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You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.12

Question 1.
Solve by the method of elimination
(i) 2x – y = 3; 3x + y = 7
(ii) x – y = 5; 3x + 2y = 25
(iii) \(\frac{x}{10}+\frac{y}{5}\) = 14; \(\frac{x}{8}+\frac{y}{6}\) = 15
(iv) 3(2x + y) = 7xy; 3(x + 3y) = 11xy
(v) \(\frac{4}{x}\) + 5y = 7; \(\frac{3}{x}\) + 4y = 5
(vi) 13x + 11y = 70; 11x + 13y = 74
Solution:
(i) 2x – y = 3 ………….. (1)
3x + y = 7 ………… (2)

Substitute x = 2 in (1)
2(2) – y = 3
4 – y = 3
-y = 3 – 4
-y = -1
∴ Solution: x = 2; y = 1
Verification:
Substitute x = 2, y = 1 in (2)
3(2) + 1 = 7 = RHS
∴ Verified.

Substitute y = 2 in (1)
x – 2 = 5
x = 5 + 2
x = 7
∴ Solution: x = 7, y = 2
Verification:
Substitute x = 7, y = 2 in (2)
3(7) + 2(2) = 21 + 4 = 25 = RHS
∴ Verified.

Substitute y = 30 in (1)
x + 2 (30) = 140
x + 60 = 140
x = 140 – 60
x = 80
∴ Solution: x = 80; y = 30
Verification:
Substitute x = 80, y = 30 in (2)
3(80) + 4(30) = 240 + 120 = 360 = RHS
∴ Verified.

(iv) 3(2x +y) = 7xy ⇒ 6x + 3y = 7xy ………. (1)
3(x + 3y) = 11xy ⇒ 3x + 9y = 11xy ………….. (2)


Substitute a = 1 in (5)
6b + 3(1) = 7
6b + 3 = 7
6b = 7 – 3
b = \(\frac{4}{6}=\frac{2}{3}\)
∴a = \(\frac{1}{x}\) = 1 ⇒ x = 1
b = \(\frac{1}{y}=\frac{2}{3}\) ⇒ y = \(\frac{3}{2}\)
∴ Solution: x = 1; y = \(\frac{3}{2}\)

Substitute y = 4 in (1)
13x + 11 (4) = 70
13x + 44 = 70
13x = 70 – 44 = 26
x = \(\frac{26}{13}\) = 2
∴ Solution: x = 2; y = 4

Question 2.
The monthly income of A and B are in the ratio 3:4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 5,000 per month, find the monthly income of each.
Solution:
Let the monthly income of A and B be 3x and 4x respectively.
Let the monthly expenditure of A and B be 5y and 7y respectively.
∴ 3x – 5y = 5000 ……… (1)
4x – 7y = 5000 ……….. (2)

Substitute y = 5000 in (1)
3x – 5 (5000) = 5000
3x – 25000 = 5000
3x = 5000 + 25000
3x = 30000
x = 10000
∴ Monthly income of A is 3x = 3 × 10000 = ₹ 30000
Monthly income of B is 4x = 4 × 10000 = ₹ 40000

Question 3.
Five years ago, a man was seven times as old as his son, while five year hence, the man will be four times as old as his son. Find their present age.
Solution:
Let the man’s present age = x
Five years ago his age is = x – 5
Let his son’s age be = y
5 years ago his son’s age = y – 5
∴ x – 5 = 7(y – 5)
x – 5 = 7y – 35
x – 7y = -35 + 5
x – 7y = – 30 ……….. (1)
After 5 years, man’s age will be = x + 5
His son’s age will be = y + 5
∴ x + 5 = 4(y + 5)
x + 5 = 4y + 20
x – 4y = 20 – 5
⇒ x – 4y = 15 ………….. (2)

Substitute y = 15 in (1)
x – 7 (15) = -30
x – 105 – 30
x = – 30 + 105
x = 75
∴ Man’s Age = 75, His son’s Age =15

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions

Exercise 5.1

Question 1.
State whether the following statements are true/false.
(i) (5, 7) is a point in the IV quadrant.
(ii) (-2, -7) is a point in the III quadrant.
(iii) (8, -7) lies below the x-axis.
(iv) (-2, 3) lies in the II quadrant.
(v) For any point on the x-axis its y-coordinate is zero.
Solution:
(i) False
(ii) True
(iii) True
(iv) True
(v) True

Question 2.
Locate the points
(i) (3, 5) and (5, 3)
(ii) (-2, -5) and (-5, -2) in the rectangular coordinate system.
Solution:

Question 3.
In which quadrant does the following points lie?
(i) (5, 2)
(ii) (-5, -8)
(iii) (-7, 1)
(iv) (8, -3)
Solution:
(i) I quadrant
(ii) III quadrant
(iii) II quadrant
(iv) IV quadrant.

