1
Chapter 1: Orienting Yourself — The Use of Coordinates
Ganita Manjari (Part I) — Exercise Sets 1.1 & 1.2 — Suggested Step-by-Step Solutions
Exercise Set 1.1 — The Cartesian Plane
Q1
In which quadrant or on which axis do the following points lie? (i) (−3, 4) (ii) (5, −2) (iii) (0, 7) (iv) (−4, −5) (v) (3, 0)
5 Marks
✓ Suggested Step-by-Step Solution
(i) (−3, 4): x negative, y positive → Quadrant II
(ii) (5, −2): x positive, y negative → Quadrant IV
(iii) (0, 7): x = 0 → lies on the positive y-axis
(iv) (−4, −5): both negative → Quadrant III
(v) (3, 0): y = 0 → lies on the positive x-axis
Suggested Marks Distribution
Each correct identification (x5)1 Mark each
Q2
Write the coordinates of a point that is 4 units to the right of the y-axis and 3 units below the x-axis.
1 Mark
✓ Suggested Step-by-Step Solution
"4 units right of y-axis" → x = +4
"3 units below x-axis" → y = −3
Answer: (4, −3) — Quadrant IV.
Suggested Marks Distribution
Correct answer (4, −3)1 Mark
Q3
Plot the following points on a Cartesian plane: A(2, 3), B(−3, 1), C(−2, −4), D(4, −1), E(0, −5), F(3, 0).
3 Marks
✓ Suggested Step-by-Step Solution
From origin (0,0): move along x-axis first, then y-axis.
A(2, 3): 2 right, 3 up → Quadrant I
B(−3, 1): 3 left, 1 up → Quadrant II
C(−2, −4): 2 left, 4 down → Quadrant III
D(4, −1): 4 right, 1 down → Quadrant IV
E(0, −5): On negative y-axis
F(3, 0): On positive x-axis
Suggested Marks Distribution
Correct axes drawn with labels½ Mark
All 6 points correctly plotted2½ Marks
Q4
Reiaan drew a map of his room using coordinates. The door is at (2, 0) and the window is at (2, 6). If each unit represents 1 metre, how far apart are the door and the window?
2 Marks
✓ Suggested Step-by-Step Solution
Both points have x = 2, so the segment is vertical.
Distance = |6 − 0| = 6 metres.
Suggested Marks Distribution
Identifying vertical segment (same x)1 Mark
Correct distance = 6 metres1 Mark
Q5
If a point P(x, y) is in Quadrant III, what can you say about the signs of x and y? Give two examples.
2 Marks
✓ Suggested Step-by-Step Solution
In Quadrant III, both x and y are negative (x < 0 and y < 0).
Examples: (−2, −3) and (−7, −1).
Suggested Marks Distribution
Both negative with reason1 Mark
Two correct examples1 Mark
Exercise Set 1.2 — Distance Formula and Applications
Q1
Find the distance between A(3, 4) and the origin B(0, 0) using the Baudhayana–Pythagoras theorem.
2 Marks
✓ Suggested Step-by-Step Solution
Horizontal leg = 3, vertical leg = 4.
AB² = 3² + 4² = 9 + 16 = 25
AB = √25 = 5 units
Suggested Marks Distribution
Correct Pythagoras application1 Mark
Correct answer: 5 units1 Mark
Q2
Find the distance between P(1, 2) and Q(4, 6).
2 Marks
✓ Suggested Step-by-Step Solution
Horizontal: 4 − 1 = 3; Vertical: 6 − 2 = 4
PQ = √(3² + 4²) = √25 = 5 units
Suggested Marks Distribution
Correct differences computed1 Mark
Correct answer 5 units1 Mark
Q3
Find the midpoint of segment joining A(2, 6) and B(8, 2).
2 Marks
✓ Suggested Step-by-Step Solution
M = ((x₁+x₂)/2, (y₁+y₂)/2) = ((2+8)/2, (6+2)/2) = (5, 4)
Answer: Midpoint = (5, 4).
Suggested Marks Distribution
Correct formula applied1 Mark
Correct answer (5, 4)1 Mark
Q4
Are points A(1, 2), B(3, 6), and C(5, 10) collinear? Justify.
3 Marks
✓ Suggested Step-by-Step Solution
AB = √((3−1)²+(6−2)²) = √(4+16) = √20 = 2√5
BC = √((5−3)²+(10−6)²) = √20 = 2√5
AC = √((5−1)²+(10−2)²) = √80 = 4√5
AB + BC = 2√5 + 2√5 = 4√5 = AC ✓
Conclusion: A, B, C are collinear.
Suggested Marks Distribution
AB and BC computed correctly1 Mark
AC computed correctly1 Mark
Correct conclusion with reason1 Mark
Q5
A city planner places a park at (0,0), school at (6,0), and hospital at (3, 3√3). Show that the three locations form an equilateral triangle and find each side.
4 Marks
✓ Suggested Step-by-Step Solution
PS = √(6²+0²) = 6
SH = √((3−6)²+(3√3)²) = √(9+27) = √36 = 6
PH = √(3²+(3√3)²) = √(9+27) = 6
PS = SH = PH = 6 → equilateral triangle, each side = 6 units.
Suggested Marks Distribution
PS = 6 computed1 Mark
SH = 6 computed1 Mark
PH = 6 computed1 Mark
Conclusion: equilateral, each side 6 units1 Mark
Q6
Find the coordinates of the reflection of the point P(−4, 5) in: (i) the x-axis, (ii) the y-axis, and (iii) the origin.
2 Marks
✓ Suggested Step-by-Step Solution
(i) Reflection in x-axis: Change sign of y-coordinate → P′(−4, −5).
(ii) Reflection in y-axis: Change sign of x-coordinate → P″(4, 5).
(iii) Reflection in origin: Change signs of both coordinates → P‴(4, −5).
Suggested Marks Distribution
Reflection in x-axis and y-axis correct1 Mark
Reflection in origin correct1 Mark
Q6
Show that the points A(1, 7), B(4, 2), C(−1, −1), and D(−4, 4) are the vertices of a square.
3 Marks
✓ Suggested Step-by-Step Solution
Using distance formula d = √[(x₂−x₁)² + (y₂−y₁)²]:
AB² = (4−1)² + (2−7)² = 9 + 25 = 34 → AB = √34
BC² = (−1−4)² + (−1−2)² = 25 + 9 = 34 → BC = √34
CD² = (−4−(−1))² + (4−(−1))² = 9 + 25 = 34 → CD = √34
DA² = (1−(−4))² + (7−4)² = 25 + 9 = 34 → DA = √34
Diagonals: AC² = (−1−1)² + (−1−7)² = 4 + 64 = 68; BD² = (−4−4)² + (4−2)² = 64 + 4 = 68.
Since all four sides are equal (√34) and both diagonals are equal (√68), ABCD is a square.
Suggested Marks Distribution
Calculation of all 4 side lengths1½ Marks
Calculation of both diagonal lengths and conclusion1½ Marks
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 1CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Smart Delivery Drone Navigation (Cartesian Coordinates): A delivery hub is located at the origin O(0, 0). Drone Alpha delivers packages to customer A at (4, 3) and then directly to customer B at (-2, 3), before returning to the hub.
(a) In which quadrant is customer A located?
(b) Calculate the straight-line distance from customer A(4, 3) to customer B(-2, 3).
(c) Find the total round-trip distance traveled by Drone Alpha.
(a) In which quadrant is customer A located?
(b) Calculate the straight-line distance from customer A(4, 3) to customer B(-2, 3).
(c) Find the total round-trip distance traveled by Drone Alpha.
(a) Customer A(4, 3) has both coordinates positive, so it lies in Quadrant I.
(b) Since both points have the same y-coordinate (y = 3), distance AB = |x₂ - x₁| = |(-2) - 4| = 6 units (or by distance formula: √((-2-4)² + (3-3)²) = √36 = 6).
(c) OA = √(4² + 3²) = √25 = 5 units. OB = √((-2)² + 3²) = √(4 + 9) = √13 ≈ 3.61 units. Total round-trip distance = OA + AB + BO = 5 + 6 + √13 = 11 + √13 units (≈ 14.61 units).
(b) Since both points have the same y-coordinate (y = 3), distance AB = |x₂ - x₁| = |(-2) - 4| = 6 units (or by distance formula: √((-2-4)² + (3-3)²) = √36 = 6).
(c) OA = √(4² + 3²) = √25 = 5 units. OB = √((-2)² + 3²) = √(4 + 9) = √13 ≈ 3.61 units. Total round-trip distance = OA + AB + BO = 5 + 6 + √13 = 11 + √13 units (≈ 14.61 units).
🧠 HOTS & Assertion-Reasoning
Assertion (A): The distance between the points P(-3, 0) and Q(5, 0) on the x-axis is 8 units.
Reason (R): The distance between two points (x₁, 0) and (x₂, 0) on the x-axis is given by |x₂ - x₁|.
Choose: (a) Both A and R are true and R is correct explanation of A. (b) Both A and R are true, but R is not the correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): The distance between two points (x₁, 0) and (x₂, 0) on the x-axis is given by |x₂ - x₁|.
Choose: (a) Both A and R are true and R is correct explanation of A. (b) Both A and R are true, but R is not the correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation of A.
Explanation: Distance PQ = |5 - (-3)| = |5 + 3| = 8 units. Reason (R) states the universal formula for points lying on the horizontal axis, which directly explains (A).
Explanation: Distance PQ = |5 - (-3)| = |5 + 3| = 8 units. Reason (R) states the universal formula for points lying on the horizontal axis, which directly explains (A).
2
Chapter 2: Introduction to Linear Polynomials
Ganita Manjari (Part I) — Exercise Sets 2.1 & 2.2 — Suggested Marks Distribution
Exercise Set 2.1 — Polynomials and Linear Expressions
Q1
Which of the following are polynomials? (i) 2x + 3 (ii) x² − 5 (iii) 3/x + 1 (iv) √x + 2 (v) 7
5 Marks
✓ Suggested Step-by-Step Solution
A polynomial has non-negative integer exponents for all variables.
(i) 2x + 3: Exponent of x is 1. Yes, polynomial (linear).
(ii) x² − 5: Exponent is 2. Yes, polynomial (quadratic).
(iii) 3/x + 1: Written as 3x⁻¹ + 1; exponent is −1 (negative). Not a polynomial.
(iv) √x + 2: Written as x¹∕²; exponent is 1/2 (fraction). Not a polynomial.
(v) 7: Constant = 7x⁰; exponent 0. Yes, polynomial (constant).
Suggested Marks Distribution
Each correct classification with reason (×5)1 Mark each
Q2
State the degree of each polynomial: (i) 5x − 3 (ii) 2x² + x − 7 (iii) 9 (iv) 4x³ − 2x + 1
2 Marks
✓ Suggested Step-by-Step Solution
(i) 5x − 3: Highest power = 1 → Degree 1 (Linear)
(ii) 2x² + x − 7: Highest power = 2 → Degree 2 (Quadratic)
(iii) 9: Constant polynomial → Degree 0
(iv) 4x³ − 2x + 1: Highest power = 3 → Degree 3 (Cubic)
Suggested Marks Distribution
All four degrees correct2 Marks
Q3
Find the value of p(x) = 3x − 7 at (i) x = 2 (ii) x = 0 (iii) x = −1.
3 Marks
✓ Suggested Step-by-Step Solution
(i) p(2) = 3(2) − 7 = 6 − 7 = −1
(ii) p(0) = 3(0) − 7 = 0 − 7 = −7
(iii) p(−1) = 3(−1) − 7 = −3 − 7 = −10
Suggested Marks Distribution
Each correct substitution and value (×3)1 Mark each
Q4
Find the zero of the linear polynomial p(x) = 4x − 12.
2 Marks
✓ Suggested Step-by-Step Solution
Set p(x) = 0: 4x − 12 = 0
4x = 12 → x = 12/4 = 3
The zero of p(x) is x = 3. Verify: p(3) = 4(3) − 12 = 0 ✓
Suggested Marks Distribution
Setting p(x) = 0 and solving1 Mark
Correct zero x = 31 Mark
Q5
A shopkeeper sells mangoes at ₹(3x + 5) each. Write a polynomial for the total cost if he sells 4 mangoes, and find the total cost when x = 10.
3 Marks
✓ Suggested Step-by-Step Solution
Step 1: Total cost polynomial = 4 × (3x + 5) = 12x + 20
Step 2: At x = 10: Total cost = 12(10) + 20 = 120 + 20 = ₹140
Suggested Marks Distribution
Correct polynomial 12x + 201 Mark
Correct evaluation at x = 101 Mark
Final answer ₹140 with units1 Mark
Exercise Set 2.2 — Graphs of Linear Polynomials
Q1
Draw the graph of p(x) = 2x + 1 by making a table of values for x = −2, −1, 0, 1, 2.
3 Marks
✓ Suggested Step-by-Step Solution
Table of values:
| x | p(x) = 2x+1 | Point |
|---|---|---|
| −2 | 2(−2)+1 = −3 | (−2, −3) |
| −1 | 2(−1)+1 = −1 | (−1, −1) |
| 0 | 2(0)+1 = 1 | (0, 1) |
| 1 | 2(1)+1 = 3 | (1, 3) |
| 2 | 2(2)+1 = 5 | (2, 5) |
Plot these points and join them with a straight line. The graph is a line crossing the y-axis at (0, 1) (y-intercept = 1) and x-axis at (−1/2, 0) (zero of p(x)).
Suggested Marks Distribution
Correct table of values1 Mark
Correct plotting of points1 Mark
Correct straight line drawn1 Mark
Q2
What does the graph of a linear polynomial always look like? How many zeros can it have?
2 Marks
✓ Suggested Step-by-Step Solution
The graph of a linear polynomial p(x) = ax + b (where a ≠ 0) is always a straight line.
A linear polynomial has exactly one zero — the x-coordinate of the point where the line crosses the x-axis (i.e., where y = 0).
Suggested Marks Distribution
Graph is a straight line1 Mark
Exactly one zero1 Mark
Q3
From the graph of p(x) = x − 4, find its zero. Verify algebraically.
