NCERT Solutions: Introduction to Linear Polynomials

Think and Reflect (Page 17)

Set 1
1. Identify the terms, variables and coefficients of the algebraic expression:
200l + 160w + 50lw

Answer::
Terms: 200l, 160w, 50lw
Variables: l, w
Coefficients:

• 200 → coefficient of l
• 160 → coefficient of w
• 50 → coefficient of lw

2. How is this expression different from the expression in Example 1 (4x + 5y + 3)?

Answer::

• The expression 200l + 160w + 50lw has a term (50lw) with two variables multiplied together, while 4x + 5y + 3 has only one variable in each term.
• The expression 4x + 5y + 3 has a constant term (3), but 200l + 160w + 50lw does not have any constant term.
• Hence, 200l + 160w + 50lw is more complex than 4x + 5y + 3.

Think and Reflect (Page 17)

Set 2

1. Identify the terms, variables and coefficients of the algebraic expression:
x(10 – x) or 10x – x²
Answer::
Terms: 10x, -x²
Variables: x
Coefficients:
10 → coefficient of x
-1 → coefficient of x²

2. Can you point out any similarity or difference between the algebraic expressions obtained in Example 1, 2 and 3?
Example 1: 4x + 5y + 3
Example 2: 200l + 160w + 50lw
Example 3: x(10 – x) or 10x – x²
Answer::
Similarity:

• All three are algebraic expressions.
• Each expression contains variables and coefficients.
• Each is made up of terms joined by addition or subtraction.

Difference:

• Example 1: Linear, has constant (3), single variables.
• Example 2: Has a term with two variables (lw), no constant.
• Example 3: Has a power (x²), so it is not linear (quadratic).

Exercise Set 2.1 (Page 18)

Q1. Find the degrees of the following polynomials:
(i) 2x² – 5x + 3   (ii) y³ + 2y – 1   (iii) -9   (iv) 4z – 3

Answer:: 

(i) 2x² – 5x + 3: The highest power of x is 2. Degree = 2 (Quadratic polynomial).

(ii) y³ + 2y – 1: The highest power of y is 3. Degree = 3 (Cubic polynomial).

(iii) -9: This is a constant, written as -9x⁰. Degree = 0 (Constant polynomial).

(iv) 4z – 3: The highest power of z is 1. Degree = 1 (Linear polynomial).

Q2. Write polynomials of degrees 1, 2 and 3.

Answer::

Degree 1 (Linear): $3x+5$
Degree 2 (Quadratic): $x^2-4x+7$
Degree 3 (Cubic): $2x^3+x^2-6x+1$

Q3. What are the coefficients of x² and x³ in the polynomial x⁴ – 3x³ + 6x² – 2x + 7?
Answer: The polynomial is: $x^4-3x^3+6x^2-2x+7$
Coefficient of x² = 6
Coefficient of x³ = -3

Q4. What is the coefficient of z in the polynomial 4z³ + 5z² – 11?

Answer: The polynomial is: $4z^3+5z^2+0\cdot z-11$
There is no explicit z term. The coefficient of z = 0.

 Q5. What is the constant term of the polynomial 9x³ + 5x² – 8x – 10?

Answer: The constant term is the term with no variable. In $9x^3+5x^2-8x-10$, the constant term = -10.

Think and Reflect (Page 19)

Q. 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:
Perimeter of a square = 4 × side

• Side = 1 cm → Perimeter = 4 cm
• Side = 1.5 cm → Perimeter = 6 cm
• Side = 2 cm → Perimeter = 8 cm
• Side = 2.5 cm → Perimeter = 10 cm
• Side = 3 cm → Perimeter = 12 cm
• When the side increases by 0.5 cm, the perimeter increases by 2 cm each time.

Q. If a player paid `750, how many matches did he play? (200 + 50m)
Answer: If the player paid ₹750, then:
200 + 50m = 750
50m = 750 – 200
50m = 550
m = 550 ÷ 50
m = 11

Think and Reflect (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 – x², 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: Yes, it can be interpreted as an input-output process where x is the input and the value of 10x – x² gives the output (area). When x = 6, the value is 10(6) – 6² = 60 – 36 = 24 cm².