Question 4.
Write down the ordinate of the following points.
(i) (7, 5)
(ii) (2, 9)
(iii) (-5, 8)
(iv) (7, -4)
Solution:
(i) 5
(ii) 9
(iii) 8
(iv) -4 (ordinate is the y-coordinate)

Exercise 5.2

Question 1.
Find the distance between the following pairs of points.
(i) (-4, 0) and (3, 0)
(ii) (-7, 2) and (5, 2)
Solution:
(i) The points (-4, 0) and (3, 0) lie on the x-axis. Hence,

(ii) The points (5,2) and (-7,2) lie on a line parallel to the x-axis. Hence the distance

Question 2.
Show that the three points (4, 2), (7, 5) and (9, 7) lie on a straight line.
Solution:
Let the points be A(4, 2), B(7, 5) and C(9, 7). By the distance formula.

Hence the points A, B and C are collinear.

Question 3.
Determine whether the points are vertices of a right triangle A(-3, -4), B(2, 6) and C (-6, 10).
Solution:

Hence ABC is a right angled triangle since the square of one side is equal to sum of the squares of the other two sides.

Question 4.
Show that the points (a, a), (-a, -a) and (\(-a \sqrt{3}, a \sqrt{3}\)) form an equilateral triangle.
Solution:
Let the points be represented by A (a, a), B(-a, -a) and C(\(-a \sqrt{3}, a \sqrt{3}\)) using the distance formula.

Since all the sides are equal the points form an equilateral triangle.

Question 5.
Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the corners of a parallelogram.
Solution:
Let A, B, C and D represent the points (-7, -3), (5, 10), (15, 8) and (3, -5) respectively.

i.e. The opposite sides are equal. Hence ABCD is a parallelogram.

Question 6.
Show that the following points A (3, 1) B(6, 4) and C(8, 6) lies on a straight line.
Solution:
Using the distance formula, we have

Therefore the points lie on a straight line.

Question 7.
If the distance between the points (5, -2), (1, a) is 5 units. Find the value of a.

Solution:

Exercise 5.3

Question 1.
A, B and C are vertices of ∆ ABC. D, E and F are mid points of sides AB, BC and AC respectively. If the coordinates of A, D and F are (-3, 5), (5, 1) and (-5, -1) respectively. Find the coordinates of B, C and E.
Solution:

Question 2.
If A(10, 11) and B(2, 3) are the coordinates of end points of diameter of circle. Then find the centre of the circle.
Solution:

Question 3.
Find the coordinates of the point which divides the line segment joining the points (3, 1) and (5, 13) internally in the ratio 3 : 5.
Solution:

Exercise 5.4

Question 1.
Using section formula, show that the points A(7, -5), B(9, -3) and C(13, 1) are collinear.
Solution:

Question 2.
A car travels at an uniform speed. At 2pm it is at a distance of 5 km at 6 pm it is at a distance of 120 km. Using section formula, find at what distance it will reach 2 mid night.
Solution:

Question 3.
Find the coordinates of the point which divides the line segment joining the point A(3, 7) and B(-11, -2) in the ratio 5 : 1.
Solution:

Exercise 5.5

Question 1.
Find the centroid of the triangle whose vertices are (2, -5), (5, 11) and (9, 9)
Solution:

Question 2.
If the centroid of a triangle is at (10, -1) and two of its vertices are (3, 2) and (5, -11). Find the third vertex of the triangle.
Solution:

Exercise 5.6

Multiple Choice Questions :

Question 1.
The point (-2, 7) lies is the quadrant
(1) I
(2) II
(3) III
(4) IV
Hint:
(-, +) lies in II^nd quadrant
Solution:
(2) II

Question 2.
The point (x, 0) where x < 0 lies on
(1) OX
(2) OY
(3) OX’
(4) OY’
Hint:
(-, 0) lies on OX’
Solution:
(3) OX’

Question 3.
For a point A(a, b) lying in quadrant III.
(1) a > 0, b < 0
(2) a < 0, b < 0
(3) a > 0, b > 0
(4) a < 0, b > 0
Hint:
(-, -) lies in III^rd quadrant
Solution:
(2) a < 0, b < 0

Question 4.
The diagonal of a square formed by the points (1, 0) (0, 1) and (-1, 0) is
(1) 2
(2) 4
(3) \(\sqrt{2}\)
(4) 8
Hint:

Solution:
(1) 2

Question 5.
The triangle obtained by joining the points A(-5, 0) B(5, 0) and C(0, 6) is
(1) an isosceles triangle
(2) right triangle
(3) scalene triangle
(4) an equilateral triangle
Hint:
Triangles having two sides equal are called isosceles.
Solution:
(a) an isosceles triangle

Text Book Activities

Activity 1.
Plot the following points on a graph sheet by taking the scale as 1cm = 1 unit. Find how far the points are from each other? A (1, 0) and D (4, 0). Find AD and also DA. Is AD = DA? You plot another set of points and verify your result.