3 Marks
✓ Suggested Step-by-Step Solution
Graphical method: Plot points: (0, −4), (2, −2), (4, 0), (6, 2). The line crosses x-axis at (4, 0).
Zero from graph: x = 4.
Algebraic verification: p(4) = 4 − 4 = 0 ✓
Suggested Marks Distribution
Correct graph with x-intercept at x = 41 Mark
Identifying zero = 41 Mark
Algebraic verification1 Mark
Q6
Find the value of k if x = 2 is a zero of the polynomial p(x) = 2x² + 3x + k.
2 Marks
✓ Suggested Step-by-Step Solution
Since x = 2 is a zero of p(x), we have p(2) = 0.
Substitute x = 2: 2(2)² + 3(2) + k = 0
2(4) + 6 + k = 0 → 8 + 6 + k = 0 → 14 + k = 0
Therefore, k = −14.
Suggested Marks Distribution
Applying condition p(2) = 0 with substitution1 Mark
Solving linear equation for k = -141 Mark
Q4
Find the coordinates where the graph of the linear equation 2x + 3y = 12 cuts: (i) the x-axis, and (ii) the y-axis.
2 Marks
✓ Suggested Step-by-Step Solution
(i) At the x-axis: y = 0.
2x + 3(0) = 12 → 2x = 12 → x = 6. Point is (6, 0).
(ii) At the y-axis: x = 0.
2(0) + 3y = 12 → 3y = 12 → y = 4. Point is (0, 4).
Suggested Marks Distribution
Correct x-intercept (6, 0)1 Mark
Correct y-intercept (0, 4)1 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 2CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
App-Based Cab Fare Modeling (Linear Polynomials): In a metropolitan city, a cab aggregator charges a fixed base booking fee of ₹40 plus ₹15 per kilometer traveled. Let x denote the distance in kilometers and C(x) denote the total fare.
(a) Formulate the fare as a linear polynomial C(x).
(b) Find the fare for a trip of 12 km.
(c) Find the zero of the polynomial C(x) and state whether it has physical significance.
(a) Formulate the fare as a linear polynomial C(x).
(b) Find the fare for a trip of 12 km.
(c) Find the zero of the polynomial C(x) and state whether it has physical significance.
(a) Linear polynomial: C(x) = 15x + 40 (degree 1).
(b) For x = 12 km: C(12) = 15(12) + 40 = 180 + 40 = ₹220.
(c) Zero of C(x): 15x + 40 = 0 ⟹ 15x = -40 ⟹ x = -40/15 = -8/3 ≈ -2.67 km. Since distance cannot be negative, this zero has mathematical validity but no physical significance for a real cab ride.
(b) For x = 12 km: C(12) = 15(12) + 40 = 180 + 40 = ₹220.
(c) Zero of C(x): 15x + 40 = 0 ⟹ 15x = -40 ⟹ x = -40/15 = -8/3 ≈ -2.67 km. Since distance cannot be negative, this zero has mathematical validity but no physical significance for a real cab ride.
🧠 HOTS & Assertion-Reasoning
Assertion (A): Every linear polynomial in one variable ax + b (where a ≠ 0) has exactly one unique zero.
Reason (R): A polynomial of degree n can have at most n distinct real zeros, and a linear polynomial has degree 1.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true but R is not the correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): A polynomial of degree n can have at most n distinct real zeros, and a linear polynomial has degree 1.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true but R is not the correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: Setting ax + b = 0 with a ≠ 0 yields x = -b/a. Since a and b are fixed constants, -b/a is unique. This is directly substantiated by the Fundamental Theorem of Algebra (Reason R).
Explanation: Setting ax + b = 0 with a ≠ 0 yields x = -b/a. Since a and b are fixed constants, -b/a is unique. This is directly substantiated by the Fundamental Theorem of Algebra (Reason R).
3
Chapter 3: The World of Numbers
Ganita Manjari (Part I) — Exercise Sets 3.1, 3.2 & 3.3 — Suggested Marks Distribution
Exercise Set 3.1 — Number Systems and Rational Numbers
Q1
Classify each number as rational or irrational: (i) √9 (ii) √7 (iii) 3/4 (iv) 0.1010010001... (v) π
5 Marks
✓ Suggested Step-by-Step Solution
(i) √9 = 3 → Can be written as 3/1. Rational.
(ii) √7 → Not a perfect square; its decimal is non-terminating non-repeating. Irrational.
(iii) 3/4 → Already in p/q form. Rational.
(iv) 0.1010010001... → Non-terminating and non-repeating pattern. Irrational.
(v) π → Famous irrational number; decimal is non-terminating non-repeating. Irrational.
Suggested Marks Distribution
Each correct classification with reason (×5)1 Mark each
Q2
Express 0.̄3̄ (0.333...) as a fraction in simplest form.
3 Marks
✓ Suggested Step-by-Step Solution
Let x = 0.3333...
Step 1: Multiply both sides by 10: 10x = 3.3333...
Step 2: Subtract: 10x − x = 3.333... − 0.333... → 9x = 3
Step 3: x = 3/9 = 1/3
Answer: 0.̄3̄ = 1/3
Suggested Marks Distribution
Setting up equation and multiplying by 101 Mark
Subtracting to eliminate repeating part1 Mark
Correct simplification: 1/31 Mark
Q3
Show that √2 is irrational.
4 Marks
✓ Suggested Step-by-Step Solution
Proof by contradiction:
Assume √2 is rational. Then √2 = p/q where p, q are integers, q ≠ 0, and HCF(p, q) = 1.
Step 1: Squaring both sides: 2 = p²/q² → p² = 2q². So p² is even → p is even.
Step 2: Let p = 2m. Then (2m)² = 2q² → 4m² = 2q² → q² = 2m². So q² is even → q is even.
Contradiction: Both p and q are even, so HCF(p,q) ≥ 2. But HCF(p,q) = 1. This contradicts our assumption.
Conclusion: √2 is irrational.
Suggested Marks Distribution
Correct assumption and setup1 Mark
Showing p is even1 Mark
Showing q is even1 Mark
Contradiction and correct conclusion1 Mark
Exercise Set 3.2 — Operations on Irrational Numbers
Q1
Simplify: (i) √5 + 2√5 (ii) √18 − √2 (iii) (√3 + 1)²
3 Marks
✓ Suggested Step-by-Step Solution
(i) √5 + 2√5 = (1+2)√5 = 3√5
(ii) √18 − √2 = √(9×2) − √2 = 3√2 − √2 = 2√2
(iii) (√3 + 1)² = (√3)² + 2(√3)(1) + 1² = 3 + 2√3 + 1 = 4 + 2√3
Suggested Marks Distribution
Each correct simplification (×3)1 Mark each
Q2
Rationalise the denominator of: (i) 1/√7 (ii) 5/(√3 + √2)
4 Marks
✓ Suggested Step-by-Step Solution
(i) 1/√7 × √7/√7 = √7/7
(ii) 5/(√3+√2) × (√3−√2)/(√3−√2)
= 5(√3−√2) / ((√3)²−(√2)²) = 5(√3−√2)/(3−2)
= 5(√3 − √2)
Suggested Marks Distribution
(i) Multiplying by √7/√7, correct result2 Marks
(ii) Conjugate multiplication, correct simplification2 Marks
Exercise Set 3.3 — Laws of Exponents for Real Numbers
Q1
Simplify using laws of exponents: (i) 2³ × 2² (ii) 5⁷ ÷ 5² (iii) (3²)³ (iv) (2/3)⁻²
4 Marks
✓ Suggested Step-by-Step Solution
(i) 2³ × 2² = 2⁻⁺² = 2⁵ = 32
(ii) 5⁷ ÷ 5² = 5⁷⁻² = 5⁵ = 3125
(iii) (3²)³ = 3²×³ = 3⁶ = 729
(iv) (2/3)⁻² = (3/2)² = 9/4 = 2.25
Suggested Marks Distribution
Each correct simplification (×4)1 Mark each
Q2
Find the value of: (i) 27⁹⁻² (ii) 64¹∕³ (iii) (125)⁻¹∕³
3 Marks
✓ Suggested Step-by-Step Solution
(i) 27⁹⁻² = (3³)⁻³∕² = 3⁻⁹∕² = 1/(3⁹∕²) = 1/(3³)¹∕² = 1/(√27) = 1/(3√3) = √3/9
(ii) 64¹∕³ = (4³)¹∕³ = 4¹ = 4
(iii) 125⁻¹∕³ = (5³)⁻¹∕³ = 5⁻¹ = 1/5
Suggested Marks Distribution
Each correct evaluation with working (×3)1 Mark each
Q4
Find five rational numbers between 3/5 and 4/5.
2 Marks
✓ Suggested Step-by-Step Solution
To find 5 rational numbers, multiply numerator and denominator by (5 + 1) = 6:
3/5 = (3 × 6) / (5 × 6) = 18/30
4/5 = (4 × 6) / (5 × 6) = 24/30
Five rational numbers between 18/30 and 24/30 are: 19/30, 20/30 (2/3), 21/30 (7/10), 22/30 (11/15), and 23/30.
Suggested Marks Distribution
Converting fractions with common larger denominator1 Mark
Listing 5 correct intermediate rational numbers1 Mark
Q3
Rationalise the denominator of (√3 − 1) / (√3 + 1).
2 Marks
✓ Suggested Step-by-Step Solution
Multiply numerator and denominator by the conjugate of (√3 + 1), which is (√3 − 1):
[(√3 − 1)(√3 − 1)] / [(√3 + 1)(√3 − 1)]
Numerator = (√3)² − 2√3 + 1 = 3 − 2√3 + 1 = 4 − 2√3
Denominator = (√3)² − 1² = 3 − 1 = 2
Result = (4 − 2√3) / 2 = 2 − √3.
Suggested Marks Distribution
Multiplying by conjugate correctly1 Mark
Simplification to 2 - √31 Mark
Q3
Evaluate: (i) (64)¹⁄³ (ii) (125)⁻¹⁄³ (iii) 16¾.
2 Marks
✓ Suggested Step-by-Step Solution
(i) (64)¹⁄³: 64 = 4³ → (4³)¹⁄³ = 4¹ = 4.
(ii) (125)⁻¹⁄³: 125 = 5³ → (5³)⁻¹⁄³ = 5⁻¹ = 1/5.
(iii) (16)¾: 16 = 2⁴ → (2⁴)¾ = 2³ = 8.
Suggested Marks Distribution
Evaluating parts (i) and (ii)1 Mark
Evaluating part (iii)1 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 3CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Precision Engineering & Real Number Expansions: A precision laser cutting machine works with tolerances measured in rational fractions and irrational square roots. An engineer models a component with dimension d = 3.142857142857... mm and diagonal D = √8 mm.
(a) Express d in p/q form.
(b) State whether diagonal D is rational or irrational and find its simplest radical form.
(c) Is the sum d + D rational or irrational?
(a) Express d in p/q form.
(b) State whether diagonal D is rational or irrational and find its simplest radical form.
(c) Is the sum d + D rational or irrational?
(a) d has a 6-digit repeating block: 3.142857... = 3 + 142857/999999 = 3 + 1/7 = 22/7 mm.
(b) D = √8 = √(4 × 2) = 2√2 mm. Since 2 is a non-square integer, √2 is irrational, so D is irrational.
(c) d is rational and D is irrational. The sum of a rational and an irrational number is always irrational. Therefore, d + D is irrational.
(b) D = √8 = √(4 × 2) = 2√2 mm. Since 2 is a non-square integer, √2 is irrational, so D is irrational.
(c) d is rational and D is irrational. The sum of a rational and an irrational number is always irrational. Therefore, d + D is irrational.
🧠 HOTS & Assertion-Reasoning
Assertion (A): Between any two distinct rational numbers, there exist infinitely many rational and irrational numbers.
Reason (R): The set of real numbers satisfies the completeness and density property on the number line.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): The set of real numbers satisfies the completeness and density property on the number line.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: Density property ensures that for any a < b, the arithmetic mean (a+b)/2 provides an intermediate rational number, which can be repeated indefinitely. The real number continuum guarantees uncountably many numbers between any interval.
Explanation: Density property ensures that for any a < b, the arithmetic mean (a+b)/2 provides an intermediate rational number, which can be repeated indefinitely. The real number continuum guarantees uncountably many numbers between any interval.
4
Chapter 4: Exploring Algebraic Identities
Ganita Manjari (Part I) — Exercise Sets 4.1 & 4.2 — Suggested Marks Distribution
Exercise Set 4.1 — Standard Algebraic Identities
Q1
Expand using appropriate identity: (i) (x + 4)² (ii) (2a − 3b)² (iii) (x + 5)(x − 5)
3 Marks
✓ Suggested Step-by-Step Solution
(i) (x+4)² = x² + 2(x)(4) + 4² = x² + 8x + 16 [Using (a+b)² = a²+2ab+b²]
(ii) (2a−3b)² = (2a)² − 2(2a)(3b) + (3b)² = 4a² − 12ab + 9b²
(iii) (x+5)(x−5) = x² − 5² = x² − 25 [Using (a+b)(a−b) = a²−b²]
Suggested Marks Distribution
Each correct expansion with identity cited (×3)1 Mark each
Q2
Evaluate using identities (without direct multiplication): (i) 102² (ii) 99² (iii) 103 × 97
3 Marks
✓ Suggested Step-by-Step Solution
(i) 102² = (100+2)² = 100² + 2(100)(2) + 2² = 10000 + 400 + 4 = 10404
(ii) 99² = (100−1)² = 100² − 2(100)(1) + 1² = 10000 − 200 + 1 = 9801
(iii) 103 × 97 = (100+3)(100−3) = 100² − 3² = 10000 − 9 = 9991
Suggested Marks Distribution
Each correct evaluation with identity (×3)1 Mark each
Q3
Expand: (2x + y + 3z)²
2 Marks
✓ Suggested Step-by-Step Solution
Using (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
Here a = 2x, b = y, c = 3z:
= (2x)² + y² + (3z)² + 2(2x)(y) + 2(y)(3z) + 2(3z)(2x)
= 4x² + y² + 9z² + 4xy + 6yz + 12xz
Suggested Marks Distribution
Correct identity applied1 Mark
All 6 terms correct1 Mark
Exercise Set 4.2 — Factorisation using Identities
Q1
Factorise: (i) x² − 9 (ii) 4a² − 25b² (iii) x² + 6x + 9 (iv) 9y² − 12y + 4
4 Marks
✓ Suggested Step-by-Step Solution
(i) x² − 9 = x² − 3² = (x+3)(x−3)
(ii) 4a² − 25b² = (2a)² − (5b)² = (2a+5b)(2a−5b)
(iii) x² + 6x + 9 = x² + 2(x)(3) + 3² = (x+3)²
(iv) 9y² − 12y + 4 = (3y)² − 2(3y)(2) + 2² = (3y−2)²
Suggested Marks Distribution
Each correct factorisation with identity (×4)1 Mark each
Q2
If x + 1/x = 5, find the value of x² + 1/x².
