Bhagya

Students can download Maths Chapter 3 Algebra Ex 3.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.3

Question 1.
Find the LCM and GCD for the following and verify that f(x) × g(x) = LCM × GCD

(i) 21x²y, 35xy²
Answer:
p(x) = 21 x²y = 3 × 7 × x² × y
g(x) = 35xy² = 5 × 7 × x × y²
G.C.D = 7 xy
L.C.M = 3 × 5 × 7 x² × y²
= 105 x²y²
L.C.M × G.C.D = 105x²y² × 7xy
= 735 x³y³ ….(1)
p(x) × g(x) = 21x²y × 35xy²
= 735x³y³ ….(2)
From (1) and (2) we get
L.C.M × G.C.D. = p(x) × g(x)

(ii) (x³ – 1)(x + 1),(x³ + 1)
Answer:
p(x) = (x³ – 1) (x + 1) = (x – 1) (x² + x + 1) (x + 1)
g(x) = x³ + 1 = (x + 1) (x² – x + 1)
G.C.D = (x + 1)
L.C.M = (x + 1) (x – 1) (x² + x + 1) (x² – x + 1)
L.C.M × G.C.D = (x + 1) (x – 1)(x² + x + 1)(x² – x + 1)x(x + 1)
= (x + 1)² (x – 1) (x² + x + 1) (x² – x + 1) ……….(1)
p(x) × g(x) = (x – 1) (x² + x + 1) (x + 1) (x + 1) (x² – x + 1)
= (x + 1)² (x – 1) (x² + x + 1) (x² – x + 1) ……….(2)
From (1) and (2) we get
L.C.M × G.C.D. = p(x) × g(x)

(iii) (x²y + xy²), (x² + xy)
Answer:
p(x) = x²y + xy² = xy(x + y)
g(x) = x² + xy = x(x + y)
G.C.D = x(x+y)
L.C.M = xy (x +y).
L.C.M × G.C.D = xy(x + y) × x(x + y)
= x²y(x + y)² …..(1)
p(x) × g(x) = xy(x + y) × x(x + y)
= x²y(x + y)²
From (1) and (2) we get
L.C.M × G.C.D. = p(x) × g(x)

Question 2.
Find the LCM of each pair of the following polynomials
(i) a² + 4a – 12, a² – 5a + 6 whose GCD is a – 2
Answer:
p(x) = a² + 4a – 12
= a² + 6a – 2a – 12
= a (a + 6) – 2(a + 6)
= (a + 6) (a – 2)

g(x) = a² – 5a + 6
= a² – 3a – 2a + 6

= a(a – 3) – 2 (a – 3)
= (a – 3) (a – 2)

(ii) x⁴ – 27a³x, (x – 3a)² whose GCD is (x – 3a)
Answer:
p(x) = x⁴ – 27a³x = x[x³ – 27a³]
= x[x³ – (3a)³]
= x(x – 3a) (x² + 3ax + 9a²)
g(x) = (x – 3a)²
G.C.D. = x – 3a

L.C.M. = x (x – 3a)² (x² + 3ax + 9a²)

Question 3.
Find the GCD of each pair of the following polynomials
(i) 12(x⁴ – x³), 8(x⁴ – 3x³ + 2x²) whose LCM is 24x³ (x – 1)(x – 2)
Answer:
p(x) = 12(x⁴ – x³)
= 12x³(x- 1)
g(x) = 8(x⁴ – 3x³ + 2x²)

= 8x²(x² – 3x + 2)
= 8x²(x – 2)(x – 1)
L.C.M. = 24x³ (x – 1) (x – 2)

(ii) (x³ + y³), (x⁴ + x²y² + y⁴) whose LCM is (x³ + y³) (x² + xy + y²)
Answer:
p(x) = x³ + y³
= (x + y)(x² – xy + y²)
g(x) = x⁴ + x²y² + y⁴ = [x² + y²]² – (xy)²
= (x² + y² + xy) (x² + y² – xy)
L.C.M. = (x³ + y³) (x² + xy + y²)
(x + y) (x² – xy + y²) (x² + xy + y²)

G.C.D. = x² – xy + y²

Question 4.
Given the L.C.M and G.C.D of the two polynomials p(x) and q(x) find the unknown polynomial in the following table

Answer:
L.C.M. = a³ – 10a² + 11a + 70
= (a – 7) (a² – 3a – 10)
= (a – 7) (a – 5) (a + 2)

G.C.D. = (a – 7)
p(x) = a² -12a + 35
= (a – 5)(a – 7)

q(x) = \(\frac{\mathrm{LCM} \times \mathrm{GCD}}{p(x)}\)

(ii) L.C.M (x² + y²)(x⁴ + x²y² + y⁴)
(x² + y²)[(x² + y²)²-(xy)²]
(x² + y²) (x² + y² + xy) (x² + y² – xy)
G.C.D. = x² – y²
(x + y)(x – y)
q(x) = (x⁴ – y⁴) (x² + y² – xy)
= [(x²)² – (y²)²](x² + y² – xy)
= (x² + y²) (x² – y²) (x² + y² – xy)
(x² + y²) (x + y) (x – y) (x² + y² – xy)
P(x) = x² + y² + xy

Students can download Maths Chapter 3 Algebra Ex 3.11 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.11

Question 1.
Solve, using the method of substitution.
(i) 2x – 3y – 7; 5x + y = 9
Solution:
2x – 3y = 7 → (1)
5x + y = 9 → (2)
Equation (2) becomes
y = 9 – 5x
Substitute the value of y in (1)
2x – 3 (9 – 5x) = 7
2x – 27 + 15x = 7
17x = 7 + 27
17x = 34
x = \( \frac{34}{17}\)
= 2
Substitute the value of x = 2 (in) (2)
y = 9 – 5 (2) = 9 – 10 = -1
∴ The value of x = 2 and y = -1

(ii) 1.5x + 0.1y = 6.2; 3x – 0.4y = 11.2
Solution:
1.5x + 0.1y = 6.2
Multiply by 10
15x + y = 62
y = 62 – 15x → (1)
3x – 0.4y = 11.2
Multiply by 10
30x – 4y = 112
divided by (2) we get
15x – 2y = 56 → (2)
Substitute the value of y in (2)
15x – 2(62 – 15x) = 56
15x – 124 + 30x = 56
45x = 56 + 124
45x = 180
x = \( \frac{180}{45}\)
= 4
Substitute the value of x = 4 in (1)
y = 62 – 15(4)
= 62 – 60
y = 2
∴ The value of x = 4 and y = 2

(iii) 10% of x + 20% of y = 24; 3x – y = 20
Solution:
\( \frac{10}{100}\) × x + \( \frac{20}{100}\) × y = 24
\( \frac{x}{10}\) + \( \frac{y}{5}\) = 24
Multiply by 10
x + 2y = 240 → (1)
3x – y = 20
– y = 20 – 3x
y = 3x – 20 → (2)
Substitute the value of y in (1)
x + 2 (3x – 20) = 240
x + 6x – 40 = 240
7x – 40 = 240
7x = 240 + 40
7x = 280
x = \( \frac{280}{7}\)
x = 40
Substitute the value of x = 40 in (2)
y = 3 (40) – 20 = 120 – 20
y = 100
∴ The value of x = 40 and y = 100