Solution:
AD = DA is correct.

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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
(ii) 1.5x + 0.1y = 6.2; 3x – 0.4y = 11.2
(iii) 10% of x + 20% of y = 24; 3x – y = 20
(iv) \(\sqrt{2} x-\sqrt{3} y=1 ; \sqrt{3} x-\sqrt{8} y=0\)
Solution:
(i) 2x – 3y = 7 ………….. (1)
5x + y = 9 ………….. (2)
Step (1)
From the equation (2)
5x+ y = 9
y = -5x + 9
Step (2)
substitute (3) in (1)
2x – 3(-5x + 9) = 7
2x + 15x – 27 = 7
17x = 7 + 27
17x = 34
x = \(\frac{34}{17}\) = 2; x = 2
Step (3)
substitute x = 2 in (3)
y = – 5(2) + 9 = -10 + 9 = -1
Solution: x = 2; y = -1

(ii) 1.5x + 0.1y = 6.2 …………. (1)
3x – 0.1y = 11.2 ………….. (2)
Multiply (1 ) x 10 15x + y = 62 ……….(1)
(2) × 10 ⇒ 30x – 4y = 112 …………. (4)
Step (1)
From equation (3)
15x + y = 62
y = -15x + 62 …………. (5)
Step (2)
substitute (5) in (4)
30x – 4 (-15x + 62) = 112
30x + 60x – 248 = 112
90x = 112 + 248
90x = 360
x = \(\frac{360}{90}\)
x = 4
Step (3)
substitute x = 4 in (5)
y = -15(4) + 62
= -60 + 62
y = 2
Solution: x = 4; y = 2

x + 2y = 240 ………. (1)
3 x – y =20 ……….. (2)
Step (1)
From equation (2)
3x – y = 20
-y = 20 – 3x
y = 3x – 20 — (3)
Step (2)
substitute (3) in (1)
x + 2(3x – 20) = 240
x + 6x – 40 = 240
7x = 240 + 40
x = \(\frac{280}{7}\)
x = 40
Step (3)
substitute x = 40 in (3)
y = 3 (40) – 20
= 120 – 20 = 100
Solution : x = 40 and y = 100

(iv) \(\sqrt{2} x-\sqrt{3} y\) = 1 ………… (1)
\(\sqrt{3} x-\sqrt{8} y\) = 0 ……….. (2)
Step (1)
From the equation (2)


Solution: x = \(\sqrt{8}\) and y = \(\sqrt{3}\)

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 = x
Let the sum of his two sons age = y
now x = 3y ⇒ x – 3y = 0 ……… (1)
After 5 years,
Step (3)
x + 5 = 2(y + 10)
x + 5 = 2y + 20
x – 2y = 20 – 5
x – 2y = 15
Step (1)
From equation (1) x = 3y
Step (2)
Substitute x = 3y in (2)
3y – 2y = 15
y = 15
Step (3)
Substitute y = 15 in (1)
x = 3y = 3 × 15
x = 45
∴ Raman’s age is 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 number be x0y
x + y = 13 …………….. (1)
If the digits are reversed the number so formed is y0x
x0y = 100x + 10 × 0 + 1 × y
y0x = 100y + 10 × 0 + 1 × x
100y + x – (100x + y) = 495
100y + x – 100x – y = 495
-99x + 99y = 495 ………….. (2)

Substitute x = 4 in (1)
4 + y = 13 = 13 – 4 = 9
The number is 409.

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.7

Multiple Choice Questions :

Question 1.
The exterior angle of a triangle is equal to the sum of two

(1) Exterior angles
(2) Interior opposite angles
(3) Alternate angles
(4) Interior angles
Hint: Exterior angle = 180°- Interior angle = sum of interior opposite angle
Solution:
(2) Interior opposite angles

Question 2.
In the quadrilateral ABCD, AB = BC and AD = DC Measure of ∠BCD is

(1) 150°
(2) 30°
(3) 105°
(4) 72°
Hint:

Solution:
(3) 105°

Question 3.
ABCD is a square, diagonals AC and BD meet at O. The number of pairs of congruent triangles are

(1) 6
(2) 8
(3) 4
(4) 12
Solution:
(1) 6

Question 4.
In the given figure CE || DB then the value of x° is

(1) 45°
(2) 30°
(3) 75°
(4) 85°
Hint: 35° + x°+ 60° = 180° ⇒ x = 85°
Solution:
(4) 85°

Question 5.
The correct statement out of the following is

(1) ∆ABC ≅ ∆DEF
(2) ∆ABC ≅ ∆DFE
(3) ∆ABC ≅ ∆FDE
(4) ∆ABC ≅ ∆FED
Hint: ∠C = ∠D; ∠B = E; ∠A = ∠F
Solution:
(4) ∆ABC = ∆FED

Question 6.
If the diagonal of a rhombus are equal, then the rhombus is a
(1) Parallelogram but not a rectangle
(2) Rectangle but not a square
(3) Square
(4) Parallelogram but not a square
Solution:
(3) Square

