Chapter 2 Introduction to linear polynomial

Class 9 Ganita Manjari Textbook solution

Exercise Set 2.1 (Pages 3-4)

1. Find the degrees of the following polynomials:

• (i) $2x^{2}-5x+3$
  • Solution: The highest power of $x$ in this polynomial is 2, so the degree is 2.
• (ii) $y^{3}+2y-1$
  • Solution: The highest power of $y$ is 3, so the degree is 3.
• (iii) -9
  • Solution: The constant -9 can be written as $-9x^{0}$ in which the power of the variable is 0. Therefore, the degree is 0.
• (iv) $4z-3$
  • Solution: The highest power of $z$ is 1, so the degree is 1.

2. Write polynomials of degrees 1, 2 and 3.

• Solution: (Note: These are just examples, as there are infinite correct answers)
  • Degree 1 (Linear): $5x-2$
  • Degree 2 (Quadratic): $x^{2}+4x+4$
  • Degree 3 (Cubic): $2y^{3}-y+8$

3. What are the coefficients of $x^{2}$ and $x^{3}$ in the polynomial $x^{4}-3x^{3}+6x^{2}-2x+7$?

• Solution: The coefficient of $x^{2}$ is 6, and the coefficient of $x^{3}$ is -3.

4. What is the coefficient of z in the polynomial $4z^{3}+5z^{2}-11$?

• Solution: Since there is no $z$ term (which means it is equivalent to $0z$), the coefficient is 0.

5. What is the constant term of the polynomial $9x^{3}+5x^{2}-8x-10$?

• Solution: The constant term is -10.

EXERCISE SET 2.2 (Page 6-7)

1. Find the value of the linear polynomial $5x-3$ if:

• (i) $x=0$
  • Solution: $5(0) – 3 = -3$
• (ii) $x=-1$
  • Solution: $5(-1) – 3 = -8$
• (iii) $x=2$
  • Solution: $5(2) – 3 = 7$

2. Find the value of the quadratic polynomial $7s^{2}-4s+6$ if:

• (i) $s=0$
  • Solution: $7(0)^{2} – 4(0) + 6 = 6$
• (ii) $s=-3$
  • Solution: $7(-3)^{2} – 4(-3) + 6 = 63 + 12 + 6 = 81$
• (iii) $s=4$
  • Solution: $7(4)^{2} – 4(4) + 6 = 112 – 16 + 6 = 102$

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 be $x$, making his mother’s age $3x$. In 5 years, their ages will be $(x + 5)$ and $(3x + 5)$. Setting up the equation: $(x + 5) + (3x + 5) = 70$, which simplifies to $4x + 10 = 70$. Solving for $x$ gives $4x = 60$, or $x = 15$. Salil is 15 years old, and his mother is 45 years old.

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

• Solution: Let the integers be $2x$ and $5x$. The equation is $5x – 2x = 63$, which simplifies to $3x = 63$, or $x = 21$. The two integers are $2(21) = 42$ and $5(21) = 105$.

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 the number of ₹5 coins be $x$. The number of ₹2 coins is $3x$. The total value equation is $5(x) + 2(3x) = 88$. This gives $11x = 88$, so $x = 8$. She has 8 five-rupee coins and 24 two-rupee coins.

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 $x$ and the longer piece be $4x$. $x + 4x = 300$, so $5x = 300$, meaning $x = 60$. The lengths are 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: Let the width be $w$, making the length $2w + 3$. The perimeter equation is $2(w + 2w + 3) = 24$. Simplifying gives $6w + 6 = 24$, so $6w = 18$, and $w = 3$. The length is $2(3) + 3 = 9$. The dimensions are 3 cm by 9 cm.

EXERCISE SET 2.3 (Pages 8-9)

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: At the end of month 1: ₹650. Month 2: ₹800. Month 3: ₹950. The linear expression representing the total amount in the $n^{th}$ month is $500 + 150n$.

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 remaining after 1 hour: 111. After 2 hours: 102. After 3 hours: 93. The linear expression is $120 – 9n$.