(iv) √2x – √3y = 1; √3x – √8y = 0
Solution:
√2x – √3y = 1
– √3y = 1 – √2x
√3y = √2x – 1
y = \(\frac{√2x-1}{√3}\) → (1)
√3x – √8y = 0 → (2)
Substitute the value of y in (2)
\(\sqrt{3x}-\frac{√8(√2x-1)}{√3}\)
multiply by √3
⇒ \(\frac{3x-√8(√2x-1)}{√3}\)
3x – √8(√2x – 1) = 0
3x – 4x + √8 = 0
-x = √8
Substitute the value of x in (1)

The value of x = √8 and y = √3

Question 2.
Raman’s age is three times the sum of the ages of his two sons. After 5 years his age will be twice the sum of the ages of his two sons. Find the age of Raman.
Solution:
Let Raman’s age be “x” years and the sum of the ages of two sons be “y” years.
By the given first condition
x = 3y
x – 3y = 0 → (1)
After 5 years Raman’s age is x + 5 years
Sum of sons age is (y + 10) years
(each son age increases by 5 years)
By the given second condition
x + 5 = 2 (y + 10)
x + 5 = 2y + 20
x – 2y = 20 – 5
x – 2y = 15 → (2)
Equation (1) becomes
x = 3y
Substitute the value of x in (2)
3y – 2y = 15
y = 15
Substitute the value of y = 15 in x = 3y
x = 3(15)
x = 45
∴ Raman’s age = 45 years

Question 3.
The middle digit of a number between 100 and 1000 is zero and the sum of the other digit is 13. If the digits are reversed, the number so formed exceeds the original number by 495. Find the number.
Solution:
Let the unit digit be and the 100 is digit be X. The number is XOY (100x + y)
By the given first condition
x + y = 13 ….(1)
If the digits are reversed the number is 100 y + x.
By the given second condition.
100y + x = 100x + y + 495
-99x + 99y = 495
-x + y = 5 …. (2)
x + y = 13 ….(1)
Add (1) and (2)
2y = 18
y = 9
Substitute the value of y = 9 in (1)
x + 9 = 13
x = 13 – 9
x = 4
∴ The number is 409

Students can download Maths Chapter 3 Algebra Ex 3.10 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.10

Question 1.
Draw the graph for the following:
(i) y = 2x
Solution:
When x = -2, y = 2 (-2) = -4
When x = 0, y = 2 (0) = 0
When x = 2, y = 2 (2) = 4
When x = 3, y = 2 (3) = 6

Plot the points (-2, -4) (0, 0) (2, 4) and (3, 6) in the graph sheet we get a straight line.

(ii) y = 4x – 1
Solution:
When x = – 1; y = 4 (-1) -1 ⇒ y = -5
When x = 0; y = 4 (0) – 1 = 0 – 1 ⇒ y = -1
When x = 2; y = 4 (2) -1 = 8 – 1 ⇒ y = l

Plot the points (-1, -5) (0, -1) and (2, 7) in the graph sheet we get a straight line. At the time of printing change the direction.

(iii) y = (\(\frac{3}{2}\))x + 3
Solution:
When x = -2;
y = \(\frac{3}{2}\)(-2) + 3
y = -3 + 3 = 0
when x = 0;
y = \(\frac{3}{2}\)(0) + 3
y = 3
when x = 2;
y = \(\frac{3}{2}\)(2) + 3
y = 3 + 3
= 6


Plot the points (-2, 0) (0, 3) and (2, 6) in the graph sheet we get a straight line.

(iv) 3x + 2y = 14
Solution:
y = \(\frac{-3x+14}{2}\)
y = – \(\frac{3}{2}\)x + 7
when x = -2
y = –\(\frac{3}{2}\)(-2) + 7 = 10
when x = 0
y = –\(\frac{3}{2}\)(0) + 7 = 7
when x = 2
y = –\(\frac{3}{2}\)(2) + 7 = 4


Plot the points (-2, 10) (0, 7) and (2, 4) in the graph sheet we get a straight line.

Question 2.
Solve graphically (i) x + y = 7, x – y = 3
Solution:
x + y = 7
y = 7 – x

Plot the points (-2, 9), (0, 7) and (3, 4) in the graph sheet
x – y = 3
-y = -x + 3
y = x – 3

Plot the points (-2, -5), (0, -3) and (4, 1) in the same graph sheet.

The point of intersection is (5, 2) of lines (1) and (2).
The solution set is (5,2).

(ii) 3x + 2y = 4; 9x + 6y – 12 = 0
Solution:
2y = -3x + 4
y = \(\frac{-3x+4}{2}\)
= \(\frac{-3}{2}\)x + 2

Plot the points (-2, 5), (0, 2) and (2, -1) in the graph sheet
9x + 6y= 12 (÷3)
3x + 2y = 4
2y = \(\frac{-3x+4}{2}\)
= \(\frac{-3}{2}\)x + 2

Plot the points (-2, 5), (0, 2) and (2, -1) the same graph sheet

Here both the equations are identical, but in different form.
Their solution is same.
This equations have an infinite number of solution.

(iii) \(\frac{x}{2}\) + \(\frac{y}{4}\) = 1: \(\frac{x}{2}\) + \(\frac{y}{4}\) = 2
Solution:
\(\frac{x}{2}\) + \(\frac{y}{4}\) = 1
multiply by 4
2x + y = 4
y = -2x + 4

Plot the points (-3, 10), (-1, 6), (0, 4) and (2, 0) in the graph sheet
\(\frac{x}{2}\) + \(\frac{y}{4}\) = 2
multiply by 4
2x + y = 8
y = -2x + 8

Plot the points (-2, 12), (-1, 10), (0, 8) and (2, 4) in the same graph sheet.

The given two lines are parallel.
∴ They do not intersect a point.
∴ There is no solution.

(iv) x – y = 0; y + 3 = 0
Solution:
x – y = 0
-y = -x
y = x

Plot the points (-2, -2), (0, 0), (1, 1) and (3, 3) in the same graph sheet.
y + 3 = 0
y = -3

Plot the points (-2, -3), (0, -3), (1, -3) and (2, -3) in the same graph sheet.

The two lines l₁ and l₂ intersect at (-3, -3). The solution set is (-3, -3).

(v) y = 2x + 1; 3x – 6 = 0
Solution:
y = 2x + 1

Plot the points (-3, -5), (-1, -1), (0, 1) and (2, 5) in the graph sheet
3x – 6 = 0
y = -3x + 6

Plot the points (-2, 12), (-1, 9), (0, 6) and (2, 0) in the same graph sheet

The two lines l₁ and l₂ intersect at (1, 3).
∴ The solution set is (1, 3).

