These solutions follow the chapter exactly as it's laid out in the textbook — each section's "Think and Reflect" boxes first, then that section's Exercise Set — with every step of working shown in full.
Key Concepts and Formulas
- Algebraic expression: combines numbers, variables and operations, e.g. \(2x^2+5xy-3y^2\).
- Univariate polynomial: an algebraic expression in one variable. Its degree is the highest power of that variable.
- Linear polynomial: a polynomial of degree 1, e.g. \(2x+3\).
- Linear growth — quantity increases by a fixed amount over equal intervals; linear decay — quantity decreases by a fixed amount over equal intervals.
- Linear pattern: a sequence where the difference between consecutive terms is constant.
- Linear relationship: \(y=ax+b\), where a is the slope and b is the y-intercept. When b = 0, the line passes through the origin.
- Parallel lines share the same slope a but have different y-intercepts b.
2.1 Introduction
Think and Reflect
TR(After Example 2: garden fencing, 200l + 160w + 50lw) 1. Identify the terms, variables and coefficients. 2. How is it different from the expression in Example 1?
Algebraic expression: \(200l + 160w + 50lw\)
The terms are 200l, 160w and 50lw.
The variables used are l and w.
Coefficient of l = 200; coefficient of w = 160; coefficient of lw = 50. There is no constant term.
Example 1 (4x + 5y + 3) has a constant term, and every term involves only one variable. Example 2 has no constant term, and its term 50lw is a product of both variables — a degree-2 term — unlike Example 1, where every term has degree 1 or 0.
TR(After Example 3: wire bent into a rectangle, 10x – x²) 1. Identify the terms, variables and coefficients. 2. Compare with Example 1.
Algebraic expression: \(10x - x^2\)
The terms are 10x and –x².
Only one variable is used: x.
Coefficient of x = 10; coefficient of x² = –1.
Example 3 involves only one variable (x), while Example 1 involves two (x and y). Example 3 also contains a degree-2 term (x²), while every term in Example 1 has degree 1 or 0.
Exercise Set 2.1
1Find the degrees of the following polynomials: (i) 2x² – 5x + 3 (ii) y³ + 2y – 1 (iii) –9 (iv) 4z – 3
The degree of a polynomial is the highest power of the variable that appears in it.
(i) In \(2x^2-5x+3\), the powers of x present are 2, 1 and 0. The highest power is 2, so the degree is 2 (a quadratic polynomial).
(ii) In \(y^3+2y-1\), the powers of y present are 3, 1 and 0. The highest power is 3, so the degree is 3 (a cubic polynomial).
(iii) \(-9\) can be written as \(-9x^0\), so the only power present is 0. The degree is 0 (a constant polynomial).
(iv) In \(4z-3\), the powers of z present are 1 and 0. The highest power is 1, so the degree is 1 (a linear polynomial).
(ii) Degree 3
(iii) Degree 0
(iv) Degree 1
2Write polynomials of degrees 1, 2 and 3.
We need one polynomial each where the highest power of the variable is 1, 2 and 3 respectively. Any correct example is acceptable.
Degree 1 (linear): \(2x + 3\) — highest power of x here is 1.
Degree 2 (quadratic): \(x^2 + 2x + 1\) — highest power of x here is 2.
Degree 3 (cubic): \(x^3 + x^2 + x + 1\) — highest power of x here is 3.
Degree 2: x² + 2x + 1
Degree 3: x³ + x² + x + 1
3What are the coefficients of x² and x³ in the polynomial x⁴ – 3x³ + 6x² – 2x + 7?
Given: \(x^4 - 3x^3 + 6x^2 - 2x + 7\)
The term containing \(x^3\) is \(-3x^3\), so its coefficient is –3.
The term containing \(x^2\) is \(6x^2\), so its coefficient is 6.
Coefficient of x³ = –3
4What is the coefficient of z in the polynomial 4z³ + 5z² – 11?