Exercise Set 2.2 (Page 21)

Q1. Find the value of the linear polynomial 5x – 3 if: (i) x = 0   (ii) x = -1   (iii) x = 2
Answer:
Let $p(x)=5x-3$
(i) $p(0)=5(0)-3=-3$
(ii) $p(-1)=5(-1)-3=-5-3=-8$
(iii) $p(2)=5(2)-3=10-3=7$

Q2. Find the value of the quadratic polynomial 7s² – 4s + 6 if: (i) s = 0   (ii) s = -3   (iii) s = 4

Answer:

Let $p(s)=7s^2-4s+6$
(i) $p(0)=7(0{)}^2-4(0)+6=6$
(ii) $p(-3)=7(-3{)}^2-4(-3)+6=7(9)+12+6=63+12+6=81$
(iii) $p(4)=7(4{)}^2-4(4)+6=7(16)-16+6=112-16+6=102$

Q3. 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.

Answer:

Let Salil’s present age = x years.
Mother’s present age = 3x years.
Setting up the equation
After 5 years: Salil’s age = x + 5, Mother’s age = 3x + 5
Given: $(x+5)+(3x+5)=70$
$4x+10=70$
$4x=60⟹x=15$
Salil’s present age = 15 years
Mother’s present age = 3 × 15 = 45 years

Verification: After 5 years: 20 + 50 = 70 

Q4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.

Answer:

Let the two integers be 2k and 5k (since their ratio is 2:5).
Setting up the equation
Difference: $5k-2k=63$
$3k=63⟹k=21$
Smaller integer = 2 × 21 = 42
Larger integer = 5 × 21 = 105
Verification: 105 – 42 = 63 
Ratio = 42:105 = 2:5 

Q5. 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?

Answer:

Let number of five-rupee coins = x.
Then number of two-rupee coins = 3x.
Setting up the equation
Total value: $5x+2(3x)=88$
$5x+6x=88$
$11x=88⟹x=8$
Five-rupee coins = 8
Two-rupee coins = 3 × 8 = 24
Verification: 5(8) + 2(24) = 40 + 48 = 88 

Q6. 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?

Answer:

Let shorter piece = x feet. Then longer piece = 4x feet.
Setting up the equation
$x+4x=300$
$5x=300⟹x=60$
Shorter piece = 60 feet
Longer piece = 4 × 60 = 240 feet
Verification: 60 + 240 = 300 

Q7. 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?

Answer:

Let width = w cm. Then length = 2w + 3 cm.
Setting up the equation
Perimeter = 2(length + width) = 24
$2[(2w+3)+w]=24$
$2[3w+3]=24$
$6w+6=24$
$6w=18⟹w=3$
Width = 3 cm
Length = 2(3) + 3 = 9 cm
Verification: Perimeter = 2(9 + 3) = 2(12) = 24 cm

Think and Reflect (Page 22)

1. Predict the number of squares in the next three stages of the pattern and write the sequence up to Stage 7, given by 2n – 1.
Answer:

For n = 1: 2(1) – 1 = 1
For n = 2: 2(2) – 1 = 3
For n = 3: 2(3) – 1 = 5
For n = 4: 2(4) – 1 = 7
For n = 5: 2(5) – 1 = 9
For n = 6: 2(6) – 1 = 11
For n = 7: 2(7) – 1 = 13
Sequence: 1, 3, 5, 7, 9, 11, 13

2. Using the expression 2n – 1, find the number of tiles in the 15th and 26th stages. Also, find which stage has 21 tiles and 47 tiles.
Answer:
For n = 15: 2(15) – 1 = 30 – 1 = 29 tiles
For n = 26: 2(26) – 1 = 52 – 1 = 51 tiles
For 21 tiles:
2n – 1 = 21
2n = 22
n = 11 → 11th stage
For 47 tiles:
2n – 1 = 47
2n = 48
n = 24 → 24th stage

Think and Reflect (Page 23)

1. What amount will be left on the 15th day? How many days will it take for the entire amount to be spent? (Bela has ₹100 and spends ₹5 every day.)
Answer: Amount left after 15 days = 100 – (5 × 15) = 100 – 75 = ₹25
Total days to spend all money = 100 ÷ 5 = 20 days

2. An auto-rickshaw fare is given by 15n – 5 (for n ≥ 2). For how many km will the fare be ₹130?
Answer:
15n – 5 = 130
15n = 135
n = 9
So, the fare will be ₹130 for 9 km.