3 Marks
✓ Suggested Step-by-Step Solution
Given: x + 1/x = 5
Step 1: Square both sides: (x + 1/x)² = 5² = 25
Step 2: x² + 2(x)(1/x) + 1/x² = 25 → x² + 2 + 1/x² = 25
Step 3: x² + 1/x² = 25 − 2 = 23
Suggested Marks Distribution
Squaring both sides correctly1 Mark
Expanding using identity1 Mark
Final answer x² + 1/x² = 231 Mark
Q3
Expand (2a + b)(2a − b)(4a² + b²).
3 Marks
✓ Suggested Step-by-Step Solution
Step 1: (2a+b)(2a−b) = (2a)² − b² = 4a² − b²
Step 2: (4a² − b²)(4a² + b²) = (4a²)² − (b²)² = 16a⁴ − b⁴
Suggested Marks Distribution
Correct first step: (2a+b)(2a−b) = 4a²−b²1 Mark
Applying difference of squares again1 Mark
Final answer 16a⁴ − b⁴1 Mark
Q4
Expand: (3a − 7b − c)² using the standard identity for trinomial square.
2 Marks
✓ Suggested Step-by-Step Solution
Use identity: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
Here x = 3a, y = −7b, z = −c:
= (3a)² + (−7b)² + (−c)² + 2(3a)(−7b) + 2(−7b)(−c) + 2(−c)(3a)
= 9a² + 49b² + c² − 42ab + 14bc − 6ca.
Suggested Marks Distribution
Correct formula and substitute signs1 Mark
Correctly expanded product1 Mark
Q4
Factorise: 27x² + y³ + z³ − 9xyz (or x³ + y³ + z³ − 3xyz form). Factorise: 8x³ + y³ + 27z³ − 18xyz.
2 Marks
✓ Suggested Step-by-Step Solution
Recognise form a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).
Here a = 2x, b = y, c = 3z. Check 3abc = 3(2x)(y)(3z) = 18xyz.
Applying identity: = (2x + y + 3z)(4x² + y² + 9z² − 2xy − 3yz − 6zx).
Suggested Marks Distribution
Identification of cubic identity components1 Mark
Complete factorised form1 Mark
Q5
Without actually calculating the cubes, find the value of (−12)³ + (7)³ + (5)³.
2 Marks
✓ Suggested Step-by-Step Solution
Let x = −12, y = 7, z = 5.
Check their sum: x + y + z = −12 + 7 + 5 = 0.
Identity: If x + y + z = 0, then x³ + y³ + z³ = 3xyz.
Therefore, (−12)³ + 7³ + 5³ = 3(−12)(7)(5) = −36 × 35 = −1260.
Suggested Marks Distribution
Verifying x + y + z = 0 and stating identity1 Mark
Correct calculation 3(-12)(7)(5) = -12601 Mark
Q6
Factorise: x³ − 23x² + 142x − 120.
2 Marks
✓ Suggested Step-by-Step Solution
Testing factors of constant 120: let x = 1:
p(1) = 1³ − 23(1) + 142(1) − 120 = 1 − 23 + 142 − 120 = 0. Thus, (x − 1) is a factor.
Dividing p(x) by (x − 1) gives quotient: x² − 22x + 120.
Splitting the middle term: x² − 12x − 10x + 120 = (x − 12)(x − 10).
Hence, p(x) = (x − 1)(x − 10)(x − 12).
Suggested Marks Distribution
Finding (x - 1) via Factor Theorem1 Mark
Factorising quadratic factor into (x - 10)(x - 12)1 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 4CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Architectural Space Planning & Identities: An architect designs a modular pavilion with square base of side (x + y). Later, a storage corridor of width 2 units is attached so total dimensions become (x + y + 2) by (x + y - 2).
(a) Use standard algebraic identity to find the expanded expression for the area.
(b) If x + y = 10 meters, find the total floor area.
(c) Verify your answer using numerical substitution.
(a) Use standard algebraic identity to find the expanded expression for the area.
(b) If x + y = 10 meters, find the total floor area.
(c) Verify your answer using numerical substitution.
(a) Using identity (A + B)(A - B) = A² - B² where A = (x + y) and B = 2:
Area = (x + y)² - 2² = x² + 2xy + y² - 4.
(b) If x + y = 10 m: Area = (10)² - 4 = 100 - 4 = 96 m².
(c) Direct substitution: (10 + 2)(10 - 2) = 12 × 8 = 96 m². The results match perfectly.
Area = (x + y)² - 2² = x² + 2xy + y² - 4.
(b) If x + y = 10 m: Area = (10)² - 4 = 100 - 4 = 96 m².
(c) Direct substitution: (10 + 2)(10 - 2) = 12 × 8 = 96 m². The results match perfectly.
🧠 HOTS & Assertion-Reasoning
Assertion (A): If a + b + c = 0, then a³ + b³ + c³ = 3abc.
Reason (R): a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: If (a + b + c) = 0, then the product (a + b + c)(a² + b² + c² - ab - bc - ca) vanishes to 0. Transposing -3abc yields a³ + b³ + c³ = 3abc.
Explanation: If (a + b + c) = 0, then the product (a + b + c)(a² + b² + c² - ab - bc - ca) vanishes to 0. Transposing -3abc yields a³ + b³ + c³ = 3abc.
5
Chapter 5: I’m Up and Down, and Round and Round
Ganita Manjari (Part I) — Exercise Sets 5.1, 5.2 & 5.3 — Suggested Marks Distribution
About this chapter: This chapter explores oscillatory and circular motion through mathematics — including sinusoidal behaviour, angles, and periodic patterns. It connects geometry with real-world motion.
Exercise Set 5.1 — Angles and Circular Motion
Q1
Convert the following angles from degrees to radians: (i) 90° (ii) 180° (iii) 270° (iv) 360°
2 Marks
✓ Suggested Step-by-Step Solution
Formula: Radians = Degrees × π/180
(i) 90° = 90 × π/180 = π/2 radians
(ii) 180° = 180 × π/180 = π radians
(iii) 270° = 270 × π/180 = 3π/2 radians
(iv) 360° = 360 × π/180 = 2π radians
Suggested Marks Distribution
Correct formula stated½ Mark
All four correct conversions1½ Marks
Q2
A wheel rotates 5 times per second. Find the angle swept per second in radians and the arc length swept by a point 30 cm from the centre.
4 Marks
✓ Suggested Step-by-Step Solution
Step 1: Angle per second
One full rotation = 2π radians.
5 rotations per second → Angle = 5 × 2π = 10π radians/second
One full rotation = 2π radians.
5 rotations per second → Angle = 5 × 2π = 10π radians/second
Step 2: Arc length
Arc length = radius × angle = 30 × 10π = 300π cm/second
Arc length = radius × angle = 30 × 10π = 300π cm/second
≈ 942.5 cm/second
Suggested Marks Distribution
Angle = 10π radians/s2 Marks
Arc length = 300π cm/s2 Marks
Exercise Set 5.2 — Periodic and Oscillatory Patterns
Q1
The height (in cm) of a child on a swing is given by h(t) = 80 + 30 sin(t), where t is time in seconds. Find the maximum and minimum heights and the height at t = 0.
3 Marks
✓ Suggested Step-by-Step Solution
Maximum: sin(t) reaches max value 1. h_max = 80 + 30(1) = 110 cm
Minimum: sin(t) reaches min value −1. h_min = 80 + 30(−1) = 50 cm
At t = 0: sin(0) = 0. h(0) = 80 + 30(0) = 80 cm
Suggested Marks Distribution
Maximum height 110 cm1 Mark
Minimum height 50 cm1 Mark
Height at t=0 is 80 cm1 Mark
Q2
What is a periodic function? Give two real-life examples and state one mathematical example.
3 Marks
✓ Suggested Step-by-Step Solution
Periodic function: A function f(x) is periodic if there exists a positive number T such that f(x + T) = f(x) for all x. T is called the period.
Real-life examples:
(1) Motion of a pendulum — it swings back and forth with equal time periods.
(2) Heart beats — the pattern of contraction and relaxation repeats regularly.
(1) Motion of a pendulum — it swings back and forth with equal time periods.
(2) Heart beats — the pattern of contraction and relaxation repeats regularly.
Mathematical example: f(x) = sin(x) has period 2π; sin(x + 2π) = sin(x).
Suggested Marks Distribution
Definition with T stated1 Mark
Two real-life examples1 Mark
Mathematical example (sin x)1 Mark
Exercise Set 5.3 — Applications
Q1
A Ferris wheel of radius 20 m completes one revolution in 40 seconds. Find: (i) Angular speed in radians per second (ii) Speed of a passenger on the rim.
4 Marks
✓ Suggested Step-by-Step Solution
(i) Angular speed: ω = 2π/T = 2π/40 = π/20 radians per second
(ii) Linear speed: v = rω = 20 × π/20 = π metres per second ≈ 3.14 m/s
Suggested Marks Distribution
Angular speed = π/20 rad/s2 Marks
Linear speed = π m/s2 Marks
Q3
Find the angle in degrees swept by the minute hand of a clock in 25 minutes.
2 Marks
✓ Suggested Step-by-Step Solution
A minute hand completes 360° in 60 minutes.
Rate of rotation = 360° / 60 = 6° per minute.
Angle swept in 25 minutes = 25 × 6° = 150°.
Suggested Marks Distribution
Rate per minute (6 degrees/min)1 Mark
Total angle calculation = 150 degrees1 Mark
Q3
The period of oscillation of a simple pendulum is T = 2 seconds. How many complete oscillations does it make in 5 minutes?
2 Marks
✓ Suggested Step-by-Step Solution
Total time t = 5 minutes = 5 × 60 = 300 seconds.
Time period for 1 oscillation T = 2 s.
Number of oscillations = Total time / T = 300 / 2 = 150 complete oscillations.
Suggested Marks Distribution
Time conversion to seconds1 Mark
Calculating 150 oscillations1 Mark
Q2
A bicycle wheel has diameter 70 cm. How many revolutions will it make to travel a distance of 1.1 km?
2 Marks
✓ Suggested Step-by-Step Solution
Radius r = 70 / 2 = 35 cm = 0.35 m.
Circumference C = 2πr = 2 × (22/7) × 35 cm = 220 cm = 2.2 m.
Total distance = 1.1 km = 1,100 m.
Number of revolutions = 1,100 / 2.2 = 500 revolutions.
Suggested Marks Distribution
Circumference = 2.2 m1 Mark
Revolutions = 5001 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 5CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Clockwork Mechanisms & Rotational Angles: In an antique clock tower, the minute hand has length 14 cm and the hour hand has length 10.5 cm.
(a) Find the angle swept by the minute hand between 3:00 PM and 3:25 PM.
(b) What angle does the hour hand rotate through in 20 minutes?
(c) At 3:00 PM, what is the measure of the angle between the two hands?
(a) Find the angle swept by the minute hand between 3:00 PM and 3:25 PM.
(b) What angle does the hour hand rotate through in 20 minutes?
(c) At 3:00 PM, what is the measure of the angle between the two hands?
(a) In 60 minutes, the minute hand covers 360°. In 25 minutes: θ = (360°/60) × 25 = 6° × 25 = 150°.
(b) The hour hand rotates 360° in 12 hours (720 minutes), which is 0.5° per minute. In 20 minutes: (0.5°) × 20 = 10°.
(c) At 3:00 PM, the hour hand points at 3 (90°) and the minute hand points at 12 (0°). The angle between them is 90° (right angle).
(b) The hour hand rotates 360° in 12 hours (720 minutes), which is 0.5° per minute. In 20 minutes: (0.5°) × 20 = 10°.
(c) At 3:00 PM, the hour hand points at 3 (90°) and the minute hand points at 12 (0°). The angle between them is 90° (right angle).
🧠 HOTS & Assertion-Reasoning
Assertion (A): A pendulum bob that oscillates through an angle of 30° on either side of the vertical sweeps a total angular span of 60° during each swing.
Reason (R): The total angular range is the sum of displacements from the central equilibrium position to both extreme positions.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): The total angular range is the sum of displacements from the central equilibrium position to both extreme positions.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: From -30° to +30° the angular displacement is 30° + 30° = 60°, which matches the physical span from one amplitude to the other.
Explanation: From -30° to +30° the angular displacement is 30° + 30° = 60°, which matches the physical span from one amplitude to the other.
6
Chapter 6: Measuring Space — Perimeter and Area
Ganita Manjari (Part I) — Exercise Sets 6.1, 6.2 & 6.3 — Suggested Marks Distribution
Exercise Set 6.1 — Perimeter of Plane Figures
Q1
Find the perimeter of: (i) A rectangle with l = 12 cm, b = 8 cm (ii) An equilateral triangle with side 9 cm (iii) A regular hexagon with side 5 cm
3 Marks
✓ Suggested Step-by-Step Solution
(i) Rectangle: P = 2(l + b) = 2(12 + 8) = 2 × 20 = 40 cm
(ii) Equilateral triangle: P = 3 × 9 = 27 cm
(iii) Regular hexagon: P = 6 × 5 = 30 cm
Suggested Marks Distribution
Each correct perimeter with formula (×3)1 Mark each
Q2
A wire of length 88 cm is bent into a circle. Find the radius and area of the circle. (Use π = 22/7)
3 Marks
✓ Suggested Step-by-Step Solution
Step 1: Circumference = 2πr = 88
r = 88/(2π) = 88 ÷ (2 × 22/7) = 88 × 7/44 = 14 cm
r = 88/(2π) = 88 ÷ (2 × 22/7) = 88 × 7/44 = 14 cm
Step 2: Area = πr² = 22/7 × 14 × 14 = 22/7 × 196 = 616 cm²
Suggested Marks Distribution
Finding radius r = 14 cm1½ Marks
Area = 616 cm²1½ Marks
Exercise Set 6.2 — Area of Triangles and Quadrilaterals
Q1
Find the area of a triangle with sides 5 cm, 12 cm, and 13 cm using Heron’s formula.