Given: \(4z^3 + 5z^2 - 11\)
Scanning the polynomial, the powers of z present are 3, 2 and 0 — there is no term with z raised to the power 1.
Since there is no \(z^1\) term, its coefficient must be 0.
5What is the constant term of the polynomial 9x³ + 5x² – 8x – 10?
Given: \(9x^3 + 5x^2 - 8x - 10\)
The constant term is the term with no variable attached to it — the term that remains if x = 0. Here it is –10.
2.2 Linear Polynomials
Think and Reflect
TR(After Example 4: perimeter 4x) Find the perimeter for sides 1, 1.5, 2, 2.5, 3 cm. What happens if the side increases by 0.5 cm each time?
Perimeter of a square of side x = \(4x\)
x = 1: \(4(1)=4\) cm
x = 1.5: \(4(1.5)=6\) cm
x = 2: \(4(2)=8\) cm
x = 2.5: \(4(2.5)=10\) cm
x = 3: \(4(3)=12\) cm
| Side (cm) | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|
| Perimeter (cm) | 4 | 6 | 8 | 10 | 12 |
Each time the side increases by 0.5 cm, the perimeter increases by \(6-4=2\), \(8-6=2\), \(10-8=2\), \(12-10=2\) — a constant 2 cm every time.
TR(After Example 5: chess club, 200 + 50m) If a player paid ₹750, how many matches did he play?
Total cost for m matches = \(200+50m\); total paid = ₹750
\(200+50m = 750\)
\(50m = 750-200 = 550\)
\(m = \dfrac{550}{50} = 11\)
TR(Input-output discussion) Interpret the rectangle's area, 10x – x², as an input-output process. What value does it take at x = 6 cm?
Here x, the length of the rectangle in cm, is the input fed into the machine. The machine computes the corresponding area using the rule \(10x - x^2\), and the area is the output.
At x = 6: \(10(6) - 6^2 = 60 - 36 = 24\)
Exercise Set 2.2
1Find the value of the linear polynomial 5x – 3 if: (i) x = 0 (ii) x = –1 (iii) x = 2
To evaluate a polynomial at a given value, substitute that value in place of the variable and simplify.
(i) At x = 0: \(5(0) - 3 = 0 - 3 = -3\)
(ii) At x = –1: \(5(-1) - 3 = -5 - 3 = -8\)
(iii) At x = 2: \(5(2) - 3 = 10 - 3 = 7\)
(ii) –8
(iii) 7
2Find the value of the quadratic polynomial 7s² – 4s + 6 if: (i) s = 0 (ii) s = –3 (iii) s = 4
(i) At s = 0: \(7(0)^2 - 4(0) + 6 = 0 - 0 + 6 = 6\)
(ii) At s = –3: \(7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81\)
(iii) At s = 4: \(7(4)^2 - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 16 + 6 = 102\)
(ii) 81
(iii) 102
3The present age of Salil's mother is three times Salil's present age. After 5 years, their ages will add up to 70 years. Find their present ages.
Let Salil's present age = x years. Then his mother's present age = 3x years.
After 5 years, Salil's age = \(x+5\) and his mother's age = \(3x+5\).
Since their ages will add up to 70:
\((x+5) + (3x+5) = 70\)
\(4x + 10 = 70\)
\(4x = 60\)
\(x = 15\)
So Salil's present age = 15 years, and his mother's present age = \(3(15) = 45\) years.
Check: after 5 years, Salil = 20, mother = 50, and \(20+50=70\), which matches.
4The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Since the ratio of the integers is 2:5, let the integers be 2k and 5k for some positive number k.
Their difference is 63:
\(5k - 2k = 63\)
\(3k = 63\)
\(k = 21\)
So the integers are \(2k = 2(21) = 42\) and \(5k = 5(21) = 105\).
Check: \(105 - 42 = 63\), and \(42:105 = 2:5\), which matches.
5Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total of ₹88, how many coins does she have of each type?
