Bhagya

Students can download Maths Chapter 3 Algebra Ex 3.7 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 3.7

Question 1.
Find the square root of the following.
(i) \(\frac{400 x^{4} y^{12} z^{16}}{100 x^{8} y^{4} z^{4}}\)
Answer:

(ii) \(\frac{7 x^{2}+2 \sqrt{14} x+2}{x^{2}-\frac{1}{2} x+\frac{1}{16}}\)
Answer:

(iii) \(\frac{121(a+b)^{8}(x+y)^{8}(b-c)^{8}}{81(b-c)^{4}(a-b)^{12}(b-c)^{4}}\)
Answer:

Question 2.
Find the square root of the following
(i) 4x² + 20x + 25
Answer:

(ii) 9x² – 24xy + 30xz – 40yz + 25z² + 16y²
Answer:

(iii) \(1+\frac{1}{x^{6}}+\frac{2}{x^{3}}\)
Answer:

(iv) (4x² – 9x + 2)(7x² – 13x – 2)(28x² – 3x – 1)
Answer:
4x² – 9x +2 = 4x² – 8x – x + 2
= 4x(x – 2)-1 (x – 2)
= (x – 2)(4x – 1)

7x² – 13x – 2 = 7x² – 14x + x – 2
= 7x (x – 2) + 1 (x – 2)
= (x – 2) (7x + 1)

28x² – 3x – 1 = 28x² – 7x + 4x – 1
= 7x (4x – 1) + 1 (4x – 1)
= (4x – 1) (7x + 1)

(v) (2x² + \(\frac { 17 }{ 6 } \)x + 1) (\(\frac { 3 }{ 2 } \) x² + 4x + 2) (\(\frac { 4 }{ 3 } \) x² + \(\frac { 11 }{ 3 } \) x + 2)
Answer:

Students can download Maths Chapter 3 Algebra Ex 3.18 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 3.18

Question 1.
Find the order of the product matrix AB if

Answer:
Given A = [a_ij]_p×q and B = [a_ij]_q×r
Order of product of AB = p × r
Order of product of BA is not defined. Number columns in r is not equal to the number of rows in P.
∴ Product BA is not defined.

Question 2.
A has ‘a’ rows and ‘a + 3 ’ columns. B has ‘6’ rows and ‘17 – b’ columns, and if both products AB and BA exist, find a, b?
Solution:
A has a rows, a + 3 columns.
B has b rows, 17 – b columns
If AB exists a × a + 3
b × 17 – b
a + 3 = 6 ⇒ a – 6 = -3 ………… (1)
If BA exists 6 × 17-6
a × a + 3
17 – 6 = a ⇒ a + 6 = 17 …………. (2)
(1) + (2) ⇒ 2a = 14 ⇒ a = 7
Substitute a = 7 in (1) ⇒ 7 – b = -3 ⇒ b = 10
a = 7, b = 10

Question 3.
A has ‘a’ rows and ‘a + 3 ’ columns. B has rows and ‘b’ columns, and if both products AB and BA exist, find a,b?
Answer:

1. Order of matrix AB = 3 × 3
2. Order of matrix AB = 4 × 2
3. Order of matrix AB = 4 × 2
4. Order of matrix AB = 4 × 1
5. Order of matrix AB = 1 × 3

Question 4.

find AB, BA and check if AB = BA?
Answer:

Question 5.
Given that


verify that A(B + C) = AB + AC
Answer:

From (1) and (2) we get
A (B + C) = AB + AC

Question 6.
Show that the matrices

satisfy commutative property AB = BA
Answer:

From (1) and (2) we get
AB = BA. It satisfy the commutative property.

Question 7.

Show that (i) A(BC) = (AB)C
(ii) (A-B)C = AC – BC
(iii) (A-B)^T = A^T – B^T
Answer:



From (1) and (2) we get
A(BC) = (AB)C

From (1) and (2) we get
(A – B) C = AC – BC


From (1) and (2) we get
(A-B)^T = A^T – B^T

Question 8.

then snow that A² + B² = I.
Answer:

Question 9.

prove that AA^T = I.
Answer:

AA^T = I
∴ L.H.S. = R.H.S.

Question 10.
Verify that A² = I when

Answer:

∴ L.H.S. = R.H.S.

Question 11.

show that A² – (a + d)A = (bc – ad)I₂.
Answer:

L.H.S. = R.H.S.
A² – (a + d) A = (bc – ad)I₂

Question 12.

verify that (AB)^T = B^T A^T
Answer:


From (1) and (2) we get, (AB)^T = B^T A^T

Question 13.

show that A² – 5A + 7I₂ = 0
Answer:


L.H.S. = R.H.S.
∴ A² – 5A + 7I₂ = 0

Students can download Maths Chapter 3 Algebra 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 3 Algebra Additional Questions

I. Multiple choice questions

Question 1.
Which of the following is a monomial?
(a) 4x²
(b) a + b
(c) a + b + c
(d) a + b + c + d
Solution:
(a) 4x²

Question 2.
Which of the following is trinomial?
(a) -7z
(b) z² – 4y²
(c) x²y – xy² + y
(d) 12a – 9ab + 5b – 3
Solution:
(c) x²y – xy² + y

Question 3.
The sum of 5x²; -7x²; 8x²; 11x² and -9x² is ………
(a) 2x²
(b) 4x²
(c) 6x²
(d) 8x²
Solution:
(d) 8x²

Question 4.
The area of a rectangle with length 2l²m and breadth 3lm² is ………
(a) 6l³m³
(b) l³m³
(c) 2l³m³
(d) 4l³m³
Solution:
(a) 6l³m³

Question 5.
The coefficient of x² and x in 2x³ – 5x² + 6x – 3 are respectively ………
(a) 2, -5
(b) 2, 6
(c) – 5, 6
(d) -5, -3
Solution:
(c) – 5, 6

Question 6.
In the system 6x -2y = 3; kx – y = 2 has a unique solution then ………
(a) k = 3
(b) k ≠ 3
(c) k = 4
(d) k ≠ 4
Solution:
(b) k ≠ 3

Question 7.
A system of two linear equation in two variables is inconsistent. If their graphs ………
(a) coincide
(b) intersect only at a point
(c) do not intersect at any point
(d) cut the x-axis
Solution:
(c) do not intersect at any point

Question 8.
The system of equation x – 4y = 8; 3x – 12y = 24 ……….
(a) has infinitely many solution
(b) has no solution
(c) has a unique solution
(d) may or may not have a solution
Solution:
(a) has infinitely many solution

Question 9.
The solution set of x – ay = 4 and x + y = 0 is (1, -1) the value of a is ………
(a) -1
(b) 1
(c) -3
(d) 3
Solution:
(d) 3

Question 10.
The solution set of x + y = 7; x – y = 3 is ………
(a) (-5, -2)
(b) (-5, 2)
(c) (5, 2)
(d) (2, 5)
Solution:
(c) (5, 2)

II. Answer following Questions

Question 1.
What must be added to x⁴ – 3x² + 2x + 6 to get x⁴ – 2x³ – x + 8?
Solution:
Let A be the required number to be added.
(x⁴ – 3x² + 2x + 6) + A = x⁴ – 2x³ – x + 8
A = x⁴ – 2x³ – x + 8 – (x⁴ – 3x² + 2x + 6)
= x⁴ – 2x³ – x + 8 – x⁴ + 3x² – 2x – 6
= -2x³ + 3x² – 3x + 2
Hence -2x³ + 3x² – 3x + 2 must be added.