(vi) x = -3; y = 3
Solution:
x = -3

Plot the points (-3, -3), (-3, -2), (-3, 2) and (-3, 3) in the graph sheet
y = 3

Plot the points (-3, 3), (-1, 3), (0, 3) and (2, 3) in the same graph sheet

The two lines l₁ and l₂ intersect at (-3, 3)
∴ The solution set is (-3, 3)

Question 3.
Two cars are 100 miles apart. If they drive towards each other they will meet in 1 hour. If they drive in the same direction they will meet in 2 hours. Find their speed by using graphical method.
Solution:
Let the speed of the two cars be “x” and “y”.
By the given first condition

x+ y = 100 → (1)
(They travel in opposite direction)
By the given second condition.
\(\)\frac{100}{x-y}= 2 [time taken in 2 hours in the same direction]
2x – 2y = 100
x – y = 50 → (2)
x + y = 100
y = 100 – x

Plot the points (30, 70), (50, 50), (60, 40) and (70,30) in the graph sheet
x – y = 50
-y = -x + 50
y = x – 50

Plot the points (40, -10), (50, 0), (60, 10) and (70, 20) in the same graph sheet

The two cars intersect at (75, 25)
The speed of the first car 75 km/hr
The speed of the second car 25 km/hr

Students can download Maths Chapter 3 Algebra Ex 3.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.2

Question 1.
Find the GCD of the given polynomials by Division Algorithm
(i) x⁴ + 3x³ – x – 3, x³ + x² – 5x + 3
Answer:
p(x) = x⁴ + 3x³ – x – 3
g(x) = x³ + x² – 5x + 3

3x² + 6x – 9 = 3(x² + 2x – 3)
Now dividing g(x) = x³ + x² – 5x + 3
by the new remainder
(leaving the constant 3)
we get x² + 2x – 3
G.C.F. = x² + 2x – 3

(ii) x⁴ – 1, x³ – 11x² + x – 11
p(x) = x⁴ – 1
g(x) = x³ – 11x² + x – 11

120x² + 120 = 120 (x² + 1)
Now dividing g(x) = x³ – 11x² + x – 11 by the new remainder (leaving the constant) we get x² + 1

G.C.D. = x² + 1

(iii) 3x⁴ + 6x³ – 12x² – 24x, 4x⁴ + 14x³ + 8x² – 8x
Answer:
p(x) = 3x⁴ + 6x³ – 12x² – 24x
= 3x (x³ + 2x² – 4x – 8)
g(x) = 4x⁴ + 14x³ + 8x² – 8x
= 2x (2x³ + 7x² + 4x – 4)
G.C.D. of 3x and 2x = x
Now g(x) is divide by p(x) we get

3x² + 12x + 12 = 3 (x² + 4x + 4)
Now dividing p(x) = x³ + 2x² – 4x – 8
by the new remainder
(leaving the constant)
x² + 4x + 4

G.C.D. = x(x² + 4x + 4) [Note x is common for p(x) and g(x)]

(iv) 3x³ + 3x² + 3x + 3, 6x³ + 12x² + 6x+12
p(x) = 3x³ + 3x² + 3x + 3
= 3(x³ + x² + x + 1)
g(x) = 6x³ + 12x² + 6x + 12
= 6(x³ + 2x² + x + 2)
G.C.D. of 3 and 6 = 3
Now g(x) is divided by p(x)

Now dividing p(x) by the remainder x² + 1
we get x + 1

∴ G.C.D. = 3(x² + 1) [3 is the G.C.D. of 3 and 6]

Question 2.
Find the LCM of the given polynomials
(i) 4x²y, 8x³y²
Answer:
4x² y = 2 × 2 × x² × y
8 x³ y² = 2 × 2 × 2 × x³ × y²
L.C.M. = 2³ × x³ × y²
= 8x³y²

Aliter: L.C.M of 4 and 8 = 8
L.C.M. of x²y and x³y² = x³y²
∴ L.C.M. = 8x³y²

(ii) -9a³b², 12a²b²c
Answer:
-9a³b² = -(3² × a³ × b²)
12a²b²c = 2² × 3 × a² × b² × c
L.C.M. = -(2² × 3² × a³ × b² × c)
= -36 a³b²c

(iii) 16m, -12m²n², 8n²
Answer:
16m = 24 × m
-12 m²n² = -(2² × 3 × m² × n²)
8n² = 2³ × n²
L.C.M. = -(2⁴ × 3 × m² × n²)
= -48 m²n²

(iv) p² – 3p + 2, p² – 4
Answer:
P² – 3p + 2 = p² – 2p – p + 2
= p(p – 2) – 1 (p – 2)
= (p – 2) (p – 1)

p² – 4 = p² – 2² (using a² – b² = (a + b) (a – b)]
= (p + 2) (p – 2)
L.C.M. = (p – 2) (p + 2) (p – 1)

(v) 2x² – 5x – 3,4.x² – 36
Answer:
2x² – 5x – 3 = 2x² – 6x + x – 3
= 2x (x – 3) + 1 (x – 3)
= (x – 3) (2x + 1)

= 4x² – 36 = 4 [x² – 9]
= 4 [x² – 3²]
= 4(x + 3) (x – 3)
L.C.M. = 4(x – 3) (x + 3) (2x + 1)

(vi) (2x² – 3xy)²,(4x – 6y)³,(8x³ – 27y³)
Answer:
(2x² – 3xy)² = x² (2x – 3y)²
(4x – 6y)³ = 2³ (2x – 3y)³
= 8 (2x – 3y)³
8x³ – 27y³ = (2x)³ – (3y)³
= (2x – 3y) [(2x)² + 2x × 3y + (3y²)]
[using a³ – b³ = (a – b) (a² + ab + b²)
(2x – 3y) (4x² + 6xy + 9y²)
L.C.M. = 8x² (2x – 3y)³ (4x² + 6xy + 9y)²

Students can download Maths Chapter 3 Algebra Ex 3.9 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Find the GCD for the following:
(i) P⁵, P¹¹, P³
Solution:
p⁵ = p⁵
p¹¹ = p¹¹
P⁹ = P⁹
G.C.D. is p⁵ (Highest common power is 5)

(ii) 4x³, y³, z³
Solution:
4x³ = 2 × 2 × x³
y³ = y³
z³ = z³
G.C.D. of 4x³, y³ and z³ = 1

(iii) 9a²b²c³, 15a³b²c⁴
Solution:
9a²b²c³ = 3 × 3 × a² × b² × c³
15a³b²c³ = 3 × 5 × a³ × b² × c⁴
G.C.D = 3 × a² × b² × c³
= 3a²b²c³

(iv) 64x⁸, 240x⁶
Solution:
64x⁸ = 2 × 2 × 2 × 2 × 2 × 2 × x⁸
= 2⁶ × x⁸

240x⁶ = 2⁴ × 3 × 5 × x⁶
G.C.D = 2⁴ × x⁶
= 16x⁶

(v) ab²c³, a²b³c, a³ac²
Solution:
ab²c³ = a × b² × c³
a²b³c = a² × b³ × c
a³bc² = a³ × b × c²
G.C.D. = abc

(vi) 35x⁵y³z⁴, 49x²yz³, 14xy²z²
Solution:
35x⁵y³z⁴ = 5 × 7 × x⁵ × y³ × z⁴
49x²yz³ = 7 × 7 × x² × z³
14xy²z² = 2 × 7 × x × y² × z²
G.C.D. = 7 × x × y × z²
= 7xyz²

