iv) have slope 2 → parallel.
*8.
The relation between Kelvin (x) and Fahrenheit (y) is y = (9/5)(x - 273) + 32.
(i) Find y when x = 313 K.
(ii) Find x when y = 158°F.
(i) Find y when x = 313 K.
(ii) Find x when y = 158°F.
Solution
(i) y = (9/5)(40) + 32 = 72 + 32 = 104°F
(ii) 158 = (9/5)(x - 273) + 32 → 126 = (9/5)(x - 273) → x - 273 = 70 → x = 343 K.
(ii) 158 = (9/5)(x - 273) + 32 → 126 = (9/5)(x - 273) → x - 273 = 70 → x = 343 K.
*9.
Work done = force × distance. Express as linear equation (force = 3 units). Find work when distance = 2 units.
Solution
w = 3d. When d = 2, w = 6 units.
10.
The graph of p(x) passes through (1,5) and (3,11). (i) Find p(x). (ii) Find where it cuts the axes.
Solution
p(x) = 3x + 2. x-intercept: (-2/3, 0). y-intercept: (0,2).
11.
Let p(x)=ax+b, q(x)=cx+d. Given: p(0)=5, p(x)-q(x) cuts x-axis at (3,0), p(x)+q(x)=6x+4. Find p and q.
Solution
p(x)=2x+5, q(x)=4x-1.
12.
Growing pattern of hexagons made using matchsticks. A new hexagon shares a side with the previous stage.
(i) Draw next two stages. How many matchsticks?
(ii) Complete table.
(iii) Find rule for nth stage.
(i) Draw next two stages. How many matchsticks?
(ii) Complete table.
(iii) Find rule for nth stage.
Solution
(i) Stage 4 = 21, Stage 5 = 26 matchsticks.
(ii) Table: 6, 11, 16, 21, 26, ...
(iii) Rule: M_n = 5n + 1
(ii) Table: 6, 11, 16, 21, 26, ...
(iii) Rule: M_n = 5n + 1
13.
p(x)=ax+b passes through (2,3) and (6,11). q(x)=cx+d passes through (4,-1) and is parallel to p(x). Find p and q. Find x-intercepts.
Solution
p(x)=2x-1, q(x)=2x-9. x-intercepts: p: (0.5,0), q: (4.5,0).
14.
What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Solution
All have positive slope, cut y-axis above origin, and pass through (-1,0).
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