Think and Reflect
Q1. Page 16: Can you identify the terms, variables and coefficients of this algebraic expression? (Reference: Example 2: 200l + 160w + 50lw)
Answer:
Example 2: 200l + 160w + 50lw
Terms: 200l, 160w, 50lw
Variables: l, w
Coefficients: 200, 160, 50
Q2. Page 16: Think and Reflect: How is it different from the algebraic expression in Example 1? (Compare 200l + 160w + 50lw from Example 2 with 4x + 5y + 3 from Example 1)
Answer:
The expression in Example 2, 200l + 160w + 50lw, has a term (50lw) where two variables (l and w) are multiplied together, making it a non-linear term. The expression in Example 1, 4x + 5y + 3, has only linear terms i.e each appearing only to the power 1 (4x, 5y) and a constant (3).
Q3. Page 17: Think and Reflect: Can you identify the terms, variables and coefficients of this algebraic expression? (Reference: Example 3: 10x - x^2)
Answer:
Terms: 10x, -x²
Variables: x
Coefficients: 10 (for the term 10x), -1 (for the term -x²)
Q4. Page 17: Think and Reflect: Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3? (Compare 4x + 5y + 3 from Example 1 with 10x - x² from Example 3)
Answer:
Similarity: Both are algebraic expressions involving variables and constants.
Difference: Example 1 has two variables (x and y) and is linear. Example 3 has only one variable (x), but its highest power is 2 (quadratic), making it non-linear.
Q5. Page 19: Think and Reflect: Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?
Answer:
Perimeters:
| Side (cm) | Perimeter (cm) |
|---|---|
| 1 | 4 |
| 1.5 | 6 |
| 2 | 8 |
| 2.5 | 10 |
| 3 | 12 |
When the side increases by 0.5 cm, the perimeter increases by a constant amount of 2 cm.
Q6. Page 19: Think and Reflect: If a player paid £750, how many matches did he play? (Reference: Chess club charges a joining fee of £200 plus £50 for every match played)
Answer:
Let the number of matches be m. The total cost is given by the linear expression: 200 + 50m.
If the player paid £750, then:
200 + 50m = 750
50m = 550
m = 11
The player played 11 matches.
Q7. Page 20: Think and Reflect: We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x - x², is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm?
Answer:
Input-Output Process: The variable x (length of the rectangle) is the input. The expression 10x - x² is the process. The resulting value is the output, which is the area of the rectangle.
Value at x = 6 cm:
Area = 10(6) - (6)² = 60 - 36 = 24 cm²
Q8. Page 22: Think and Reflect: Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern. (Reference: Growing pattern of square tiles: 1, 3, 5, 7, 9...)
Answer:
The pattern increases by 2 each time (constant difference of 2). The sequence up to Stage 7 is:
1, 3, 5, 7, 9, 11, 13
The next three stages (Stages 8, 9, 10) would have:
15, 17, 19 squares respectively.
Q9. Page 22: Think and Reflect: Using the expression 2n - 1, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?
Answer:
15th Stage: 2(15) - 1 = 30 - 1 = 29 tiles
26th Stage: 2(26) - 1 = 52 - 1 = 51 tiles
For 21 tiles: 2n - 1 = 21 => 2n = 22 => n = 11. So, the 11th stage has 21 tiles.
For 47 tiles: 2n - 1 = 47 => 2n = 48 => n = 24. So, the 24th stage has 47 tiles.
Q10. Page 23: Think and Reflect: What amount will be left on the 15th day? How many days will it take for the entire amount to be spent? (Reference: Bela has ₹100 pocket money and spends ₹5 every day)
Answer:
Amount left on the 15th day:
Amount = 100 - 5n = 100 - 5(15) = 100 - 75 = ₹25
Days to spend entire amount:
100 - 5n = 0 => 5n = 100 => n = 20 days
Q11. Page 23: Think and Reflect: For how many km will the auto-rickshaw fare be ₹130? (Reference: Auto-rickshaw fare: ₹25 for first 2 km, then ₹15 per km)
Answer:
The fare for n km (n ≥ 2) is given by: 25 + 15(n - 2).
To find n when fare is ₹130:
25 + 15(n - 2) = 130
15(n - 2) = 105
n - 2 = 7
n = 9 km
The fare will be ₹130 for a travel of 9 km.
Q12. Page 24: Think and Reflect: What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be 700? (Reference: Cost function C(d) = 100 + 60d)
Answer:
Cost for 15 km:
C(15) = 100 + 60(15) = 100 + 900 = ₹1000
Kilometres for ₹700:
100 + 60d = 700
60d = 600
d = 10 km
The cost will be ₹700 for a journey of 10 km.
Q13. Page 25: Think and Reflect: What will be the height of the water at the end of 5 months? (Reference: Height function h(t) = 3 - 0.5t)
Answer:
h(5) = 3 - 0.5(5) = 3 - 2.5 = 0.5 m
The height of the water at the end of 5 months will be 0.5 meters.
Q14. Page 27: Think and Reflect: Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent? (Reference: Telecom bill example)
Answer:
20 (the slope 'a'): This represents the additional cost per GB of data used. In this context, it costs ₹20 per GB.
150 (the y-intercept 'b'): This represents the fixed monthly fee charged by the company, regardless of data usage.
Q15. Page 28: Think and Reflect: Identify other points on the line by completing the following table. (Reference: Table for y = 2x + 1)
| x | 1 | 2 | 5 | 7 | 9 | 12 | 20 |
|---|---|---|---|---|---|---|---|
| y | 3 | 5 | 11 | 15 | 19 | 25 | 41 |
Answer:
The completed table is:
| x | 1 | 2 | 5 | 7 | 9 | 12 | 20 |
|---|---|---|---|---|---|---|---|
| y | 3 | 5 | 11 | 15 | 19 | 25 | 41 |
For each value of x, y is calculated as 2x + 1.
Q16. Page 34: Think and Reflect: Differentiate between the graphs of the equations y = 3x + 1, and y = -3x + 1.
Answer:
Similarities: Both lines have the same y-intercept (b = 1). Therefore, both lines cut the y-axis at the same point (0, 1).
Differences:
- Slope (a): y = 3x + 1 has a positive slope (a = 3), meaning it rises from left to right. It represents linear growth.
- y = -3x + 1 has a negative slope (a = -3), meaning it falls from left to right. It represents linear decay.
Q17. Page 35: Think and Reflect: Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies? (Reference: Graphs of y = 2x - 1, y = 2x + 1, y = 2x + 5)
Answer:
When the slope 'a' is fixed (e.g., a = 2) and only the y-intercept 'b' varies, the lines have the same steepness and are parallel to each other. They are vertical translations of each other, with each line shifting up or down along the y-axis depending on the value of b.