(vii) 25ab³c, 100a²bc, 125 ab
Solution:
25ab³c = 5 × 5 × a × b³ × c
100a²be = 2 × 2 × 5 × 5 × a² × b × c
125ab = 5 × 5 × 5 × a × b
G.C.D. = 5 × 5 × a × b
= 25ab

(viii) 3abc, 5xyz, 7pqr
Solution:
3abc = 3 × a × b × c
5xyz = 5 × x × y × z
7pqr = 7 × p × q × r
G.C.D. = 1

Question 2.
Find the GCD for the following:
(i) (2x + 5), (5x + 2)
(ii) a^m+1, a^m+2, a^m+3
(iii) 2a² + a, 4a² – 1
(iv) 3a², 5b³, 7c⁴
(v) x⁴ – 1, x² – 1
(vi) a³ – 9ax², (a – 3x)²
Solution:
(i) (2x + 5) = 2x + 5
5x + 2 = 5x + 2
G.C.D. = 1

(ii) a^m+1 = a^m × a¹
a^m+2 = a^m × a²
a^m+3 = a^m × a³
G.C.D.= a^m × a
= a^m+1

(iii) 2a² + a = a(2a + 1)
4a² – 1 = (2a)2 – 1
(Using a² – b² = (a + b)(a – b)
= (2a + 1)(2a – 1)
G.C.D. = 2a + 1

(iv) 3a² = 3 × a²
5b³ = 5 × b³
7c⁴ = 7 × c⁴
G.C.D. = 1

(v) x⁴ – 1 = (x²)² – 1
= (x² + 1 ) (x² – 1)
= (x² + 1 ) (x + 1 ) (x – 1 )
x² – 1 = (x + 1 ) (x – 1 )
G.C.D. = (x + 1 ) (x – 1 )

(vi) a³ – 9ax² = a(a² – 9x²)
= a[a² – (3x)²]
= a(a + 3x)(a – 3x)
(a – 3x)² = (a – 3x)²
G.C.D. = a – 3x

Students can download Maths Chapter 3 Algebra Ex 3.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.1

Question 1.
Solve the following system of linear equations in three variables
(i) x + y + z = 5
2x – y + z = 9
x – 2y + 3z = 16
Answer:
x + y + z = 5 ….(1)
2x – y + z = 9 ….(2)
x – 2y + 3z = 16 ….(3)
by adding (1) and (2)

Substituting z = 4 (4)
3x + 2(-4) = 14
3x – 8 = 14
3x = 14 – 8
3x = 6
x = \(\frac { 6 }{ 3 } \) = 2

Substituting x = 2 and z = 4 in (1)
2 + y + 4 = 5
y + 6 = 5
y = 5 – 6
= -1
∴ The value of x = 2, y = -1 and z = 4

(ii) \(\frac { 1 }{ x } \) – \(\frac { 2 }{ y } \) + 4 = 0, \(\frac { 1 }{ y } \) – \(\frac { 1 }{ z } \) + 1 = 0, \(\frac { 2 }{ z } \) + \(\frac { 3 }{ x } \) = 14
Answer:
Let \(\frac { 1 }{ x } \) = p, \(\frac { 1 }{ y } \) = q and \(\frac { 1 }{ z } \) = r
p – 2q + 4 = 0
p – 2q = -4 ……(1)
q – r + 1 = 0
q – r = -1 ……(2)
3p + 2r = 14 ……(3)

Substituting the value of p = 2 in (1)
2 – 2q = -4
-2q = – 4 – 2
-2q = -6
q = \(\frac { 6 }{ 2 } \) = 3
Substituting the value of q = 3 in (2)
3 – r = 1
– r = – 1 – 3
r = 4

The value of x = \(\frac { 1 }{ 2 } \), y = \(\frac { 1 }{ 3 } \) and z = \(\frac { 1 }{ 4 } \)

(iii) x + 20 = \(\frac { 3y }{ 2 } \) + 10 = 2z + 5 = 110 – (y + z)
Answer:
x + 20 = \(\frac { 3y }{ 2 } \) + 10
Multiply by 2
2x + 40 = 3y + 20
2x – 3y = -40 + 20
2x – 3y = -20 ……(1)

\(\frac { 3y }{ 2 } \) + 10 = 2z + 5
Multiply by 2
3y + 20 = 4z + 10
3y – 4z = 10 – 20
3y – 4z = -10 ……(2)

2z + 5 = 110 – (y + z)
2z + 5 = 110 – y – z
y + 3z = 110 – 5
y + 3z = 105 ….(3)

Substitute the value of z = 25 in (2)
3y – 4(25) = -10
3y – 100 = – 10
3y = – 10 + 100
3y = 90
y = \(\frac { 90 }{ 3 } \) = 30
∴ The value of x = 35, y = 30 and z = 25

Substitute the value of y = 30 in (1)
2x – 3(30) = -20
2x – 90 = -20
2x = -20 + 90
2x = 70
x = \(\frac { 70 }{ 2 } \) = 35

Question 2.
Discuss the nature of solutions of the following system of equations
(i) x + 2y – z = 6, – 3x – 2y + 5z = -12 , x – 2z = 3
Answer:
x + 2y – z = 6 …..(1)
-3x – 2y + 5z = -12 …..(2)
x – 2z = 3 ……(3)
Adding (1) and (2) we get

Adding (3) and (4) we get

The above statement tells us that the system has an infinite number of solutions.

(ii) 2y + z = 3(- x + 1) ,-x + 3y – z = -4, 3x + 2y + z = –\(\frac { 1 }{ 2 } \)
2y + z = 3 (- x + 1)
2y + z = -3x + 3 ……(1)
3x + 2y + z = –\(\frac { 1 }{ 2 } \)

-x + 3y – z = – 4
x – 3y + z = 4 …..(2)

3x + 2y + z = – \(\frac { 1 }{ 2 } \)

Hence we arrive at a contradiction as 0 ≠ 7
This means that the system is inconsistent and has no solution.

(iii) \(\frac { y+z }{ 4 } \) = \(\frac { z+x }{ 3 } \) = \(\frac { x+y }{ 2 } \), x + y + z = 27
Answer:
\(\frac { y+z }{ 4 } \) = \(\frac { z+x }{ 3 } \)
3y + 3z = 4z + 4x
-4x + 3y + 3z – 4z = 0
-4x + 3y – z = 0
4x – 3y + z = 0 ………(1)

\(\frac { z+x }{ 3 } \) = \(\frac { x+y }{ 2 } \)
2z + 2x = 3x + 3y
-3x + 2x – 3y + 2z = 0
-x – 3y + 2z = 0
x + 3y – 2z = 0 ………(2)

Substituting the value of x in (5)
6 + 5z = 81
5z = 81 – 6
5z = 75
z = \(\frac { 75 }{ 5 } \) = 15
Substituting the value of x = 3
and z = 15 in (3)
3 + y + 15 = 27
y + 18 = 27
y = 27 – 18
= 9
The value of x = 3, y = 9 and z = 15
This system of equations have unique solution.