Let the number of ₹5 coins = x. Then the number of ₹2 coins = 3x.
Total value of all coins is ₹88:
\(5x + 2(3x) = 88\)
\(5x + 6x = 88\)
\(11x = 88\)
\(x = 8\)
So she has 8 coins of ₹5, and \(3(8)=24\) coins of ₹2.
Check: \(5(8) + 2(24) = 40 + 48 = 88\), which matches.
6A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?
Let the shorter piece = x feet. Then the longer piece = 4x feet.
Together they make up the full fence:
\(x + 4x = 300\)
\(5x = 300\)
\(x = 60\)
So the shorter piece = 60 feet, and the longer piece = \(4(60) = 240\) feet.
Check: \(60+240=300\), which matches.
7If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?
Let the width = w cm. Then the length = \(2w+3\) cm.
Using the perimeter formula, Perimeter = \(2(\text{length}+\text{width})\):
\(2(w + 2w+3) = 24\)
\(2(3w+3) = 24\)
\(6w+6 = 24\)
\(6w = 18\)
\(w = 3\)
So the width = 3 cm, and the length = \(2(3)+3 = 9\) cm.
Check: Perimeter = \(2(9+3) = 24\), which matches.
2.3 Exploring Linear Patterns
Think and Reflect
TRPredict the number of squares in the next three stages of the tile pattern, and write the sequence up to Stage 7.
Stages 1 to 4 give 1, 3, 5, 7 tiles — each stage adds 2 more tiles than the previous one.
Stage 5 = 7 + 2 = 9. Stage 6 = 9 + 2 = 11. Stage 7 = 11 + 2 = 13.
TRUsing 2n – 1, find the tiles at the 15th and 26th stage. Which stage has 21 tiles? Which has 47 tiles?
Stage 15: \(2(15)-1 = 30-1 = 29\) tiles
Stage 26: \(2(26)-1 = 52-1 = 51\) tiles
For 21 tiles: \(2n-1=21 \Rightarrow 2n = 22 \Rightarrow n = 11\)
For 47 tiles: \(2n-1=47 \Rightarrow 2n = 48 \Rightarrow n = 24\)
Stage 26 → 51 tiles
21 tiles occurs at Stage 11
47 tiles occurs at Stage 24
TR(Bela's pocket money, 100 – 5n) What amount is left on the 15th day? How many days until the entire amount is spent?
Amount left on day 15: \(100 - 5(15) = 100 - 75 = 25\)
The amount becomes 0 when: \(100 - 5n = 0 \Rightarrow 5n = 100 \Rightarrow n = 20\)
TR(Auto-rikshaw fare, 15n – 5) For how many km will the fare be ₹130?
\(15n - 5 = 130\)
\(15n = 135\)
\(n = 9\)
Exercise Set 2.3
1A student has ₹500 in her savings account. She gets ₹150 every month as pocket money. Find a linear expression for the amount in the nth month.
Starting amount = ₹500. After n months, she has received \(150n\) rupees in pocket money.
Amount at the end of the nth month \(= 500 + 150n\)
| Month, n | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Amount (₹) | 650 | 800 | 950 | 1100 |
2A rally starts with 120 members. Each hour, 9 members drop out. Find a linear expression for the nth hour.
Starting members = 120. After n hours, \(9n\) members have dropped out.
Members remaining after n hours \(= 120 - 9n\)
| Hour, n | 1 | 2 | 3 |
|---|---|---|---|
| Members | 111 | 102 | 93 |
3Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm (ii) 10 cm (iii) 8 cm. Find the linear pattern.
Area = length × breadth = \(13 \times b\), where b is the breadth.
(i) At b = 12: \(13 \times 12 = 156\) cm²
(ii) At b = 10: \(13 \times 10 = 130\) cm²
(iii) At b = 8: \(13 \times 8 = 104\) cm²
(i) 156 cm²
(ii) 130 cm²
(iii) 104 cm²
4Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm (ii) 9 cm (iii) 13 cm. Find the linear pattern.
