EXERCISE SET 2.1
1. Find the degrees of the following polynomials:
Definition: The degree of a polynomial is the highest power of the variable in the polynomial expression.
(i) 2x² – 5x + 3
The highest exponent of the variable x is 2. Therefore, the Degree = 2 (Quadratic Polynomial).
(ii) y³ + 2y – 1
The highest exponent of the variable y is 3. Therefore, the Degree = 3 (Cubic Polynomial).
(iii) -9
This is a non-zero constant polynomial. It can be written as -9x⁰. Therefore, the Degree = 0.
(iv) 4z – 3
The highest exponent of the variable z is 1. Therefore, the Degree = 1 (Linear Polynomial).
2. Write polynomials of degrees 1, 2 and 3.
- Polynomial of Degree 1 (Linear): P(x) = 3x + 5
- Polynomial of Degree 2 (Quadratic): P(x) = 2x² – 4x + 7
- Polynomial of Degree 3 (Cubic): P(x) = x³ + 3x² – 2x + 1
3. What are the coefficients of x² and x³ in the polynomial x⁴ – 3x³ + 6x² – 2x + 7?
Rewriting the polynomial terms to clearly identify coefficients:
1x⁴ + (-3)x³ + (6)x² – 2x + 7
- The coefficient of x³ is -3.
- The coefficient of x² is 6.
4. What is the coefficient of z in the polynomial 4z³ + 5z² – 11?
Since there is no linear term containing ‘z’, the polynomial can be written as:
4z³ + 5z² + 0z – 11
Therefore, the coefficient of z is 0.
5. What is the constant term of the polynomial 9x³ + 5x² – 8x – 10?
The constant term is the term independent of the variable x.
Therefore, the constant term is -10.
EXERCISE SET 2.2
1. Find the value of the linear polynomial 5x – 3 if:
Let P(x) = 5x – 3
(i) x = 0
P(0) = 5(0) – 3 = 0 – 3 = -3
(ii) x = -1
P(-1) = 5(-1) – 3 = -5 – 3 = -8
(iii) x = -3
P(-3) = 5(-3) – 3 = -15 – 3 = -18
(iv) x = 2
P(2) = 5(2) – 3 = 10 – 3 = 7
2. Find the value of the quadratic polynomial 7s² – 4s + 6 if:
Let P(s) = 7s² – 4s + 6
(i) s = 0
P(0) = 7(0)² – 4(0) + 6 = 6
(ii) s = 4
P(4) = 7(4)² – 4(4) + 6 = 7(16) – 16 + 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.
Step 1: Define variables based on conditions.
Let Salil’s present age = x years.
Then, Salil’s mother’s present age = 3x years.
Step 2: Set up equations for ages after 5 years.
Salil’s age after 5 years = (x + 5) years.
Mother’s age after 5 years = (3x + 5) years.
According to the problem statement:
(x + 5) + (3x + 5) = 70
4x + 10 = 70
4x = 60
x = 15
Conclusion:
Salil’s present age = 15 years.
Salil’s mother’s present age = 3(15) = 45 years.
4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Let the common ratio variable be x.
Therefore, the two integers are 2x and 5x.
Given difference = 63:
5x – 2x = 63
3x = 63
x = 21
The integers are:
First integer = 2(21) = 42
Second integer = 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?
Let the number of ₹5 coins = x.
Then, the number of ₹2 coins = 3x.
Total value equation:
5(x) + 2(3x) = 88
5x + 6x = 88
11x = 88
x = 8
Therefore:
Number of ₹5 coins = 8
Number of ₹2 coins = 3(8) = 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?
Let the length of the shorter piece = x feet.
Then, the length of the longer piece = 4x feet.
Total length equation:
x + 4x = 300
5x = 300
x = 60
Therefore:
Shorter piece = 60 feet
Longer piece = 4(60) = 240 feet
7. If the length of a rectangle is more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?
Let the width of the rectangle = w cm.
Then, the length (l) = (2w + 3) cm.
Formula for Perimeter = 2(l + w)
2[(2w + 3) + w] = 24
2[3w + 3] = 24
6 w + 6 = 24
6w = 18
w = 3
Dimensions:
Width = 3 cm
Length = 2(3) + 3 = 9 cm
EXERCISE SET 2.3
1. Pocket money linear expression model:
Initial deposit = ₹500. Monthly addition = ₹150.
Let n represent the number of months passed.
- End of month 2: 500 + 150(2) = ₹800
- Linear Expression for the nth month: Amount = 500 + 150n
2. Rally members dropout model:
Initial members = 120. Dropout rate = 9 members per hour.