Question 2.
What must be subtracted to y⁴ + 2y³ – 3y + 8 to get y⁴ – 2y³ + 6?
Solution:
Let A be the required number to be subtracted.
(y⁴ + 2y³ – 3y² + 8) – A = y⁴ – 2y³ + 6
y⁴ + 2y³ – 3y² + 8 – (y⁴ – 2y³ + 6) = A
y⁴ + 2y³ – 3y² + 8 – y⁴ + 2y³ – 6 = A
4y³ – 3y² + 2 = A
Hence 4y³ – 3y² + 2 must be subtracted.

Question 3.
The area of a rectangle is x⁴ + 9x² + 20 sq.units and its length is x² + 4 units. Find its breadth in term of x.
Solution:
Let the breadth of a rectangle be “b”
Length of the rectangle = x² + 4
Area of the rectangle = x⁴ + 9x² + 20
Length × Breadth = x⁴ + 9x² + 20
(x² + 4) × b = x⁴ + 9x² + 20
b = \(\frac{x^{4}+9x^{2}+20}{x^{2}+4}\)
= x² + 5
breadth of a rectangle = x² + 5

Question 4.
Solve 3x + 4y = 24; 20x – 11y = 47 using cross multiplication method.
Solution:
3x + 4y – 24 = 0 → (1)
20x – 11y – 47 = 0 → (2)

\(\frac{x}{-452}\) = \(\frac{1}{-113}\)
-113 = -452
x = \(\frac{452}{113}\)
= 4
But \(\frac{y}{-339}\) = \(\frac{1}{-113}\)
-113y = -339
y = \(\frac{339}{113}\)
= 3
∴ The solution set is (4, 3)

Question 5.
A fraction such that if the numerator is multiplied by 3 and the denominator is reduced by 3, we get \(\frac{18}{11}\), but if the numerator is increased by 8 and the denominator is doubled, we get \(\frac{2}{5}\). Find the fraction.
Solution:
Let the numerator be x and the denominator be y
∴ The fraction is \(\frac{x}{y}\)
According to the given condition
\(\frac{3x}{y-3}\) = \(\frac{18}{11}\)
33x = 18(y – 3)
33x = 18y – 54
33x – 18y – 54 = 0
11x – 6y – 18 = 0 ……. (1)
According to the second condition
\(\frac{x+8}{2y}\) = \(\frac{2}{5}\)
5x + 40 = 4y
5x – 4y + 40 = 0 ……..(2)

∴ The fraction is = \(\frac{12}{25}\)

Question 6.
One number is greater than the thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the number.
Solution:
Let the greater number be x and the smaller number be “y” By the given first condition
x = 3y + 2
x – 3y = 2 ……(1)
Again by the given second condition
4y = x + 5
-x + 4y = 5 …….(2)
Add (1), (2) ⇒ y = 7
Substitute the value of y = 7 in (1)
x – 3(7) = 2
x = 2 + 21
= 23
The greater number is 23 and the smaller number is 7.

Question 7.
The cost of 11 pencils and 3 erasers is Rs 50 and the cost of 8 pencils and 3 erasers is Rs 38. Find the cost of 5 pencils and 5 erasers.
Solution:
Let the cost of a pencil be Rs x and the cost of an eraser be Rs y. According to the first condition.
11x + 3y = 50 …….(1)
According to the second condition
8x + 3y = 38 ……..(2)
(1) – (2) ⇒ 3x = 12
x = \(\frac{12}{3}\)
= 4
Substitute the value of x = 4 in (1)
11 (4) + 3y = 50
3y = 50 – 44
3y = 6
y = \(\frac{6}{3}\)
= 2
Cost of 5 pencils + 5 erasers = 5(4) + 5(2)
= 20 + 10
= 30
The required cost is Rs 30

Students can download Maths Chapter 3 Algebra Ex 3.15 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 3.15

Question 1.
Graph the following quadratic equations and state their nature of solutions.
(i) x² – 9x + 20 = 0
(ii) x² – 4x + 4 = 0
(iii) x² + x + 7 = 0
(iv) x² – 9 = 0
(v) x² – 6x + 9 = 0
(vi) (2x – 3) (x + 2) = 0

(i) x² – 9x + 20 = 0
Answer:
Let y = x² – 9x + 20
(i) Prepare the table of values for y = x² – 9x + 20

(ii) Plot the points (-1, 30) (0,20) (1, 12) (2, 6) (3,2), (4, 0), (5, 0), (6,2) (omit the high value)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (4, 0) and (5, 0)
There are two points of intersection with the X-axis at 4 and 5. The solution set is 4 and 5. The quadratic equation has real and unequal roots.
(v) Since there is two point of intersection with X-axis (different solution)
∴ The equation x² – 9x + 20 = 0 has real and unequal roots.

(ii) x² – 4x + 4 = 0
Answer:
Let y = x² – 4x + 4
(i) Prepare the table of values for y = x² – 4x + 4

(ii) Plot the points (-3,25) (-2,16) (-1, 9) (0,4) (1,-1) (2, 0), (3,1) and (4, 4)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (2, 0) which is 2.
(v) Since there is only one point of intersection with X-axis (2, 0).
∴ The solution set is 2.
The Quadratic equation x² – Ax + 4 = 0 has real and equal roots.

(iii) x² + x + 7 = 0
Answer:
Let y = x² + x + 7
(i) Prepare the table of values for y = x² + x + 7

(ii) Plot the points (-4,19) (-3,13) (-2, 9) (-1, 7) (0, 7) (1, 9), (2,13) (3,19) and (4,27)
(iii) Join the points by a free hand smooth curve.
(iv) The solution of the given quadratic equation are the X-coordinates of the intersecting points of the parabola with the X-axis.
(v) The curve does not intersecting the X-axis. There is no solution set.
The equation x² + x + 7 = 0 has no real roots.

(iv) x² – 9 = 0
Answer:
Let y = x² – 9
(i) Prepare the table of values for y = x² – 9

(ii) Plot the points (-4, 7) (-3, 0) (-2, -5) (-1, -8) (0, -9) (1, -8), (2, -5) (3, 0) (4, 7)
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X-axis at -3 and 3.
The solution is (-3, 3).
(v) Since there are two points of intersection -3 and 3 with the X-axis the quadratic equation has real and unequal roots.