Question 7.
If bisectors of ∠A and ∠B of a quadrilateral ABCD meet at O, then ∠AOB is

Hint:

Solution:
(2) \(\frac{1}{2}(\angle \mathbf{C}+\angle \mathbf{D})\)

Question 8.
The interior angle made by the side in a parallelogram is 90° then the parallelogram
is a
(1) rhombus
(2) rectangle
(3) trapezium
(4) Kite
Hint:
If one angle of a parallelogram is 90°, then it is a rectangle
Solution:
(2) rectangle

Question 9.
Which of the following statement is correct?
(1) Opposite angles of a parallelogram are not equal.
(2) Adjacent angles of a parallelogram are complementary.
(3) Diagonals of a parallelogram are always equal.
(4) Both pairs of opposite side of a parallelogram are always equal.
Hint:
Opposite sides of a parallelogram are equal.
Solution:
(4) Both pairs of opposite side of a parallelogram are always equal

Question 10.
The angles of the triangle are 3x – 40, x + 20 and 2x – 10 then the value of x is
(1) 40°
(2) 35°
(3) 50°
(4) 45°
Hint: 3x – 40 + x + 20 + 2x- 10 – 180° ⇒ 6x = 210 ⇒ x = 35°
Solution:
(2) 35

Question 11.
PQ and RS are two equal chords of a circle with centre O such that ∠POQ = 70°, then ORS =
(1) 60°
(2) 70°
(3) 55°
(4) 80°
Solution:
(3) 55°
Hint:

Question 12.
A chord is at a distance of 15cm from the centre of the circle of radius 25cm. The length of the chord is
(1) 25 cm
(2) 20 cm
(3) 40 cm
(4) 18 cm
Hint:

Solution:
(3) 40 cm

Question 13.
In the figure, O is the centre of the circle and ∠ACB = 40° then ∠AOB =
(1) 80°
(2) 85°
(3) 70°
(4) 65°

Hint:

Solution:
(1) 80°

Question 14.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is

(1) 30°
(2) 20°
(3) 15°
(4) 25°
Hint:

Solution:
(1) 30°

Question 15.
In the figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. The radius of the circle is
(1) 8 cm
(2) 4 cm
(3) 6 cm
(4) 10 cm
Hint:

Solution:
(4) 10 cm

Question 16.
In the figure, PQRS and PTVS are two cyclic quadrilaterals, if ∠QRS = 80°, then ∠TVS =
(1) 80°
(2) 100°
(3) 70°
(4) 90°

Hint:

Solution:
(1) 80°

Question 17.
If one angle of a cyclic quadrilateral is 75°, then the opposite angle is
(1) 100°
(2) 105°
(3) 85°
(4) 90°
Hint: 180° – 75° =105°
Solution:
(2) 105°

Question 18.
In the figure, ABCD is a cyclic quadrilateral in which DC produced to E and CF is drawn parallel to AB such that ∠ADC = 80° and ∠ECF = 20°, then ∠BAD = ?
(1) 100°
(2) 20°
(3) 120°
(4) 110°

Hint:

Solution:
(3) 120°

Question 19.
AD is a diameter of a circle and AB is a chord. If AD = 30 cm and AB = 24 cm then the distance of AB from the centre of the circle is
(1) 10 cm
(2) 9 cm
(3) 8 cm
(4) 6 cm
Hint:

Question 20.
In the given figure, If OP = 17cm PQ = 30 cm and OS is perpendicular to PQ, then RS is
(1) 10 cm
(2) 6 cm
(3) 7 cm
(4) 9 cm

Hint:

Solution:
(4) 9 cm

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Additional Questions

Answer the following questions.

Question 1.
How many thousands are there in 1 lakhs?
Solution:
\(\frac{1,00,0000}{1000}\) = 100 Thousands

Question 2.
The difference between successor and predecessor of any number is 2. Is it true? Justify your answer.
Solution:
It is true that the difference between successor and predecessor of any number is 2.
Because the difference between any number and its predecessor is 1.
Also the difference between the number and its successor is 1.
The total difference is 2.

Question 3.
The expanded form of the number 6,00,001 is given as 6 × 100000 + 1 × 1. Can you write like this Comment.
Solution:
Yes. We can write the expansion of the number 600001 as 6 × 100000 + 1 × 1.
Because 6 × 100000 + 1 × 1 = 600000 + 1 = 600001

Question 4.
Write the greatest five digit number using the digits 2, 3, 4, 0 and 7.
Solution:
Greatest five digit number = 74320

Question 5.
Can you write the least five digit number using the digits 2,3,4,0 and 7 as 02347. Why? What will be the correct number?
Solution:
No, we cannot write the least five digit number using the digits 2, 3, 4, 0 and 7 as 02347. If it is 02347, the left most zero has no value. It becomes 4 digit number 2347.
The correct number will be 20347.