3 Marks
✓ Suggested Step-by-Step Solution
Step 1: Semi-perimeter s = (5 + 12 + 13)/2 = 30/2 = 15 cm
Step 2: Heron’s formula
Area = √[s(s−a)(s−b)(s−c)] = √[15(15−5)(15−12)(15−13)]
Area = √[s(s−a)(s−b)(s−c)] = √[15(15−5)(15−12)(15−13)]
= √[15 × 10 × 3 × 2] = √900 = 30 cm²
Suggested Marks Distribution
s = 15 cm calculated1 Mark
Heron’s formula applied correctly1 Mark
Area = 30 cm²1 Mark
Q2
A rhombus has diagonals of lengths 16 cm and 12 cm. Find its area and perimeter.
4 Marks
✓ Suggested Step-by-Step Solution
Area = (1/2) × d₁ × d₂ = (1/2) × 16 × 12 = 96 cm²
Side of rhombus: Half-diagonals are 8 and 6.
Side = √(8² + 6²) = √(64+36) = √100 = 10 cm
Side = √(8² + 6²) = √(64+36) = √100 = 10 cm
Perimeter = 4 × 10 = 40 cm
Suggested Marks Distribution
Area = 96 cm²1 Mark
Side using Pythagoras = 10 cm2 Marks
Perimeter = 40 cm1 Mark
Exercise Set 6.3 — Areas of Composite Figures
Q1
A rectangular park 60 m × 40 m has a semicircular flowerbed of diameter 14 m at one end. Find the area of the park excluding the flowerbed. (π = 22/7)
4 Marks
✓ Suggested Step-by-Step Solution
Step 1: Rectangle area = 60 × 40 = 2400 m²
Step 2: Semicircle area (r = 7 m) = (1/2)πr² = (1/2) × (22/7) × 49 = 11 × 7 = 77 m²
Step 3: Park area (excluding flowerbed) = 2400 − 77 = 2323 m²
Suggested Marks Distribution
Rectangle area = 2400 m²1 Mark
Semicircle area = 77 m²2 Marks
Final answer = 2323 m²1 Mark
Q3
A rectangular field is 75 m long and 45 m broad. A runner runs around it 4 times. Find the total distance covered by the runner.
2 Marks
✓ Suggested Step-by-Step Solution
Perimeter of rectangular field = 2 × (length + breadth) = 2 × (75 + 45) = 2 × 120 = 240 m.
Distance in 1 round = 240 m.
Total distance in 4 rounds = 4 × 240 = 960 m.
Suggested Marks Distribution
Perimeter calculation = 240 m1 Mark
Total distance in 4 rounds = 960 m1 Mark
Q3
Find the area of a triangular park ABC with sides 120 m, 80 m, and 50 m using Heron's formula.
3 Marks
✓ Suggested Step-by-Step Solution
Semi-perimeter s = (a + b + c) / 2 = (120 + 80 + 50) / 2 = 250 / 2 = 125 m.
s − a = 125 − 120 = 5 m
s − b = 125 − 80 = 45 m
s − c = 125 − 50 = 75 m
Area = √[s(s−a)(s−b)(s−c)] = √[125 × 5 × 45 × 75]
= √[(5³) × 5 × (3² × 5) × (3 × 5²)] = √[5⁷ × 3³] = 5³ × 3 × √(5 × 3) = 375√15 m² (≈ 1452.37 m²).
Suggested Marks Distribution
Calculating semi-perimeter s = 125 m1 Mark
Applying Heron formula and exact simplified square root2 Marks
Q2
Find the area of a trapezium whose parallel sides are 25 cm and 13 cm, and the distance between them is 8 cm.
2 Marks
✓ Suggested Step-by-Step Solution
Formula for area of trapezium = ½ × (sum of parallel sides) × height.
Area = ½ × (25 + 13) × 8 = ½ × 38 × 8 = 19 × 8 = 152 cm².
Suggested Marks Distribution
Trapezium area formula1 Mark
Correct evaluation 152 cm^21 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 6CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Agricultural Land Surveying (Heron's Formula): A farmer owns a triangular plot of land with sides a = 35 m, b = 53 m, and c = 66 m. The farmer wants to sow wheat which costs ₹12 per square meter for seeds and tilling.
(a) Calculate the semi-perimeter s of the triangular field.
(b) Calculate the total area using Heron's formula.
(c) Determine the total cost of cultivation.
(a) Calculate the semi-perimeter s of the triangular field.
(b) Calculate the total area using Heron's formula.
(c) Determine the total cost of cultivation.
(a) Semi-perimeter: s = (a + b + c)/2 = (35 + 53 + 66)/2 = 154/2 = 77 m.
(b) s - a = 77 - 35 = 42; s - b = 77 - 53 = 24; s - c = 77 - 66 = 11.
Area = √(77 × 42 × 24 × 11) = √(11 × 7 × 7 × 6 × 6 × 4 × 11) = 11 × 7 × 6 × 2 = 924 m².
(c) Cost of cultivation = 924 × ₹12 = ₹11,088.
(b) s - a = 77 - 35 = 42; s - b = 77 - 53 = 24; s - c = 77 - 66 = 11.
Area = √(77 × 42 × 24 × 11) = √(11 × 7 × 7 × 6 × 6 × 4 × 11) = 11 × 7 × 6 × 2 = 924 m².
(c) Cost of cultivation = 924 × ₹12 = ₹11,088.
🧠 HOTS & Assertion-Reasoning
Assertion (A): Among all triangles with a given fixed perimeter, the equilateral triangle encloses the maximum area.
Reason (R): In Heron's formula √[s(s-a)(s-b)(s-c)], the product (s-a)(s-b)(s-c) is maximized when the factors are equal (AM = GM inequality).
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): In Heron's formula √[s(s-a)(s-b)(s-c)], the product (s-a)(s-b)(s-c) is maximized when the factors are equal (AM = GM inequality).
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: By the AM-GM inequality, the product of terms with constant sum is maximized when all terms are equal: s - a = s - b = s - c ⟹ a = b = c.
Explanation: By the AM-GM inequality, the product of terms with constant sum is maximized when all terms are equal: s - a = s - b = s - c ⟹ a = b = c.
7
Chapter 7: The Mathematics of Maybe — Introduction to Probability
Ganita Manjari (Part I) — Exercise Sets 7.1, 7.2 & 7.3 — Suggested Marks Distribution
Exercise Set 7.1 — Basic Concepts of Probability
Q1
Define: (i) Sample space (ii) Event (iii) Probability of an event. Write the range of probability values.
4 Marks
✓ Suggested Step-by-Step Solution
Sample Space (S): The set of all possible outcomes of a random experiment. E.g., for a coin toss: S = {H, T}.
Event (E): Any subset of the sample space. E.g., getting a head: E = {H}.
Probability: P(E) = (Number of favourable outcomes) / (Total number of outcomes) = n(E)/n(S)
Range: 0 ≤ P(E) ≤ 1. P(impossible event) = 0; P(certain event) = 1.
Suggested Marks Distribution
Sample space definition with example1 Mark
Event definition1 Mark
Probability formula1 Mark
Range 0 ≤ P ≤ 11 Mark
Q2
A die is rolled once. Find the probability of getting: (i) An even number (ii) A number greater than 4 (iii) A multiple of 3 (iv) The number 7
4 Marks
✓ Suggested Step-by-Step Solution
Sample space S = {1, 2, 3, 4, 5, 6}; n(S) = 6.
(i) Even numbers: {2, 4, 6}; P = 3/6 = 1/2
(ii) Greater than 4: {5, 6}; P = 2/6 = 1/3
(iii) Multiple of 3: {3, 6}; P = 2/6 = 1/3
(iv) Number 7: Impossible (7 not on die); P = 0
Suggested Marks Distribution
Each correct probability with working (×4)1 Mark each
Exercise Set 7.2 — Experimental and Theoretical Probability
Q1
A bag contains 3 red, 5 blue, and 2 green balls. A ball is drawn at random. Find the probability of drawing: (i) A red ball (ii) A blue ball (iii) Not a green ball
3 Marks
✓ Suggested Step-by-Step Solution
Total balls = 3 + 5 + 2 = 10.
(i) Red: P = 3/10 = 0.3
(ii) Blue: P = 5/10 = 1/2
(iii) Not green: Balls that are not green = 3 + 5 = 8. P = 8/10 = 4/5
Suggested Marks Distribution
Each correct probability (×3)1 Mark each
Q2
Two coins are tossed simultaneously. Find the probability of getting: (i) Exactly one head (ii) At least one head (iii) No head
3 Marks
✓ Suggested Step-by-Step Solution
S = {HH, HT, TH, TT}; n(S) = 4.
(i) Exactly one H: {HT, TH}; P = 2/4 = 1/2
(ii) At least one H: {HH, HT, TH}; P = 3/4 = 3/4
(iii) No H: {TT}; P = 1/4 = 1/4
Suggested Marks Distribution
Each correct probability with sample space (×3)1 Mark each
Exercise Set 7.3 — Applications and Word Problems
Q1
In a class of 40 students, 18 like cricket, 15 like football, and 7 like both. A student is chosen at random. Find the probability that the student likes neither cricket nor football.
3 Marks
✓ Suggested Step-by-Step Solution
Step 1: n(Cricket ∪ Football) = 18 + 15 − 7 = 26
Step 2: Students who like neither = 40 − 26 = 14
P(neither) = 14/40 = 7/20
Suggested Marks Distribution
Union formula applied: 261 Mark
Neither = 141 Mark
P = 7/201 Mark
Q3
What is the sum of probabilities of all elementary events of an experiment? If P(E) = 0.05, what is the probability of 'not E'?
2 Marks
✓ Suggested Step-by-Step Solution
The sum of the probabilities of all elementary events of an experiment is always equal to 1.
We know that P(E) + P(not E) = 1.
Therefore, P(not E) = 1 − 0.05 = 0.95.
Suggested Marks Distribution
Stating total probability = 11 Mark
Calculation P(not E) = 0.951 Mark
Q3
A card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting: (i) a king of red colour, (ii) a face card.
2 Marks
✓ Suggested Step-by-Step Solution
Total number of outcomes = 52.
(i) King of red colour: 2 cards (King of hearts, King of diamonds). Probability = 2/52 = 1/26.
(ii) Face card: 4 Kings + 4 Queens + 4 Jacks = 12 cards. Probability = 12/52 = 3/13.
Suggested Marks Distribution
Probability of red king = 1/261 Mark
Probability of face card = 3/131 Mark
Q2
In a lottery of 100 tickets numbered 1 to 100, one ticket is drawn at random. What is the probability that the ticket drawn bears a perfect square number?
2 Marks
✓ Suggested Step-by-Step Solution
Total outcomes = 100.
Perfect squares between 1 and 100 are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 (10 numbers).
Favourable outcomes = 10.
Probability = 10/100 = 1/10 (or 0.1).
Suggested Marks Distribution
Listing perfect squares (10 numbers)1 Mark
Correct probability = 1/101 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 7CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Quality Control in Electronics (Probability): A batch of 500 LED computer monitors was tested at a manufacturing plant. 465 passed all quality checks on the first attempt, 25 had minor display defects that could be repaired, and 10 were completely discarded.
(a) What is the probability that a randomly chosen monitor passes without defect?
(b) What is the probability that a monitor requires repair?
(c) Verify that the sum of probabilities of all elementary events equals 1.
(a) What is the probability that a randomly chosen monitor passes without defect?
(b) What is the probability that a monitor requires repair?
(c) Verify that the sum of probabilities of all elementary events equals 1.
(a) P(Pass) = 465 / 500 = 93/100 = 0.93.
(b) P(Repair) = 25 / 500 = 1/20 = 0.05.
(c) P(Discard) = 10 / 500 = 0.02. Sum = 0.93 + 0.05 + 0.02 = 1.00. This verifies the fundamental probability law.
(b) P(Repair) = 25 / 500 = 1/20 = 0.05.
(c) P(Discard) = 10 / 500 = 0.02. Sum = 0.93 + 0.05 + 0.02 = 1.00. This verifies the fundamental probability law.
🧠 HOTS & Assertion-Reasoning
Assertion (A): If the probability of winning a chess game is 0.68, then the probability of not winning is 0.32.
Reason (R): For any event E, P(E) + P(not E) = 1.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): For any event E, P(E) + P(not E) = 1.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: The complementary event of winning is 'not winning'. By rule P(not E) = 1 - P(E) = 1 - 0.68 = 0.32, which directly validates both statement and reason.
Explanation: The complementary event of winning is 'not winning'. By rule P(not E) = 1 - P(E) = 1 - 0.68 = 0.32, which directly validates both statement and reason.
8
Chapter 8: Predicting What Comes Next — Exploring Sequences
Ganita Manjari (Part I) — Exercise Sets 8.1, 8.2 & 8.3 — Suggested Marks Distribution
Exercise Set 8.1 — Arithmetic Progressions
Q1
Identify whether the following are APs. If yes, find the common difference: (i) 2, 5, 8, 11, ... (ii) 1, 4, 9, 16, ... (iii) 10, 7, 4, 1, ...
3 Marks
✓ Suggested Step-by-Step Solution
(i) 2, 5, 8, 11: Differences = 3, 3, 3 (constant). Yes, AP. Common difference d = 3
(ii) 1, 4, 9, 16: Differences = 3, 5, 7 (not constant). Not an AP.
(iii) 10, 7, 4, 1: Differences = −3, −3, −3 (constant). Yes, AP. Common difference d = −3
Suggested Marks Distribution
Each correct identification with reason (×3)1 Mark each
Q2
Find the 20th term of the AP: 3, 7, 11, 15, ...
