End-of-Chapter Exercises – Solutions
Q1. Write a polynomial of degree 3 in the variable , in which the coefficient of the term is .
A polynomial of degree 3 must contain the term .
The coefficient of the term should be .
One such polynomial is:
Hence, the required polynomial is:
Q2. Find the values of the following polynomials at the indicated values of the variables.
(i) if
Given:
Substitute
(ii)
Given:
Substitute
Q3. If we multiply a number by 5/2 and add 2/3 to the product, we get -7/12. Find the number.
Let the number be x.
According to the question:
(5/2)x + 2/3 = -7/12
(5/2)x = -7/12 – 2/3
Take LCM 12:
(5/2)x = -7/12 – 8/12
(5/2)x = -15/12
(5/2)x = -5/4
Multiply both sides by 2/5:
x = (-5/4) × (2/5)
x = -1/2
Answer: -1/2
Q4. 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?
Let the smaller number be x.
Then the larger number is 5x.
After adding 21:
Smaller number = x + 21
Larger number = 5x + 21
According to the question:
5x + 21 = 2(x + 21)
5x + 21 = 2x + 42
5x – 2x = 42 – 21
3x = 21
x = 7
Larger number = 5 × 7 = 35
Answer: 7 and 35
Q5. If you have Rs 800 and you save Rs 250 every month, find the amount you have after:
(i) 6 months
Savings in 6 months = 250 × 6 = 1500
Total amount = 800 + 1500
Total amount = 2300
Answer: Rs 2300
(ii) 2 years
2 years = 24 months
Savings in 24 months = 250 × 24 = 6000
Total amount = 800 + 6000
Total amount = 6800
Answer: Rs 6800
Linear Pattern
If the number of months is x,
Amount = 800 + 250x
Answer: 800 + 250x
Q6. 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.
Let the tens digit be x.
Then the ones digit is x – 3.
Original number = 10x + (x – 3)
Original number = 11x – 3
Interchanged number = 10(x – 3) + x
Interchanged number = 11x – 30
According to the question:
(11x – 3) + (11x – 30) = 143
22x – 33 = 143
22x = 176
x = 8
Ones digit = 8 – 3 = 5
Original number = 85
Interchanged number = 58
Answer: 85 and 58
Q7. Draw the graph of the following equations and identify their slopes and y-intercepts.
(i) y = -3x + 4
Slope = -3
y-intercept = 4
Point where the line cuts the y-axis = (0, 4)
(ii) 2y = 4x + 7
Divide by 2:
y = 2x + 7/2
Slope = 2
y-intercept = 7/2
Point where the line cuts the y-axis = (0, 7/2)
(iii) 5y = 6x – 10
Divide by 5:
y = (6/5)x – 2
Slope = 6/5
y-intercept = -2
Point where the line cuts the y-axis = (0, -2)
(iv) 3y = 6x – 11
Divide by 3:
y = 2x – 11/3
Slope = 2
y-intercept = -11/3
Point where the line cuts the y-axis = (0, -11/3)
Are any of the lines parallel?
Lines are parallel if their slopes are equal.
Slope of equation (ii) = 2
Slope of equation (iv) = 2
Therefore, equations (ii) and (iv) are parallel.
Answer: Lines (ii) and (iv) are parallel.
Q8. The relation between Kelvin and Fahrenheit temperature is:
y = (9/5)(x – 273) + 32
where x is temperature in Kelvin and y is temperature in Fahrenheit.
(i) Find the temperature in Fahrenheit if the temperature is 313 K.
Substitute x = 313:
y = (9/5)(313 – 273) + 32
y = (9/5)(40) + 32
y = 72 + 32
y = 104
Answer: 104°F
(ii) If the temperature is 158°F, find the temperature in Kelvin.
Substitute y = 158:
158 = (9/5)(x – 273) + 32
158 – 32 = (9/5)(x – 273)
126 = (9/5)(x – 273)
Multiply both sides by 5/9:
70 = x – 273
x = 343
Answer: 343 K
Q9. The work done by a body on the application of a constant force is the product of force and distance travelled.
Work = Force × Distance
Let:
Work = w
Distance = d
Given force = 3 units
Therefore:
w = 3d
This is the required linear equation.
Table for graph
d = 0, w = 0
d = 1, w = 3
d = 2, w = 6
d = 3, w = 9
Plot the points:
(0, 0), (1, 3), (2, 6), (3, 9)
and join them to get a straight line.
Work done when distance travelled is 2 units
w = 3 × 2
w = 6
Answer: 6 units
Q10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).
(i) Find the polynomial p(x)
Let:
p(x) = ax + b
Since the graph passes through (1, 5):
a + b = 5 …..(1)
Since the graph also passes through (3, 11):
3a + b = 11 …..(2)
Subtract equation (1) from equation (2):
3a + b – (a + b) = 11 – 5
2a = 6
a = 3
Put a = 3 in equation (1):
3 + b = 5
b = 2
Therefore:
p(x) = 3x + 2
Answer: p(x) = 3x + 2
(ii) Find the coordinates where the graph of p(x) cuts the axes
x-axis
For x-axis:
y = 0
3x + 2 = 0
3x = -2
x = -2/3
Point on x-axis = (-2/3, 0)
y-axis
For y-axis:
x = 0
p(0) = 3(0) + 2
p(0) = 2
Point on y-axis = (0, 2)
Answer:
x-axis point = (-2/3, 0)
y-axis point = (0, 2)
(iii) Verification by graph
Draw the graph of y = 3x + 2 which passes through:
(1, 5)
(3, 11)
(0, 2)
and cuts the x-axis at (-2/3, 0).