Question 6.
Write the relation between Largest two digit number and Smallest three digit number.
Solution:
Largest two digit number + 1 = Smallest three digit number.
99 + 1 = 100

Question 7.
Name the property being illustrated in each of the cases.

1. (30 + 20) + 10 = 30 + (20 + 10)
2. 10 × 35 = (10 × 30) + (10 × 5)

Solution:

1. Associativity
2. Distribution of multiplication over addition.

Question 8.
10 crore = ____
Solution:
100 million

Question 9.
The heights of five boys in class VI are 135, 141, 129, 132, 145 (in centimetres) in height. Arrange their heights as how they stand in the assembly?
Solution:
129 cm < 132 cm < 135 cm < 141 cm < 145 cm

Question 10.
The number lock has the password number with 3 digits. The number is the least even number and less than 200. Middle digit has no value separately. Find the password. The digits are used only once.
Solution:
102

Question 11.
Arrange in ascending order. 123456, 123546, 123623, 123511
Solution:
123456 < 123511 < 123546 < 123623 Question 12. Arrange in descending order. 8461, 7535, 2943, 6214 Solution: 8461 > 7535 > 6214 > 2943

Question 13.
Find the numbers between 572634 and 562634 which is approximated to ten thousand place.
Solution:
562634 < 570000 < 572634

Question 14.
Evaluate the following:
(a) 44 ÷ 2 + (7 + 80 ÷ 10) – 14 + 23
(b) 17 × 6 – 4 – 2 + 20 – (22 + 18)
(c) 16 × 144 ÷ 16 ÷ 9 + 16 + 15 – 20
(d) 12 × 36 ÷ 12 ÷ 3 + 5 + 6 – 2
(e) 15 – [17 + 30 ÷ 6 – (6 + 6) + 7]
Solution:
(a) 44 ÷ 2 + (7 + 80 ÷ 10) – 14 + 23 (Given)
= 44 ÷ 2 + (7 + 8) – 14 + 23 (To complete the bracket ÷ done first)
= 44 ÷ 2 + 15 – 14 + 23 (Bracket completed second)
= 22 + 15 – 14 + 23 (÷ completed third)
= 37 – 37 (+ completed fourth)
= 0 (- completed last)
∴ 44 ÷ 2 + (7 + 80 ÷ 10) – 14+ 23 = 0.

(b) 17 × 6 – 4 – 2 + 20 – (22 + 18) (Given)
= 17 × 6 – 4 – 2 + 20 – 40 (Bracket completed first)
= 102 – 4 – 2 + 20 – 40 (× completed second)
= 102 – 4 – 22 – 40 (+ completed third)
= 98 – 22 – 40 (÷ completed one by one)
= 76 – 40
= 36
∴ 17 × 6 – 4 – 2+ 20 – (22 + 18) = 36

(c) 16 × 144 ÷ 16 ÷ 9 + 16 + 15 – 20 (Given)
= 16 × 9 ÷ 9+16 + 15 – 20 (÷ completed first)
= 16 × 1 + 16 + 15 – 20 (÷ completed second)
= 16 + 16 + 15 – 20 (× completed third)
= 32 + 15 – 20 (+ completed fourth)
= 47 – 20 (+ completed fifth)
= 27 (- completed last)
∴ 16 × 144 ÷ 16 ÷ 9 + 16 + 15 – 20 = 27

(d) 12 × 36 ÷ 12 ÷ 3 + 5 + 6 – 2 (Given)
= 12 × 3 ÷ 3 + 5 + 6 – 2 (÷ completed first)
= 12 × 1 + 5 + 6 – 2 (÷ completed second)
= 12 + 5 + 6 – 2 (× completed third)
= 17 + 6 – 2 (+ completed forth)
= 23 – 2 (+ completed fifth)
= 21 (- completed last)
∴ 12 × 36 ÷ 12 ÷ 3 + 5 + 6 – 2 = 21

(e) 15 – [17 + 30 ÷ 6 – (6 + 6) + 7] (Given)
= 15 – [17 + 30 ÷ 6 – 12 + 7] (Inner bracket completed first)
= 15 – [17 + 5 – 12 + 7] (÷ completed second)
= 15 – [22 – 19] (+ completed third)
= 15 – 3 (bracket completed forth)
= 12 (- completed last)
∴ 15 – [17 + 30 ÷ 6 – (6 + 6) + 7] = 12.

