CG Board Class 10 Maths Chapter 7 Solutions: Graph (आलेख) (2026-27)

(i)** What conclusions can you draw from graph (A)?

• (ii) How are these conclusions different from those drawn from graph (B)?
• (i) Conclusion from Graph (A):

Since Graph (A) is a straight line, it implies that the ratio of change in Y to change in X, given by (Y₂ – Y₁)/(X₂ – X₁), is constant. This demonstrates a direct linear relationship where Y changes at a uniform rate with respect to X.

• (ii) Difference from Graph (B):

Graph (B) is a curve, which means that the ratio (Y₂ – Y₁)/(X₂ – X₁) is not constant. In Graph (B), the rate of change between the two quantities is non-uniform, unlike the uniform rate observed in Graph (A).

Step 1: Identify Variables:**

• Independent Variable (X-axis): Number of Days.
• Dependent Variable (Y-axis): Petrol Remaining (in Liters).
• Step 2: Table of Coordinates (X, Y):
• Day 1: (1, 4)
• Day 2: (2, 3)
• Day 3: (3, 2)
• Day 4: (4, 1)
• Day 5: (5, 0)
• Step 3: Select Scale:
• On X-axis: 1 big square = 1 Day.
• On Y-axis: 1 big square = 1 Liter.
• Step 4: Verification of Linearity:

(Y₂ – Y₁)/(X₂ – X₁) = (3 – 4)/(2 – 1) = -1/1 = -1

(Y₃ – Y₂)/(X₃ – X₂) = (2 – 3)/(3 – 2) = -1/1 = -1

Since the ratio is constant (-1), plotting and joining these points (1,4), (2,3), (3,2), (4,1), (5,0) produces a straight line slanting downwards from left to right, reaching zero on Day 5.

Step 1: Formula and Given Data:**

• Principal (P) = Rs. 300
• Rate (R) = 5% per annum
• Formula: Simple Interest (S.I.) = (P × R × T)/(100)
• Step 2: Calculation for different time periods (T):
• For T = 0 year: S.I. = (300 × 5 × 0)/(100) = Rs. 0
• For T = 1 year: S.I. = (300 × 5 × 1)/(100) = Rs. 15
• For T = 2 years: S.I. = (300 × 5 × 2)/(100) = Rs. 30
• For T = 3 years: S.I. = (300 × 5 × 3)/(100) = Rs. 45
• For T = 4 years: S.I. = (300 × 5 × 4)/(100) = Rs. 60
• Step 3: Observation Table:
• Time T in years (X-axis): 0, 1, 2, 3, 4
• Simple Interest in Rs. (Y-axis): 0, 15, 30, 45, 60
• Step 4: Scale & Plotting:
• X-axis: 1 big square = 1 Year.
• Y-axis: 1 big square = Rs. 15.
• Points to plot: (0,0), (1,15), (2,30), (3,45), (4,60).

Joining these points yields a straight line passing through the origin (0,0).

Step 1: Value Calculation Table (y = x²):**

• When x = -4 ⇒ y = (-4)² = 16
• When x = -3 ⇒ y = (-3)² = 9
• When x = -2 ⇒ y = (-2)² = 4
• When x = -1 ⇒ y = (-1)² = 1
• When x = 0 ⇒ y = (0)² = 0
• When x = 1 ⇒ y = (1)² = 1
• When x = 2 ⇒ y = (2)² = 4
• When x = 3 ⇒ y = (3)² = 9
• When x = 4 ⇒ y = (4)² = 16
• Step 2: Table of Values:

| x (X-axis) | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |

| y = x² (Y-axis) | 16 | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |

• Step 3: Scale & Graph Nature:
• X-axis: 1 big square = 1 unit.
• Y-axis: 1 big square = 2 units.
• Plotting points (-4,16), (-3,9), (-2,4), (-1,1), (0,0), (1,1), (2,4), (3,9), (4,16) and joining them with a smooth curve produces a symmetric U-shaped parabola curve opening upwards with vertex at origin (0,0).

Step 1: Assign Axes:**

• X-axis: Week (1, 2, 3, 4, 5)
• Y-axis: Quantity of onion in kg (1, 2, 3, 4, 5)
• Step 2: Scale:
• X-axis: 1 big square = 1 Week.
• Y-axis: 1 big square = 1 kg.
• Step 3: Coordinates:

Points to plot: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5).

• Step 4: Verification of Linearity:

(Y₂ – Y₁)/(X₂ – X₁) = (2 – 1)/(2 – 1) = 1

Joining these coordinates forms a straight line inclination at 45° passing through (0,0).

Calculations using equation s = (t – 20)/(5):**

• For t = 25°C ⇒ s = (25 – 20)/(5) = 5/5 = 1 cm/sec
• For t = 30°C ⇒ s = (30 – 20)/(5) = 10/5 = 2 cm/sec
• For t = 35°C ⇒ s = (35 – 20)/(5) = 15/5 = 3 cm/sec
• For t = 40°C ⇒ s = (40 – 20)/(5) = 20/5 = 4 cm/sec
• Table of Values:

| Temperature t (°C) | 25 | 30 | 35 | 40 |

| Speed s (cm/sec) | 1 | 2 | 3 | 4 |

• Answers to Sub-questions:

(i)** What should be the length of pipe to obtain a pitch of 160 cycles per second?

• (ii) What will be the pitch when the length of pipe is 60 cm?
• (i) Length of pipe for pitch = 160 cycles/sec:

Locating 160 cycles/sec on the Y-axis and moving horizontally to intersect the graph curve, the corresponding X-axis reading is 30 cm (approx. 30 cm to 31 cm).

• (ii) Pitch for pipe length = 60 cm:

Locating 60 cm on the X-axis and moving vertically to intersect the graph curve, the corresponding Y-axis reading is 80 cycles per second.

Verification for Table – A:**

• Point 1 to 2: (Y₂ – Y₁)/(X₂ – X₁) = (8π – 4π)/(4 – 2) = (4π)/(2) = 2π
• Point 2 to 3: (Y₃ – Y₂)/(X₃ – X₂) = (12π – 8π)/(6 – 4) = (4π)/(2) = 2π
• Point 3 to 4: (Y₄ – Y₃)/(X₄ – X₃) = (16π – 12π)/(8 – 6) = (4π)/(2) = 2π
• Point 4 to 5: (Y₅ – Y₄)/(X₅ – X₄) = (20π – 16π)/(10 – 8) = (4π)/(2) = 2π

(i)** What was the city temperature at 6 am?

• (ii) What was the city temperature at 3 pm in the afternoon?
• (iii) At what time was the temperature 30°C?
• (iv) What would have been the temperature at 12 midnight?
• (v) What was the temperature at 9 pm in the night?

Reading values directly from the textbook temperature graph (Graph-7)

• (ii) Temperature at 3 pm: From the graph peak at 3 pm, the reading is 39°C.
• (iv) Temperature at 12 Midnight: Extrapolating the downward evening curve after 6 pm (30°C at 6 pm), the estimated temperature at 12 midnight is approximately 18°C – 20°C.

(i)** What was the pollution level at 8 in the morning?

• (ii) At what time was the highest pollution level recorded?

Reading from the textbook pollution level graph

• (ii) Time of highest pollution level: The peak point of the pollution curve occurs at 1 pm.