Volume = length × breadth × height = \(7 \times 11 \times h = 77h\)
(i) At h = 5: \(77 \times 5 = 385\) cm³
(ii) At h = 9: \(77 \times 9 = 693\) cm³
(iii) At h = 13: \(77 \times 13 = 1001\) cm³
(i) 385 cm³
(ii) 693 cm³
(iii) 1001 cm³
5Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express as a linear pattern.
Pages left after n days \(= 500 - 20n\)
At n = 15: \(500 - 20(15) = 500 - 300 = 200\)
2.4 Linear Growth and Decay
Think and Reflect
TR(Example 9: journey cost, C = 100 + 60d) What is the cost for 15 km? For how many km will the cost be ₹700?
At d = 15: \(C = 100 + 60(15) = 100 + 900 = 1000\)
Solving \(100 + 60d = 700\): \(60d = 600 \Rightarrow d = 10\)
TR(Example 10: water tank, h = 3 – 0.5t) What is the height of the water at the end of 5 months?
At t = 5: \(h(5) = 3 - 0.5(5) = 3 - 2.5 = 0.5\)
Exercise Set 2.4
1A plant has height 1.75 feet and grows 0.5 feet each month. (i) height after 7 months (ii) table for t = 0 to 10 (iii) expression and why it's linear growth.
(i) At t = 7: \(h = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25\) feet
(ii) Table of values:
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| h | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |
(iii) \(h(t) = 1.75 + 0.5t\). This represents linear growth because h increases by a fixed amount, 0.5 feet, for every one-month increase in t — a constant rate of increase, i.e. a positive slope.
(iii) h(t) = 1.75 + 0.5t — linear growth (positive slope of 0.5)
2A mobile phone bought for ₹10,000 loses ₹800 in value every year. (i) value after 3 years (ii) table for t = 0 to 8 (iii) expression and why it's linear decay.
(i) At t = 3: \(v = 10000 - 800(3) = 10000 - 2400 = 7600\)
(ii) Table of values:
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| v (₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
(iii) \(v(t) = 10000 - 800t\). This represents linear decay because v decreases by a fixed amount, ₹800, every year — a constant rate of decrease, i.e. a negative slope.
(iii) v(t) = 10000 – 800t — linear decay (negative slope of –800)
3Initial population 750. Every year 50 people move in. (i) population after 6 years (ii) table for t = 0 to 10 (iii) expression and why it's linear growth.
(i) At t = 6: \(P = 750 + 50(6) = 750 + 300 = 1050\)
(ii) Table of values:
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| P | 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |
(iii) \(P(t) = 750 + 50t\). This is linear growth because P increases by the fixed amount 50 every year — a positive slope of 50.
(iii) P(t) = 750 + 50t — linear growth (positive slope of 50)
4Telecom recharge ₹600 reduces by ₹15 per day. (i) equation for b(x) (ii) when will it run out? (iii) table for x = 1 to 10.
(i) \(b(x) = 600 - 15x\). This is linear decay because the balance decreases by the fixed amount ₹15 every day — a negative slope of –15.
(ii) Solving \(600 - 15x = 0\): \(15x = 600 \Rightarrow x = 40\)
(iii) Table of values:
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| b(x) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |
2.5 Linear Relationships
Think and Reflect
TR(Example 11: telecom bill, y = 20x + 150) What do the numbers 20 and 150 represent?
In \(y = 20x + 150\), x is the amount of data used (GB) and y is the total bill (₹).
20 is the slope — the extra cost charged for every additional GB of data used.
150 is the y-intercept — the fixed monthly fee the student pays even if she uses 0 GB of data.
Exercise Set 2.5
1A learning platform's bill: 10 modules → ₹400; 14 modules → ₹500. Find a and b in y = ax + b.