Question 15.
An export company produced 235219 shirts, 158342 trousers and 11704 jackets in a year. What is the total production of all the three items in that year?
Solution:
Number of shirts produced = 235219
Number of trousers produced = 158342
Number of jackets produced = 11704
Total production of all items = 405265
Total production of all items in that year = 4,05,265

Question 16.
India’s population has been steadily increasing from 439 million in 1961 to 1028 million in 2001. Find the total increase in population from 1961 to 2001. Write the increase in population in the Indian system of Numeration using commas suitably.
Solution:
Population of India in 1961 = 439 millions = 439,000,000
Population of India in 2001 = 1028 millions = 1,028,000,000
Increase in population from 1961 to 2001 = Population in 2001 – Population in 1961
= 1028000000 – 439000000
= 589000000
= 589 million.
Increase in population in Indian System = 58,90,00,000

Question 17.
A person had ₹ 10,00,000 with him. He purchased a flat for ₹ 8,70,000. With the remaining money, he has to buy a T.V. for 1 lakh. How much money was left with him to buy a T.V?
Solution:
Total money the person had = ₹ 10,00,000
Cost of flat = ₹ 8,70,000
Remaining money = ₹ 1,30,000
Now he has ₹ 1,30,000. So it is enough to buy a TV for ₹ 1,00,000.

Question 18.
A box contains 50 packets of biscuits, each weighing 120g. How many such boxes can be loaded in a van, which cannot carry more than 900 kg?
Solution:
Given: Total number of packets = 50.
Weight of each packet = 120 g
Weight of a box = 50 × 120 g = 6000 g = 6 kg [∵ 1000 g = 1 kg]
Required number of boxes = \(\frac{900}{6}\) = 150.
150 boxes are required.

Question 19.
How much money was collected from 5342 students for a charity show, if each student contributed ₹ 670?
Solution:
Total number of students = 5342
Contribution of each student = ₹ 670
Total money collected = 5342 × 670 = ₹ 35,79,140
Total money collected = ₹ 35,79,140

Question 20.
Estimate the following to the nearest hundreds
(a) 439 + 334 + 4317
(b) 1,08,734 – 47,599
(c) 8325 – 491
(d) 4,89,348 – 48,365
Solution:
(a) 439 + 334 + 4317
439 ⇒ 400
334 ⇒ 300
4317 ⇒ 4300
Sum = 5,000

(b) 1,08,734 – 47,599
1,08,734 ⇒ 1,08,700
47,599 ⇒ 47,600
Difference = 61,100

(c) 8325 – 491
8325 ⇒ 8300
491 ⇒ 500
Differences = 7,800

(d) 4,89,348 – 48,365
4,89,348 ⇒ 4,89,300
48,365 ⇒ 48,400
Difference = 4,40,900

Question 21.
Estimate the following products:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Solution:
(a) 578 × 161
578 ⇒ 600
161 ⇒ 200
Estimated product is 600 × 200 = 1,20,000

(b) 5281 × 3491
5281 ⇒ 5000
3491 ⇒ 3500
Estimated Product = 5000 × 3500 = 1,75,00,000

(c) 1291 × 592
1291 ⇒ 1300
592 ⇒ 600
Estimated Product is = 1300 × 600 = 7,80,000

(d) 9250 × 29
9250 ⇒ 9000
29 ⇒ 30
Estimated Product is 9000 × 30 = 2,70,000

Question 22.
Are all whole numbers are natural numbers? Justify your answer?
Solution:
No, all whole numbers are not natural numbers.
Because ‘0’ belongs to the whole number system. But it is not in a natural number system.
All whole numbers except ‘0’ are natural numbers.

Question 23.
Use associative property of addition to add 847 + 306 + 453
Solution:
847 + 306 + 453
= (847 + 453) + 306
= 1300 + 306
= 1606
∴ 847 + 306 + 453 = 1606

Question 24.
Find the value of (1063 × 127) – (1063 × 27)
Solution:
(1063 × 127) – (1063 × 27)
= 1063 (127 – 27) [Taking 1063 as common]
= 1063 × 100
= 106300.
i.e (1063 × 127) – (1063 × 27) = 106300

Question 25.
Find the product using suitable properties
(a) 738 × 103
(b) 1005 × 168
Solution:
(a) We have 738 × 103
= 738 × (100 + 3)
= 738 × 100 + 738 × 3 [By distributive property of multiplication over addition]
= 73800 + 2214
= 76014

(b) 1005 × 168
= (1000 + 5) × 168
= (168 × (1000 + 5) (By commutative property)
= (168 × 1000) + (168 × 5)
= 1,68,000 + 840
= 1,68,840

Question 26.
Write the largest six-digit number and write the number names in words using the Indian and International system.
Solution:
The largest six-digit number is 999999
Number names are nine lakh ninety-nine thousand nine hundred and ninety-nine

Question 27.
In a mobile store, the number of mobiles sold during a month is 1250, Assuming that the same number of mobiles are sold every month, find the number of mobiles sold in 2 years.
Solution:
Number of mobiles sold in 1 month = 1250
1 year = 12 months
2 years = 2 × 12 = 24 months
Number of mobiles sold in 24 months = 1250 × 24= 30,000
Number of mobiles sold in 2 years = 30,000

Question 28.
Simplify 24 + 2 × 8 ÷ 2 – 1
Solution:
24 + 2 × 8 ÷ 2 – 1
= 24 + 2 × 4 – 1
= 24 + 8 – 1
= 32 – 1
= 31

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 2 Introduction to Algebra Additional

Questions

Question 1.
Additive identity ____
Solution:
0

Question 2.
Multiplicative identity ____
Solution:
1

Question 3.
Express to an algebraic statement.
(i) ‘t’ is added to 100
(ii) 4 less to 9 times of y.
Solution:
(i) t + 100
(ii) 9y – 4

Question 4.
Find the rule which gives the number of sticks in the following pattern.