Linear expression after n hours: Remaining members = 120 – 9n
3. Area of a rectangle with constant length = 13 cm:
Formula: Area = Length × 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
Linear pattern matching expression: Area(b) = 13b
4. Volume of a box with length = 7 cm, breadth = 11 cm:
Base Area = 7 × 11 = 77 cm². Volume = Base Area × 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
Linear pattern matching expression: Volume(h) = 77h
5. Sarita reading progress model:
Total pages = 500. Reading speed = 20 pages/day.
Pages left after 15 days = 500 – (20 × 15) = 500 – 300 = 200 pages.
Linear pattern expression after d days: Pages Left = 500 – 20d
EXERCISE SET 2.4
1. Plant growth problem
(i) Height after 7 months: 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25 feet.
(ii) Values Table:
| Months (t) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Height (h) | 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 it expands at a constant addition rate of 0.5 feet per month.
2. Telecom problem
(i) Equation & Linear Decay Explanation
Equation: b(x) = 600 – 15x
Why it represents linear decay:
- Constant Rate of Decrease: The balance drops by the exact same amount (15) each day. A constant rate of change is the key feature of a linear function.
- Linear Form: The equation matches the standard slope-intercept form y = mx + c, where:
- Slope (m) = -15 (negative shows a steady decrease)
- y-intercept (c) = 600 (starting balance at x = 0)
(ii) Days Until the Balance Runs Out
Set b(x) = 0:
600 – 15x = 0 15x = 600 x = 600 / 15 x = 40
The balance will run out after 40 days.
(iii) Table of Values (x = 1 to 10)
Here is how the balance b(x) reduces day by day for the first 10 days:
| Day (x) | Calculation | Remaining Balance (b(x)) |
| 1 | 600 – 15(1) | 585 |
| 2 | 600 – 15(2) | 570 |
| 3 | 600 – 15(3) | 555 |
| 4 | 600 – 15(4) | 540 |
| 5 | 600 – 15(5) | 525 |
| 6 | 600 – 15(6) | 510 |
| 7 | 600 – 15(7) | 495 |
| 8 | 600 – 15(8) | 480 |
| 9 | 600 – 15(9) | 465 |
| 10 | 600 – 15(10) | 450 |
EXERCISE SET 2.5
1. Digital learning platform charges model: y = ax + b
Given data pairs: (10, 400) and (14, 500)
Equation 1: 10a + b = 400
Equation 2: 14a + b = 500
Subtracting Equation 1 from Equation 2:
4a = 100 ⇒ a = 25
Substitute a = 25 back into Equation 1:
10(25) + b = 400 ⇒ 250 + b = 400 ⇒ b = 150
3. Temperature scale conversion constants
Given formula structure: °C = a(°F) + b
Condition 1: C = 0, F = 32 ⇒ 0 = 32a + b ⇒ b = -32a
Condition 2: C = 100, F = 212 ⇒ 100 = 212a + b
Substitute value of b into Condition 2:
100 = 212a – 32a
100 = 180a ⇒ a = 5/9
Find b using standard substitution:
b = -32(5/9) = -160/9
Linear formula: °C = (5/9)(°F – 32)
END-OF-CHAPTER EXERCISES
3. Rational number configuration puzzle:
Let the target unknown number be x.
According to the condition statement:
(5/2)x + 2/3 = -7/12
(5/2)x = -7/12 – 2/3
(5/2)x = -7/12 – 8/12
(5/2)x = -15/12 = -5/4
x = (-5/4) × (2/5)
x = -1/2
6. Two-digit number value configurations:
Let the tens digit be x and unit digit be y. Number format value = 10x + y.
Sum of number + reversed combination: (10x + y) + (10y + x) = 11x + 11y = 143
Dividing by 11 gives: x + y = 13
Given digit variance condition: |x – y| = 3
- Case A (x – y = 3): Adding the equations gives 2x = 16 ⇒ x = 8, y = 5. The number is 85.
- Case B (y – x = 3): Adding the equations gives 2y = 16 ⇒ y = 8, x = 5. The number is 58.
11. Linear Polynomial parameters p(x) & q(x):
Let p(x) = ax + b and q(x) = cx + d
Condition 1: p(0) = 5 ⇒ b = 5.
Condition 3: p(x) + q(x) = 6x + 4 ⇒ (a + c)x + (b + d) = 6x + 4
Comparing variables constants:
a + c = 6
b + d = 4 ⇒ 5 + d = 4 ⇒ d = -1
Condition 2: p(x) – q(x) cuts x-axis at (3,0), meaning value at x=3 equals 0.
p(x) – q(x) = (a – c)x + (b – d) = (a – c)x + (5 – (-1)) = (a – c)x + 6
At x = 3:
3(a – c) + 6 = 0 ⇒ 3(a – c) = -6 ⇒ a – c = -2
Solving the system equations:
a + c = 6
a – c = -2
Adding both formulas gives: 2a = 4 ⇒ a = 2, then c = 4.
Final Polynomial Equations:
p(x) = 2x + 5
q(x) = 4x – 1