When x = 10, y = 400: \(400 = 10a+b\) ... (i)
When x = 14, y = 500: \(500 = 14a+b\) ... (ii)
Subtracting (i) from (ii): \(4a = 100 \Rightarrow a = 25\)
Substituting a = 25 into (i): \(400 = 10(25)+b = 250+b \Rightarrow b = 150\)
2A gym's badminton court bill: 10 hours → ₹800; 15 hours → ₹1100. Find a and b.
When x = 10, y = 800: \(800 = 10a+b\) ... (i)
When x = 15, y = 1100: \(1100 = 15a+b\) ... (ii)
Subtracting (i) from (ii): \(5a = 300 \Rightarrow a = 60\)
Substituting a = 60 into (i): \(800 = 10(60)+b = 600+b \Rightarrow b = 200\)
3Given °C = a°F + b, ice melts at 0°C/32°F and water boils at 100°C/212°F. Find a and b.
When °F = 32, °C = 0: \(0 = 32a+b\) ... (i)
When °F = 212, °C = 100: \(100 = 212a+b\) ... (ii)
Subtracting (i) from (ii): \(180a = 100 \Rightarrow a = \dfrac{100}{180} = \dfrac{5}{9}\)
Substituting \(a = \dfrac{5}{9}\) into (i): \(0 = 32\left(\dfrac{5}{9}\right)+b \Rightarrow b = -\dfrac{160}{9}\)
2.6 Visualising Linear Relationships
Think and Reflect
TR(Line y = 2x + 1) Complete the table for x = 1, 2, 5, 7, 9, 12, 20.
Substituting each x-value into \(y = 2x + 1\):
x = 1: \(2(1)+1=3\)
x = 2: \(2(2)+1=5\)
x = 5: \(2(5)+1=11\)
x = 7: \(2(7)+1=15\)
x = 9: \(2(9)+1=19\)
x = 12: \(2(12)+1=25\)
x = 20: \(2(20)+1=41\)
| x | 1 | 2 | 5 | 7 | 9 | 12 | 20 |
|---|---|---|---|---|---|---|---|
| y | 3 | 5 | 11 | 15 | 19 | 25 | 41 |
y = 2x + 1, marking the points (1,3) and (7,15) from the table.
TR(Fig. 2.9: lines y = ax, a > 0) What happens as a varies? What happens when a > 1 and when a < 1?
Every line of the form y = ax passes through the origin (0, 0), since substituting x = 0 gives y = 0 regardless of a.
When a > 1, the line is steeper than y = x. When 0 < a < 1, the line is flatter (less steep) than y = x. When a = 1, the line coincides exactly with y = x.
y = x/2, y = x, y = 2x — all through the origin, steepness increasing with a.
TR(Fig. 2.11: lines y = –ax, a > 0) What happens as a varies?
These lines also pass through the origin, but slope downward since the coefficient of x is negative.
When a > 1, the line falls more steeply than y = –x. When 0 < a < 1, the line falls more gently than y = –x.
y = -x/3, y = -x, y = -3x — all through the origin, sloping downward.
TRDifferentiate between the graphs of y = 3x + 1 and y = –3x + 1.
Both lines have the same y-intercept, since at x = 0 both give y = 1 — they cross the y-axis at the same point, (0, 1).
y = 3x + 1 has slope +3 (rises left to right). y = –3x + 1 has slope –3, equal in magnitude but opposite in sign (falls left to right).
y = 3x + 1 and y = -3x + 1, meeting at their common y-intercept (0, 1).
TR(Fig. 2.13: y = 2x – 1, y = 2x + 1, y = 2x + 5) What do you conclude when a is fixed but b varies?
All three lines have the same slope, a = 2, so they are equally steep. Only b, the y-intercept, changes: –1, 1, and 5.
y = 2x - 1, y = 2x + 1, y = 2x + 5 — parallel lines with the same slope, different y-intercepts.
Exercise Set 2.6
For each part, two points are calculated and plotted, and the line is drawn through them.