Solution:
Let ‘x’ be the no. of R’s formed.

The rule is 6x.
Let y ’ be the no. of S’s formed.

The rule is 5y.

Question 5.
How old was Suja 6 years from now?
Solution:
Let Suja’s present age be ‘a’ years.
6 years from now Suja will be (a + 6) years old.

Question 6.
Price of Apple per kg is ₹ 50 more than price of orange per kg. What is the cost of Apple per kg?
Solution:
Let the price of orange be ₹ b
Price of Apple will be ₹ (b + 50)

Question 7.
Given ‘n’ students like ice cream. What may 2n show?
Solution:
2n shows double the number of students who like ice cream.

Question 8.
Price of oil per litre is ₹ 5 more than three times the price of cool drinks ₹ ‘p’ Express algebraically.
Solution:
Price of cool drinks per kg = ₹ p
Three times = 3p
5 Rs. more = 3p + 5
Price of oil per kg = ₹ (3p + 5)

Question 9.
Complete the table and by inspection of the table find the value of m when m + 10 = 16.

Solution:

From the table m+ 10 = 16 when m = 6.

Question 10.
Express algebraically (a) y divided by r (b) double times x is subtracted from 10
Solution:
(a) \(\frac{y}{r}\)
(b) 10 – 2x

Question 11.
Give verbal expression of
(a) 7x + 18
(b) \(\frac{4 x}{3}\)
Solution:
(a) 18 added to 7 times x
(b) 4 times x divided by 3.

Question 12.
Rajini’s Father’s age is 5 years more than 3 times Rajini’s age. What is her father’s age?
Solution:
3x + 5

Question 13.

Find the rule for the above pattern.
Solution:
2p

Question 14.
Prepare a table for 3x + 10. From the table find the value of x when 3x + 10 = 25.
Solution:
5

Question 15.
Complete the table and find the solution of the equation \(\frac{z}{3}=4\) using the table.

Solution:

Question 16.
Form the expression for which Ramu is 3 years younger than Mathu.
Solution:
m – 3

Question 17.
A tap is to be pasted along the edges of a square shaped gift box. Its length is 4 cm. What is the length of tap needed for one side.
Solution:
\(\frac{4 p}{4}=p\)

Question 18.
The value of y in 7y – 20 = 99.
Solution:
y = 17

Question 19.
Nine added to two times x gives 301. Find the value of x.
Solution:
x = 146

Question 20.
Aarthi is 3 years younger to Harini. If the sum of their ages is 23, how old is Harini?
Solution:
Let Harini’s age be x years
Aarthi’s age is x – 3 years
Given sum of their ages is 23.
i.e., x + (x – 3) = 23

Students can Download Maths Chapter 5 Information Processing Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Intext Questions

Try these (TextBook Page No 122)

Question 1.
Find the number of all possible triangles that can be formed from the triangle given below.
Solution:

Question 2.
Use the given figure to form a 3 × 3 magic square.

Solution:

Question 3.
Convert the tree diagram into a numeric expression

Solution:
(10 × 5) + (9 × 4)

Question 4.
(i) Find out the total time taken by the bus to reach from A to E via B , C and D.
(ii) Find which is the shortest route from A to E.
Solution:

(i) Route ⇒ 2A ➝ B ➝ C ➝ D ➝ E
Time taken ⇒ (7 + 5 + 3 + 6) hrs = 21 hrs

Try this (TextBook Page No 127)

Question 1.
Kumar has four different hats. He always wears a hat. Sometimes he wears the same hat more than once in a week as in the figure.

In how many different ways might he decide to wear his hats in one week?
Solution:

H1, H2, H3, H4 be Hat 1, Hat 2, Hat 3, Hat 4 respectively.
∴ In 7 × 4 = 28 different ways kumar may decide to wear his hat in one week.

Exercise 5.1

Try these (TextBook Page No 134)

Question 1.
Find any four SETs among these set of cards.

Solution:

Question 2.
This is an example for a magic square in SETs, can you make another one?

Solution:

Try this (TextBook PageNo 137)

Question 1.
Think why this graph colouring is invalid?

Solution:
The objective of graph colouring is to assign minimum number of colours to the verties do not have the same colour.
Here adjacent edges have the same colour.
∴ It is invalid.