Exercise Set 2.3 (Page 23)

Q1. 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.
Answer:
Initial amount = ₹500. She adds ₹150 each month.

Month (n)	1	2	3	4
Amount (₹)	650	800	950	1100

Amount at end of n^th month:
$$A(n)=500+150n$$
This is a linear polynomial in n. The amount increases by a constant ₹150 each month (linear growth).

Q2. 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 n^th hour.

Answer:

Initial members = 120. Each hour, 9 members leave.

Hour (n)	1	2	3
Members	111	102	93

Members at end of n^th hour:
$$M(n)=120-9n$$
This is a linear polynomial with negative slope (linear decay). Members decrease by 9 each hour.

Q3. 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.

Answer:

Length = 13 cm. Area = 13 × breadth.
(i) Breadth = 12 cm: Area = 13 × 12 = 156 cm²
(ii) Breadth = 10 cm: Area = 13 × 10 = 130 cm²
(iii) Breadth = 8 cm: Area = 13 × 8 = 104 cm²
If breadth = b cm, then:
$$\text{Area}=13b{\text{ cm}}^2$$
This is a linear pattern: Area decreases by 26 cm² for every 2 cm decrease in breadth (constant difference = 26 when breadth changes by 2).

Q4. 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.

Answer:

Length = 7 cm, Breadth = 11 cm. Volume = 7 × 11 × height = 77 × height.
(i) Height = 5 cm: Volume = 77 × 5 = 385 cm³
(ii) Height = 9 cm: Volume = 77 × 9 = 693 cm³
(iii) Height = 13 cm: Volume = 77 × 13 = 1001 cm³
If height = h cm:
$$V=77h{\text{ cm}}^3$$
This is a linear pattern. Volume increases by 77 × 4 = 308 cm³ for every 4 cm increase in height (constant difference per unit = 77).

Q5. 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.

Answer:

Pages remaining after n days:$$P(n)=500-20n$$
After 15 days: $P(15)=500-20(15)=500-300=\mathbf{200}\text{ pages}$
This is a linear pattern (linear decay). Pages decrease by 20 each day (constant difference = 20).

Think and Reflect (Page 24)

The cost of travel is given by C=100+60d, where d is the distance in km. What is the cost for travelling 15 km? For how many kilometres will the cost be ₹700?Answer: 
For d = 15:
C = 100 + 60(15) = 100 + 900 = ₹1000
For C = 700:
100 + 60d = 700
60d = 600
d = 10
So, the cost for 15 km is ₹1000 and the cost will be ₹700 for 10 km.

Think and Reflect (Page 25)

The height of water is given by h(t)=3-0.5t.  What will be the height at the end of 5 months?
Answer: h(5) = 3 – 0.5(5) = 3 – 2.5 = 0.5 m

Exercise Set 2.4 (Page 25)

Q1. 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 and show how the height, h, increases every month.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.

Answer:

(i) Height after 7 months:
$h=1.75+0.5\times 7=1.75+3.5=\mathbf{5.25}\text{ feet}$

(ii) Table of values:

Month (t)	0	1	2	3	4	5	6	7	8	9	10
Height h (ft)	1.75	2.25	2.75	3.25	3.75	4.25	4.75	5.25	5.75	6.25	6.75

(iii) Expression:
$$h(t)=1.75+0.5t$$
This represents linear growth because the height increases by a fixed amount (0.5 feet) over equal intervals (every month). The slope is +0.5 (positive), confirming growth.

Q2. 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.
(ii) Make a table of values for t varying from 0 to 8 years.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.

Answer:

(i) Value after 3 years:
$v=10000-800\times 3=10000-2400=\mathbf{\text{₹}}\mathbf{7600}$

(ii) Table of values:

Year (t)	0	1	2	3	4	5	6	7	8
Value v (₹)	10000	9200	8400	7600	6800	6000	5200	4400	3600

(iii) Expression:

$$v(t)=10000-800t$$

This represents linear decay because the value decreases by a fixed amount (₹800) over equal intervals (every year). The slope is -800 (negative), confirming decay.

Q3. 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.
(ii) Make a table of values for t varying from 0 to 10 years.
(iii) Find an expression that relates P and t, and explain why it represents linear growth.