1(i)y = 4x, y = 2x, y = x
All three lines pass through (0, 0) since b = 0 in each case. A second point for each:
y = 4x → (1, 4)
y = 2x → (1, 2)
y = x → (1, 1)
y = 4x, y = 2x, y = x — all through the origin, steepness increasing with a.
1(ii)y = –6x, y = –3x, y = –x
All three lines pass through (0, 0). A second point for each:
y = –6x → (1, –6)
y = –3x → (1, –3)
y = –x → (1, –1)
y = -6x, y = -3x, y = -x — all through the origin, sloping downward.
1(iii)y = 5x, y = –5x
y = 5x passes through (0, 0) and (1, 5).
y = –5x passes through (0, 0) and (1, –5).
y = 5x and y = -5x — mirror images of each other across the x-axis.
1(iv)y = 3x – 1, y = 3x, y = 3x + 1
y = 3x – 1 passes through (0, –1) and (1, 2).
y = 3x passes through (0, 0) and (1, 3).
y = 3x + 1 passes through (0, 1) and (1, 4).
y = 3x - 1, y = 3x, y = 3x + 1 — parallel lines, same slope, different y-intercepts.
1(v)y = –2x – 3, y = –2x, y = 2x + 3
y = –2x – 3 passes through (0, –3) and (1, –5).
y = –2x passes through (0, 0) and (1, –2).
y = 2x + 3 passes through (0, 3) and (1, 5).
y = -2x - 3, y = -2x, y = 2x + 3 — first two are parallel; the third is not.
End-of-Chapter Exercises
1Write a polynomial of degree 3 in x, in which the coefficient of the x² term is –7.
We need the highest power of x to be 3 (a cubic polynomial), and the term with x² must have coefficient –7.
\(x^3 - 7x^2 + 2x + 5\) satisfies both conditions: highest power is 3, and the coefficient of x² is –7.
2Find the values: (i) 5x² – 3x + 7 if x = 1 (ii) 4t³ – t² + 6 if t = a
(i) At x = 1: \(5(1)^2 - 3(1) + 7 = 5 - 3 + 7 = 9\)
(ii) Replacing t with a in the expression directly gives \(4a^3 - a^2 + 6\).
(ii) 4a³ – a² + 6
3If we multiply a number by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, we get \(-\frac{7}{12}\). Find the number.
Let the number be x.
\(\dfrac{5}{2}x + \dfrac{2}{3} = -\dfrac{7}{12}\)
\(\dfrac{5}{2}x = -\dfrac{7}{12} - \dfrac{2}{3}\)
Taking LCM 12: \(-\dfrac{7}{12} - \dfrac{8}{12} = -\dfrac{15}{12} = -\dfrac{5}{4}\)
\(\dfrac{5}{2}x = -\dfrac{5}{4}\)
\(x = -\dfrac{5}{4} \times \dfrac{2}{5} = -\dfrac{10}{20} = -\dfrac{1}{2}\)
4A positive number is 5 times another number. If 21 is added to both, one new number becomes twice the other. Find the numbers.
Let the smaller number = x. Then the larger number = 5x.
Adding 21 to both: new smaller number = x + 21; new larger number = 5x + 21.
Since 5x + 21 was already the larger quantity, it becomes twice the other new number:
\(5x + 21 = 2(x+21)\)
\(5x+21 = 2x+42\)
\(3x = 21\)
\(x = 7\)
So the smaller number = 7, and the larger number = \(5(7) = 35\).
Check: new numbers are 28 and 56, and indeed \(56 = 2 \times 28\).
5If you have ₹800 and save ₹250 every month, find the amount after (i) 6 months (ii) 2 years. Express as a linear pattern.
Amount after n months \(= 800 + 250n\)
(i) At n = 6: \(800 + 250(6) = 800 + 1500 = 2300\)
(ii) 2 years = 24 months. At n = 24: \(800 + 250(24) = 800 + 6000 = 6800\)
(i) ₹2,300
(ii) ₹6,800
6*The digits of a two-digit number differ by 3. Interchanging and adding gives 143. Find both numbers.