(v) x² – 6x + 9 = 0
Answer:
Let y = x² – 6x + 9
(i) Prepare a table of values for y = x² – 6x + 9

(ii) Plot the points (-2,25) (-1,16) (0,9) (1,4) (2,1) (3,0), (4,1) and (5,4) on the graph using suitable scale (omit the points (-4, 49) and (-3, 36)
(iii) Join the points by a free hand smooth curve.
(iv) The X – coordinates of the point of intersection of the curve with X-axis are the roots of the , given equation, provided they intersect.
The solution is 3.
(v) Since there is only one point of intersection with X-axis the quadratic equation x² – 6x + 9 = 0 has real and equal roots.

(vi) (2x – 3) (x + 2) = 0
Answer:
y = (2x – 3) (x + 2)
= 2x² + 4x – 3x – 6
= 2x² + x – 6

(i) Prepare a table of values for y from x – 4 to 4

(ii) Plot the points (-4, 22) (-3, 9) (-2, 0) (-1, -5) (0, -6) (1, -3), (2, 4), (3, 15) and (4, 30).
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X – axis at (-2, 0) and (1\(\frac { 1 }{ 2 } \), 0)
∴ The solution set is (-2,1\(\frac { 1 }{ 2 } \))
(v) Since there are two points of intersection with X – axis, the quadratic equation has real and un – equal roots.

Question 2.
Draw the graph of y = x² – 4 and hence solve x² – x – 12 = 0
Answer:
(i) Draw the graph of y = x² – 4 by preparing the table of values as below.

(ii) Plot the points for the ordered pairs (-4, 12) (-3, 5) (-2, 0) (-1, -3) (0, -4) (1, -3), (2, 0), (3, 5) and (4, 12). Draw the curve with the suitable scale.
(iii) To solve x² – x – 12 = 0 subtract x² – x – 12 from y = x² – 4

The equation y = x + 8 represents a straight line. Prepare a table for y = x + 8

(iv) Mark the point of intersection of the curve and the straight line is (-3, 5) and 4, 12)
∴ The solution set is (-3, 4) for x² – x – 12 = 0.

Question 3.
Draw the graph of y = x² + x and hence solve x² + 1 = 0
Answer:
Let y = x² + x
(i) Draw the graph of y = x² + x by preparing the table.

(ii) Plot the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12) and (4, 20).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² + 1 = 0, subtract x² + 1 = 0 from x² + x we get.

The equation represent a straight line. Draw a line y = x – 1

Observe the graph of y = x² + 1 does not interset the parabola y = x² + x.
This x² + 1 has no real roots.

Question 4.
Draw the graph of y = x² + 3x + 2 and use it to solve x² + 2x + 1 = 0.
(i) Draw the graph of y = x² + 3x + 2 preparing the table of values as below.

(ii) Plot the points (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20) (4, 30).
(iii) To solve x² + 2x + 1 = 0 subtract x² + 2x + 1 = 0 from y = x² + 3x + 2

(iv) Draw the graph of y = x + 1 from the table

The equation y = x + 1 represent a straight line.
This line intersect the curve at only one point (-1, 0). The solution set is (-1).

Question 5.
Draw the graph of y = x² + 3x – 4 and hence use it to solve x² + 3x – 4 = 0
Answer:
Let y = x² + 3x – 4
(i) Draw the graph of y = x² + 3x – 4

(ii) Plot the points (-5, 6), (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14) on the graph using suitable scale.
(iii) Join the points by a free hand smooth curve.
The smooth curve is the graph of y = x² – 4x + 4
(iv) To solve x + 3x – 4 = 0, subtract x² + 3x – 4 = 0 from y = x² + 3x – 4.
y = 0
∴ The point of intersection with the x – axis is the solution set.
The solution set is -4 and 1.

Question 6.
Draw the graph of y = x² – 5x – 6 and hence solve x², – 5x – 14 = 0
Answer:
Let y = x² – 5x – 6
(i) Draw the graph of y = x² – 5x – 6 by preparing the table of values as below.

(ii) Plot the points (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5,-6), (6, 0) and (7, 8).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² – 5x – 14 = 0, subtract x² – 5x – 14 = 0 from y = x² – 5x – 6.

The equation y = 8 represent a straight line draw a straight line through y = 8 intersect the curve at two places. From the two points draw perpendicular line to the X – axis it will intersect at -2 and 7.
The solution is -2 and 7

Question 7.
Draw the graph of y = 2x² – 3x – 5 and hence solve 2x² – 4x – 6 = 0
Answer:
(i) Draw the graph of y = 2x² – 3x – 5 by preparing the table of values given below.

(ii) Plot the points (-3, 22), (-2, 9), (-1, 0), (0, -5), (1,-6), (2, -3), (3, 4), (4, 15) on the graph sheet using suitable scale.
(iii) To solve 2x² – 4x – 6 = 0 subtract 2x² – 4x – 6 = 0 from y = 2x² – 3x – 5

(iv) y = x + 1 represent a straight line.

The straight line intersect the curve at (-1, 0) and (3, 4). From the two point draw perpendicular lines to the X – axis it will intersect at -1 and 3.
The solution set is (-1, 3)

Question 8.
Draw the graph of y = (x – 1) (x + 3) and hence solve x² – x – 6 = 0
Answer:
y = (x – 1) (x + 3)
y = x² + 2x – 3
(i) Draw the graph of y = x² + 2x – 3 by preparing the table of values given below

(ii) Plot the points (-4, 5), (-3, 0), (-2, -3), (-1, -4), (0, -3), (1, 0), (2, 5), (3, 12) and (4, 21) on the graph sheet using suitable scale.
(iii) To solve x² – x – 6 = 0 subtract x² – x – 6 = 0 from y = x2 + 2x – 3
(iv) Draw the graph of y = 3x + 3 by preparing the table.

(v) The straight line cuts the curve at (-2, -3) and (3, 12). Draw perpendicular lines from the point to X – axis.
The line cut the X – axis at -2 and 3.
The solution set is (-2, 3)

Students can download Maths Chapter 3 Algebra Ex 3.15 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 3.15

Multiple Choice Questions.