Question 3.
Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. If 4 years ago Vani’s grandfather was four times as old as Vani then how old are they all now?
Answer:
Let the age of Vani be”x” years
Vani father age = “y” years
Vani grand father = “z” years
By the given first condition.
\(\frac { x+y+z }{ 3 } \) = 53
x + y + z = 159 ….(1)
By the given 2^nd Condition.
\(\frac { 1 }{ 2 } \) z + \(\frac { 1 }{ 3 } \)y + \(\frac { 1 }{ 4 } \)x = 65
Multiply by 12
6z + 4y + 3x = 780
3x + 4y + 6z = 780 ….(2)
By the given 3^rd condition
z – 4 = 4 (x – 4) ⇒ z – 4 = 4x – 16
– 4x + z = – 16 + 4
4x – z = 12 ……(3)

Vani age = 24 years
Vani’s father age = 51 years
Vani grand father age = 84 years
Substitute the value of x = 24 in (3)
4 (24) – z = 12
96 – z = 12
-z = 12 – 96
z = 84
Substitute the value of
x = 24 and z = 84 in (1)
24 + y + 84 = 159
y + 108 = 159
y = 159 – 108
= 51

Question 4.
The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the old number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number ?
Answer:
Let the hundreds digit be x and the tens digit be ”y” and the unit digit be “z”
∴ The number is 100x + 10y + z
If the digits are reversed the new number is 100z + 10y + x
By the given first condition
x + y + z = 11 ….(1)
By the given second condition
100z + 10y + x = 5 (100x + 10y + z) + 46
= 500x + 50y + 5z + 46
x – 500x + 10y – 50y + 100z – 5z = 46
– 499x – 40y + 95z = 46
499x + 40y – 95z = -46 ….(2)
By the given third condition
x + 2y = z
x + 2y – z = 0 ….(3)

∴ The number is 137
Subtituting the value of y = 3 in (5)
2x + 3(3) = 11
2x = 11 – 9
2x = 2
x = \(\frac { 2 }{ 2 } \) = 1
Subtituting the value of x = 1, y = 3 in (1)
1 + 3 + z = 11
z = 11 – 4
= 7

Question 5.
There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. But when first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort.
Answer:
Let the number of ₹5 currencies be “x”
Let the number of ₹10 currencies be “y”
and the number of ₹20 currencies be “z”
By the given first condition
x + y + z = 12 ………(1)
By the given second condition
5x + 10y + 20z = 105
x + 2y + 4z = 21 (÷5) ……….(2)
By the given third condition
10x + 5y + 20z = 105 + 20
10x + 5y + 20z = 125
2x + y + 4z = 25 ………..(3)


Substituting the value of x = 7 in (5)
7 – y = 4 ⇒ -y = 4 – 7
-y = -3 ⇒ y = 3
Substituting the value of x = 7, y = 3 in …. (1)
7 + 3 + z = 12
z = 12 – 10 = 2
x = 7, y = 3, z = 2
Number of currencies in ₹ 5 = 7
Number of currencies in ₹ 10 = 3
Number of currencies in ₹ 20 = 2

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Factorise each of the following polynomials using synthetic division:
(i) x³ – 3x² – 10x + 24
Solution:
p(x) – x³ – 3x² – 10x + 24
p(1) = 1³ – 3(1)² – 10(1) + 24
= 1 – 3 – 10 + 24
= 25 – 13
≠ 0
x – 1 is not a factor

p(-1) = (-1)³ – 3(-1)² – 10(-1) + 24
= – 1 – 3(1) + 10 + 24
= -1 – 3 + 10 + 24
= 34 – 4
= 30
≠ 0
x + 1 is not a factor

p(2) = 2³ – 3(2)² – 10(2) + 24
= 8 – 3(4) – 20 + 24
= 8 – 12 – 20 + 24
= 32 – 32
= 0
∴ x – 2 is a factor

x² – x – 12 = x² – 4x + 3x – 12

= x(x – 4) + 3 (x – 4)
= (x – 4) (x + 3)
∴ The factors of x³ – 3x² – 10x + 24 = (x – 2) (x – 4) (x + 3)

(ii) 2x³ – 3x² – 3x + 2
Solution:
p(x) = 2x³ – 3x² – 3x + 2
P(1) = 2(1)³ – 3(1)² – 3(1) + 2
= 2 – 3 – 3 + 2
= 2 – 6
= -4
≠ 0
x – 1 is not a factor

P(-1) = 2(-1)³ – 3(-1)² – 3(-1) + 2
= -2 – 3 + 3 + 2
= 5 – 5
= 0
∴ x + 1 is a factor

2x² – 5x + 2 = 2x² – 4x – x + 2

= 2x(x – 2) – 1 (x – 2)
= (x – 2) (2x – 1)
∴ The factors of 2x³ – 3x² – 3x + 2 = (x + 1) (x – 2) (2x – 1)

(iii) – 7x + 3 + 4x³
Solution:
p(x) = – 7x + 3 + 4x³
= 4x³ – 7x + 3
P(1) = 4(1)³ – 7(1) + 3
4 – 7 + 3
= 7 – 7
= 0
∴ x – 1 is a factor

4x² + 4x – 3 = 4x² + 6x – 2x – 3

= 2x(2x + 3) – 1 (2x + 3)
= (2x + 3) (2x – 1)
∴ The factors of – 7x + 3 + 4x³ = (x – 1) (2x + 3) (2x – 1)

(iv) x³ + x² – 14x – 24
Solution:
p(x) = x³ + x² – 14x – 24
p(1) = (1)³ + (1)² – 14 (1) – 24
= 1 + 1 – 14 – 24
= -36
≠ 0
x + 1 is not a factor.

p(-1) = (-1)³ + (-1)² – 14(-1) – 24
= -1 + 1 + 14 – 24
= 15 – 25
≠ 0
x – 1 is not a factor.

p(2) = (-2)³ + (-2)² – 14 (-2) – 24
= -8 + 4 + 28 – 24
= 32 – 32
= 0
∴ x + 2 is a factor

x² – x – 12 = x² – 4x + 3x – 12

= x(x – 4) + 3 (x – 4)
= (x – 4) (x + 3)
This (x + 2) (x + 3) (x – 4) are the factors.
x³ + x² – 14x – 24 = (x + 2) (x + 3) (x – 4)

(v) x³ – 7x + 6
Solution:
p(x) = x³ – 7x + 6
P( 1) = 1³ – 7(1) + 6
= 1 – 7 + 6
= 7 – 7
= 0
∴ x – 1 is a factor

x² + x – 6 = x² + 3x – 2x – 6

= x(x + 3) – 2 (x + 3)
= (x + 3) (x – 2)
This (x – 1) (x – 2) (x + 3) are factors.
∴ x³ – 7x + 6 = (x – 1) (x – 2) (x + 3)

(vi) x³ – 10x² – x + 10
p(x) = x³ – 10x² – x + 10
= 1 – 10 – 1 + 10
= 11 – 11
= 0
∴ x – 1 is a factor

x² – 9x – 10 = x² – 10x + x – 10

= x(x – 10) + 1 (x – 10)
= (x – 10) (x + 1)
This (x – 1) (x + 1) (x – 10) are the factors.
∴ x³ – 10x² – x + 10 = (x – 1) (x – 10) (x + 1)