Let the tens digit = x and the units digit = y. Original number \(= 10x+y\). Interchanged number \(= 10y+x\).
Sum of the two numbers is 143:
\((10x+y)+(10y+x) = 143\)
\(11x+11y = 143\)
\(11(x+y) = 143\)
\(x+y = 13\) ... (i)
The digits differ by 3:
\(x-y = 3\) ... (ii)
Adding (i) and (ii): \(2x = 16 \Rightarrow x=8\)
From (i): \(y = 13-8 = 5\)
Original number = \(10(8)+5 = 85\); interchanged number = \(10(5)+8 = 58\)
Check: \(85+58=143\), and \(8-5=3\), both matching.
7*Graph: (i) y = –3x + 4 (ii) 2y = 4x + 7 (iii) 5y = 6x – 10 (iv) 3y = 6x – 11. Identify slopes, y-intercepts, and check for parallels.
Rewriting each equation in the form y = ax + b:
(i) \(y=-3x+4\) is already in this form; slope = –3, y-intercept = 4
(ii) \(2y=4x+7 \Rightarrow y=2x+3.5\); slope = 2, y-intercept = 3.5
(iii) \(5y=6x-10 \Rightarrow y=1.2x-2\); slope = 1.2, y-intercept = –2
(iv) \(3y=6x-11 \Rightarrow y=2x-\dfrac{11}{3}\); slope = 2, y-intercept = \(-\dfrac{11}{3} \approx -3.67\)
Points where each line cuts the y-axis:
(i) (0, 4)
(ii) (0, 3.5)
(iii) (0, –2)
(iv) \((0, -\dfrac{11}{3})\)
All four lines with their y-intercepts marked. Lines (ii) and (iv) are parallel.
Comparing the slopes: lines (ii) and (iv) both have slope 2, so these two lines are parallel to each other.
8*\(y = \frac{9}{5}(x - 273) + 32\), x in Kelvin, y in Fahrenheit. (i) y at x = 313 K. (ii) x at y = 158°F.
(i) At x = 313: \(y = \dfrac{9}{5}(313-273)+32 = \dfrac{9}{5}(40)+32 = 72+32 = 104\)
(ii) At y = 158: \(158 = \dfrac{9}{5}(x-273)+32\)
\(126 = \dfrac{9}{5}(x-273)\)
\(x-273 = 126 \times \dfrac{5}{9} = 70\)
\(x = 343\)
(ii) 343 K
9*Work = force × distance. Express as a linear equation, taking force = 3 units. Find w at d = 2.
Work \(w = \) force \(\times\) distance \(d\). With force = 3 units: \(w = 3d\)
This passes through the origin (0, 0) with slope 3; a second point is (2, 6).
w = 3d, passing through the origin and (2, 6).
At d = 2: \(w = 3(2) = 6\)
10*p(x) passes through (1, 5) and (3, 11). (i) Find p(x). (ii) Where does it cut the axes? (iii) Draw and verify.
Let p(x) = ax + b.
At (1, 5): \(5=a+b\) ... (i)
At (3, 11): \(11=3a+b\) ... (ii)
Subtracting (i) from (ii): \(2a=6 \Rightarrow a=3\)
Substituting into (i): \(5=3+b \Rightarrow b=2\)
y-intercept: at x = 0, \(p(0)=2\), so the line cuts the y-axis at (0, 2).
x-intercept: \(0=3x+2 \Rightarrow x=-\dfrac{2}{3}\), so the line cuts the x-axis at \((-\dfrac{2}{3}, 0)\).
p(x) = 3x + 2, passing through (1,5) and (3,11); cuts axes at (0,2) and (-2/3, 0).
The graph confirms the line passes through (1, 5) and (3, 11), and crosses the axes at (0, 2) and \((-\dfrac{2}{3}, 0)\) as calculated.
