📚 Question Paper
Linear Equations in Two Variables & 3D Geometry
Total Marks: 80 | Time: 3 Hours | Number of Questions: 28
📌 SECTION A: Multiple Choice Questions (1 Mark Each)
Choose the correct option. Total: 10 Marks
Q1.1M
The standard form of a linear equation in two variables is:
(a) x + y = 5
(b) ax + by + c = 0
(c) x² + y² = 1
(d) x/2 + y/3 = 1
Q2.1M
How many solutions does the equation x + 2y = 5 have?
(a) One unique solution
(b) Two solutions
(c) Infinitely many solutions
(d) No solution
Q3.1M
Which of the following is a solution of 2x + 3y = 12?
(a) (0, 4)
(b) (3, 2)
(c) (6, 0)
(d) All of the above
Q4.1M
The curved surface area of a cone is:
(a) πrl
(b) πr²h
(c) 2πrh
(d) 4πr²
Q5.1M
For a cone, the relationship between radius (r), height (h), and slant height (l) is:
(a) l = r + h
(b) l² = r² + h²
(c) l = rh
(d) l = √(r + h)
Q6.1M
The surface area of a sphere with radius r is:
(a) 2πr²
(b) 3πr²
(c) 4πr²
(d) πr²
Q7.1M
The volume of a hemisphere is:
(a) (4/3)πr³
(b) (2/3)πr³
(c) (1/3)πr³
(d) πr³
Q8.1M
In the equation 2x + 3y - 4 = 0, the value of a, b, and c are:
(a) a=2, b=3, c=-4
(b) a=2, b=3, c=4
(c) a=-2, b=-3, c=4
(d) a=3, b=2, c=-4
Q9.1M
The total surface area of a cone is:
(a) πrl
(b) πr² + πrl
(c) πr(l + r)
(d) Both (b) and (c)
Q10.1M
The volume of a cone is:
(a) πr²h
(b) (1/3)πr²h
(c) (2/3)πr²h
(d) 3πr²h
📌 SECTION B: Short Answer Questions (2 Marks Each)
Answer in 30-50 words. Total: 16 Marks
Q11.2M
Write the equation 3x - 4y = 7 in the standard form ax + by + c = 0 and identify the values of a, b, and c.
Q12.2M
Find two solutions of the linear equation x + y = 4.
Q13.2M
Check whether (2, 3) is a solution of the equation 2x + 3y = 13.
Q14.2M
If a cone has radius 5 cm and height 12 cm, find the slant height.
Q15.2M
Write one real-life situation that can be expressed as a linear equation in two variables.
Q16.2M
Why does a linear equation in two variables have infinitely many solutions? Explain briefly.
Q17.2M
A sphere has a radius of 7 cm. Find its surface area. (Use π = 22/7)
Q18.2M
Differentiate between curved surface area and total surface area of a cone.
📌 SECTION C: Medium Answer Questions (3 Marks Each)
Answer in 80-100 words. Total: 18 Marks
Q19.3M
Find four different solutions of the equation 2x + y = 6.
Q20.3M
Rearrange the equation y = 2x - 3 into the standard form ax + by + c = 0 and identify a, b, c.
Q21.3M
A cone has a slant height of 13 cm and radius of 5 cm. Find its curved surface area. (Use π = 22/7)
Q22.3M
Find the total surface area of a cone with radius 3.5 cm and slant height 10 cm. (Use π = 22/7)
Q23.3M
A hemisphere has a radius of 7 cm. Find its total surface area and volume. (Use π = 22/7)
Q24.3M
Express the following statement as a linear equation in two variables: "The cost of 2 notebooks and 3 pens is ₹50."
📌 SECTION D: Long Answer Questions (4-5 Marks Each)
Answer in 150-200 words. Total: 28 Marks
Q25.4M
Verify that (3, 2) and (0, 5) are solutions of the equation 3x + y = 11, but (1, 8) is not. Also, find two more solutions.