Question 1.
If x³ + 6x² + kx + 6 is exactly divisible by x + 2, then k = 2
(a) 6
(b) -7
(c) -8
(d) 11
Solution:
(d) 11
Hint:
p(x) = x³ + 6x² + kx + 6
Given p(-2) = 0
(-2)³ + 6(-2)² + k(-2) + 6 = 0
-8 + 24 – 2k + 6 = 0
22 – 2k = 0
k = \(\frac{22}{2}\)
= 11

Question 2.
The root of the polynomial equation 2x + 3 = 0 is…….
(a) \(\frac{1}{3}\)
(b) –\(\frac{1}{3}\)
(c) –\(\frac{3}{2}\)
(d) –\(\frac{2}{3}\)
Solution:
(c) –\(\frac{3}{2}\)
Hint:
2x + 3 = 0
2x = – 3 ⇒ x = –\(\frac{3}{2}\)

Question 3.
The type of the polynomial 4 – 3x³ is ……..
(a) constant polynomial
(b) linear polynomial
(c) quadratic polynomial
(d) cubic polynomial
Solution:
(d) cubic polynomial

Question 4.
If x⁵¹ + 51 is divided by x + 1, then the remainder is …….
(a) 0
(b) 1
(c) 49
(d) 50
Solution:
(d) 50
Hint:
p(x) = x⁵¹ + 51
p(-1)= (-1)⁵¹ + 51
= -1 + 51
= 50

Question 5.
The zero of the polynomial 2x + 5 is ……..
(a) \(\frac{5}{2}\)
(b) –\(\frac{5}{2}\)
(c) \(\frac{2}{5}\)
(d) –\(\frac{2}{5}\)
Solution:
(b) –\(\frac{5}{2}\)
Hint:
2x + 5 = 0 ⇒ 2x = -5 ⇒ x = –\(\frac{5}{2}\)

Question 6.
The sum of the polynomials p(x) = x³ – x² – 2, q(x) = x² – 3x + 1
(a) x³ – 3x – 1
(b) x³ + 2x² – 1
(c) x³ – 2x² – 3x
(d) x³ – 2x² + 3x – 1
Solution:
(a) x³ – 3x – 1
Hint:
p(x) + q(x) = (x³ – x² – 2) + (x² – 3x + 1) = x³ – x² – 2 + x² – 3x + 1
= x³ – 3x – 1

Question 7.
Degree of the polynomial (y³ – 2) (y³ + 1) is
(a) 9
(b) 2
(c) 3
(d) 6
Solution:
(d) 6
(y³ – 2) (y³ + 1) = y⁶ + y³ – 2y³ – 2
= y⁶ – y³ – 2

Question 8.
Let the polynomials be
(A) -13q⁵ + 4q² + 12q
(B) (x² + 4)(x² + 9)
(C) 4q⁸ – q⁶ + q²
(D) –\(\frac{5}{7}\) y¹² + y³ + y⁵.
Then ascending order of their degree is
(a) A, B, D, C
(b) A, B, C, D
(c) B, C, D, A
(d) B, A, C, D
Solution:
(d) B, A, C, D

Question 9.
If p(a) = 0 then (x – a) is a …….. of p(x)
(a) divisor
(b) quotient
(c) remainder
(d) factor
Solution:
(d) factor

Question 10.
Zeros of (2 – 3x) is ……..
(a) 3
(b) 2
(c) \(\frac{2}{3}\)
(d) \(\frac{3}{2}\)
Solution:
(c) \(\frac{2}{3}\)

Question 11.
Which of the following has x -1 as a factor?
(a) 2x – 1
(b) 3x – 3
(c) 4x – 3
(d) 3x – 4
Solution:
(b) 3x – 3

Question 12.
If x – 3 is a factor of p(x), then the remainder is ……..
(a) 3
(b) -3
(c) p(3)
(d) p(-3)
Solution:
(c) p(3)

Question 13.
(x +y)(x² – xy + y²) is equal to ……..
(a) (x + y)³
(b) (x – y)³
(c) x³ + y³
(d) x³ – y³
Solution:
(c) x³ + y³

Question 14.
(a + b – c)² is equal to ……..
(a) (a – b + c)²
(b) (-a – b + c)²
(c) (a + b + c)²
(d) (a – b – c)²
Solution:
(b) (-a – b + c)²
Hint:
(a + b – c)² = a² + b² + c² + 2ab – 2bc – 2ac
(- a – b + c)² = a² + b² + c² + 2ab – 2bc – 2ac
(OR)
(- a – b + c)² = (-1)² (a + b + c)² (taking – 1 as common)
= (a + b – c)²

Question 15.
In an expression ax² + bx + c the sum and product of the factors respectively ……..
(a) a, bc
(b) b, ac
(c) ac, b
(d) bc, a
Solution:
(b) b, ac

Question 16.
If (x + 5) and (x – 3) are the factors of ax² + bx + c, then values of a, b and c are ………
(a) 1, 2, 3
(b) 1, 2, 15
(c) 1, 2, -15
(d) 1, -2, 15
Solution:
(c) 1, 2, -15
Hint:
(x + 5) (x – 3) = x² + (5 – 3) x + (5) (-3)
= x² + 2x – 15
compare with ax² + bx + c
a = 1, b = 2 and c = -15

Question 17.
Cubic polynomial may have maximum of ……… linear factors.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 18.
Degree of the constant polynomial is ……..
(a) 3
(b) 2
(c) 1
(d) 0
Solution:
(d) 0

Question 19.
Find the value of m from the equation 2x + 3y = m. If its one solution is x = 2 and y = -2.
(a) 2
(b) -2
(c) 10
(d) 0
Solution:
(b) – 2
Hint:
The equation is 2x + 3y = m
Substitute x – 2 and y = -2 we get
2(2) + 3(-2) = m ⇒ 4 – 6 = m ⇒ -2 = m

Question 20.
Which of the following is a linear equation?
(a) x + \(\frac{1}{2}\) = 2
(b) x (x – 1) = 2
(c) 3x + 5 = \(\frac{2}{3}\)
(d) x³ – x = 5
Solution:
(c) 3x + 5 = \(\frac{2}{3}\)

Question 21.
Which of the following is a solution of the equation 2x – y = 6?
(a) (2, 4)
(b) (4, 2)
(c) (3, -1)
(d) (0, 6)
Solution:
(b) (4, 2)
Hint:
2x – y = 6
Substitute x – 4 and y = 2 we get
2(4) – 2 = 6 ⇒ 8 – 2 = 6 ⇒ 6 = 6
∴ (4, 2) is the solution

Question 22.
If (2, 3) is a solution of linear equation 2x + 3y = k then, the value of k is ……..
(a) 12
(b) 6
(c) 0
(d) 13
Solution:
(d) 13
Hint:
The equation is 2x + 3y = k
Substitute x = 2 and y = 3 we get,
2(2) + 3(3) = k ⇒ 4 + 9 = k ⇒ 13 = k

Question 23.
Which condition does not satisfy the linear equation ax + by + c = 0 ……..
(a) a ≠ 0, b = 0
(b) a = 0, b ≠ 0
(c) a = 0, b = 0, c ≠ 0
(d) a ≠ 0, b ≠ 0
Solution:
(c) a = 0, b = 0, c ≠ 0

Question 24.
Which of the following is not a linear equation in two variable?
(a) ax + by + c = 0
(b) 0x + 0y + c = 0
(c) 0x + by + c = 0
(d) ax + 0y + c = 0
Solution:
(b) 0x + 0y + c = 0
Hint:
0x + 0y + c = 0
0 + 0 + c = 0 ⇒ c = 0
There is no variable.
∴ It is not a linear equation

Question 25.
The value of k for which the pair of linear equations 4x + 6y – 1 = 0 and 2x + ky – 1 = 0 represents parallel lines is ……..
(a) k = 3
(b) k = 2
(c) k = 4
(d) k = -3
Solution:
(a) k = 3
Hint:
Slope of 4x + 6y – 1 = 0 is
6y = -4x + 1 ⇒ y = \(\frac{-4}{6}\) x + \(\frac{1}{6}\)
Slope = \(\frac{-4}{6}\) = \(\frac{-2}{3}\)
Slope of 2x + ky – 7 = 0
ky = -2x + 7
y = \(\frac{-2}{k}\)x + \(\frac{7}{k}\)
Slope of a line = \(\frac{-2}{k}\)
Since the lines are parallel
\(\frac{-2}{3}\) = \(\frac{-2}{k}\)
-2k = – 6
k = \(\frac{6}{2}\)
= 3

Question 26.
A pair of linear equations has no solution then the graphical representation is ……..