Students can download Maths Chapter 3 Algebra Ex 3.13 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.13

Question 1.
Solve by cross-multiplication method.
(i) 8x – 3y = 12; 5x = 2y + 7
Solution:
8x – 3y – 12 = 0 → (1)
5x – 2y – 7 = 0 → (2)
Use the coefficients for cross multiplication

\(\frac{x}{-3}\) = -1 ⇒ x = 3
\(\frac{y}{-4}\) = -1 ⇒ y = 4
∴ The value of x = 3 and y = 4

(ii) 6x + 7y – 11 = 0; 5x + 2y = 13
Solution:
6x + 7y – 11 = 0 → (1)
5x + 2y = 13 → (2)
Use the coefficient for cross multiplication

-23x = -69
∴ 23x = 69
x= \(\frac{69}{23}\)
= 3
\(\frac{y}{23}\) = \(\frac{1}{-23}\)
-23y = 23
23y = -23
y = –\(\frac{23}{23}\)
y = -1
∴ The value of x = 3 and y = -1

(iii) \(\frac{2}{x}\) + \(\frac{3}{y}\) =5; \(\frac{3}{x}\) – \(\frac{1}{y}\) + 9 = 0
Solution:
\(\frac{1}{x}\) = a; \(\frac{1}{y}\) = b
2a + 3b – 5 = 0 → (1)
3a – b + 9 = 0 → (2)
Using the coefficients for cross multiplication

-11b = -33
11b = 33
b = \(\frac{33}{11}\) = 3
But \(\frac{1}{x}\) = a ⇒ \(\frac{1}{x}\) = -2
-2x = 1 ⇒ 2x = -1
x = –\(\frac{1}{2}\)
but \(\frac{1}{y}\) = b
\(\frac{1}{y}\) = 3 ⇒ 3y = 1
y = \(\frac{1}{3}\)
∴ The value of x = –\(\frac{1}{2}\) and y = \(\frac{1}{3}\)

Question 2.
Akshaya has 2 rupee coins and 5 rupee coins in her purse. If in all she has 80 coins totalling Rs 220, how many coins of each kind does she have.
Solution:
Let the number of 2 rupee coins be “x” and the number of 5 rupee coins be “y”.
By the given first condition
x + y = 80 → (1)
Again by the given second condition
2x + 5y = 220 → (2)
x + y – 80 = 0 → (3)
2x + 5y – 220 = 0 → (4)
Using the coefficients for cross multiplication

\(\frac{x}{180}\) = \(\frac{1}{3}\)
3x = 180
x = \(\frac{180}{3}\)
= 60
But \(\frac{y}{60}\) = \(\frac{1}{3}\)
3y = 60
y = \(\frac{6}{30}\)
= 20
Number of 2 rupee coins = 60
Number of 5 rupee coins = 20

Question 3.
It takes 24 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 8 hours and the pipe of the smaller diameter is used for 18 hours. Only half of the pool is filled. How long would each pipe take to fill the swimming pool.
Solution:
Let the time taken by the larger diameter pipe be “x” hours and the time taken by the smaller diameter pipe be “y” hours.
By the given first condition
\(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{24}\) → (1)
Also
In 8 hours the large pipe fill \(\frac{8}{x}\)
In 18 hours the smaller pipe fill \(\frac{18}{y}\)
By the given second condition ( \(\frac{1}{2}\) of the tank)
\(\frac{8}{x}\) + \(\frac{18}{y}\) = \(\frac{1}{2}\) → (2)
Solve (1) and (2) we get
Let \(\frac{1}{x}\) = a; \(\frac{1}{y}\) = b
a + b = \(\frac{1}{24}\)
Multiply by 24
24a + 24b = 1
24a + 24b – 1 = 0 → (3)
8a + 18b = \(\frac{1}{2}\)
Multiply by 2
16a + 36b = 1
16a + 36b – 1 = 0 → (4)

x = 40
\(\frac{1}{y}\) = b ⇒ \(\frac{1}{y}\) = \(\frac{1}{60}\)
y = 60
To fill the remaining half of the pool.
Time taken by larger pipe = \(\frac{1}{2}\) × 40 = 20 hours
Time taken by smaller pipe = \(\frac{1}{2}\) × 60 = 30 hours

Students can download Maths Chapter 2 Real Numbers Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Additional Questions

I. Multiple choice question

Question 1.
The decimal form of –\(\frac{3}{4}\) is ………
(a) – 0.75
(b) – 0.50
(c) – 0.25
(d) – 0.125
Solution:
(a) – 0.75

Question 2.
If a number has a non-terminating and non-recurring decimal expansion, then it is……….
(a) a rational number
(b) a natural number
(c) an irrational number
(d) an integer
Solution:
(c) an irrational number

Question 3.
Which one of the following has terminating decimal expansion?
(a) \(\frac{7}{9}\)
(b) \(\frac{8}{15}\)
(c) \(\frac{1}{12}\)
(d) \(\frac{5}{32}\)
Solution:
(d) \(\frac{5}{32}\)

Question 4.
Which of the following are irrational numbers?
(i) \(\sqrt{2+\sqrt3}\)
(ii) \(\sqrt{4+\sqrt25}\)
(iii) \(\sqrt[3]{5+\sqrt7}\)
(iv) \(\sqrt{8-\sqrt[3]8}\)
(a) (ii), (iii) and (iv)
(b) (i), (iii) and (iv)
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(d) (i), (iii) and (iv)

Question 5.
Irrational number has a
(a) terminating decimal
(b) no decimal part
(c) non-terminating and recurring decimal
(d) non-terminating and non-recurring decimal
Solution:
(d) non-terminating and non-recurring decimal

Question 6.
If \(\frac{1}{7}\) = 0.142857, then the value of \(\frac{3}{7}\) is……..
(a) 0.285741
(b) 0.428571
(c) 0.285714
(d) 0.574128
Solution:
(b) 0.428571

Question 7.
Which of the following are not rational numbers?
(a) 7√5
(b) \(\frac{7}{\sqrt{5}}\)
(c) \(\sqrt{36}\) – 9
(d) π + 2
Solution:
(c) \(\sqrt{36}\) – 9

Question 8.
The product of 2√5 and 6√5 is……….
(a) 12√5
(b) 60
(c) 40
(d) 8√5
Solution:
(b) 60

Question 9.
The rational number lying between \(\frac{1}{5}\) and \(\frac{1}{2}\)
(a) \(\frac{7}{20}\)
(b) \(\frac{2}{10}\)
(c) \(\frac{2}{7}\)
(d) \(\frac{3}{10}\)
Solution:
(a) \(\frac{7}{20}\)

Question 10.
The value of 0.03 + 0.03 is ……….
(a) 0.\(\overline { 09 }\)
(b) 0.\(\overline { 0303 }\)
(c) 0.\(\overline { 06 }\)
(d) 0
Solution:
(c) 0.06

Question 11.
The sum of \(\sqrt{343}\) + \(\sqrt{567}\) is
(a) 18√3
(b) 16√7
(c) 15√3
(d) 14√7
Solution:
(b) 16√7

Question 12.
If \(\sqrt{363}\) = x√3 then x = ………
(a) 8
(b) 9
(c) 10
(d) 11
Solution:
(d) 11

Question 13.
The rationalising factor of \(\frac{1}{\sqrt{7}}\) is ……….
(i) 7
(b) √7
(c) \(\frac{1}{7}\)
(d) \(\frac{1}{\sqrt{7}}\)
Solution:
(b) √7

Question 14.
The value of \((\frac{1}{3^5})^4\) is ……..
(a) 3²⁰
(b) 3^-20
(c) \(\frac{1}{3^{-20}}\)
(d) \(\frac{1}{3^{9}}\)
Solution:
(b) 3^-20

Question 15.
What is 3.976 × 10^-4 written in decimal form?
(a) 0.003976
(b) 0.0003976
(c) 39760
(d) 0.03976
Solution:
(b) 0.0003976

II. Answer the following Questions.

Question 1.
Find any seven rational numbers between \(\frac{5}{8}\) and –\(\frac{5}{6}\)
Solution:
Let us convert the given rational numbers having the same denominators.
L.C.M of 8 and 6 is 24.