Q26.4M
A cone has a radius of 6 cm, height of 8 cm. Find: (a) Slant height (b) Curved surface area (c) Total surface area (d) Volume. (Use π = 22/7)
Q27.5M
A motorcyclist performs stunts on a hollow sphere. The sphere has a diameter of 14 m. Calculate: (a) Radius (b) Surface area available for riding (c) Volume of the sphere. Also, explain why a linear equation in two variables always has infinitely many solutions. (Use π = 22/7)
Q28.5M
In a One-Day International cricket match, two batsmen together scored 315 runs. Express this as a linear equation in two variables. Find all possible solutions where both batsmen score more than 50 runs. Also, explain: (a) What is a solution to this equation? (b) Why can there be multiple solutions? (c) What are the constraints in a real cricket scenario?
✅ ANSWER KEY WITH EXPLANATIONS
Click on each answer to reveal the solution and explanation
SECTION A: Answer Key (Multiple Choice)
Q1.1M
✓ Answer: (b) ax + by + c = 0
The standard form of a linear equation in two variables is ax + by + c = 0, where a, b, and c are real numbers, and a and b are not both zero. Option (a) is already in this form. Options (c) and (d) are not linear equations in standard form.
Q2.1M
✓ Answer: (c) Infinitely many solutions
Linear equations in two variables always have infinitely many solutions because you can choose any value for x, and by substituting it, you'll find a corresponding value of y. For example: If x=0, y=5; If x=1, y=4; If x=2, y=3, etc.
Q3.1M
✓ Answer: (d) All of the above
Let's verify each: (a) 2(0)+3(4)=0+12=12 ✓ | (b) 2(3)+3(2)=6+6=12 ✓ | (c) 2(6)+3(0)=12+0=12 ✓ All three are solutions of the equation 2x + 3y = 12.
Q4.1M
✓ Answer: (a) πrl
The curved surface area (CSA) of a cone is πrl, where r is the radius and l is the slant height. This is the area of the curved surface, not including the circular base.
Q5.1M
✓ Answer: (b) l² = r² + h²
This is the Pythagorean theorem applied to the cone. The slant height, radius, and height form a right-angled triangle where l is the hypotenuse. Therefore: l² = r² + h²
Q6.1M
✓ Answer: (c) 4πr²
The surface area of a sphere is 4πr². This is the total outer surface area. A hemisphere has curved surface area 2πr² (half of the sphere).
Q7.1M
✓ Answer: (b) (2/3)πr³
A hemisphere is half of a sphere. The volume of a sphere is (4/3)πr³, so a hemisphere's volume is (4/3)πr³ ÷ 2 = (2/3)πr³.
Q8.1M
✓ Answer: (a) a=2, b=3, c=-4
In the equation 2x + 3y - 4 = 0, comparing with ax + by + c = 0: a (coefficient of x) = 2, b (coefficient of y) = 3, c (constant term) = -4.
Q9.1M
✓ Answer: (d) Both (b) and (c)
TSA = Curved Surface Area + Base Area = πrl + πr² = πr(l + r). Both expressions are equivalent and correct.
Q10.1M
✓ Answer: (b) (1/3)πr²h
The volume of a cone is one-third the volume of a cylinder with the same radius and height. Volume = (1/3)πr²h, where r is radius and h is height.
SECTION B: Answer Key (Short Answer)
Q11.2M
✓ Answer: 3x - 4y - 7 = 0
Original: 3x - 4y = 7
Standard form: 3x - 4y - 7 = 0
Where: a = 3, b = -4, c = -7
Standard form: 3x - 4y - 7 = 0
Where: a = 3, b = -4, c = -7
Q12.2M
✓ Answer: (0, 4) and (4, 0)
For x + y = 4:
Solution 1: Set x = 0 → 0 + y = 4 → y = 4, so (0, 4) ✓
Solution 2: Set y = 0 → x + 0 = 4 → x = 4, so (4, 0) ✓
(Other valid solutions: (1, 3), (2, 2), (3, 1), etc.)