Solution:

Hint:
Since there is no solution the two lines are parallel. (l11m)

Question 27.
If \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\) where a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 then the given pair of linear equation has …….. solution(s).
(a) no solution
(b) two solutions
(c) unique
(d) infinite
Solution:
(c) unique
Hint:
Since it has unique solution
\(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\)

Question 28.
\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\) where a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 then the given pair of linear equation has …….. solution(s).
(a) no solution
(b) two solutions
(c) infinite
(d) unique
Solution:
(a) no solution
Hint:
\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\) the linear equation has no solution.

Question 29.
GCD of any two prime numbers is …….
(a) -1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

Question 30.
The GCD of x⁴ – y⁴ and x² – y² is ……..
(a) x⁴ – y⁴
(b) x² – y²
(c) (x + y)²
(d) (x + y)⁴
Solution:
(b) x² – y²
Hint:
x⁴ – y⁴ = (x²)² – (y²)²
= (x² + y²)(x² – y²)
x² – y² = (x² – y²)
G.C.D. = x² – y²

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.4 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 5 Coordinate Geometry Ex 5.4

Question 1.
Find the coordinates of the point which divides the line segment joining the points A (4,-3) and B (9,7) in the ratio 3 : 2.
Solution:
A line divides internally in the ratio m : n

Question 2.
In what ratio does the point P(2, -5) divide the line segment joining A(-3, 5) and B(4, -9).
Solution:
A line divides internally in the ratio m : n

\(\frac{4m-3n}{m+n}\)
= 2
4m – 3n = 2m + 2n
4m – 2m = 3n + 2n
2m = 5n
\(\frac{m}{n}\) = \(\frac{5}{2}\)
m : n = 5 : 2
The ratio is 5 : 2.
and
\(\frac{-9m+5n}{m+n}\)
= -5
-9m + 5 n = -5(m + n)
-9m + 5 n = -5m – 5n
-9m + 5 m = -5n – 5n
-4m = -10
\(\frac{m}{n}\) = \(\frac{10}{4}\) ⇒ \(\frac{m}{n}\) = \(\frac{5}{2}\)
m : n = 5 : 2

Question 3.
Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in such a way that AP = \(\frac{2}{5}\) AB.
Solution:
Let the point A (1, 2) and B (6, 7)
AP = \(\frac{2}{5}\) AB
\(\frac{AP}{PB}\) = \(\frac{2}{5}\)
∴ AP = 2; PB = 5 – 2 = 3
A line divides internally in the ratio m : n

Question 4.
Find the coordinates of the points of trisection of the line segment joining the points A (-5, 6) and B (4, -3).
Solution:
Let P and Q be the point of trisection
so that AP = PB = QB

(\(\frac{3}{3}\), \(\frac{0}{3}\)) = (1, 0)
The Point P is (-2, 3), The Point Q is (1, 0)

Question 5.
The line segment joining A(6, 3) and B(-1, -4) is doubled in length by adding half of AB to each end. Find the coordinates of the new end points.
Solution:

m : n = 3 : 1
A line divides externally in the ratio m : n

∴ BA’ divides in the ratio 2 : 1
A line divides internally in the ratio m : n the point is \(\frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}\)
Let the point A’ be (a, b)

\(\frac{2a-1}{3}\) = 6
2a – 1 = 18
2a = 19
a = \(\frac{19}{2}\)
and
\(\frac{2b-4}{3}\) = 3
2b – 4 = 9
2b = 13
b = \(\frac{13}{2}\)

The point A’ is (\(\frac{19}{2}\), \(\frac{13}{2}\))
To find B’
Let B’ be (a, b)
AB = 7√2
BB’ = \(\frac{1}{2}\) × 7√2 = \(\frac{7√2}{2}\)
\(\frac{AB}{BB’}\) = 7√2 ÷ \(\frac{7√2}{2}\) = \(\frac{7√2×2}{7√2}\) = 2
AB’ divides in the ratio 2 : 1
(-1, -4) = \(\frac{2a+6}{3}\), \(\frac{2b+3}{3}\)
\(\frac{2a+6}{3}\) = -1
2a + 6 = -3
2a = -3 – 6
2a = -9
a = –\(\frac{9}{2}\)
and
\(\frac{2b+3}{3}\) = -4
2b + 3 = -12
2b = -12 – 3
2b = -15
b = –\(\frac{15}{2}\) = -1
The point B’ is (-\(\frac{9}{2}\), –\(\frac{15}{2}\))

Question 6.
Using section formula, show that the points A (7, -5), B (9, -3) and C (13, 1) are collinear.
Solution:
If three points are collinear, then one of the points divide the line segment joining the other points in the ratio r : 1. If P is between A and B and \(\frac{AP}{PB}\) = r

The line divides in the ratio 1 : 2
A line divides internally in the ratio m : n
The point P = (\(\frac{mx_{2}+nx_{1}}{m+n}\), \(\frac{my_{2}+ny_{1}}{m+n}\))
m = 1, n = 2, x₁ = 7, x₂ = 13, y₁ = – 5, y₂ = 1
By the given equation,
The Point B = (\(\frac{13+14}{3}\), \(\frac{1-10}{3}\))
= (\(\frac{27}{3}\), \(\frac{-9}{3}\))
= (9, -3)
∴ The three points A, B and C are collinear.