11*p(x) = ax + b, q(x) = cx + d. Given: p(0) = 5; p(x) – q(x) cuts the x-axis at (3, 0); p(x) + q(x) = 6x + 4. Find p(x) and q(x).
From p(0) = 5: \(p(0)=a(0)+b=b\), so \(b=5\)
From \(p(x)+q(x) = 6x+4\): \((ax+b)+(cx+d) = (a+c)x+(b+d)\)
Comparing with 6x + 4: \(a+c=6\) and \(b+d=4\)
Since b = 5: \(5+d=4 \Rightarrow d=-1\)
From p(x) – q(x) cutting the x-axis at (3, 0): \(p(x)-q(x) = (a-c)x+(b-d)\), which equals 0 at x = 3:
\(3(a-c)+(5-(-1))=0\)
\(3(a-c)+6=0 \Rightarrow a-c=-2\)
Solving \(a+c=6\) and \(a-c=-2\) together: adding gives \(2a=4 \Rightarrow a=2\), then \(c=6-2=4\)
Check: \(p(x)-q(x) = -2x+6\), which is 0 at x = 3, and \(p(x)+q(x)=6x+4\), both matching.
12*Hexagon matchstick pattern: a new hexagon shares a side with the previous one at each stage.
Stage 1 is a single hexagon, using all 6 of its sides = 6 matchsticks. Every new hexagon added shares exactly 1 side (1 matchstick) with the hexagon before it, so it only needs 5 new matchsticks.
(i) Next two stages:
Stage 4 — 4 hexagons, 21 matchsticks.
Stage 5 — 5 hexagons, 26 matchsticks.
Stage 4 = 3(5) + 6 = 21 matchsticks. Stage 5 = 4(5) + 6 = 26 matchsticks.
(ii) Table of values:
| Stage, n | 1 | 2 | 3 | 4 | 5 | … | n |
|---|---|---|---|---|---|---|---|
| Matchsticks | 6 | 11 | 16 | 21 | 26 | … | 5n+1 |
(iii) The first hexagon needs 6 matchsticks, and each of the remaining (n – 1) hexagons adds 5 more:
Matchsticks \(= 6 + 5(n-1) = 6+5n-5 = 5n+1\)
(iv) At n = 15: \(5(15)+1 = 75+1 = 76\)
(v) Solving \(5n+1=200\): \(5n=199 \Rightarrow n=39.8\). Since n must be a whole number, this is not possible.
Stage 4 = 21, Stage 5 = 26
15th stage = 76 matchsticks
200 matchsticks cannot form any stage since n = 39.8 is not a whole number
13*p(x) passes through (2, 3) and (6, 11). q(x) passes through (4, –1) and is parallel to p(x). Find both and their x-intercepts.
Let p(x) = ax + b.
At (2, 3): \(3=2a+b\) ... (i)
At (6, 11): \(11=6a+b\) ... (ii)
Subtracting (i) from (ii): \(4a=8 \Rightarrow a=2\)
Substituting into (i): \(3=2(2)+b=4+b \Rightarrow b=-1\)
Since q(x) is parallel to p(x), their slopes are equal, so c = a = 2. Let q(x) = 2x + d.
q passes through (4, –1): \(-1=2(4)+d=8+d \Rightarrow d=-9\)
x-intercept of p(x): \(2x-1=0 \Rightarrow x=\dfrac{1}{2}\), giving \((\dfrac{1}{2}, 0)\)
x-intercept of q(x): \(2x-9=0 \Rightarrow x=\dfrac{9}{2}\), giving \((\dfrac{9}{2}, 0)\)
p(x) = 2x - 1 and q(x) = 2x - 9, parallel lines with x-intercepts at (1/2,0) and (9/2,0).
14*What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
\(f(x) = ax+a = a(x+1)\)
Testing x = –1: \(f(-1) = a(-1+1) = a(0) = 0\), true for every value of a.
Frequently Asked Questions
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