Solution 1: Set x = 0 → 0 + y = 4 → y = 4, so (0, 4) ✓
Solution 2: Set y = 0 → x + 0 = 4 → x = 4, so (4, 0) ✓
(Other valid solutions: (1, 3), (2, 2), (3, 1), etc.)
Q13.2M
✓ Answer: YES, (2, 3) is a solution
Substitute x = 2, y = 3 in 2x + 3y = 13:
2(2) + 3(3) = 4 + 9 = 13 ✓
Since both sides are equal, (2, 3) is a solution.
2(2) + 3(3) = 4 + 9 = 13 ✓
Since both sides are equal, (2, 3) is a solution.
Q14.2M
✓ Answer: l = 13 cm
Using l² = r² + h²:
l² = 5² + 12²
l² = 25 + 144 = 169
l = √169 = 13 cm
l² = 5² + 12²
l² = 25 + 144 = 169
l = √169 = 13 cm
Q15.2M
✓ Answer: Examples
Example 1: The cost of notebooks and pens: 5x + 3y = 100 (where x = cost of notebook, y = cost of pen)
Example 2: Age relationship: 2a + b = 50 (where a and b are ages)
Example 3: Speed-distance: 60t + 40s = 300 (where t and s are times)
Example 2: Age relationship: 2a + b = 50 (where a and b are ages)
Example 3: Speed-distance: 60t + 40s = 300 (where t and s are times)
Q16.2M
✓ Answer: Explanation
A linear equation in two variables has infinitely many solutions because:
• For any chosen value of x, we can calculate a corresponding value of y
• Since x can be any real number, y has infinitely many values
• Each (x, y) pair forms a solution
Example: x + y = 5 has solutions (0,5), (1,4), (2,3), (3,2), ...
• For any chosen value of x, we can calculate a corresponding value of y
• Since x can be any real number, y has infinitely many values
• Each (x, y) pair forms a solution
Example: x + y = 5 has solutions (0,5), (1,4), (2,3), (3,2), ...
Q17.2M
✓ Answer: Surface Area = 616 cm²
Given: r = 7 cm, π = 22/7
Formula: Surface Area = 4πr²
SA = 4 × (22/7) × 7 × 7
SA = 4 × 22 × 7
SA = 88 × 7 = 616 cm²
Formula: Surface Area = 4πr²
SA = 4 × (22/7) × 7 × 7
SA = 4 × 22 × 7
SA = 88 × 7 = 616 cm²
Q18.2M
✓ Answer: Difference
Curved Surface Area (CSA): Only the curved part of the cone = πrl
Total Surface Area (TSA): Curved part + circular base = πrl + πr² = πr(l + r)
The difference is that TSA includes the circular base area πr², while CSA does not.
Total Surface Area (TSA): Curved part + circular base = πrl + πr² = πr(l + r)
The difference is that TSA includes the circular base area πr², while CSA does not.
SECTION C: Answer Key (Medium Answer)
Q19.3M
✓ Answer: Four Solutions
For equation 2x + y = 6:
Solution 1: x = 0 → y = 6 → (0, 6)
Solution 2: x = 1 → 2(1) + y = 6 → y = 4 → (1, 4)
Solution 3: x = 2 → 2(2) + y = 6 → y = 2 → (2, 2)
Solution 4: x = 3 → 2(3) + y = 6 → y = 0 → (3, 0)
Solution 1: x = 0 → y = 6 → (0, 6)
Solution 2: x = 1 → 2(1) + y = 6 → y = 4 → (1, 4)
Solution 3: x = 2 → 2(2) + y = 6 → y = 2 → (2, 2)
Solution 4: x = 3 → 2(3) + y = 6 → y = 0 → (3, 0)
Q20.3M
✓ Answer: 2x - y - 3 = 0
Original: y = 2x - 3
Rearrange: y - 2x + 3 = 0
Standard form: 2x - y - 3 = 0
Where: a = 2, b = -1, c = -3
Rearrange: y - 2x + 3 = 0
Standard form: 2x - y - 3 = 0
Where: a = 2, b = -1, c = -3