2 Marks
✓ Suggested Step-by-Step Solution
a = 3, d = 4. Formula: aₙ = a + (n−1)d
a₂₀ = 3 + (20−1)×4 = 3 + 19×4 = 3 + 76 = 79
Suggested Marks Distribution
Correct formula applied1 Mark
Correct answer a₀ = 791 Mark
Q3
Find the sum of first 15 terms of the AP: 5, 8, 11, 14, ...
3 Marks
✓ Suggested Step-by-Step Solution
a = 5, d = 3, n = 15.
Sₙ = (n/2)[2a + (n−1)d] = (15/2)[2×5 + 14×3]
= (15/2)[10 + 42] = (15/2) × 52 = 15 × 26 = 390
Suggested Marks Distribution
Correct sum formula used1 Mark
Correct substitution1 Mark
Sₙ = 3901 Mark
Exercise Set 8.2 — Geometric and Other Sequences
Q1
Find the next three terms of the geometric sequence: 2, 6, 18, ...
2 Marks
✓ Suggested Step-by-Step Solution
Common ratio r = 6/2 = 3. Each term is multiplied by 3.
4th term = 18 × 3 = 54; 5th = 162; 6th = 486.
Next three terms: 54, 162, 486
Suggested Marks Distribution
Correct common ratio r = 31 Mark
Three correct terms: 54, 162, 4861 Mark
Exercise Set 8.3 — Real-life Applications of Sequences
Q1
A savings plan starts with ₹500 in month 1 and increases by ₹200 each month. How much will be saved in month 12? What is the total savings over 12 months?
4 Marks
✓ Suggested Step-by-Step Solution
a = 500, d = 200, n = 12.
Month 12: a₁₂ = 500 + (12−1)×200 = 500 + 2200 = ₹2700
Total savings: S₁₂ = (12/2)[2×500 + 11×200] = 6[1000 + 2200] = 6×3200 = ₹19,200
Suggested Marks Distribution
Month 12 amount = ₹27002 Marks
Total S₁₂ = ₹19,2002 Marks
Q4
Which term of the AP: 21, 18, 15, ... is −81? Also, is any term 0?
2 Marks
✓ Suggested Step-by-Step Solution
Here a = 21, d = 18 − 21 = −3. Let aₙ = −81.
aₙ = a + (n − 1)d → −81 = 21 + (n − 1)(−3)
−102 = −3(n − 1) → n − 1 = 34 → n = 35. (35th term is −81).
For aₙ = 0: 0 = 21 + (n − 1)(−3) → 3(n − 1) = 21 → n − 1 = 7 → n = 8 (8th term is 0).
Suggested Marks Distribution
Finding n = 35 for term -811 Mark
Finding n = 8 for term 01 Mark
Q2
In a Geometric Progression (GP), the first term is 3 and the common ratio is 2. Write down the first five terms of this GP.
2 Marks
✓ Suggested Step-by-Step Solution
General term of GP: aₙ = a × rⁿ⁻¹. Here a = 3, r = 2.
Term 1: 3
Term 2: 3 × 2 = 6
Term 3: 6 × 2 = 12
Term 4: 12 × 2 = 24
Term 5: 24 × 2 = 48
The first 5 terms are: 3, 6, 12, 24, 48.
Suggested Marks Distribution
Formula for GP terms1 Mark
Listing first five terms correctly1 Mark
Q2
A bacteria culture doubles every hour. If there were initially 100 bacteria, write the sequence of bacteria population for the first 5 hours.
2 Marks
✓ Suggested Step-by-Step Solution
Initial count (Hour 0) = 100. Growth factor = 2 per hour.
Hour 1: 100 × 2 = 200
Hour 2: 200 × 2 = 400
Hour 3: 400 × 2 = 800
Hour 4: 800 × 2 = 1600
Hour 5: 1600 × 2 = 3200
The population sequence is: 200, 400, 800, 1600, 3200.
Suggested Marks Distribution
Identifying geometric doubling model1 Mark
Listing correct sequence up to hour 51 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 8CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Solar Energy Deployment Trends (Sequences): A clean energy initiative installs 120 rooftop solar units in Month 1. Due to increased subsidies, they install 25 additional units each consecutive month (145 in Month 2, 170 in Month 3, and so on).
(a) Identify the type of sequence and state its first term a and common difference d.
(b) How many solar units will be installed in Month 12?
(c) Find the total number of solar units installed during the first year (12 months).
(a) Identify the type of sequence and state its first term a and common difference d.
(b) How many solar units will be installed in Month 12?
(c) Find the total number of solar units installed during the first year (12 months).
(a) It forms an Arithmetic Progression (AP) with first term a = 120 and common difference d = 25.
(b) Month 12: a₁₂ = a + (12 - 1)d = 120 + 11(25) = 120 + 275 = 395 units.
(c) Total units in year 1: S₁₂ = (12/2)[2a + (12 - 1)d] = 6[2(120) + 275] = 6[240 + 275] = 6 × 515 = 3,090 solar units.
(b) Month 12: a₁₂ = a + (12 - 1)d = 120 + 11(25) = 120 + 275 = 395 units.
(c) Total units in year 1: S₁₂ = (12/2)[2a + (12 - 1)d] = 6[2(120) + 275] = 6[240 + 275] = 6 × 515 = 3,090 solar units.
🧠 HOTS & Assertion-Reasoning
Assertion (A): The n-th term of any Arithmetic Progression is a linear expression in n of the form an = pn + q.
Reason (R): In an AP, an = a + (n - 1)d = dn + (a - d), where the coefficient of n is the common difference d.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): In an AP, an = a + (n - 1)d = dn + (a - d), where the coefficient of n is the common difference d.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: Expanding an = a + (n-1)d yields dn + (a-d). Here d and (a-d) are constants, making an a linear polynomial in n.
Explanation: Expanding an = a + (n-1)d yields dn + (a-d). Here d and (a-d) are constants, making an a linear polynomial in n.
9
Chapter 9: Propositions and Their Converses
Ganita Manjari (Part II) — Exercise Sets 9.1, 9.2 & 9.3 — Suggested Marks Distribution
Exercise Set 9.1 — If–Then Statements and Logic
Q1
Write the converse of each statement: (i) If it rains, then the ground is wet. (ii) If a triangle is equilateral, then all its angles are 60°. (iii) If a number is divisible by 6, then it is divisible by 2.
3 Marks
✓ Suggested Step-by-Step Solution
The converse of "If P, then Q" is "If Q, then P".
(i) If the ground is wet, then it rained.
(ii) If all angles of a triangle are 60°, then the triangle is equilateral.
(iii) If a number is divisible by 2, then it is divisible by 6. (Note: The converse is NOT always true — e.g., 4 is divisible by 2 but not 6.)
Suggested Marks Distribution
Each correct converse (×3)1 Mark each
Q2
For the statement "If a quadrilateral is a square, then it is a rectangle": (i) Write the converse. (ii) Is the original statement true? (iii) Is the converse true? Justify.
4 Marks
✓ Suggested Step-by-Step Solution
(i) Converse: If a quadrilateral is a rectangle, then it is a square.
(ii) Original is TRUE: A square has all right angles and equal sides → it is a rectangle.
(iii) Converse is FALSE: A rectangle has all right angles but sides need not be equal. Example: A 4 cm × 6 cm rectangle is NOT a square. Counterexample disproves the converse.
Suggested Marks Distribution
Correct converse written1 Mark
Original is true with reason1 Mark
Converse is false with counterexample2 Marks
Exercise Set 9.2 — Proofs in Geometry
Q1
Prove: The base angles of an isosceles triangle are equal.
4 Marks
✓ Suggested Step-by-Step Solution
Given: Triangle ABC with AB = AC. To prove: ∠B = ∠C.
Construction: Draw AD, the bisector of ∠A, meeting BC at D.
In △ABD and △ACD:
AB = AC (given)
AD = AD (common)
∠BAD = ∠CAD (AD bisects ∠A)
AB = AC (given)
AD = AD (common)
∠BAD = ∠CAD (AD bisects ∠A)
By SAS congruence: △ABD ≅ △ACD
Therefore: ∠ABD = ∠ACD (CPCT), i.e., ∠B = ∠C □
Suggested Marks Distribution
Correct construction stated1 Mark
Three correct pairs for congruence2 Marks
SAS, CPCT and conclusion1 Mark
Exercise Set 9.3 — Counterexamples
Q1
Disprove the following using a counterexample: "If n is a prime number, then n is odd."
2 Marks
✓ Suggested Step-by-Step Solution
Counterexample: n = 2.
2 is a prime number (divisible only by 1 and itself).
But 2 is even, not odd.
Conclusion: The statement is FALSE. n = 2 is a prime that is not odd. One counterexample is sufficient to disprove a universal statement.
Suggested Marks Distribution
Correct counterexample n = 21 Mark
Explanation why it disproves the statement1 Mark
Q3
Write the converse, inverse, and contrapositive of: "If two angles are vertical angles, then they are equal."
2 Marks
✓ Suggested Step-by-Step Solution
Let p = "two angles are vertical angles", q = "they are equal".
Converse (q → p): "If two angles are equal, then they are vertical angles" (Note: false).
Inverse (∼p → ∼q): "If two angles are not vertical angles, then they are not equal."
Contrapositive (∼q → ∼p): "If two angles are not equal, then they are not vertical angles" (True).
Suggested Marks Distribution
Correct converse statement1 Mark
Correct contrapositive statement1 Mark
Q2
Prove that the sum of the angles of a triangle is 180°.
2 Marks
✓ Suggested Step-by-Step Solution
Given: A triangle ABC.
Construction: Draw line XY through A parallel to BC.
Since XY || BC and AB is transversal: ∠XAB = ∠B (alternate interior angles).
Similarly, AC is transversal: ∠YAC = ∠C (alternate interior angles).
Since XAY is a straight line: ∠XAB + ∠A + ∠YAC = 180°.
Substituting: ∠B + ∠A + ∠C = 180° → ∠A + ∠B + ∠C = 180°. Hence proved.
Suggested Marks Distribution
Construction of parallel line and alternate angles1 Mark
Straight line property and conclusion1 Mark
Q3
State Euclid's Fifth Postulate in simple words.
2 Marks
✓ Suggested Step-by-Step Solution
Euclid's Fifth Postulate (Parallel Postulate):
If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles (180°), then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.
Suggested Marks Distribution
Accurate and clear statement of Euclid 5th postulate2 Marks
Q2
Provide a counterexample to disprove the statement: "Every prime number is an odd number."
2 Marks
✓ Suggested Step-by-Step Solution
Consider the number 2.
2 is a prime number because its only positive divisors are 1 and 2.
However, 2 is an even number, not an odd number.
Thus, 2 serves as a valid counterexample disproving the general statement.
Suggested Marks Distribution
Identifying 2 as the counterexample1 Mark
Justifying primality and even nature1 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 9CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Logical Deductions & Geometry Theorems: Consider the geometric proposition: 'If two lines intersect, then the vertically opposite angles are equal.'
(a) Identify the hypothesis (P) and the conclusion (Q) in the proposition.
(b) Write the converse of the statement.
(c) Is the converse always true? Explain with a counterexample or diagrammatic explanation.
(a) Identify the hypothesis (P) and the conclusion (Q) in the proposition.
(b) Write the converse of the statement.
(c) Is the converse always true? Explain with a counterexample or diagrammatic explanation.
(a) Hypothesis (P): 'Two straight lines intersect.' Conclusion (Q): 'The vertically opposite angles are equal.'
(b) Converse: 'If two angles formed by intersecting lines are equal, then they are vertically opposite angles.'
(c) The converse is not necessarily true in general without specifying geometric orientation: two non-vertically opposite angles can also be equal (for example, alternate interior angles or adjacent supplementary angles each equal to 90°).
(b) Converse: 'If two angles formed by intersecting lines are equal, then they are vertically opposite angles.'
(c) The converse is not necessarily true in general without specifying geometric orientation: two non-vertically opposite angles can also be equal (for example, alternate interior angles or adjacent supplementary angles each equal to 90°).
🧠 HOTS & Assertion-Reasoning
Assertion (A): A single counterexample is sufficient to disprove a universally quantified mathematical conjecture.
Reason (R): For a universal statement 'For all x, P(x)' to be true, P(x) must hold for every member of the domain without exception.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): For a universal statement 'For all x, P(x)' to be true, P(x) must hold for every member of the domain without exception.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: The negation of '∀x, P(x)' is '∃x such that ¬P(x)'. Finding even one instance where the conjecture fails proves the negation and invalidates the claim.
Explanation: The negation of '∀x, P(x)' is '∃x such that ¬P(x)'. Finding even one instance where the conjecture fails proves the negation and invalidates the claim.
10
Chapter 10: How Quantities Combine — Understanding Data
Ganita Manjari (Part II) — Exercise Sets 10.1, 10.2 & 10.3 — Suggested Marks Distribution
Exercise Set 10.1 — Weighted Averages and Mixtures
Q1
A student scored 72 in Maths (out of 100), 68 in Science (out of 80), and 55 in English (out of 60). Find the weighted average percentage, where the weights are the total marks.
4 Marks
✓ Suggested Step-by-Step Solution
Total marks obtained = 72 + 68 + 55 = 195
Total possible marks = 100 + 80 + 60 = 240
Weighted average percentage = (195/240) × 100 = (195 × 100)/240 = 81.25%
This differs from the simple average: (72% + 85% + 91.67%)/3 ≈ 82.9% — illustrating why weights matter.
Suggested Marks Distribution
Total marks obtained = 1951 Mark
Total possible = 2401 Mark
Weighted % = 81.25%2 Marks
Q2
20 litres of 30% salt solution is mixed with 30 litres of 50% salt solution. Find the concentration (%) of salt in the mixture.
3 Marks
✓ Suggested Step-by-Step Solution
Salt in first solution = 30% of 20 = 6 litres
Salt in second solution = 50% of 30 = 15 litres
Concentration = (6+15)/(20+30) × 100 = 21/50 × 100 = 42%
Suggested Marks Distribution
Salt in each solution1 Mark
Total salt = 21 L1 Mark
Concentration = 42%1 Mark
Exercise Set 10.2 — Data Visualisation
Q1
The following data shows the number of books read per month by 5 students: 4, 7, 3, 9, 5. Find the mean, median, and mode.