Now the rational numbers between

We can take any seven of them.

Question 2.
Find any three rational numbers between \(\frac{1}{2}\) and \(\frac{1}{5}\)
Solution:

Thus the three rational numbers are \(\frac{7}{20}\), \(\frac{17}{40}\) and \(\frac{37}{80}\)

Question 3.
Represent \(-\frac{2}{11}\), \(-\frac{5}{11}\) and \(-\frac{9}{11}\) on the number lines.
Solution:

To Represent \(-\frac{2}{11}\), \(-\frac{5}{11}\) and \(-\frac{9}{11}\) on the number line we make 11 markings each being equal distence \(\frac{1}{11}\) on the left of 0.
The point A represent \((-\frac{2}{11})\), the point B represents \((-\frac{5}{11})\) and the point C represents \((-\frac{9}{11})\)

Question 4.
Express the following in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0.
(i) 0.\(\overline { 47 }\)
Solution:
Let x = 0.474747…….. →(1)
100 x = 47.4747…….. →(2)
(2) – (1) ⇒ 100x – x = 47.4747……..
(-) 0.4747……..
99 x = 47.0000
x = \(\frac{47}{99}\)
∴ 0.\(\overline { 47 }\) = \(\frac{47}{99}\)

(ii) 0.\(\overline { 57 }\)
Solution:
Let x = 0.57777…….. →(1)
10 x = 5.77777…….. →(2)
100 x = 57.7777…….. →(3)
(3) – (2) ⇒ 100 x – 10 x = 57.7777……..
(-) 5.7777……..
99 x = 52.0000
x = \(\frac{52}{90}\) = \(\frac{26}{45}\)
∴ 0.\(\overline { 57 }\) = \(\frac{26}{45}\)

(iii) 0.\(\overline { 245 }\)
Solution:
Let x = 0.2454545…….. →(1)
10 x = 2.454545…….. →(2)
1000 x = 245.4545…….. →(3)
(3) – (2) ⇒ 1000 x – 10 x = 245.4545
(-) 2.4545………
990 x = 243.00000
x = \(\frac{243}{990}\) (or) \(\frac{27}{110}\)
∴ 0.\(\overline { 245 }\) = \(\frac{27}{110}\)

Question 5.
Without actual division classify the decimal expansion of the following numbers as terminating or non-terminating and recurring.
(i) \(\frac{7}{16}\)
(ii) \(\frac{13}{150}\)
(ii) –\(\frac{11}{75}\)
(iv) \(\frac{17}{200}\)
Solution:
(i) \(\frac{7}{16}\) = \(\frac{7}{2^4}\) = \(\frac{7}{2^{4} \times 5^{0}}\)
∴ \(\frac{7}{16}\) has a terminating decimal expansion.

(ii) \(\frac{13}{150}=\frac{13}{2 \times 3 \times 5^{2}}\)
Since it is not in the form of \(\frac{P}{2^{m} \times 5^{n}}\)
∴ \(\frac{13}{150}\) as non-terminating and recurring decimal expansion.

(iii) \(-\frac{11}{75}=-\frac{11}{3 \times 5^{2}}\)
Since it is not in the form of \(\frac{P}{2^{m} \times 5^{n}}\)
∴ –\(\frac{11}{75}\) as non-terminating and recurring decimal expansion.

(iv) \(\frac{17}{200}=\frac{17}{2^{3} \times 5^{2}}\)
∴ \(\frac{17}{200}\) has a terminating decimal expansion.

Question 6.
Find the value of \(\sqrt{27}\) + \(\sqrt{75}\) – \(\sqrt{108}\) + \(\sqrt{48}\)
Solution:

= 3√3 + 5√3 – 6√3 + 4√3
= 12√3 – 6√3
= 6√3
= 6 × 1.732
= 10.392

Question 7.
Evaluate \(\frac{\sqrt{2}+1}{\sqrt{2-1}}\)
Solution:

= 2√2 + 3
= 2 × 1.414 + 3
= 2.828 + 3
= 5.828

Question 8.

Solution:

= 69984 × 1021^-21-20+9
= 69984 × 10^-32
= 6.9984 × 10⁴ × 10^-32
= 6.9984 × 10^-32+4
= 6.9984 × 10^-28

Question 9.
Write
(a) 9.87 × 10⁹
(b) 4.134 × 10^-4 and
(c) 1.432 × 10^-9 in decimal form.
Solution:
(a) 9.87 × 10⁹ = 9870000000
(b) 4.134 × 10^-4 = 0.0004134
(c) 1.432 × 10^-9 = 0.000000001432

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.6 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.6

Question 1.
Find the sum of the following
(i) 3, 7, 11,… up to 40 terms.
Answer:
3,7,11,… up to 40 terms
First term (a) = 3
Common difference (d) = 7 – 3 = 4
Number of terms (n) = 40
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1) d]
S₄₀ = \(\frac { 40 }{ 2 } \) [6 + 39 × 4] = 20 [6 + 156]
= 20 × 162
S₄₀ = 3240

(ii) 102,97, 92,… up to 27 terms.
Answer:
Here a = 102, d = 97 – 102 = -5
n = 27
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S₂₇ = \(\frac { 27 }{ 2 } \) [2(102) + 26(-5)]
= \(\frac { 27 }{ 2 } \) [204 – 130]
= \(\frac { 27 }{ 2 } \) × 74
= 27 × 37 = 999
S₂₇ = 999

(iii) 6 + 13 + 20 + …. + 97
Answer:
Here a = 6, d = 13 – 6 = 7, l = 97
n = \(\frac { l-a }{ d } \) + 1
= \(\frac { 97-6 }{ 7 } \) + 1
= \(\frac { 91 }{ 7 } \) + 1 =
13 + 1 = 14
S_n = \(\frac { n }{ 2 } \) (a + l)
S_n = \(\frac { 14 }{ 2 } \) (a + l)
S_n = \(\frac { 14 }{ 2 } \) (6 + 97)
= 7 × 103
S_n = 721