Hence verified.
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Q11. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) p(0) = 5
(ii) p(x) – q(x) cuts the x-axis at (3, 0)
(iii) p(x) + q(x) = 6x + 4
Find p(x) and q(x).
From p(0) = 5:
b = 5
So:
p(x) = ax + 5
Given:
p(x) + q(x) = 6x + 4
(ax + 5) + (cx + d) = 6x + 4
(a + c)x + (5 + d) = 6x + 4
Comparing coefficients:
a + c = 6 …..(1)
5 + d = 4
d = -1
So:
q(x) = cx – 1
Now:
p(x) – q(x) = (ax + 5) – (cx – 1)
= (a – c)x + 6
Since it cuts the x-axis at (3, 0):
0 = (a – c)(3) + 6
3(a – c) = -6
a – c = -2 …..(2)
Now solve equations (1) and (2):
a + c = 6
a – c = -2
Add both equations:
2a = 4
a = 2
Put a = 2 in equation (1):
2 + c = 6
c = 4
Therefore:
p(x) = 2x + 5
q(x) = 4x – 1
Answer:
p(x) = 2x + 5
q(x) = 4x – 1
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Q12. Look at the first three stages of a growing pattern of hexagons made using matchsticks.
(i) Draw the next two stages. How many matchsticks will be required?
Stage 1 uses 6 matchsticks.
Each new hexagon shares one side, so every new stage adds 5 matchsticks.
Stage 2 = 11 matchsticks
Stage 3 = 16 matchsticks
Stage 4 = 21 matchsticks
Stage 5 = 26 matchsticks
Answer:
Stage 4 = 21 matchsticks
Stage 5 = 26 matchsticks
(ii) Complete the table
Stage Number:
1 2 3 4 5 n
Number of Matchsticks:
6 11 16 21 26 5n + 1
(iii) Find a rule for the nth stage
First stage uses 6 matchsticks.
Each next stage adds 5 matchsticks.
Therefore:
Number of matchsticks = 6 + (n – 1) × 5
= 6 + 5n – 5
= 5n + 1
Answer:
Rule = 5n + 1
(iv) How many matchsticks are required for the 15th stage?
Using the formula:
5n + 1
For n = 15:
5(15) + 1
= 75 + 1
= 76
Answer: 76 matchsticks
(v) Can 200 matchsticks form a stage in this pattern?
Using the formula:
5n + 1 = 200
5n = 199
n = 199/5
n = 39.8
Since n is not a whole number, 200 matchsticks cannot form a stage in this pattern.
Answer: No, 200 matchsticks cannot form a stage in this pattern.
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Q13. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) The graph of p(x) passes through (2, 3) and (6, 11)
(ii) The graph of q(x) passes through (4, -1)
(iii) The graph of q(x) is parallel to the graph of p(x)
Find p(x), q(x), and the coordinates where they meet the x-axis.
Finding p(x)
Slope of p(x):
m = (11 – 3)/(6 – 2)
m = 8/4
m = 2
So:
p(x) = 2x + b
Using point (2, 3):
3 = 2(2) + b
3 = 4 + b
b = -1
Therefore:
p(x) = 2x – 1
Finding q(x)
Since q(x) is parallel to p(x), its slope is also 2.
So:
q(x) = 2x + d
Using point (4, -1):
-1 = 2(4) + d
-1 = 8 + d
d = -9
Therefore:
q(x) = 2x – 9
Points where the lines meet the x-axis
For p(x)
2x – 1 = 0
2x = 1
x = 1/2
Point = (1/2, 0)
For q(x)
2x – 9 = 0
2x = 9
x = 9/2
Point = (9/2, 0)
Answer:
p(x) = 2x – 1
q(x) = 2x – 9
x-axis point of p(x) = (1/2, 0)
x-axis point of q(x) = (9/2, 0)
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Q14. What do all linear functions of the form f(x) = ax + a, where a > 0, have in common?
Given:
f(x) = ax + a
Take a common:
f(x) = a(x + 1)
For x-intercept:
f(x) = 0
a(x + 1) = 0
Since a > 0:
x + 1 = 0
x = -1
Therefore, all such functions cut the x-axis at the same point.
Answer:
All the graphs pass through the point (-1, 0).
Students can also check all other exercise solutions of the chapter “Introduction to Linear Polynomial” for complete preparation. Visit Exercise 2.1 Solutions, Exercise 2.2 Solutions, Exercise 2.3 Solutions, Exercise 2.4 Solutions, Exercise 2.5 Solutions, and Exercise 2.6 Solutions to understand every concept step by step and score better in exams.
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