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) Area = $13 \times 12 = 156$ cm²
  • (ii) Area = $13 \times 10 = 130$ cm²
  • (iii) Area = $13 \times 8 = 104$ cm²
  • The linear pattern is $13b$ (where $b$ is the 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: Base Area = $7 \times 11 = 77$ cm².
  • (i) Volume = $77 \times 5 = 385$ cm³
  • (ii) Volume = $77 \times 9 = 693$ cm³
  • (iii) Volume = $77 \times 13 = 1001$ cm³
  • The linear pattern is $77h$ (where $h$ is the 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: After 15 days, she will have read $15 \times 20 = 300$ pages. The remaining pages will be $500 – 300 = 200$. The linear expression for the $n^{th}$ day is $500 – 20n$.

EXERCISE SET 2.4 (Pages 10-11)

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.
  • Solution: $1.75 + 7(0.5) = 5.25$ feet.
• (ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
  • Solution: $t=0$: 1.75, $t=1$: 2.25, $t=2$: 2.75, $t=3$: 3.25… up to $t=10$: 6.75.
• (iii) Find an expression that relates h and t, and explain why it represents linear growth.
  • Solution: The relationship is $h(t) = 1.75 + 0.5t$. This represents linear growth because the height $h$ increases by a constant amount (0.5 feet) over equal intervals (every 1 month).

2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.

• (i) Find the value of the phone after 3 years.
  • Solution: $10000 – 3(800) = 7600$.
• (ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.
  • Solution: $t=0$: 10,000, $t=1$: 9,200, $t=2$: 8,400… up to $t=8$: 3,600.
• (iii) Find an expression that relates v and t, and explain why it represents linear decay.
  • Solution: The relationship is $v(t) = 10000 – 800t$. It represents linear decay because the value decreases by a constant ₹800 over equal one-year intervals.

3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.

• (i) Find the population of the village after 6 years.
  • Solution: $750 + 6(50) = 1050$.
• (ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
  • Solution: $t=0$: 750, $t=1$: 800, $t=2$: 850… up to $t=10$: 1250.
• (iii) Find an expression that relates P and t, and explain why it represents linear growth.
  • Solution: $P(t) = 750 + 50t$. It shows linear growth as the population increases by a constant 50 people per year.

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 that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
  • Solution: $b(x) = 600 – 15x$. It represents linear decay because the quantity (balance) drops by a fixed amount (₹15) over equal time intervals (days).
• (ii) After how many days will the balance run out?
  • Solution: Set $600 – 15x = 0$. $15x = 600$, resulting in $x = 40$ days.
• (iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.
  • Solution: $x=1$: 585, $x=2$: 570, $x=3$: 555… up to $x=10$: 450.

EXERCISE SET 2.5 (Page 12)

1. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation $y=ax+b$ find the values of a and b.

• Solution: Set up a system of equations: $10a + b = 400$ and $14a + b = 500$. Subtract the first from the second: $4a = 100 \Rightarrow a = 25$. Substitute $a$ back into the first equation: $10(25) + b = 400 \Rightarrow b = 150$. So, $a = 25$ and $b = 150$.

2. A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation $y=ax+b$ find the values of a and b.

• Solution: $10a + b = 800$ and $15a + b = 1100$. Subtracting gives $5a = 300$, so $a = 60$. Using the first equation to solve for b: $10(60) + b = 800$, which gives $b = 200$. So, $a = 60$ and $b = 200$.

3. Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by $°C = a°F + b$. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.

• Solution: Using the hint, when $°F = 32$, $°C = 0$. So, $0 = 32a + b$. When $°F = 212$, $°C = 100$. So, $100 = 212a + b$. Subtracting the first from the second gives $100 = 180a \Rightarrow a = 100/180 = 5/9$. Substituting back gives $b = -32(5/9) = -160/9$.

EXERCISE SET 2.6 (Pages 21-22)

1. Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’. (Note: Graphical descriptions provided instead of drawn images)

• (i) $y=4x$, $y=2x$, $y=x$
  • Solution: Here, the y-intercept $b$ is 0, so all lines pass through the origin. The value of $a$ changes the slope. As $a$ increases (from 1 to 2 to 4), the line becomes steeper.
• (ii) $y=-6x$, $y=-3x$, $y=-x$
  • Solution: Again, $b=0$ so lines pass through the origin. These represent linear decay. As the negative value of $a$ becomes larger in magnitude, the line slopes downward more steeply.
• (iii) $y=5x$, $y=-5x$
  • Solution: Both lines pass through the origin and have the same steepness, but $y=5x$ rises (positive slope) while $y=-5x$ falls (negative slope).
• (iv) $y=3x-1$, $y=3x$, $y=3x+1$
  • Solution: Since $a=3$ is fixed, these three lines are parallel to each other. The values of $b$ (-1, 0, 1) shift the lines up or down on the y-axis (these are the y-intercepts).
• (v) $y=-2x-3$, $y=-2x$, $y=2x+3$
  • Solution: $y=-2x-3$ and $y=-2x$ are parallel descending lines with different y-intercepts (-3 and 0). $y=2x+3$ is an ascending line intersecting the y-axis at 3.