Answer:

(i) Population after 6 years:
$P=750+50\times 6=750+300=\mathbf{1050}$

(ii) Table of values:

Year (t)	0	1	2	3	4	5	6	7	8	9	10
Population P	750	800	850	900	950	1000	1050	1100	1150	1200	1250

(iii) Expression:
$$P(t)=750+50t$$
This represents linear growth because the population increases by a fixed amount (50 people) each year. The slope is +50 (positive).

Q4. 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.
(ii) After how many days will the balance run out?
(iii) Make a table of values for x varying from 1 to 10 days.

Answer:

(i) Balance after x days:
$$b(x)=600-15x$$
This represents linear decay: the balance decreases by ₹15 (a fixed amount) each day. Slope = -15 (negative).

(ii) Balance runs out when b(x) = 0:
$600-15x=0⟹15x=600⟹x=40$
Balance runs out after 40 days.
(iii) Table for x = 1 to 10:

Day (x)	1	2	3	4	5	6	7	8	9	10
Balance b(x) (₹)	585	570	555	540	525	510	495	480	465	450

Think and Reflect (Page 27)

In the equation y = 20x + 150, what do the numbers 20 and 150 represent?Answer: The number 20 represents the cost per GB of data (₹20 per GB), and 150 represents the fixed monthly fee (₹150).

Exercise Set 2.5 (Page 27)

Q1. 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.

Answer:

Given: y = ax + b
Step 1 – Form equations
When x = 10, y = 400: $10a+b=400\cdots (1)$
When x = 14, y = 500: $14a+b=500\cdots (2)$
Step 2 – Subtract (1) from (2)
$4a=100⟹a=25$
Step 3 – Substitute back
$b=400-10(25)=400-250=150$
a = 25, b = 150
The relation is: $y=25x+150$
Interpretation: Fixed monthly fee = ₹150, Cost per module = ₹25

Q2. 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.

Answer:

Given: y = ax + b
Step 1 – Form equations
When x = 10, y = 800: $10a+b=800\cdots (1)$
When x = 15, y = 1100: $15a+b=1100\cdots (2)$
Step 2 – Subtract (1) from (2)
$5a=300⟹a=60$
Step 3 – Substitute back
$b=800-10(60)=800-600=200$
a = 60, b = 200
The relation is: $y=60x+200$
Interpretation: Fixed monthly fee = ₹200, Cost per hour = ₹60

Q3. 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.

Answer:

Given: °C = a(°F) + b
Step 1 – Form equations
When °C = 0, °F = 32: $0=32a+b\cdots (1)$
When °C = 100, °F = 212: $100=212a+b\cdots (2)$
Step 2 – Subtract (1) from (2)
$100=180a⟹a=\frac{100}{180}=\frac{5}{9}$
Step 3 – Find b
From (1): $b=-32a=-32\times \frac{5}{9}=-\frac{160}{9}$
a=59,b=−1609
The relation is:
 $${}^{\circ }C=\frac{5}{9}({}^{\circ }F)-\frac{160}{9}=\frac{5}{9}({}^{\circ }F-32)$$
This is the well-known Celsius-Fahrenheit conversion formula.

Think and Reflect (Page 28)

Identify other points on the line y=2x+1 by completing the table.

Using $y=2x+1$:

• x = 2 → y = 5
• x = 5 → y = 11
• x = 9 → y = 19
• x = 12 → y = 25
• x = 20 → y = 41x125791220y351115192541

Think and Reflect (Page 33)

Differentiate between the graphs of the equations y = 3x + 1 and y = -3x + 1.
Answer: Both graphs are straight lines and have the same y-intercept (1). However, their slopes are different. The line y = 3x + 1 has a positive slope, so it rises from left to right. The line y = -3x + 1 has a negative slope, so it falls from left to right.

Think and Reflect (Page 35)

Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?
Answer: Yes. When a is fixed and b varies, the graphs are straight lines with the same slope, so they are parallel to each other. Changing b only shifts the line up or down without changing its slope.