Question 2.
How many consecutive odd integers beginning with 5 will sum to 480?
Answer:
First term (a) = 5
Common difference (d) = 2
(consecutive odd integer)
S_n = 480
\(\frac { n }{ 2 } \) [2a + (n-1) d] = 480
\(\frac { n }{ 2 } \) [10 + (n-1)2] = 480

n + 24 = 0 or n – 20 = 0
n = -24 or n = 20
[number of terms cannot be negative]
∴ Number of consecutive odd integers is 20

Question 3.
Find the sum of first 28 terms of an A.P. whose n^th term is 4n – 3.
Solution:
n = 28
t_n = 4n – 3
t₁ = 4 × 1 – 3 = 1
t₂ = 4 × 2 – 3 = 5
t₂₈ = 4 × 28 – 3
= 112 – 3 = 109
∴ a = 1, d = t₂ – t₁ = 5 – 1 = 4
l = 109.
S_n = \(\frac{n}{2}\) (2a+(n – 1)d)
S₂₈ = \(\frac{28}{2}\) (2 × 1 + 27 × 4)
= 14(2 + 108)
= 14 × 110
= 1540

Question 4.
The sum of first n terms of a certain series is given as 2n² – 3n . Show that the series is an A.P.
Answer:
Let tn be nth term of an A.P.
t_n = Sn – S_n-1
= 2n² – 3n – [2(n – 1)² – 3(n – 1)]
= 2n² – 3n – [2(n² – 2n + 1) – 3n + 3]
= 2n² – 3n – [2n² – 4n + 2 – 3n + 3]
= 2n² – 3n – [2n² – 7n + 5]
= 2n² – 3n – 2n² + 7n – 5
t_n = 4n – 5
t₁ = 4(1) – 5 = 4 – 5 = -1
t₂ = 4(2) -5 = 8 – 5 = 3
t₃ = 4(3) – 5 = 12 – 5 = 7
t₄ = 4(4) – 5 = 16 – 5 = 11
The A.P. is -1, 3, 7, 11,….
The common difference is 4
∴ The series is an A.P.

Question 5.
The 104th term and 4^th term of an A.P are 125 and 0. Find the sum of first 35 terms?
Answer:
104th term of an A.P = 125
t₁₀₄ = 125
[t_n = a + (n – 1) d]
a + 103d = 125 …..(1)
4^th term = 0
t₄ = 0
a + 3d = 0 …..(2)


Sum of 35 terms = 612.5

Question 6.
Find the sum of ail odd positive integers less than 450.
Answer:
Sum of odd positive integer less than 450
1 + 3 + 5 + …. 449
Here a = 1, d = 3 – 1 = 2,l = 449

Aliter: Sum of all the positive odd intergers
= n²
= 225 × 225
= 50625
Sum of the odd integers less than 450
= 50625

Question 7.
Find the sum of all natural numbers between 602 and 902 which are not divisible by 4?
Answer:
First find the sum of all the natural’s number between 602 and 902
Here a = 603, d = 1, l = 901

Find the sum of all the numbers between 602 and 902 which are divisible by 4.
Here a = 604; l = 900; d = 4

Sum of the numbers which are not divisible
by 4 = S_n1 – S_n2
= 224848 – 56400
= 168448
Sum of the numbers = 168448

Question 8.
Raghu wish to buy a laptop. He can buy it by paying ₹40,000 cash or by giving it in 10 installments as ₹4800 in the first month, ₹4750 in the second month, ₹4700 in the third month and so on. If he pays the money in this fashion, find
(i) total amount paid in 10 installments.
(ii) how much extra amount that he has to pay than the cost?
Solution:
4800 + 4750 + 4700 + … 10 terms
Here a = 4800
(i) d = t₂ – t₁ = 4750 – 4800 = -50
n = 10
S_n = \(\frac{n}{2}\) (2a + (n – 1)d)
S₁₀ = \(\frac{10}{2}\)  (2 × 4800 + 9 × -50)
= 5 (9600 – 450)
= 5 × 9150 = 45750
Total amount paid in 10 installments = ₹ 45750.
(ii) The extra amount he pays in installments
= ₹ 45750 – ₹ 40,000
= ₹ 5750

Question 9.
A man repays a loan of ₹ 65,000 by paying ₹ 400 in the first month and then increasing the payment by ₹ 300 every month. How long will it take for him to clear the loan?
Answer:
(i) Total loan amount = ₹ 65,000
S_n = 65,000
First month payment (a) = 400
Every month increasing ₹ 300
d = 300
S_n = \(\frac { n }{ 2 } \) [2a + (n-1)d]
65000 = \(\frac { n }{ 2 } \) [2(400) + (n – 1)300]
130000 = n [800 + 300n – 300]
= n [500 + 300n]
13000 = 500n + 300n²
Dividing by (100)

Number of installments will not be negative
∴ Time taken to pay the loan is 20 months.

Question 10.
A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step.
(i) How many bricks are required for the top most step?
(ii) How many bricks are required to build the stair case?
Solution:
100 + 98 + 96 + 94 + … 30 steps.
Here
a = 100
d = -2
n = 30
∴ S_n = \(\frac{n}{2}\)  (2a + (n – 1)d)
S₃₀ = \(\frac{30}{2}\)  (2 × 100 + 29 × -2)
= 15(200 – 58)
= 15 × 142
= 2130
t₃₀ = a + (n – 1)d
= 100 + 29 × -2
= 100 – 58
= 42
(i) No. of bricks required for the top step are 42.
(ii) No. of bricks required to build the stair case are 2130.

Question 11.
If S₁, S₂ , S₃, ….S_m are the sums of n terms of m A.P.,s whose first terms are 1,2, 3…… m and whose common differences are 1,3,5,…. (2m – 1) respectively, then show that (S₁ + S₂ + S₃ + ……. + S_m) = \(\frac { 1 }{ 2 } \) mn(mn + 1)
Answer:
First terms of an A.P are 1, 2, 3,…. m
The common difference are 1, 3, 5,…. (2m – 1)


By adding (1) (2) (3) we get
S₁ + S₂ + S₃ + …… + S_m = \(\frac { n }{ 2 } \) (n + 1) + \(\frac { n }{ 2 } \) (3n + 1) + \(\frac { n }{ 2 } \) (5n + 1) + ….. + \(\frac { n }{ 2 } \) [n(2m – 1 + 1)]
= \(\frac { n }{ 2 } \) [n + 1 + 3n + 1 + 5n + 1 ……. + n (2m – 1) + m)]
= \(\frac { n }{ 2 } \) [n + 3n + 5n + ……. n(2m – 1) + m]
= \(\frac { n }{ 2 } \) [n (1 + 3 + 5 + ……(2m – 1)) + m
= \(\frac { n }{ 2 } \) [n(\(\frac { m }{ 2 } \)) (2m) + m]
= \(\frac { n }{ 2 } \) [nm² + m]
S₁ + S₂ + S₃ + ……….. + S_m = \(\frac { mn }{ 2 } \) [mn + 1]
Hint:
1 + 3 + 5 + ……. + 2m – 1
S_n = \(\frac { n }{ 2 } \) (a + l)
= \(\frac { m }{ 2 } \) (1 + 2m -1)
= \(\frac { m }{ 2 } \) (2m)

Question 12.
Find the sum

Answer: