Question 1

Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.

(i) Find the height after 7 months.
(ii) Make a table of values for $t$ varying from 0 to 10 months and show how the height, $h$, increases every month.
(iii) Find an expression that relates $h$ and $t$, and explain why it represents linear growth.

Accepted Answer

(i) Initial height = 1.75 feet, growth per month = 0.5 feet
Height after 7 months = Initial height + 7 × growth per month = 1.75 + 7 × 0.5 = 1.75 + 3.5 = 5.25 feet

(ii) Table of values:
| t (months) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|------------|---|---|---|---|---|---|---|---|---|---|----|
| h (feet)   |1.75|2.25|2.75|3.25|3.75|4.25|4.75|5.25|5.75|6.25|6.75|

(iii) Expression relating height h and time t:

h = 1.75 + 0.5t

This is a linear expression because h changes by a constant amount (0.5 feet) for every unit increase in t (month). The graph of h versus t is a straight line with slope 0.5 and intercept 1.75. — Step-by-step:
- Initial height is 1.75 feet.
- Each month, height increases by 0.5 feet.
- After t months, height = initial height + (growth per month) × t = 1.75 + 0.5t.
- This is linear because the rate of change is constant.
- The table confirms the height increases by 0.5 feet each month.

Question 2

A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.

(i) Find the value of the phone after 3 years.
(ii) Make a table of values for $t$ varying from 0 to 8 years and show how the value of the phone, $
u$, depreciates with time.
(iii) Find an expression that relates $
u$ and $t$, and explain why it represents linear decay.

Accepted Answer

(i) Initial value = ₹10,000
Depreciation per year = ₹800
Value after 3 years = Initial value - 3 × depreciation = 10000 - 3 × 800 = 10000 - 2400 = ₹7600

(ii) Table of values:
| t (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|-----------|----|----|----|----|----|----|----|----|----|
| ν (₹)     |10000|9200|8400|7600|6800|6000|5200|4400|3600|

(iii) Expression relating value ν and time t:

ν = 10000 - 800t

This represents linear decay because the value decreases by a fixed amount (₹800) every year. The graph of ν versus t is a straight line with negative slope -800 and intercept 10000. — Step-by-step:
- Initial value is ₹10,000.
- Each year, value decreases by ₹800.
- After t years, value = 10000 - 800t.
- This is linear decay because the value decreases by a constant amount each year.
- The table confirms the value reduces by ₹800 every year.

Question 3

The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.

(i) Find the population of the village after 6 years.

(ii) Make a table of values for $t$ varying from 0 to 10 years and show how the population, $P$, increases every year.
(iii) Find an expression that relates $P$ and $t$, and explain why it represents linear growth.

Accepted Answer

(i) Initial population = 750
Increase per year = 50
Population after 6 years = 750 + 6 × 50 = 750 + 300 = 1050

(ii) Table of values:
| t (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|-----------|----|----|----|----|----|----|----|----|----|----|----|
| P         | 750| 800| 850| 900| 950|1000|1050|1100|1150|1200|1250|

(iii) Expression relating population P and time t:

P = 750 + 50t

This represents linear growth because the population increases by a fixed number (50) every year. The graph of P versus t is a straight line with slope 50 and intercept 750. — Step-by-step:
- Initial population is 750.
- Each year, population increases by 50.
- After t years, population = 750 + 50t.
- This is linear growth because the population increases by a constant amount each year.
- The table confirms the population increases by 50 every year.

Question 4

A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.

(i) Write an equation that models the remaining balance $b(x)$ after using the scheme for $x$ days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for $x$ varying from 1 to 10 days and show how the balance $b(x)$, reduces with time.

Accepted Answer

(i) Initial balance = ₹600
Daily reduction = ₹15
Equation for balance after x days:

b(x) = 600 - 15x

This represents linear decay because the balance decreases by a fixed amount (₹15) each day.

(ii) To find when balance runs out, set b(x) = 0:

0 = 600 - 15x
15x = 600
x = 600 / 15 = 40 days

(iii) Table of values:
| x (days) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|----------|----|----|----|----|----|----|----|----|----|----|
| b(x) (₹) |585 |570 |555 |540 |525 |510 |495 |480 |465 |450 | — Step-by-step:
- Initial balance is ₹600.
- Each day, balance reduces by ₹15.
- After x days, balance = 600 - 15x.
- This is linear decay because balance decreases by a constant amount daily.
- Balance runs out when 600 - 15x = 0, so x = 40 days.
- The table confirms the balance decreases by ₹15 each day.

Question 5

A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill $y$ depends on the number of modules accessed, $x$, according to the relation $y = ax + b$, find the values of $a$ and $b$.

Accepted Answer

Let the monthly bill be y and the number of modules accessed be x. Given y = ax + b.

From the first observation:
When x = 10, y = 400
=> 400 = 10a + b  ...(1)

From the second observation:
When x = 14, y = 500
=> 500 = 14a + b  ...(2)

Subtract (1) from (2):
500 - 400 = 14a - 10a + b - b
100 = 4a
=> a = 25

Substitute a = 25 in (1):
400 = 10*25 + b
400 = 250 + b
=> b = 150

Therefore, a = 25 and b = 150. — We form two linear equations from the given data points and solve them simultaneously to find the values of a and b. The difference of the two equations eliminates b, allowing us to find a. Substituting back gives b.

Question 6

A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill $y$ depends on the hours of the use of the badminton court, $x$, according to the relation $y = ax + b$, find the values of $a$ and $b$.

Accepted Answer

Let the monthly bill be y and the hours of use be x. Given y = ax + b.

From the first observation:
When x = 10, y = 800
=> 800 = 10a + b  ...(1)

From the second observation:
When x = 15, y = 1100
=> 1100 = 15a + b  ...(2)

Subtract (1) from (2):
1100 - 800 = 15a - 10a + b - b
300 = 5a
=> a = 60

Substitute a = 60 in (1):
800 = 10*60 + b
800 = 600 + b
=> b = 200

Therefore, a = 60 and b = 200. — Two linear equations are formed from the given data points. Subtracting one from the other eliminates b, allowing us to find a. Substituting a back into one equation gives b.

Question 7

Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by $^\circ \mathrm{C} = a^\circ \mathrm{F} + b$. Find $a$ and $b$, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.

(Hint: When $^\circ \mathrm{C} = 0$, $^\circ \mathrm{F} = 32$ and when $^\circ \mathrm{C} = 100$, $^\circ \mathrm{F} = 212$. Use this information to find $a$ and $b$, and thus, the linear relationship between $^\circ \mathrm{C}$ and $^\circ \mathrm{F}$.)

Accepted Answer

Given the relation: °C = a °F + b.

Using the two points:
When °C = 0, °F = 32
=> 0 = a*32 + b  ...(1)

When °C = 100, °F = 212
=> 100 = a*212 + b  ...(2)

From (1): b = -32a

Substitute b in (2):
100 = 212a - 32a
100 = 180a
=> a = 100 / 180 = 5/9

Then b = -32 * (5/9) = -160/9

Therefore, the relation is:
°C = (5/9) °F - 160/9

Alternatively, rearranged to express °F in terms of °C:
°F = (9/5) °C + 32. — Two points are used to form simultaneous linear equations. Solving these equations yields the values of a and b, giving the linear conversion formula between Celsius and Fahrenheit.

Question 8

Draw the graphs of the following sets of lines. In each case, reflect on the role of $a$ and $b$.

(i) $y = 4x, y = 2x, y = x$

(ii) $y = -6x, y = -3x, y = -x$

(iii) $y = 5x, y = -5x$

(iv) $y = 3x - 1, y = 3x, y = 3x + 1$

(v) $y = -2x - 3, y = -2x, y = 2x + 3$

Accepted Answer

To draw the graphs of the given lines, plot points for each equation and join them to form straight lines. Then analyze the role of 'a' (slope) and 'b' (y-intercept) in each case.

(i) Lines: y=4x, y=2x, y=x
- All lines pass through origin (0,0) since b=0.
- Slopes are 4, 2, and 1 respectively.
- Higher slope means steeper line.
- Graphs show lines with increasing steepness as slope increases.

(ii) Lines: y=-6x, y=-3x, y=-x
- All pass through origin.
- Negative slopes indicate lines slope downwards from left to right.
- Slopes -6, -3, -1 show decreasing steepness.

(iii) Lines: y=5x, y=-5x
- Both pass through origin.
- Slopes are positive 5 and negative 5.
- One line slopes steeply upwards, the other steeply downwards.

(iv) Lines: y=3x -1, y=3x, y=3x +1
- All have same slope 3, so lines are parallel.
- Different y-intercepts (-1, 0, 1) shift lines vertically.

(v) Lines: y=-2x -3, y=-2x, y=2x +3
- First two lines have slope -2, third has slope 2.
- First two lines are parallel, third line slopes upward.
- Different intercepts shift lines vertically.

Thus, 'a' determines slope (steepness and direction), 'b' determines vertical shift (y-intercept). — Step-by-step:
1. For each equation, identify slope (a) and intercept (b).
2. Plot points by substituting values of x.
3. Draw lines through plotted points.
4. Compare slopes to understand steepness and direction.
5. Compare intercepts to understand vertical shifts.
6. Conclude role of 'a' and 'b' accordingly.

Question 9

Write a polynomial of degree 3 in the variable $x$, in which the coefficient of the $x^2$ term is $-7$.

Accepted Answer

A polynomial of degree 3 in variable x with the coefficient of x^2 term as -7 can be written as: p(x) = ax^3 - 7x^2 + bx + c, where a, b, c are any real numbers and a ≠ 0. For example, p(x) = 2x^3 - 7x^2 + 3x + 5. — The degree of the polynomial is the highest power of x, which is 3 here. The coefficient of x^2 is given as -7. The other coefficients can be any real numbers. So, the polynomial can be constructed accordingly.

Question 10

Find the values of the following polynomials at the indicated values of the variables.