Exercise Set 2.6 (Page 36)

Q 1 Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’.
(i) y = 4x, y = 2x, y = x
(ii) y = -6x, y = -3x, y = -x
(iii) y = 5x, y = -5x
(iv) y = 3x – 1, y = 3x, y = 3x + 1
(v) y = -2x – 3, y = -2x, y = 2x + 3

Answer:

(i) y = 4x, y = 2x, y = x
All three lines pass through the origin (0, 0) since b = 0. As the slope a increases (1 → 2 → 4), the line becomes steeper. All have positive slope → linear growth.(ii) y = -6x, y = -3x, y = -x
All three lines pass through the origin. As |a| increases (1 → 3 → 6), the line becomes steeper but falls from left to right. All have negative slope → linear decay.(iii) y = 5x, y = -5x
Both pass through the origin. y = 5x has slope +5 (growth, rises left to right); y = -5x has slope -5 (decay, falls left to right). They are reflections of each other across the x-axis.(iv) y = 3x – 1, y = 3x, y = 3x + 1
All three lines have slope a = 3 (same steepness). They are parallel to each other. The y-intercepts are -1, 0, and 1 respectively, shifting the lines up or down. Changing b shifts the line vertically.

(v) y = -2x – 3, y = -2x, y = 2x + 3
y = -2x – 3 and y = -2x are parallel (same slope -2, different intercepts). y = 2x + 3 has slope +2 – same magnitude but opposite sign to the other two. y = -2x – 3 and y = 2x + 3 have the same y-intercept only in magnitude but opposite slopes.

End-of-Chapter Exercises (Page 36)

Q1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is -7.

Answer: $x^3-7x^2+2x+5$
Degree = 3  |  Coefficient of $x^2$ = -7 

Q2. Find the values of the following polynomials at the indicated values of the variables.
(i) 5x² – 3x + 7 if x = 1
(ii) 4t³ – t² + 6 if t = a

Answer: 
(i) $5(1{)}^2-3(1)+7=5-3+7=\mathbf{9}$
(ii) Substituting $t=a$:$$4a^3-a^2+6$$

Q3. If we multiply a number by 5/2 and add 2/3 to the product, we get -7/12. Find the number.

Answer: Let the number be $x$.
$$\frac{5}{2}x+\frac{2}{3}=\frac{-7}{12}$$
$$\frac{5}{2}x=\frac{-7}{12}-\frac{8}{12}=\frac{-15}{12}=\frac{-5}{4}$$
$$x=\frac{-5}{4}\times \frac{2}{5}=-\frac{\mathbf{1}}{\mathbf{2}}$$

Q4. 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?

Answer: Let the smaller number = $x$. Larger = $5x$.
$$5x+21=2(x+21)⟹5x+21=2x+42⟹3x=21⟹x=7$$
The two numbers are 7 and 35.
$7+21=28$; $35+21=56=2\times 28$ 

Q5. 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.

Answer: Linear expression:$$A(n)=800+250n$$
(i) After 6 months: $A(6)=800+250(6)=800+1500=\mathbf{\text{₹}}\mathbf{2300}$
(ii) After 2 years = 24 months: $A(24)=800+250(24)=800+6000=\mathbf{\text{₹}}\mathbf{6800}$
Sequence: 1050, 1300, 1550, … – consecutive terms differ by constant 250 → linear pattern.

*Q6. 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.

Answer: Let tens digit = $x$, units digit = $y$, with $x-y=3$.
Original number $=10x+y$; reversed $=10y+x$.
$$(10x+y)+(10y+x)=143⟹11(x+y)=143⟹x+y=13$$
Solving $x+y=13$ and $x-y=3$:
$$2x=16⟹x=8,y=5$$
The numbers are 85 and 58.
 $85+58=143$ 

*Q7. 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?

Answer:Rewrite each in slope-intercept form $y=ax+b$:

Equation	Slope ($a$)	y-intercept ($b$)	Cuts y-axis at
(i) $y=-3x+4$	-3	4	$(0,4)$
(ii) $y=2x+{\displaystyle \frac{7}{2}}$	2	${\displaystyle \frac{7}{2}}$	$(0,{\displaystyle \frac{7}{2}})$
(iii) $y={\displaystyle \frac{6}{5}}x-2$	${\displaystyle \frac{6}{5}}$	-2	$(0,-2)$
(iv) $y=2x-{\displaystyle \frac{11}{3}}$	2	$-{\displaystyle \frac{11}{3}}$	$(0,-{\displaystyle \frac{11}{3}})$

Parallel lines: Lines (ii) and (iv) both have slope $a=2$ but different y-intercepts → they are parallel to each other.