Question 7.
A line segment AB is increased along its length by 25% by producing it to C on the side of B. If A and B have the coordinates (-2, -3) and (2, 1) respectively, then find the coordinates of C.
Solution:
Let the point C be (a, b)

The ratio is 4 : 1 (m : n)
A line divides internally in the ratio m : n

\(\frac{4a-2}{5}\) = 2
4a – 2 = 10
4a = 12
a = \(\frac{12}{4}\) = 3
and
\(\frac{4b-3}{5}\) = 1
4b – 3 = 5
4b = 8
b = \(\frac{8}{4}\) = 2
The co-ordinate of C is (3, 2)

Students can download Maths Chapter 3 Algebra Ex 3.6 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 3 Algebra Ex 3.6

Question 1.
Simplify
(i) \(\frac{x(x+1)}{x-2}+\frac{x(1-x)}{x-2}\)
Answer:
\(\frac{x(x+1)}{x-2}+\frac{x(1-x)}{x-2}\) = \(\frac{x(x+1)+x(1-x)}{x-2}\)
= \(\frac{x^{2}+x+x-x^{2}}{x-2}\)
= \(\frac { 2x }{ x-2 } \)

(ii) \(\frac { x+2 }{ x+3 } \) + \(\frac { x-1 }{ x-2 } \)
Answer:
\(\frac { x+2 }{ x+3 } \) + \(\frac { x-1 }{ x-2 } \) = \(\frac{(x+2)(x-2)+(x-1)(x+3)}{(x+3)(x-2)}\)

(iii) \(\frac{x^{3}}{x-y}+\frac{y^{3}}{y-x}\) = \(\frac{x^{3}}{x-y}+\frac{y^{3}}{-1(x-y)}\)
Answer:
= \(\frac{x^{3}}{x-y}-\frac{y^{3}}{x-y}\)
= \(\frac{x^{3}-y^{3}}{x-y}\) (using a³ – b³ = (a – b) (a² + ab + b²))
= \(\frac{(x-y)\left(x^{2}+x y+y^{2}\right)}{x-y}\)
= x² + xy + y²

Question 2.
Simplify
(i) \(\frac{(2 x+1)(x-2)}{x-4}\) – \(\frac{\left(2 x^{2}-5 x+2\right)}{x-4}\)
Answer:

(ii) \(\frac{4 x}{x^{2}-1}-\frac{x+1}{x-1}\)
Answer:

Question 3.
Subtract \(\frac{1}{x^{2}+2}\) from \(\frac{2 x^{3}+x^{2}+3}{\left(x^{2}+2\right)^{2}}\)
Answer:

Question 4.
Which rational expression should be subtracted from \(\frac{x^{2}+6 x+8}{x^{3}+8}\)
Answer:

Question 5.
If A = \(\frac{2x+1}{2 x-1}\), B = \(\frac{2x-1}{2x+1}\) find \(\frac{1}{A-B}\) – \(\frac{2 \mathbf{B}}{\mathbf{A}^{2}-\mathbf{B}^{2}}\)
Answer:

Question 6.
If A = \(\frac { x }{ x+1 } \) B = \(\frac { 1 }{ x+1 } \) prove that \(\frac{(\mathbf{A}+\mathbf{B})^{2}+(\mathbf{A}-\mathbf{B})^{2}}{\mathbf{A}+\mathbf{B}}=\frac{2\left(x^{2}+1\right)}{x(x+1)^{2}}\)
Answer:

Question 7.
Pari needs 4 hours to complete a work. His friend Yuvan needs 6 hours to complete the same work. How long will it take to complete if they work together?
Solution:
Pari: time required to complete the work = 4 hrs.
∴ In 1 hr. he will complete = \(\frac{1}{4}\) of the work.
= \(\frac{1}{4}\) w.
Yuvan: Time required to complete the work = 6 hrs.
∴ In 1 hr. he will complete the \(\frac{1}{6}\) of the work
= \(\frac{1}{6}\) w
Working together, in 1 hr. they will complete \(\frac{w}{4}+\frac{w}{6}\) of the work.
= \(\frac{6 w+4 w}{24}=\frac{5}{12} w\)
∴ To complete the total work time taken
= \(\frac{w}{\frac{5}{12} w}=\frac{12}{5}\) = 2.4 hrs. [∵ (4) hrs = 4 × 60 = 24 min]
= 2 hrs 24 minutes.

Question 8.
Iniya bought 50 kg of fruits consisting of apples and bananas. She paid twice as much per kg for the apple as she did for the banana. If Iniya bought ? 1800 worth of apples and ₹ 600 worth bananas, then how many kgs of each fruit did she buy?
Let the weight of applies be a kg.
Let the weight of bananas be b kg.
a + b = 50
ax = ₹ 1800 ………….. (1)
by = ₹ 600 ………… (2)
x = 2y …………… (3)
Use (3) in (1) ⇒ a(2y) = 1800
y = \(\frac{900}{a}\) ………….. (4)
Use (4) in (2) ⇒ \(b\frac{900}{a}\) = 600
∵ 3b = 2a …………….. (5)
∵ a + b = 50
a + \(\frac{2 a}{3}\) = 50 ⇒ \(\frac{5 a}{3}\) = 50
⇒ a = 50 × \(\frac{3}{5}\)
= 30
∴ b = 20
∴ Iniya bought 30 kg of applies and 20 kg of bananas
= 30
.’. b = 20.
.’. Iniya bought 30 kg of applies and 20 kg of bananas.

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.2 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 5 Coordinate Geometry Ex 5.2

Question 1.
Find the distance between the following pairs of points.
(i) (1, 2) and (4, 3)
Solution:
Distance between the points (1, 2) and (4, 3)

(ii) (3, 4) and (-7, 2)
Solution:
Distance between the points (3,4) and (-7, 2)

(iii) (a, b) and (c, b)
Solution:
Distance between the two points (a, b) and (c, b)

= c – a units

(iv) (3,- 9) and (-2, 3)
Solution:
Distance between the two points (3, -9) and (-2, 3)

= 13 units

Question 2.
Determine whether the given set of points in each case are collinear or not.
(i) (7, -2), (5, 1), (3, 4)
Solution:
To prove that three points are collinear, sum of the distance between two pairs of points is equal to the third pair of points.

AB + BC = AC
\(\sqrt{13}\) + \(\sqrt{13}\) = 2\(\sqrt{13}\) ⇒ 2\(\sqrt{13}\) = 2\(\sqrt{13}\)
∴ The given three points are collinear.

(ii) (a, -2), (a, 3), (a, 0)
Solution:
A (a, -2) B (a, 3) C (a, 0)

√4
= 2
AC + BC = AB ⇒ 2 + 3 = 5
∴ The given three points are collinear.

Question 3.
Show that the following points taken in order to form an isosceles triangle.
(i) A (5, 4), B(2, 0), C (-2, 3)
Solution:

= 5√2
AB = BC = 5. (Two sides are equal)
∴ ABC is an isosceles triangle.

(ii) A (6, -1), B (-2, -4), C (2, 10)
Solution:

BC = AC = \(\sqrt{212}\) (TWO sides are equal)
∴ ABC is an isosceles triangle.

Question 4.
Show that the following points taken in order to form an equilateral triangle in each case.
(i) A(2, 2), B(-2, -2), C(-2√3, 2√3)
Solution:

AB = BC = AC (Three sides are equal)
∴ ABC is an equilateral triangle.

(ii) A(√3, 2), B (0, 1), C(0, 3)
Solution:

= √4
= 2
AB = BC = AC (Three sides are equal)
∴ ABC is an equilateral triangle.