3 Marks
✓ Suggested Step-by-Step Solution
Mean = (4+7+3+9+5)/5 = 28/5 = 5.6
Median: Arranged order: 3, 4, 5, 7, 9. Middle value = 5
Mode: All values appear once → No mode
Suggested Marks Distribution
Mean = 5.61 Mark
Median = 5 (with sorted order)1 Mark
Mode = none (or all)1 Mark
Q2
What is a stacked bar chart? When is it more useful than a simple bar chart? Give one example scenario.
3 Marks
✓ Suggested Step-by-Step Solution
Stacked bar chart: A bar chart where each bar is divided into segments representing different sub-categories, stacked on top of each other.
More useful when: You want to compare both the total value and the composition (proportions of sub-categories) across multiple groups simultaneously.
Example: Monthly sales data for a shop selling fruits, vegetables, and dairy — a stacked bar shows both total sales and the contribution of each product category per month.
Suggested Marks Distribution
Definition of stacked bar chart1 Mark
When it is more useful1 Mark
Relevant example1 Mark
Exercise Set 10.3 — Interpreting Data
Q1
The marks scored by 8 students in a test are: 45, 67, 89, 34, 72, 67, 56, 89. Find the range and identify the outlier(s), if any.
3 Marks
✓ Suggested Step-by-Step Solution
Sorted data: 34, 45, 56, 67, 67, 72, 89, 89
Range = Maximum − Minimum = 89 − 34 = 55
Mean ≈ (34+45+56+67+67+72+89+89)/8 = 519/8 ≈ 64.9. The value 34 is significantly below the mean. It could be considered an outlier (isolated low score).
Suggested Marks Distribution
Sorted data and range = 551 Mark
Mean calculated ≈ 64.91 Mark
Identifying 34 as possible outlier with reasoning1 Mark
Q3
Find the mean of the first 10 natural numbers.
2 Marks
✓ Suggested Step-by-Step Solution
First 10 natural numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Sum = n(n + 1) / 2 = 10 × 11 / 2 = 55.
Mean = Sum / Total count = 55 / 10 = 5.5.
Suggested Marks Distribution
Sum of first 10 natural numbers = 551 Mark
Mean = 5.51 Mark
Q3
What is a frequency polygon? When is it preferred over a histogram?
2 Marks
✓ Suggested Step-by-Step Solution
A frequency polygon is a graphical representation of grouped data formed by joining the midpoints (class marks) of the tops of the bars of a histogram with straight line segments.
It is preferred when comparing two or more data distributions on the same axes.
Suggested Marks Distribution
Definition of frequency polygon using class marks1 Mark
Application for comparing distributions1 Mark
Q2
Find the median and mode of the following data set: 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
2 Marks
✓ Suggested Step-by-Step Solution
Arrange in ascending order: 14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28 (n = 12 terms).
Median: Average of (12/2)th and (12/2 + 1)th terms = 6th and 7th terms = (18 + 18) / 2 = 18.
Mode: Number with highest frequency: 14 occurs 4 times. Mode = 14.
Suggested Marks Distribution
Arranging data and calculating median = 181 Mark
Mode = 141 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 10CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Environmental Science & Weighted Air Quality Index: An environmental research station records the particulate matter (PM2.5) concentrations at four different monitoring points with weights based on population exposure:
Point 1: 45 µg/m³ (weight 3), Point 2: 60 µg/m³ (weight 2), Point 3: 80 µg/m³ (weight 4), Point 4: 110 µg/m³ (weight 1).
(a) Find the simple arithmetic mean of the four readings.
(b) Calculate the weighted mean concentration of PM2.5.
(c) Why is the weighted mean a more reliable representation than the simple mean?
Point 1: 45 µg/m³ (weight 3), Point 2: 60 µg/m³ (weight 2), Point 3: 80 µg/m³ (weight 4), Point 4: 110 µg/m³ (weight 1).
(a) Find the simple arithmetic mean of the four readings.
(b) Calculate the weighted mean concentration of PM2.5.
(c) Why is the weighted mean a more reliable representation than the simple mean?
(a) Simple Mean = (45 + 60 + 80 + 110) / 4 = 295 / 4 = 73.75 µg/m³.
(b) Weighted Mean = Σ(w · x) / Σw = [3(45) + 2(60) + 4(80) + 1(110)] / (3 + 2 + 4 + 1) = [135 + 120 + 320 + 110] / 10 = 685 / 10 = 68.5 µg/m³.
(c) The weighted mean gives greater importance to heavily populated zones (Points 1 & 3), giving a truer measure of actual public health exposure.
(b) Weighted Mean = Σ(w · x) / Σw = [3(45) + 2(60) + 4(80) + 1(110)] / (3 + 2 + 4 + 1) = [135 + 120 + 320 + 110] / 10 = 685 / 10 = 68.5 µg/m³.
(c) The weighted mean gives greater importance to heavily populated zones (Points 1 & 3), giving a truer measure of actual public health exposure.
🧠 HOTS & Assertion-Reasoning
Assertion (A): The median is a more robust measure of central tendency than the mean when a dataset contains extreme outliers.
Reason (R): The mean utilizes the numerical values of all observations, whereas the median depends only on the positional rank of observations.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): The mean utilizes the numerical values of all observations, whereas the median depends only on the positional rank of observations.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: A single extreme value can shift the mean drastically, whereas the middle rank (median) remains stable. Reason (R) correctly explains this structural robustness.
Explanation: A single extreme value can shift the mean drastically, whereas the middle rank (median) remains stable. Reason (R) correctly explains this structural robustness.
11
Chapter 11: The World of Algorithms
Ganita Manjari (Part II) — Exercise Sets 11.1, 11.2 & 11.3 — Suggested Marks Distribution
Exercise Set 11.1 — What is an Algorithm?
Q1
Define an algorithm. Write an algorithm to find the largest of three numbers A, B, and C.
4 Marks
✓ Suggested Step-by-Step Solution
Algorithm: A finite, step-by-step procedure to solve a problem or accomplish a task.
Algorithm to find largest of A, B, C:
Step 1: Input A, B, C.
Step 2: If A > B and A > C, then Largest = A. Go to Step 5.
Step 3: If B > C, then Largest = B. Go to Step 5.
Step 4: Largest = C.
Step 5: Output Largest.
Step 6: Stop.
Step 1: Input A, B, C.
Step 2: If A > B and A > C, then Largest = A. Go to Step 5.
Step 3: If B > C, then Largest = B. Go to Step 5.
Step 4: Largest = C.
Step 5: Output Largest.
Step 6: Stop.
Suggested Marks Distribution
Definition of algorithm1 Mark
Correct logical steps covering all cases3 Marks
Exercise Set 11.2 — GCD and Euclid’s Algorithm
Q1
Use Euclid’s algorithm to find HCF(252, 105).
4 Marks
✓ Suggested Step-by-Step Solution
Euclid’s Algorithm: HCF(a, b) = HCF(b, a mod b) until remainder = 0.
Step 1: 252 = 105 × 2 + 42 → HCF(252,105) = HCF(105, 42)
Step 2: 105 = 42 × 2 + 21 → HCF(105, 42) = HCF(42, 21)
Step 3: 42 = 21 × 2 + 0 → HCF(42, 21) = 21
HCF(252, 105) = 21
Suggested Marks Distribution
Algorithm stated correctly1 Mark
Steps 1 and 2 correct2 Marks
Final HCF = 211 Mark
Q2
Use Euclid’s algorithm to find HCF(455, 42).
3 Marks
✓ Suggested Step-by-Step Solution
455 = 42 × 10 + 35 → HCF(455,42) = HCF(42,35)
42 = 35 × 1 + 7 → HCF(42,35) = HCF(35,7)
35 = 7 × 5 + 0 → HCF = 7
Suggested Marks Distribution
Each correct step (×3)1 Mark each
Exercise Set 11.3 — Addition Algorithms and Patterns
Q1
The Fibonacci sequence starts: 1, 1, 2, 3, 5, 8, ... Each term is the sum of the two before it. Write the next 5 terms and find the ratio of consecutive terms to 2 decimal places for the last 3 pairs.
4 Marks
✓ Suggested Step-by-Step Solution
Next 5 terms: 13, 21, 34, 55, 89
Ratios (next/previous):
34/21 ≈ 1.62
55/34 ≈ 1.62
89/55 ≈ 1.62
34/21 ≈ 1.62
55/34 ≈ 1.62
89/55 ≈ 1.62
These ratios approach the Golden Ratio φ ≈ 1.618, a famous mathematical constant.
Suggested Marks Distribution
5 correct terms: 13, 21, 34, 55, 892 Marks
Three ratios calculated ≈ 1.621 Mark
Mention of Golden Ratio1 Mark
Q2
Write an algorithm (step-by-step procedure) to find whether a given positive integer n is even or odd.
2 Marks
✓ Suggested Step-by-Step Solution
Step 1: Start.
Step 2: Input a positive integer n.
Step 3: Compute remainder r = n % 2.
Step 4: If r == 0, display "n is Even"; Else display "n is Odd".
Step 5: Stop.
Suggested Marks Distribution
Clear input, remainder check, conditional branch2 Marks
Q3
Use Euclid's Division Algorithm to find the HCF of 135 and 225.
2 Marks
✓ Suggested Step-by-Step Solution
Since 225 > 135, apply division: 225 = 135 × 1 + 90
Remainder 90 ≠ 0, apply to 135 and 90: 135 = 90 × 1 + 45
Remainder 45 ≠ 0, apply to 90 and 45: 90 = 45 × 2 + 0
Since remainder is now 0, divisor is 45. Hence, HCF(135, 225) = 45.
Suggested Marks Distribution
Steps of Euclid division lemma shown1 Mark
Final HCF = 451 Mark
Q4
Find the HCF of 867 and 255 using Euclid's algorithm.
2 Marks
✓ Suggested Step-by-Step Solution
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
The remainder is 0, so HCF(867, 255) = 51.
Suggested Marks Distribution
Division steps correctly written1 Mark
HCF = 511 Mark
Q2
Write an algorithm to compute the factorial of a positive integer n (n!).
2 Marks
✓ Suggested Step-by-Step Solution
Step 1: Start.
Step 2: Input integer n. Set fact = 1, i = 1.
Step 3: Repeat while i ≤ n: fact = fact × i; i = i + 1.
Step 4: Output fact.
Step 5: Stop.
Suggested Marks Distribution
Initialization and loop condition correctly specified2 Marks
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 11CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Computer Science & Cryptography (Euclidean Algorithm): A cybersecurity protocol requires computing the greatest common divisor of two large encryption keys, 1147 and 899, to establish a coprime shared key.
(a) Apply Euclid's Division Algorithm to find gcd(1147, 899).
(b) State the number of division steps taken to reach the remainder zero.
(c) Express the final gcd as a linear combination of 1147 and 899 (Bézout's identity).
(a) Apply Euclid's Division Algorithm to find gcd(1147, 899).
(b) State the number of division steps taken to reach the remainder zero.
(c) Express the final gcd as a linear combination of 1147 and 899 (Bézout's identity).
(a) Step 1: 1147 = 899 × 1 + 248
Step 2: 899 = 248 × 3 + 155
Step 3: 248 = 155 × 1 + 93
Step 4: 155 = 93 × 1 + 62
Step 5: 93 = 62 × 1 + 31
Step 6: 62 = 31 × 2 + 0.
The last non-zero remainder is 31. Thus, gcd(1147, 899) = 31.
(b) It took 6 division steps.
(c) Back substitution: 31 = 93 - 62 = 93 - (155 - 93) = 2(93) - 155 = 2(248 - 155) - 155 = 2(248) - 3(155) = 2(248) - 3(899 - 3×248) = 11(248) - 3(899) = 11(1147 - 899) - 3(899) = 11(1147) - 14(899) = 31.
Step 2: 899 = 248 × 3 + 155
Step 3: 248 = 155 × 1 + 93
Step 4: 155 = 93 × 1 + 62
Step 5: 93 = 62 × 1 + 31
Step 6: 62 = 31 × 2 + 0.
The last non-zero remainder is 31. Thus, gcd(1147, 899) = 31.
(b) It took 6 division steps.
(c) Back substitution: 31 = 93 - 62 = 93 - (155 - 93) = 2(93) - 155 = 2(248 - 155) - 155 = 2(248) - 3(155) = 2(248) - 3(899 - 3×248) = 11(248) - 3(899) = 11(1147 - 899) - 3(899) = 11(1147) - 14(899) = 31.
🧠 HOTS & Assertion-Reasoning
Assertion (A): The Euclidean algorithm for finding the GCD of any two positive integers always terminates in a finite number of steps.
Reason (R): The sequence of remainders r₁, r₂, r₃... forms a strictly decreasing sequence of non-negative integers bounded below by zero.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): The sequence of remainders r₁, r₂, r₃... forms a strictly decreasing sequence of non-negative integers bounded below by zero.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: Since b > r₁ > r₂ > ... ≥ 0, each remainder is strictly smaller than the preceding one. Any strictly decreasing sequence of non-negative integers must reach zero in at most b steps.
Explanation: Since b > r₁ > r₂ > ... ≥ 0, each remainder is strictly smaller than the preceding one. Any strictly decreasing sequence of non-negative integers must reach zero in at most b steps.
12
Chapter 12: Quadrilaterals
Ganita Manjari (Part II) — Exercise Sets 12.1, 12.2 & 12.3 — Suggested Marks Distribution
Exercise Set 12.1 — Properties of Parallelograms
Q1
Prove that opposite sides of a parallelogram are equal.
4 Marks
✓ Suggested Step-by-Step Solution
Given: ABCD is a parallelogram (AB ∥ CD and AD ∥ BC). To prove: AB = CD and AD = BC.
Construction: Draw diagonal AC.