*Q8.  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 = 9/5 (x – 273) + 32.
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.

Answer: (i)$$y=\frac{9}{5}(313-273)+32=\frac{9}{5}(40)+32=72+32=\mathbf{104}\mathbf{°}\mathbf{F}$$

(ii)$$158=\frac{9}{5}(x-273)+32⟹126=\frac{9}{5}(x-273)⟹x-273=70⟹x=\mathbf{343}\mathbf{K}$$

*Q9. 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? Verify it by plotting it on the graph.

Answer: 

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. Express this as a linear equation in two variables (work w and distance d), taking the constant force as 3 units. What is the work done when the distance travelled is 2 units?

Answer:
Work done = force × distance
So, w = 3d

When d = 2:
w = 3 × 2 = 6

Thus, the work done is 6 units.

*Q10. 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.

Answer: 
(i) Let p(x) = ax + b
Using (1, 5): a + b = 5
Using (3, 11): 3a + b = 11
Subtracting: 2a = 6 → a = 3
Then b = 5 – 3 = 2
So, p(x) = 3x + 2

(ii) To find intercepts:
y-intercept: x = 0 → y = 2 → (0, 2)
x-intercept: 3x + 2 = 0 → x = -2/3 → (-2/3, 0)*Q11. 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).
Answer: 

From (i): p(0) = b = 5 → p(x) = ax + 5
From (iii): p(x) + q(x) = 6x + 4
(ax + 5) + (cx + d) = 6x + 4
(a + c)x + (5 + d) = 6x + 4
So, a + c = 6 and 5 + d = 4 → d = -1
From (ii): p(x) – q(x) = 0 at x = 3
p(3) = q(3)
3a + 5 = 3c – 1
3a – 3c = -6
a – c = -2
Now solve:
a + c = 6
a – c = -2
Adding: 2a = 4 → a = 2
Then c = 4
Thus,
p(x) = 2x + 5
q(x) = 4x – 1

*Q12. 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?
(ii) Complete the following table.
Stage: 1, 2, 3, 4, 5, …, n
Number of matchsticks:
(iii) Find a rule to determine the number of matchsticks required for the nth stage.
(iv) How many matchsticks will be required for the 15th stage of the pattern?
(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.

Answer:  (i) Stage 4 = $6+3\times 5=\mathbf{21}$  |  Stage 5 = $6+4\times 5=\mathbf{26}$

(ii)

Stage ($n$)	1	2	3	4	5	…	$n$
Matchsticks	6	11	16	21	26	…	$5n+1$

(iii) Rule:$$\text{Matchsticks}=5n+1$$

(iv) Stage 15: $5(15)+1=\mathbf{76}$ matchsticks

(v) If $5n+1=200⟹n={\displaystyle \frac{199}{5}}=39.8$, which is not a whole number. Therefore, 200 matchsticks cannot form a stage in this pattern.

*Q13. 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.

Answer: Finding p(x):

Slope $a={\displaystyle \frac{11-3}{6-2}}={\displaystyle \frac{8}{4}}=2$

Using $(2,3)$: $2(2)+b=3⟹b=-1$   →   $\mathbf{p}(\mathbf{x})=\mathbf{2}\mathbf{x}-\mathbf{1}$

Finding q(x): Since parallel to $p(x)$, slope $c=2$.

Using $(4,-1)$: $2(4)+d=-1⟹d=-9$   →   $\mathbf{q}(\mathbf{x})=\mathbf{2}\mathbf{x}-\mathbf{9}$

x-intercepts (set y=0):

• $p(x)=0$: $2x-1=0⟹x={\displaystyle \frac{1}{2}}$ → $({\displaystyle \frac{1}{2}},0)$
• $q(x)=0$: $2x-9=0⟹x={\displaystyle \frac{9}{2}}$ → $({\displaystyle \frac{9}{2}},0)$

*Q14. What do all linear functions of the form f(x)=ax+a, a>0 have in common?

Answer: We can write $f(x)=ax+a=a(x+1)$.
Setting $f(x)=0$: $a(x+1)=0⟹x=-1$
All such lines pass through the fixed point (−1, 0) on the x-axis, regardless of the value of $a$.
Since both the slope and y-intercept equal $a$, as $a$ changes, the line pivots about the point $(-1,0)$.