Question 5.
Show that the following points taken in order to form the vertices of a parallelogram.
(i) A(-3, 1), B(-6, -7), C (3, -9) and D(6, -1)
Solution:

AB = CD = \(\sqrt{73}\) and BC = AD = \(\sqrt{85}\) (Opposite sides are equal)
∴ ABCD is a parallelogram.

(ii) A (-7, -3), B(5, 10), C(15, 8) and D(3, -5)
Solution:

AB = CD = \(\sqrt{313}\) and BC = AD = \(\sqrt{104}\) (Opposite sides are equal)
∴ ABCD is a parallelogram.

Question 6.
Verify that the following points taken in order to form the vertices of a rhombus.
(i) A(3, -2), B (7, 6),C (-1, 2) and D (-5, -6)
Solution:

AB = BC = CD = AD = \(\sqrt{80}\). All the four sides are equal.
∴ ABCD is a rhombus.

(ii) A (1, 1), B (2, 1),C (2, 2) and D (1, 2)
Solution:

AB = BC = CD = AD = 1. All the four sides are equal.
∴ ABCD is a rhombus.

Question 7.
A (-1, 1), B (1, 3) and C (3, a) are points and if AB = BC, then find ‘a’.
Solution:

4 + (a – 3)² = 8
(a – 3)² = 8 – 4
(a – 3)² = 4
a – 3 = √4
= ± 2
a – 3 = 2 (or) a – 3 = -2
a = 2 + 3 (or) a = 3 – 2
a = 5 (or) a = 1
∴ The value of a = 5 or a = 1.

Question 8.
The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3) is 10 units, What are the coordinates of A?
Solution:
Let the point A be (a, a) B is (1, 3)
Distance AB = 10 (Given)
By distance formula \(\sqrt{(a – 1 )² + (a – 3)²}\) = 10
Simplifying 2a² – 8a + 10 = 100
a² – 4a – 45 = 0
(a – 9)(a + 5) = 0
⇒ a = – 5; A = (-5, -5)
a = 9; A = (9, 9)

Question 9.
The point (x, y) is equidistant from the points (3, 4) and (-5, 6). Find a relation between x and y.
Solution:
Let the point O be (x, y), A be (3, 4) and B be (-5, 6).

Distance = \(\sqrt{(x_{2} – x_{1})² + (y_{2} – y_{1})²}\)
Given ,OA = OB
\(\sqrt{(x – 3 )² + (y – 4)²}\) = \(\sqrt{(x + 5 )² + (y – 6)²}\)
Squaring on both sides
(x – 3)² + (y – 4)² = (x + 5)² + (y – 6)²
x² – 6x + 9 + y² – 8y + 16 = x² + 10x + 25 + y² – 12y + 36
x² + y² – 6x – 8y + 25 = x² + y² + 10x – 12y + 61
6x – 10x – 8y + 12y = 61 – 25 ⇒ -16x + 4y = 36
÷ 4 ⇒ -4x + y = 9
∴ The relation between x and y is y = 4x + 9

Question 10.
Let A(2,3) and B(2, -4) be two points. If P lies on the x-axis, such that AP = \(\frac{3}{7}\) AB, find the coordinates of P.
Solution:
Given points are A(2, 3) and B(2, -4)
The point P lies on the x-axis.
∴ The point P is (x, 0)

16x² – 64x + 208 = 9x² – 36x + 180
16x² – 9x² – 64x + 36x + 208 – 180 = 0
7x² – 28x + 28 = 0
x² – 4x + 4 = 0
(x – 2)² = 0
x – 2 = 0
x = 2
∴ The point P is (2, 0)

Question 11.
Show that the point (11, 2) is the centre of the circle passing through the points (1, 2), (3, -4) and (5, -6)
Solution:

\(\sqrt{100}\)
= 10
OA = OB = OC = 10
O is the centre of the circle passing through A, B and C.

Question 12.
The radius of a circle with centre at origin is 30 units. Write the coordinates of the points where the circle intersects the axes. Find the distance between any two such points.
Solution:

Radius of the circle = 30 units. The point O is (0, 0).
Let a intersect the x-axis and b intersect the y-axis.
∴ The point A is (a, 0) and B is (0, b)

Squaring on both sides
30² = a²
∴ a = 30
The point A is (30, 0)
OB = \(\sqrt{(0 – 0)² + (b – 0)²}\)
= \(\sqrt{0² + b²}\)
30 = \(\sqrt{b²}\)
Squaring on both sides
30² = b²
∴ b = 30
The point B is (0, 30)

= 30√2
∴ Distance between the two points = 30√2

Students can download Maths Chapter 3 Algebra Ex 3.17 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 3.17

Question 1.
If then

verify that (i) A + B = B + A
(ii) A + (-A) = (-A) + A = O.
Answer:

From (1) and (2) we get A + B = B + A

From (1) and (2) we get
A + (-A) = (-A) + A = 0

Question 2.
If
then verify that
A + (B + C) = (A + B) + C.
Answer:


From (1) and (2) we get
A + (B + C) = (A + B) + C

Question 3.
Find X and Y if and
Answer:

Question 4.
If
find the value of (i) B – 5A (ii) 3A – 9B
Answer:

Question 5.
Find the values of x, y, z if

Answer:
(i) x – 3 = 1 ⇒ x = 1 + 3 ⇒ x = 4
3x – z = 0 (substitute the value of x)
3(4) – z = 0
12 – z = 0
∴ z = 12
x + y + z = 6
4 + y + 12 = 0
y + 16 = 6
y = 6- 16
∴ y = -10
The value of x = 4, y = -10 and z = 12

(ii) [x y – z z + 3] + [y 4 3] = [4 8 16]
x + y = 4 ….(1)
y – z + 4 = 8
Substitute the value
of z in (2)
(2) ⇒ y – 10 = 4
Substitute the value of y in (1)
z + 3 + 3 = 16
z + 6 = 16
z = 16 – 16 = 10
y = 14
(1) ⇒ x + 14 = 4
x – 4 – 14 = -10
The value of x = -10, y = 14 and z = 10

Question 6.
Find x and y if x
Answer:

4x – 2y = 4
(1) ⇒ 2x – y = 2
(2) ⇒ 3x – y = 2
– x + y = 2
Add (1) and (2)

Substitute the value of x = 4 in (2)
– 4 + y = 2
y = 2 + 4 = 6
The value of x = 4 and y = 6

Question 7.
Find the non-zero values of x satisfying the matrix equation

Answer:

The value of x = 4

Question 8.
Solve for x,y :

Answer:

x² – 4x = -5
x² – 4x + 5 = 0
(x – 5) (x + 1) = 0
x – 5 = 0 or x + 1 = 0
x = 5 or x = – 1

The value of x = -1 and 5
y² – 2y = 8
y² – 2y – 8 = 0
(y – 4) (y + 2) = 0
y – 4 = 0 or y + 2 = 0

y = 4 or y = -2
The value of y = -2 and 4

Students can download Maths Chapter 3 Algebra Ex 3.14 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 3.14