Q21.3M
✓ Answer: CSA = 204.29 cm² or 1430/7 cm²
Given: l = 13 cm, r = 5 cm, π = 22/7
Formula: CSA = πrl
CSA = (22/7) × 5 × 13
CSA = (22 × 5 × 13)/7
CSA = 1430/7 = 204.29 cm²
Formula: CSA = πrl
CSA = (22/7) × 5 × 13
CSA = (22 × 5 × 13)/7
CSA = 1430/7 = 204.29 cm²
Q22.3M
✓ Answer: TSA = 148.5 cm²
Given: r = 3.5 cm, l = 10 cm, π = 22/7
Formula: TSA = πr(l + r)
TSA = (22/7) × 3.5 × (10 + 3.5)
TSA = (22/7) × 3.5 × 13.5
TSA = 22 × 0.5 × 13.5
TSA = 11 × 13.5 = 148.5 cm²
Formula: TSA = πr(l + r)
TSA = (22/7) × 3.5 × (10 + 3.5)
TSA = (22/7) × 3.5 × 13.5
TSA = 22 × 0.5 × 13.5
TSA = 11 × 13.5 = 148.5 cm²
Q23.3M
✓ Answer: TSA = 462 cm², Volume = 718.67 cm³
Given: r = 7 cm, π = 22/7
TSA = 3πr²
TSA = 3 × (22/7) × 7 × 7 = 3 × 22 × 7 = 462 cm²
Volume = (2/3)πr³
V = (2/3) × (22/7) × 7 × 7 × 7
V = (2/3) × 22 × 49 = 718.67 cm³
TSA = 3πr²
TSA = 3 × (22/7) × 7 × 7 = 3 × 22 × 7 = 462 cm²
Volume = (2/3)πr³
V = (2/3) × (22/7) × 7 × 7 × 7
V = (2/3) × 22 × 49 = 718.67 cm³
Q24.3M
✓ Answer: 2x + 3y = 50
Let x = cost of one notebook (in ₹)
Let y = cost of one pen (in ₹)
Cost of 2 notebooks = 2x
Cost of 3 pens = 3y
Total cost = 2x + 3y
Linear equation: 2x + 3y = 50
Let y = cost of one pen (in ₹)
Cost of 2 notebooks = 2x
Cost of 3 pens = 3y
Total cost = 2x + 3y
Linear equation: 2x + 3y = 50
SECTION D: Answer Key (Long Answer)
Q25.4M
✓ Answer: Verification and Solutions
Verification:
• For (3, 2): 3(3) + 2 = 9 + 2 = 11 ✓ (Solution)
• For (0, 5): 3(0) + 5 = 0 + 5 = 5 ✗ (NOT a solution)
• For (1, 8): 3(1) + 8 = 3 + 8 = 11 ✓ (Solution)
Correction: (0, 5) is NOT a solution, but (1, 8) IS
Two more solutions:
• Set x = 2: 3(2) + y = 11 → y = 5 → (2, 5) ✓
• Set x = 4: 3(4) + y = 11 → y = -1 → (4, -1) ✓
• For (3, 2): 3(3) + 2 = 9 + 2 = 11 ✓ (Solution)
• For (0, 5): 3(0) + 5 = 0 + 5 = 5 ✗ (NOT a solution)
• For (1, 8): 3(1) + 8 = 3 + 8 = 11 ✓ (Solution)
Correction: (0, 5) is NOT a solution, but (1, 8) IS
Two more solutions:
• Set x = 2: 3(2) + y = 11 → y = 5 → (2, 5) ✓
• Set x = 4: 3(4) + y = 11 → y = -1 → (4, -1) ✓
Q26.4M
✓ Answer: All Four Measurements
Given: r = 6 cm, h = 8 cm, π = 22/7
(a) Slant Height:
l² = r² + h² = 36 + 64 = 100
l = 10 cm
(b) Curved Surface Area:
CSA = πrl = (22/7) × 6 × 10 = 1320/7 ≈ 188.57 cm²
(c) Total Surface Area:
TSA = πr(l + r) = (22/7) × 6 × 16 = 2112/7 ≈ 301.71 cm²
(d) Volume:
V = (1/3)πr²h = (1/3) × (22/7) × 36 × 8 = 6336/7 ≈ 905.14 cm³
(a) Slant Height:
l² = r² + h² = 36 + 64 = 100
l = 10 cm
(b) Curved Surface Area:
CSA = πrl = (22/7) × 6 × 10 = 1320/7 ≈ 188.57 cm²
(c) Total Surface Area:
TSA = πr(l + r) = (22/7) × 6 × 16 = 2112/7 ≈ 301.71 cm²
(d) Volume:
V = (1/3)πr²h = (1/3) × (22/7) × 36 × 8 = 6336/7 ≈ 905.14 cm³
Q27.5M
✓ Answer: Sphere Calculations and Explanation
Given: Diameter = 14 m, π = 22/7
(a) Radius:
r = Diameter/2 = 14/2 = 7 m
(b) Surface Area:
SA = 4πr² = 4 × (22/7) × 7 × 7
SA = 4 × 22 × 7 = 616 m²