In △ABC and △CDA:
∠BAC = ∠DCA (alternate interior angles, AB∥CD)
∠BCA = ∠DAC (alternate interior angles, AD∥BC)
AC = AC (common)
∠BAC = ∠DCA (alternate interior angles, AB∥CD)
∠BCA = ∠DAC (alternate interior angles, AD∥BC)
AC = AC (common)
By ASA congruence: △ABC ≅ △CDA
By CPCT: AB = CD and BC = AD. □
Suggested Marks Distribution
Construction½ Mark
Three pairs for congruence2 Marks
ASA + CPCT + conclusion1½ Marks
Q2
In parallelogram ABCD, if ∠A = 70°, find all other angles.
2 Marks
✓ Suggested Step-by-Step Solution
∠A = ∠C = 70° (opposite angles of parallelogram are equal)
∠A + ∠B = 180° (co-interior angles) → ∠B = 180° − 70° = 110°
∠D = ∠B = 110° (opposite angles)
Suggested Marks Distribution
∠C = 70° with reason1 Mark
∠B = ∠D = 110° with reason1 Mark
Exercise Set 12.2 — Midpoint Theorem and Varignon’s Theorem
Q1
State and prove the Midpoint Theorem for a triangle.
5 Marks
✓ Suggested Step-by-Step Solution
Statement: The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
Given: In △ABC, M and N are midpoints of AB and AC. To prove: MN ∥ BC and MN = BC/2.
Construction: Extend MN to point D such that MN = ND. Join CD.
In △AMN and △CDN:
AN = NC (N is midpoint of AC)
MN = ND (by construction)
∠ANM = ∠CND (vertically opposite)
AN = NC (N is midpoint of AC)
MN = ND (by construction)
∠ANM = ∠CND (vertically opposite)
By SAS: △AMN ≅ △CDN → AM = CD and CD ∥ AB
Also AM = MB → CD = MB and CD ∥ MB. So MBCD is a parallelogram → MN ∥ BC.
MD = BC (opposite sides of parallelogram); MN = MD/2 → MN = BC/2 □
Suggested Marks Distribution
Correct statement1 Mark
Correct construction1 Mark
Congruence established2 Marks
Conclusion MN∥BC and MN=BC/21 Mark
Exercise Set 12.3 — Centroids and Tiling
Q1
Find the centroid of triangle with vertices A(2, 4), B(6, 2), and C(4, 8).
2 Marks
✓ Suggested Step-by-Step Solution
Centroid G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
G = ((2+6+4)/3, (4+2+8)/3) = (12/3, 14/3) = (4, 14/3)
Suggested Marks Distribution
Correct formula applied1 Mark
Centroid = (4, 14/3)1 Mark
Q3
Show that the diagonals of a rhombus are perpendicular to each other.
2 Marks
✓ Suggested Step-by-Step Solution
Let ABCD be a rhombus with diagonals AC and BD intersecting at O.
Since a rhombus is a parallelogram, its diagonals bisect each other: OA = OC and OB = OD.
In ΔAOD and ΔCOD: OA = OC, OD = OD (common), AD = CD (sides of rhombus are equal).
Therefore, ΔAOD ≅ ΔCOD by SSS congruence.
∠AOD = ∠COD (by CPCT).
Since AC is a line, ∠AOD + ∠COD = 180° → 2∠AOD = 180° → ∠AOD = 90°. Hence proved.
Suggested Marks Distribution
Congruence of adjacent triangles using SSS1 Mark
Linear pair leading to 90 degrees1 Mark
Q2
ABC is a triangle right-angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that: (i) D is the midpoint of AC, (ii) MD ⊥ AC.
2 Marks
✓ Suggested Step-by-Step Solution
(i) In ΔABC, M is the midpoint of AB and MD || BC.
By the Converse of Midpoint Theorem, a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. Therefore, D is the midpoint of AC.
(ii) Since MD || BC and AC is transversal: ∠ADM = ∠C (corresponding angles).
Since ∠C = 90°, ∠ADM = 90° → MD ⊥ AC.
Suggested Marks Distribution
Converse of Midpoint Theorem justification1 Mark
Corresponding angles argument for perpendicularity1 Mark
Q3
State Varignon's Theorem regarding the midpoints of sides of any quadrilateral.
2 Marks
✓ Suggested Step-by-Step Solution
Varignon's Theorem:
The figure formed by connecting the consecutive midpoints of the sides of any arbitrary quadrilateral is always a parallelogram.
Furthermore, the area of the Varignon parallelogram is equal to half the area of the original quadrilateral.
Suggested Marks Distribution
Accurate statement of Varignon theorem and parallelogram property2 Marks
Q2
Find the centroid of a triangle whose vertices are P(3, −5), Q(−7, 4), and R(10, −2).
2 Marks
✓ Suggested Step-by-Step Solution
Centroid formula: G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3).
x-coordinate = (3 − 7 + 10) / 3 = 6 / 3 = 2.
y-coordinate = (−5 + 4 − 2) / 3 = −3 / 3 = −1.
Therefore, centroid G is (2, −1).
Suggested Marks Distribution
Centroid formula with correct substitution1 Mark
Coordinates (2, -1)1 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 12CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Civil Engineering & Structural Midpoints (Varignon's Theorem): Four structural columns A, B, C, D form an arbitrary non-planar quadrilateral ABCD on a construction site. Surveyors mark the midpoints of the four sides: P on AB, Q on BC, R on CD, and S on DA.
(a) State Varignon's Theorem concerning the quadrilateral PQRS formed by joining the midpoints.
(b) Prove that PQRS is always a parallelogram using the Midpoint Theorem on ΔABC and ΔADC.
(c) If the diagonals AC = 18 m and BD = 24 m intersect at right angles, what special shape does PQRS become?
(a) State Varignon's Theorem concerning the quadrilateral PQRS formed by joining the midpoints.
(b) Prove that PQRS is always a parallelogram using the Midpoint Theorem on ΔABC and ΔADC.
(c) If the diagonals AC = 18 m and BD = 24 m intersect at right angles, what special shape does PQRS become?
(a) Varignon's Theorem: The figure formed by connecting the midpoints of the sides of any quadrilateral is always a parallelogram.
(b) In ΔABC, P and Q are midpoints of AB and BC. By the Midpoint Theorem: PQ || AC and PQ = ½ AC. In ΔADC, S and R are midpoints of AD and CD. By Midpoint Theorem: SR || AC and SR = ½ AC. Hence, PQ || SR and PQ = SR. Since one pair of opposite sides is equal and parallel, PQRS is a parallelogram.
(c) When diagonals AC ⊥ BD, the adjacent sides PQ (parallel to AC) and QR (parallel to BD) are perpendicular. A parallelogram with a right angle is a rectangle.
(b) In ΔABC, P and Q are midpoints of AB and BC. By the Midpoint Theorem: PQ || AC and PQ = ½ AC. In ΔADC, S and R are midpoints of AD and CD. By Midpoint Theorem: SR || AC and SR = ½ AC. Hence, PQ || SR and PQ = SR. Since one pair of opposite sides is equal and parallel, PQRS is a parallelogram.
(c) When diagonals AC ⊥ BD, the adjacent sides PQ (parallel to AC) and QR (parallel to BD) are perpendicular. A parallelogram with a right angle is a rectangle.
🧠 HOTS & Assertion-Reasoning
Assertion (A): The diagonals of a rectangle are equal in length and bisect each other.
Reason (R): A rectangle is a parallelogram with all four interior angles equal to 90°.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): A rectangle is a parallelogram with all four interior angles equal to 90°.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: Since it is a parallelogram, its diagonals bisect each other. Considering right triangles with common base and equal heights, hypotenuses (the diagonals) are congruent by SAS.
Explanation: Since it is a parallelogram, its diagonals bisect each other. Considering right triangles with common base and equal heights, hypotenuses (the diagonals) are congruent by SAS.
13
Chapter 13: Two Variables, One Line
Ganita Manjari (Part II) — Exercise Sets 13.1, 13.2 & 13.3 — Suggested Marks Distribution
Exercise Set 13.1 — Linear Equations in Two Variables
Q1
Write two solutions each for: (i) 2x + y = 8 (ii) x − 3y = 6
4 Marks
✓ Suggested Step-by-Step Solution
(i) 2x + y = 8:
Put x = 0: y = 8 → Solution (0, 8)
Put x = 2: y = 8−4 = 4 → Solution (2, 4)
Put x = 0: y = 8 → Solution (0, 8)
Put x = 2: y = 8−4 = 4 → Solution (2, 4)
(ii) x − 3y = 6:
Put y = 0: x = 6 → Solution (6, 0)
Put y = −1: x = 6−3 = 3 → Solution (3, −1)
Put y = 0: x = 6 → Solution (6, 0)
Put y = −1: x = 6−3 = 3 → Solution (3, −1)
Suggested Marks Distribution
Two correct solutions for (i)2 Marks
Two correct solutions for (ii)2 Marks
Q2
Express y in terms of x for 3x + 2y = 12. Use this to check whether (2, 3) is a solution.
3 Marks
✓ Suggested Step-by-Step Solution
3x + 2y = 12 → 2y = 12 − 3x → y = (12 − 3x)/2
At x = 2: y = (12 − 6)/2 = 6/2 = 3
Since y = 3 when x = 2, the point (2, 3) satisfies the equation. ✓ It is a solution.
Suggested Marks Distribution
y expressed correctly1 Mark
Substituting x = 2 gives y = 31 Mark
Correct conclusion1 Mark
Exercise Set 13.2 — Graphical Representation
Q1
Draw the graph of x + 2y = 6. Find the x-intercept and y-intercept from the graph.
3 Marks
✓ Suggested Step-by-Step Solution
x + 2y = 6 → y = (6 − x)/2
| x | y = (6−x)/2 | Point |
|---|---|---|
| 0 | 3 | (0, 3) |
| 2 | 2 | (2, 2) |
| 6 | 0 | (6, 0) |
x-intercept (where y=0): x = 6, point (6, 0)
y-intercept (where x=0): y = 3, point (0, 3)
Suggested Marks Distribution
Table of values correct1 Mark
Correct graph (straight line through points)1 Mark
Intercepts: x = 6, y = 31 Mark
Exercise Set 13.3 — Pair of Linear Equations (Graphical Method)
Q1
Solve graphically: x + y = 5 and x − y = 1.
4 Marks
✓ Suggested Step-by-Step Solution
Line 1: x + y = 5 → points: (0,5), (5,0), (2,3)
Line 2: x − y = 1 → points: (0,−1), (1,0), (3,2)
Plot both lines. They intersect at a common point.
Verification: At (3, 2): 3+2=5 ✓ and 3−2=1 ✓
Solution: x = 3, y = 2
Suggested Marks Distribution
Correct table/points for each line1 Mark
Correct graphs drawn1 Mark
Intersection point identified1 Mark
Verification: x=3, y=21 Mark
Q3
Find the value of k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
2 Marks
✓ Suggested Step-by-Step Solution
Substitute x = 2 and y = 1 into the equation:
2(2) + 3(1) = k
4 + 3 = k → k = 7.
Suggested Marks Distribution
Substitution of coordinates1 Mark
k = 71 Mark
Q2
Give the geometric representations of 2x + 9 = 0 as an equation: (i) in one variable, (ii) in two variables.
2 Marks
✓ Suggested Step-by-Step Solution
2x + 9 = 0 → 2x = −9 → x = −4.5.
(i) In one variable: It represents a single point at x = −4.5 on the number line.
(ii) In two variables: It can be written as 2x + 0y + 9 = 0, which represents a straight line parallel to the y-axis passing through (−4.5, 0).
Suggested Marks Distribution
One variable representation (point on number line)1 Mark
Two variable representation (line parallel to y-axis)1 Mark
Q2
The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement and express it in standard form ax + by + c = 0.
2 Marks
✓ Suggested Step-by-Step Solution
Let the cost of a notebook be ₹x and cost of a pen be ₹y.
Given: Cost of notebook = 2 × Cost of pen → x = 2y.
In standard form: 1x − 2y + 0 = 0 (where a = 1, b = −2, c = 0).
Suggested Marks Distribution
Forming relation x = 2y1 Mark
Standard form 1x - 2y + 0 = 01 Mark
Q3
Find four different solutions of the equation x + 2y = 6.
2 Marks
✓ Suggested Step-by-Step Solution
Rewrite as: x = 6 − 2y.
If y = 0: x = 6 → (6, 0)
If y = 1: x = 6 − 2(1) = 4 → (4, 1)
If y = 2: x = 6 − 2(2) = 2 → (2, 2)
If y = 3: x = 6 − 2(3) = 0 → (0, 3)
Suggested Marks Distribution
Systematic substitution method1 Mark
Listing 4 valid coordinate pairs1 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 13CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Student Startup Economics (Break-Even Line): A student startup manufactures recycled notebooks. Their fixed monthly operational cost is ₹3,000, and variable production cost is ₹25 per notebook. They sell each notebook for ₹65.
(a) Write the total cost equation C(x) and total revenue equation R(x) for x notebooks.
(b) Set up the linear equation in two variables for the break-even point where Revenue = Cost.
(c) How many notebooks must they sell to break even?
(a) Write the total cost equation C(x) and total revenue equation R(x) for x notebooks.
(b) Set up the linear equation in two variables for the break-even point where Revenue = Cost.
(c) How many notebooks must they sell to break even?
(a) Total Cost: C(x) = 25x + 3000. Total Revenue: R(x) = 65x.
(b) Break-even occurs when R(x) = C(x) ⟹ 65x = 25x + 3000 ⟹ 40x - 3000 = 0 (or 40x - y = 0 where y is cost/revenue).
(c) Solving: 40x = 3000 ⟹ x = 3000 / 40 = 75 notebooks. They must sell 75 notebooks to break even.
(b) Break-even occurs when R(x) = C(x) ⟹ 65x = 25x + 3000 ⟹ 40x - 3000 = 0 (or 40x - y = 0 where y is cost/revenue).
(c) Solving: 40x = 3000 ⟹ x = 3000 / 40 = 75 notebooks. They must sell 75 notebooks to break even.
🧠 HOTS & Assertion-Reasoning
Assertion (A): The graph of every linear equation ax + by + c = 0 (where a and b are not both zero) is a straight line.