END-OF-CHAPTER EXERCISES (Pages 21-25)

1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the $x^{2}$ term is -7.

• Solution: Example: $x^{3} – 7x^{2} + 4x – 1$.

2. Find the values of the following polynomials at the indicated values of the variables.

• (i) $5x^{2}-3x+7$ if $x=1$
  • Solution: $5(1)^{2} – 3(1) + 7 = 5 – 3 + 7 = 9$.
• (ii) $4t^{3}-t^{2}+6$ if $t=a$
  • Solution: Substituting $a$ in place of $t$ gives $4a^{3} – a^{2} + 6$.

3. If we multiply a number by 5/2 and add 2/3 to the product, we get -7/12. Find the number.

• Solution: $\frac{5}{2}x + \frac{2}{3} = \frac{-7}{12}$. Move $\frac{2}{3}$ (which is $\frac{8}{12}$) to the right side: $\frac{5}{2}x = \frac{-7}{12} – \frac{8}{12} = \frac{-15}{12} = \frac{-5}{4}$. Then, $x = \frac{-5}{4} \times \frac{2}{5} = \frac{-1}{2}$.

4. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

• Solution: Let the smaller number be $x$, and the larger be $5x$. Adding 21 gives $(x+21)$ and $(5x+21)$. The equation is $5x+21 = 2(x+21)$. This simplifies to $5x+21 = 2x+42 \Rightarrow 3x = 21 \Rightarrow x = 7$. The two numbers are 7 and 35.

5. If you have ₹800 and you save ₹250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.

• Solution: The linear pattern is $Amount = 800 + 250m$.
  • (i) After 6 months: $800 + 250(6) = 2300$.
  • (ii) After 2 years (24 months): $800 + 250(24) = 6800$.

6. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.

• Solution: Let the digits be $x$ and $y$. Original number is $10x + y$; interchanged number is $10y + x$. The sum is $11x + 11y = 143$, so $x + y = 13$. Since the digits differ by 3, either $x – y = 3$ or $y – x = 3$. Solving the system $x + y = 13$ and $x – y = 3$ yields $2x = 16 \Rightarrow x = 8$, which means $y = 5$. The possible numbers are 85 and 58.

7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis. (i) $y=-3x+4$ (ii) $2y=4x+7$ (iii) $5y=6x-10$ (iv) $3y=6x-11$. Are any of the lines parallel?

• Solution: * (i) Slope = -3, y-intercept = 4. Cuts y-axis at (0, 4).
  • (ii) Simplify to $y = 2x + \frac{7}{2}$. Slope = 2, y-intercept = 3.5. Cuts y-axis at (0, 3.5).
  • (iii) Simplify to $y = \frac{6}{5}x – 2$. Slope = 1.2, y-intercept = -2. Cuts y-axis at (0, -2).
  • (iv) Simplify to $y = 2x – \frac{11}{3}$. Slope = 2, y-intercept = $\frac{-11}{3}$. Cuts y-axis at (0, $\frac{-11}{3}$).
  • Parallel: Yes, lines (ii) and (iv) are parallel because they share the same slope ($a=2$).

8. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation $y = \frac{9}{5}(x – 273) + 32$.

• (i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
  • Solution: $y = \frac{9}{5}(313 – 273) + 32 = \frac{9}{5}(40) + 32 = 72 + 32 = 104$ °F.
• (ii) If the temperature is 158 °F, then find the temperature in Kelvin.
  • Solution: $158 = \frac{9}{5}(x – 273) + 32 \Rightarrow 126 = \frac{9}{5}(x – 273)$. Multiply both sides by 5/9: $70 = x – 273$, resulting in $x = 343$ K.

9. The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units?

• Solution: The equation is $w = 3d$. When the distance $d=2$, the work done is $w = 3(2) = 6$ units.