(i) $5x^{2} - 3x + 7$ if $x = 1$

(ii) $4t^{3} - t^{2} + 6$ if $t = a$

Accepted Answer

(i) For x = 1, p(x) = 5(1)^2 - 3(1) + 7 = 5 - 3 + 7 = 9.

(ii) For t = a, p(t) = 4a^3 - a^2 + 6. — Substitute the given values into the polynomial and simplify.
(i) Substitute x=1: 5*1^2 - 3*1 + 7 = 5 - 3 + 7 = 9.
(ii) Substitute t=a: 4a^3 - a^2 + 6 (expression in terms of a).

Question 11

If we multiply a number by $\frac{5}{2}$ and add $\frac{2}{3}$ to the product, we get $\frac{-7}{12}$. Find the number.

Accepted Answer

Let the number be x.
Then, (5/2)x + 2/3 = -7/12.
Multiply both sides by 12 to clear denominators:
12 * (5/2)x + 12 * (2/3) = 12 * (-7/12)
=> 6 * 5x + 4 * 2 = -7
=> 30x + 8 = -7
=> 30x = -7 - 8 = -15
=> x = -15 / 30 = -1/2. — Set up the equation as per the problem statement. Multiply both sides by the least common denominator to eliminate fractions. Solve the resulting linear equation for x.

Question 12

A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Accepted Answer

Let the smaller number be x.
Then the positive number = 5x.
After adding 21 to both numbers:
New numbers are x + 21 and 5x + 21.
Given one new number is twice the other:
Case 1: 5x + 21 = 2(x + 21)
=> 5x + 21 = 2x + 42
=> 5x - 2x = 42 - 21
=> 3x = 21
=> x = 7
Numbers are 7 and 35.
Check: 35 + 21 = 56, 7 + 21 = 28, 56 = 2*28 (True).
Case 2: x + 21 = 2(5x + 21)
=> x + 21 = 10x + 42
=> x - 10x = 42 - 21
=> -9x = 21
=> x = -21/9 = -7/3 (negative, discard as positive number required).
Hence, numbers are 7 and 35. — Assign variables, translate the problem into equations, solve the linear equations, and verify the solution.

Question 13

If you have ₹ 800 and you save ₹ 250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.

Accepted Answer

(i) Amount after 6 months = 800 + 250 × 6 = 800 + 1500 = ₹ 2300.
(ii) 2 years = 24 months.
Amount after 24 months = 800 + 250 × 24 = 800 + 6000 = ₹ 6800.
Linear pattern: Amount after n months = 800 + 250n. — Initial amount is ₹800, monthly saving is ₹250. Total amount after n months is initial amount plus n times monthly saving. Substitute n = 6 and n = 24 to find amounts.

Question 14

The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.

Accepted Answer

Let the digits be x and y, where x is the tens digit and y is the units digit.
Given: |x - y| = 3.
Original number = 10x + y.
Interchanged number = 10y + x.
Sum = (10x + y) + (10y + x) = 11(x + y) = 143.
=> x + y = 13.
From |x - y| = 3, two cases:
Case 1: x - y = 3
Add with x + y = 13:
2x = 16 => x = 8
Then y = 5.
Number: 85 and 58.
Case 2: y - x = 3
Add with x + y = 13:
2y = 16 => y = 8
Then x = 5.
Number: 58 and 85.
Both pairs satisfy the conditions.
Hence, the numbers are 85 and 58. — Assign variables to digits, use given conditions to form equations, solve simultaneous equations, and find digits. Construct numbers from digits.

Question 15

Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.

(i) $y = -3x + 4$
(ii) $2y = 4x + 7$
(iii) $5y = 6x - 10$
(iv) $3y = 6x - 11$

Are any of the lines parallel?

Accepted Answer

First, rewrite each equation in the form y = mx + c:
(i) y = -3x + 4
Slope (m) = -3, y-intercept (c) = 4
Cuts y-axis at (0,4)

(ii) 2y = 4x + 7 => y = 2x + 7/2
Slope = 2, y-intercept = 3.5
Cuts y-axis at (0, 3.5)

(iii) 5y = 6x - 10 => y = (6/5)x - 2
Slope = 6/5, y-intercept = -2
Cuts y-axis at (0, -2)

(iv) 3y = 6x - 11 => y = 2x - 11/3
Slope = 2, y-intercept = -11/3 ≈ -3.67
Cuts y-axis at (0, -3.67)

Lines with same slope are parallel.
Slopes:
(i) -3
(ii) 2
(iii) 6/5 = 1.2
(iv) 2
Lines (ii) and (iv) have slope 2, so they are parallel.

Graphing these lines will show their intercepts and slopes as above. — Convert each equation to slope-intercept form y = mx + c to identify slope and y-intercept. Lines with equal slopes are parallel. Plot points to draw graphs.

Introduction to

Introduction to — Study Notes

2.1 INTRODUCTION

2.2 LINEAR POLYNOMIALS

2.3 EXPLORING LINEAR PATTERNS

Practice Questions — Introduction to

All 8 Chapters in Ganita Manjari (English)