Reason (R): Every point (x, y) whose coordinates satisfy the equation lies on this line, and every point on the line satisfies the equation.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): Every point (x, y) whose coordinates satisfy the equation lies on this line, and every point on the line satisfies the equation.
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: A linear algebraic relation between two variables defines a 1-dimensional manifold with constant slope, corresponding to Euclid's geometric definition of a straight line.
Explanation: A linear algebraic relation between two variables defines a 1-dimensional manifold with constant slope, corresponding to Euclid's geometric definition of a straight line.
14
Chapter 14: Surface Area and Volume
Ganita Manjari (Part II) — Exercise Sets 14.1, 14.2, 14.3 & 14.4 — Suggested Marks Distribution
Exercise Set 14.1 — Surface Area of Cuboids and Cubes
Q1
Find the total surface area and lateral surface area of a cuboid with length 12 cm, breadth 8 cm, and height 5 cm.
3 Marks
✓ Suggested Step-by-Step Solution
Given: l = 12 cm, b = 8 cm, h = 5 cm
Total Surface Area (TSA) = 2(lb + bh + hl)
= 2(12×8 + 8×5 + 5×12) = 2(96 + 40 + 60) = 2 × 196 = 392 cm²
= 2(12×8 + 8×5 + 5×12) = 2(96 + 40 + 60) = 2 × 196 = 392 cm²
Lateral Surface Area (LSA) = 2(l + b) × h
= 2(12 + 8) × 5 = 2 × 20 × 5 = 200 cm²
= 2(12 + 8) × 5 = 2 × 20 × 5 = 200 cm²
Suggested Marks Distribution
Correct formula for TSA½ Mark
TSA = 392 cm²1 Mark
LSA = 200 cm²1½ Marks
Q2
Find the surface area of a cube whose edge is 7 cm. How many square tiles of side 1 cm are needed to cover it?
3 Marks
✓ Suggested Step-by-Step Solution
TSA of cube = 6a² = 6 × 7² = 6 × 49 = 294 cm²
Area of each tile = 1 × 1 = 1 cm²
Number of tiles needed = 294 ÷ 1 = 294 tiles
Suggested Marks Distribution
TSA formula and TSA = 294 cm²2 Marks
Number of tiles = 2941 Mark
Q3
A room is 10 m long, 7 m wide, and 4 m high. Find the cost of whitewashing the four walls and ceiling at ₹12 per m². (Ignore doors and windows.)
4 Marks
✓ Suggested Step-by-Step Solution
Step 1: Area of four walls (LSA) = 2(l + b) × h = 2(10 + 7) × 4 = 2 × 17 × 4 = 136 m²
Step 2: Area of ceiling = l × b = 10 × 7 = 70 m²
Step 3: Total area to whitewash = 136 + 70 = 206 m²
Step 4: Cost = 206 × 12 = ₹2,472
Suggested Marks Distribution
LSA of walls = 136 m²1 Mark
Ceiling area = 70 m²1 Mark
Total area = 206 m²1 Mark
Cost = ₹2,4721 Mark
Exercise Set 14.2 — Surface Area of Cylinders and Cones
Q1
Find the curved surface area and total surface area of a cylinder with radius 7 cm and height 20 cm. (Use π = 22/7)
3 Marks
✓ Suggested Step-by-Step Solution
CSA = 2πrh = 2 × (22/7) × 7 × 20 = 2 × 22 × 20 = 880 cm²
TSA = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 20) = 44 × 27 = 1188 cm²
Suggested Marks Distribution
CSA formula and CSA = 880 cm²1½ Marks
TSA formula and TSA = 1188 cm²1½ Marks
Q2
A cone has radius 6 cm and slant height 10 cm. Find its curved surface area and total surface area. (π = 3.14)
3 Marks
✓ Suggested Step-by-Step Solution
CSA = πrl = 3.14 × 6 × 10 = 188.4 cm²
TSA = πr(l + r) = 3.14 × 6 × (10 + 6) = 3.14 × 6 × 16 = 3.14 × 96 = 301.44 cm²
Suggested Marks Distribution
CSA = πrl = 188.4 cm²1½ Marks
TSA = 301.44 cm²1½ Marks
Q3
A cylindrical pipe is open at both ends. Its radius is 3.5 cm and length is 80 cm. Find the area of metal sheet used to make it. (π = 22/7)
2 Marks
✓ Suggested Step-by-Step Solution
Open at both ends → Only curved surface area needed.
CSA = 2πrh = 2 × (22/7) × 3.5 × 80 = 2 × 22 × 0.5 × 80 = 1760 cm²
Suggested Marks Distribution
Recognising only CSA needed1 Mark
Area = 1760 cm²1 Mark
Exercise Set 14.3 — Volume of Solids
Q1
Find the volume of: (i) A cuboid with l = 15 cm, b = 10 cm, h = 6 cm (ii) A cube with edge 9 cm (iii) A cylinder with r = 7 cm, h = 10 cm. (π = 22/7)
3 Marks
✓ Suggested Step-by-Step Solution
(i) Cuboid: V = l × b × h = 15 × 10 × 6 = 900 cm³
(ii) Cube: V = a³ = 9³ = 729 cm³
(iii) Cylinder: V = πr²h = (22/7) × 49 × 10 = 22 × 7 × 10 = 1540 cm³
Suggested Marks Distribution
Each correct volume (×3)1 Mark each
Q2
Find the volume of a cone with radius 5 cm and height 12 cm. Also find its slant height. (π = 3.14)
3 Marks
✓ Suggested Step-by-Step Solution
Volume = (1/3)πr²h = (1/3) × 3.14 × 25 × 12 = (1/3) × 942 = 314 cm³
Slant height: l = √(r² + h²) = √(25 + 144) = √169 = 13 cm
Suggested Marks Distribution
Volume = 314 cm³2 Marks
Slant height l = 13 cm1 Mark
Q3
Find the volume and surface area of a sphere of radius 6 cm. (π = 3.14)
3 Marks
✓ Suggested Step-by-Step Solution
Volume = (4/3)πr³ = (4/3) × 3.14 × 216 = (4/3) × 678.24 = 904.32 cm³
Surface Area = 4πr² = 4 × 3.14 × 36 = 452.16 cm²
Suggested Marks Distribution
Volume = 904.32 cm³1½ Marks
Surface area = 452.16 cm²1½ Marks
Exercise Set 14.4 — Mixed Applications and Word Problems
Q1
A metallic sphere of radius 4.2 cm is melted and recast into a cylinder of radius 6 cm. Find the height of the cylinder. (π = 22/7)
4 Marks
✓ Suggested Step-by-Step Solution
Volume of sphere = (4/3)πr³ = (4/3) × (22/7) × (4.2)³
= (4/3) × (22/7) × 74.088 = (4 × 22 × 74.088)/(3 × 7) = 6511.75/21 ≈ 310.27 cm³
= (4/3) × (22/7) × 74.088 = (4 × 22 × 74.088)/(3 × 7) = 6511.75/21 ≈ 310.27 cm³
Volume of cylinder = πR²H = (22/7) × 36 × H
Equating: (22/7) × 36 × H = (4/3) × (22/7) × (4.2)³
36H = (4/3) × (4.2)³ = (4 × 74.088)/3 = 98.784
H = 98.784/36 = 2.744 cm ≈ 2.74 cm
36H = (4/3) × (4.2)³ = (4 × 74.088)/3 = 98.784
H = 98.784/36 = 2.744 cm ≈ 2.74 cm
Suggested Marks Distribution
Volume of sphere calculated1 Mark
Volume of cylinder set up correctly1 Mark
Equation formed correctly1 Mark
H ≈ 2.74 cm1 Mark
Q2
A conical tent is 10 m high and the radius of its base is 24 m. Find: (i) The slant height (ii) Curved surface area of the tent (iii) Cost of canvas at ₹70 per m². (π = 3.14)
5 Marks
✓ Suggested Step-by-Step Solution
(i) Slant height l = √(r² + h²) = √(576 + 100) = √676 = 26 m
(ii) CSA of cone = πrl = 3.14 × 24 × 26 = 3.14 × 624 = 1959.36 m²
(iii) Cost of canvas = 1959.36 × 70 = ₹1,37,155.20
Suggested Marks Distribution
Slant height l = 26 m2 Marks
CSA = 1959.36 m²2 Marks
Cost = ₹1,37,155.201 Mark
Q3
A hemispherical bowl of radius 10 cm is filled with water. Find the volume of water in the bowl. Also find how many cylindrical cups of radius 2 cm and height 5 cm can be filled from this bowl. (π = 3.14)
5 Marks
✓ Suggested Step-by-Step Solution
Volume of hemisphere = (2/3)πr³ = (2/3) × 3.14 × 1000 = (2/3) × 3140 = 2093.33 cm³
Volume of each cup = πr²h = 3.14 × 4 × 5 = 62.8 cm³
Number of cups = 2093.33 ÷ 62.8 ≈ 33 cups (approx.)
Suggested Marks Distribution
Hemisphere volume formula correct1 Mark
Volume = 2093.33 cm³2 Marks
Cup volume = 62.8 cm³1 Mark
Number of cups ≈ 331 Mark
Q4
A rectangular tank 80 cm long, 50 cm wide, and 40 cm deep is filled with water. How many litres of water does it hold? If water is pumped out at 2 litres per minute, how long will it take to empty it?
4 Marks
✓ Suggested Step-by-Step Solution
Volume of tank = 80 × 50 × 40 = 160,000 cm³
Convert to litres = 160,000 ÷ 1000 = 160 litres (since 1 litre = 1000 cm³)
Time to empty = 160 ÷ 2 = 80 minutes
Suggested Marks Distribution
Volume = 160,000 cm³1 Mark
Conversion to litres = 160 L2 Marks
Time = 80 minutes1 Mark
Q4
A conical tent is 10 m high and the radius of its base is 24 m. Find: (i) slant height of the tent, (ii) cost of canvas required to make the tent at ₹70 per m².
2 Marks
✓ Suggested Step-by-Step Solution
(i) Slant height l = √(r² + h²) = √(24² + 10²) = √(576 + 100) = √676 = 26 m.
(ii) Curved surface area = πrl = (22/7) × 24 × 26 = 13728 / 7 m².
Cost = Area × Rate = (13728 / 7) × 70 = 13728 × 10 = ₹1,37,280.
Suggested Marks Distribution
Slant height l = 26 m1 Mark
Cost = Rs 1,37,2801 Mark
Q4
Find the volume of a sphere whose surface area is 154 cm².
2 Marks
✓ Suggested Step-by-Step Solution
Surface area of sphere = 4πr² = 154 cm².
4 × (22/7) × r² = 154 → (88/7)r² = 154 → r² = (154 × 7) / 88 = 49 / 4.
r = √(49/4) = 7/2 = 3.5 cm.
Volume = (4/3)πr³ = (4/3) × (22/7) × (7/2) × (7/2) × (7/2)
= (11 × 49) / 3 = 539 / 3 = 179⅓ cm² (≈ 179.67 cm³).
Suggested Marks Distribution
Radius r = 3.5 cm from surface area1 Mark
Volume = 179.67 cm^31 Mark
🎯 Competency-Based Questions (CBQ) & HOTS — Chapter 14CBSE / NEP 2020
📝 Case Study & Real-Life Modeling
Sustainable Packaging Optimization (Surface Area & Volume): A juice manufacturer needs to pack 1 litre (1,000 cm³) of fruit juice. They are comparing two options: Option A is a cuboidal tetra-pack of base 10 cm × 5 cm and height 20 cm. Option B is a cylindrical tin can of radius r = 5 cm and height h = 12.73 cm.
(a) Verify that both containers have approximately 1,000 cm³ volume.
(b) Calculate the total surface area (packaging material required) for Option A.
(c) Calculate the total surface area for Option B and determine which packaging is more eco-friendly (uses less material).
(a) Verify that both containers have approximately 1,000 cm³ volume.
(b) Calculate the total surface area (packaging material required) for Option A.
(c) Calculate the total surface area for Option B and determine which packaging is more eco-friendly (uses less material).
(a) Volume A = 10 × 5 × 20 = 1,000 cm³. Volume B = πr²h = (22/7) × 25 × 12.73 ≈ 1,000 cm³.
(b) Surface Area A = 2(lb + bh + hl) = 2(10×5 + 5×20 + 20×10) = 2(50 + 100 + 200) = 2(350) = 700 cm².
(c) Surface Area B = 2πr(r + h) = 2 × (22/7) × 5 × (5 + 12.73) = (220/7) × 17.73 ≈ 557.2 cm².
Conclusion: The cylindrical can requires 142.8 cm² less material (≈ 20.4% less), making Option B significantly more eco-friendly and resource-efficient.
(b) Surface Area A = 2(lb + bh + hl) = 2(10×5 + 5×20 + 20×10) = 2(50 + 100 + 200) = 2(350) = 700 cm².
(c) Surface Area B = 2πr(r + h) = 2 × (22/7) × 5 × (5 + 12.73) = (220/7) × 17.73 ≈ 557.2 cm².
Conclusion: The cylindrical can requires 142.8 cm² less material (≈ 20.4% less), making Option B significantly more eco-friendly and resource-efficient.
🧠 HOTS & Assertion-Reasoning
Assertion (A): When all linear dimensions of a cube are tripled, its surface area increases 9-fold, while its volume increases 27-fold.
Reason (R): Surface area scales proportionally with the square of the linear scale factor (k²), while volume scales with the cube of the linear factor (k³).
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Reason (R): Surface area scales proportionally with the square of the linear scale factor (k²), while volume scales with the cube of the linear factor (k³).
Choose: (a) Both A and R are true and R is the correct explanation. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.
Correct Choice: (a) Both A and R are true and R is the correct explanation.
Explanation: For scaling factor k = 3: Area' = 6(3a)² = 9(6a²) = 9 × Area. Volume' = (3a)³ = 27a³ = 27 × Volume. The geometric square-cube scaling law perfectly explains this.
Explanation: For scaling factor k = 3: Area' = 6(3a)² = 9(6a²) = 9 × Area. Volume' = (3a)³ = 27a³ = 27 × Volume. The geometric square-cube scaling law perfectly explains this.