(c) Volume:
V = (4/3)πr³ = (4/3) × (22/7) × 7 × 7 × 7
V = (4/3) × 22 × 49 = 4312/3 ≈ 1437.33 m³
Why Linear Equations have Infinite Solutions:
Linear equations in two variables have infinitely many solutions because for any value of one variable, we can find a corresponding value of the other. Example: x + y = 10 has solutions (0,10), (1,9), (2,8), etc. We can choose x to be any real number and find a matching y.
(a) Radius:
r = Diameter/2 = 14/2 = 7 m
(b) Surface Area:
SA = 4πr² = 4 × (22/7) × 7 × 7
SA = 4 × 22 × 7 = 616 m²
(c) Volume:
V = (4/3)πr³ = (4/3) × (22/7) × 7 × 7 × 7
V = (4/3) × 22 × 49 = 4312/3 ≈ 1437.33 m³
Why Linear Equations have Infinite Solutions:
Linear equations in two variables have infinitely many solutions because for any value of one variable, we can find a corresponding value of the other. Example: x + y = 10 has solutions (0,10), (1,9), (2,8), etc. We can choose x to be any real number and find a matching y.
Q28.5M
✓ Answer: Cricket Problem Complete Solution
Linear Equation:
Let x = runs by first batsman, y = runs by second batsman
x + y = 315
Solutions where both score more than 50 runs:
We need: x > 50, y > 50, and x + y = 315
Valid range: 50 < x < 265 and y = 315 - x
Infinite solutions exist in this range!
Examples: (100, 215), (150, 165), (157, 158), (200, 115), etc.
(a) What is a solution?
A solution is a pair of values (x, y) where x + y = 315. Each pair represents the runs scored by each batsman.
(b) Why multiple solutions?
Because for any first batsman's score, there's a corresponding second batsman's score that satisfies the equation. Since there's no constraint limiting the first batsman to one specific score, infinitely many combinations are possible.
(c) Real cricket constraints:
• Both batsmen must score ≥ 0 runs
• Practical limit: both typically < 315 runs
• Each batsman's score is usually < 315
• In reality, only a few specific scores are likely (statistical constraint)
Let x = runs by first batsman, y = runs by second batsman
x + y = 315
Solutions where both score more than 50 runs:
We need: x > 50, y > 50, and x + y = 315
Valid range: 50 < x < 265 and y = 315 - x
Infinite solutions exist in this range!
Examples: (100, 215), (150, 165), (157, 158), (200, 115), etc.
(a) What is a solution?
A solution is a pair of values (x, y) where x + y = 315. Each pair represents the runs scored by each batsman.
(b) Why multiple solutions?
Because for any first batsman's score, there's a corresponding second batsman's score that satisfies the equation. Since there's no constraint limiting the first batsman to one specific score, infinitely many combinations are possible.
(c) Real cricket constraints:
• Both batsmen must score ≥ 0 runs
• Practical limit: both typically < 315 runs
• Each batsman's score is usually < 315
• In reality, only a few specific scores are likely (statistical constraint)