10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11). (i) Find the polynomial p(x). (ii) Find the coordinates where the graph of p(x) cuts the axes. (iii) Draw the graph of p(x) and verify your answers.

• Solution:
  • (i) Use $p(x) = ax + b$. Point (1,5) yields $a + b = 5$. Point (3,11) yields $3a + b = 11$. Subtracting gives $2a = 6 \Rightarrow a = 3$. Therefore $b = 2$. The polynomial is $p(x) = 3x + 2$.
  • (ii) Cuts the y-axis where $x=0$, so (0, 2). Cuts the x-axis where $p(x)=0 \Rightarrow 3x+2=0 \Rightarrow x = \frac{-2}{3}$, so coordinates are ($\frac{-2}{3}$, 0).

11. Let $p(x)=ax+b$ and $q(x)=cx+d$ be two linear polynomials such that: (i) $p(0)=5$. (ii) The polynomial $p(x)-q(x)$ cuts the x-axis at (3, 0). (iii) The sum $p(x)+q(x)$ is equal to $6x+4$ for all real x. Find the polynomials p(x) and q(x).

• Solution:
  • From (i), $p(0) = 5$, so $b = 5$.
  • From (iii), $(a+c)x + (b+d) = 6x + 4$. Matching coefficients, $a+c = 6$ and $b+d = 4$. Since $b=5$, then $d = -1$.
  • From (ii), $p(3) – q(3) = 0 \Rightarrow p(3) = q(3)$. Thus, $3a + b = 3c + d$. Substitute $b=5$ and $d=-1$: $3a + 5 = 3c – 1 \Rightarrow 3a – 3c = -6 \Rightarrow a – c = -2$.
  • Solving the system $a+c=6$ and $a-c=-2$ gives $2a=4 \Rightarrow a=2$. So $c=4$.
  • Therefore, $p(x) = 2x + 5$ and $q(x) = 4x – 1$.

12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.

• (i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?
  • Solution: Stage 1 uses 6 matchsticks. Stage 2 uses 11 (adds 5). Stage 3 uses 16 (adds 5). Stage 4 will use 21. Stage 5 will use 26.
• (ii) Complete the following table.
  • Solution: Stage 1: 6, Stage 2: 11, Stage 3: 16, Stage 4: 21, Stage 5: 26.
• (iii) Find a rule to determine the number of matchsticks required for the $n^{th}$ stage.
  • Solution: The rule is $5n + 1$.
• (iv) How many matchsticks will be required for the 15th stage of the pattern?
  • Solution: $5(15) + 1 = 76$ matchsticks.
• (v) Can 200 matchsticks form a stage in this pattern? Justify your answer.
  • Solution: If $5n + 1 = 200$, then $5n = 199$, which means $n = 39.8$. Since the stage number $n$ must be an integer, 200 matchsticks cannot form a stage.

13. Let $p(x)=ax+b$ and $q(x)=cx+d$ be two linear polynomials such that: (i) The graph of p(x) passes through the points (2, 3) and (6, 11). (ii) The graph of q(x) passes through the point (4, -1). (iii) The graph of q(x) is parallel to the graph of p(x). Find the polynomials p(x) and q(x) Also, find the coordinates of the point where these lines meet the x-axis.

• Solution: * Find $p(x)$: $2a + b = 3$ and $6a + b = 11$. Subtracting gives $4a = 8 \Rightarrow a = 2$. Using $2(2) + b = 3$, we get $b = -1$. So, $p(x) = 2x – 1$.
  • Find $q(x)$: Parallel means $c = a = 2$. So $q(x) = 2x + d$. It passes through (4, -1), meaning $2(4) + d = -1 \Rightarrow 8 + d = -1 \Rightarrow d = -9$. So, $q(x) = 2x – 9$.
  • X-axis intercepts occur where $y=0$. For $p(x)$, $2x – 1 = 0 \Rightarrow x = 0.5$, coordinate is (0.5, 0). For $q(x)$, $2x – 9 = 0 \Rightarrow x = 4.5$, coordinate is (4.5, 0).

14. What do all linear functions of the form $f(x)=ax+a$ have in common? ($a>0$)

• Solution: If you factor the equation, $f(x) = a(x+1)$. For any value of $a$, setting $x = -1$ yields $f(-1) = a(-1+1) = a(0) = 0$. Therefore, all such lines pass through the shared coordinate point (-1, 0).