Question 1.
The sum of a two digit number and the number formed by interchanging the digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number. Find the first number.
Solution:
Let the ten’s digit be x and the unit digit be y.
The number is 10x + y
If the digits are interchanged
The new number is 10y + x
By the given first condition
10x + y + 10y + x = 110
11x + 11y = 110
x + y = 10 → (1) (Divided by 11)
Again by the given second condition
10x + y – 10 = 5(x + y ) + 4
10x + y – 10 = 5x + 5y + 4
5x – 4y = 14 → (2)
(1) × 5 ⇒ 5x + 5y = 50 → (3)
(2) × 1 ⇒ 5x – 4y = 14 → (2)
(3) – (2) ⇒ 9y = 36
y = 36/9
= 4
Substitute the value of y = 4 in (1)
x + y = 10
x + 4 = 10
x = 10 – 4
= 6
∴ The number is (10 × 6 + 4) = 64

Question 2.
The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \(\frac{1}{2}\). Find the fraction.
Solution:
Let the numerator be “x” and the denominator be “y”
∴ The fraction is \(\frac{x}{y}\)
By the given first condition
x + y = 12 → (1)
Again by the second condition
\(\frac{x}{y+3}\) = \(\frac{1}{2}\)
2x = y + 3
2x – y = 3 → (2)
(1) + (2) ⇒ 3x = 15
x = \(\frac{15}{3}\) = 5
Substitute the value of x = 5 in (1)
5 + y = 12
y = 12 – 5
= 7
∴ The fraction is \(\frac{5}{7}\)

Question 3.
ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y -5)°, ∠C = (4x)° and ∠D = (7x + 5)°. Find the four angles.
Solution:

ABCD is a cyclic quadrilateral ∠A + ∠C = 180°
(Sum of the opposite angles of a cyclic quadrilateral is 180°)
(4y + 20)° + (4x)° = 180°
4y + 20 + 4x = 180
4x + 4y = 180 – 20
4x + 4y = 160
x + y = 40 → (1) (divided by 4)
∠B + ∠D = 180° (Sum of the opposite angles of a cyclic quadrilateral)
(3y – 5)° + (7x + 5)° = 180°
3y – 5 + 7x + 5 = 180
7x + 3y = 180 → (2)
(1) × 3 ⇒ 3x + 3y = 120 → (3)
(3) – (2) ⇒ -4x = – 60
4x = 60
x = \(\frac{60}{4}\)
Substitute the value of x = 15 in (1)
15 + y = 40
y = 40 – 15 = 25
∠A = 4y + 20 = 4(25) + 20 = 100 + 20 = 120°
∴ ∠A = 120°
∠B = 3y – 5 = 3(25) – 5 = 75 – 5 = 70
∴ ∠B = 70°
∠C = 4x = 4(15) = 60
∴ ∠C = 60°
∠D = 7x + 5 = 7(15) + 5
∠D = 105 + 5 = 110°
∴ ∠A= 120°, ∠B = 70°, ∠C = 60° and ∠D = 110°

Question 4.
On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains Rs 2000. But if he sells the T.V. at 10% gain and the fridge at 5% loss, he gains Rs.1500 on the transaction. Find the actual price of the T.V. and the fridge.
Solution:
Let the cost price of the TV be Rs “x” and the cost price of the fridge be Rs “y”.
By the given condition

Multiply by 20
x + 2y = 40000 → (1)
Again by the given second condition

Multiply by 20
2x – y = 30000 → (2)
(2) × 2 ⇒ 4x – 2y = 60000 → (3)
(1) + (3) ⇒ 5x + 0 = 100000
x = \(\frac{100000}{5}\)
= 20000
Substitute the value of x = 20000 in (1)
20000 + 2y = 40000
2y = 40000 – 20000
= 20000
y = \(\frac{20000}{2}\)
= 10000
Cost price of a TV = Rs 20,000
Cost price of a fridge = Rs 10,000

Question 5.
Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.
Solution:
Let the two numbers be x and y.
By the given first condition
x : y = 5 : 6
6x = 5y (Product of the extreme is equal to the product of the means)
6x – 5y = 0 → (1)
Again by the given second condition
x – 8 : y – 8 = 4 : 5
5(x – 8) = 4(y – 8)
5x – 40 = 4y – 32
5x – 4y = – 32 + 40
5x – 4y = 8 → (2)
(1) × 4 ⇒ 24x – 20y = 0 → (3)
(2) × 5 ⇒ 25x – 20y = 40 → (4)
(3) – (4) ⇒ – x + 0 = -40
∴ x = 40
Substitute the value of x = 40 in (1)
6(40) – 5y = 0
240 – 5y = 0 ⇒ – 5y = -240
5y = 240
y = \(\frac{240}{5}\)
= 48
The two numbers are 40 and 48 [∴ The ratio of the number = 40 : 48 are 5 : 6]

Question 6.
4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indian and 5 Chinese can finish it in 4 days. How long would it take for 1 Indian to do it? How long would it ‘ take for 1 Chinese to do it?
Solution:
Let the time taken by a Indian be “x”
Time taken by a Chinese be “y”
Work done by a Indian in one day = \(\frac{1}{x}\)
Work done by a Chinese in one day = \(\frac{1}{y}\)
By the given first condition
(4 Indian + 4 Chinese) finish the work in 3 days
\(\frac{4}{x}\) + \(\frac{4}{y}\) = \(\frac{1}{3}\) → (1)
Again by the given second condition
(2 Indian + 5 Chinese) finish the work in 4 days
\(\frac{2}{x}\) + \(\frac{5}{y}\) = \(\frac{1}{4}\) → (2)
Solve the equation (1) and (2)
Let \(\frac{1}{x}\) = a; \(\frac{1}{y}\) = b
4a + 4b = \(\frac{1}{3}\)
12a + 12b = 1 → (3) (Multiply by 3)
2a + 5b = \(\frac{1}{4}\)
8a + 20b = 1 → (4) (Multiply by 4)
(3) × (2) ⇒ 24a + 24b = 2 → (5)
(4) × (3) ⇒ 24a + 60b = 3 → (6)
(5) – (6) ⇒ -36b = -1
b = \(\frac{1}{36}\)
Substitute the value of b = \(\frac{1}{36}\) in (3)
12a + 12(\(\frac{1}{36}\)) = 1
12a + \(\frac{1}{3}\) = 1
36a + 1 = 3
36a = 2
a = \(\frac{2}{36}\) = \(\frac{1}{18}\)
But \(\frac{1}{x}\) = a ⇒ \(\frac{1}{x}\) = \(\frac{1}{18}\)
x = 18
\(\frac{1}{y}\) = b ⇒ \(\frac{1}{y}\) = \(\frac{1}{36}\)
y = 36
∴ Time taken by a 1 Indian is 18 days
Time taken by a 1 Chinese is 36 days