Students can Download Maths Chapter 1 Rational Numbers Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Rational Numbers Intext Questions

Exercise 1.1
Try This (Textbook Page No. 6)
Question 1.
Is the number \(\frac{7}{-5}\) a fraction or a rational number ? Why?
Solution:
\(\frac{7}{-5}\) is a rational number
Because of rational number is a rational number which is of the form \(\frac{p}{q}\) , q ≠ 0 and p and q are integers.
But fraction is part of a whole.

Question 2.
Write any 6 rational numbers of your choice.
Solution:
0, \(\frac{-1}{2}\), \(\frac{1}{2}\), \(\frac{3}{4}\), \(\frac{6}{7}\), -5, 6

Try This (Textbook Page No. 7)

Question 1.
Explain why the following statements are true?
(i) 0.8 = \(\frac{4}{5}\)
(ii) 1.4 > \(\frac{1}{4}\)
(iii) 0.74 < \(\frac{3}{4}\) (iv) 0.4 > 0.386
(v) 0.096 < 0.24
(vi) 1.128 = 0.1280
Solution:


(vi) 0.128 = \(\frac{128}{1000}=\frac{1280}{10000}\)
0.1280 = \(\frac{1280}{10000}\)
∴ Both are equal. i.e., 0.128 = 0.1280

Try This (Textbook Page No. 9)

Question 1.
Which of the pairs are equivalent rational numbers?

Solution:

Question 2.
Find the standard form of
(i) \(\frac{36}{-96}\)
(ii) \(\frac{-56}{-72}\)
(iii) \(\frac{27}{18}\)
Solution:

Question 3.
Mark the following rational numbers on a number line.
(i) \(\frac{-2}{3}\)
(ii) \(\frac{-8}{-5}\)
(iii) \(\frac{5}{-4}\)
Solution:
(i) \(\frac{-2}{3}\) lies between 0 and -1.
The unit part between 0 and -1 is divided into 3 equal parts and second part is taken.

(ii) \(\frac{-8}{-5}=1 \frac{3}{5}\)
\(1 \frac{3}{5}\) lies between 1 and 2. The unit part between 1 and 2 is divided into 5 equal parts and the third part is taken.

(iii) \(\frac{5}{-4}=-\frac{5}{4}=-1 \frac{1}{4}\)
\(-1 \frac{1}{4}\) lies between -1 and The unit part between -1 and -2 is divided into four equal parts and the first part is taken.

Try This (Textbook Page No. 15)

Question 1.
Are there any rational numbers between \(\frac{-7}{11}\) and \(\frac{6}{-11}\) ?
Solution:

Try This (Textbook Page No. 19)

Question 1.
Divide: (i) 5 by \(\frac{-7}{3}\)
(ii) \(\frac{-7}{3}\) by 5
Solution:

Exercise 1.2
Try This (Textbook Page No. 25)

Question 1.
The closure property on integers holds for subtraction and not for division. What about rational numbers? Verify.
Solution:
Let 0 and \(\frac{1}{2}\) be two rational numbers 0 – \(\frac{1}{2}\) = \(\frac{1}{2}\) is a rational number
∴ Closure property for subtraction holds for rational numbers.
But consider the two rational number \(\frac{5}{2}\) and 0.
\(\frac{5}{2} \div 0=\frac{5}{2 \times 0}=\frac{5}{0}\)
Here denominator = 0 and it is not a rational number.
∴ Closure property is not true for division of rational numbers.

Question 2.
Check whether \(\frac{3}{5}-\frac{7}{8}=\frac{7}{8}-\frac{3}{5}\)
Solution:

∴ Subtraction of rational numbers is not commutative

Question 3.
Is \(\frac{3}{5} \div \frac{7}{8}=\frac{7}{8} \div \frac{5}{3}\)? So, what do you conclude?
Solution:

∴ Commutative property not hold good for division of rational numbers.

Try This (Textbook Page No. 26)

Question 1.
Check whether the associative property holds for subtraction and division.
Solution:
Consider the rational numbers \(\frac{2}{3}\), \(\frac{1}{2}\) and \(\frac{3}{4}\)


∴ Associative property does not hold for division of rational numbers

Question 2.
Observe that

Use your reasoning skills, to find the sum of the first 7 numbers in the pattern given above.
Solution:

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

Exercise 3.1

Question 1.
Classify the following polynomials based on number of terms.
(i) x³ – x²
(ii) 5x
(iii) 4x⁴ + 2x³ + 1
(iv) 4.x³
(v) x + 2
(vi) 3x²
(vii) y⁴ + 1
(viii) y²⁰ + y¹⁸ + y²
(ix) 6
(x) 2u³ + u² + 3
(xi) u²³ – u⁴
(xii) y
Solution:
5x, 3x², 4x³, y and 6 are monomials because they have only one term.
x³ – x², x + 2, y⁴ + 1 and u²³ – u⁴ are binomials as they contain only two terms.
4x⁴ + 2x³ + 1 , y²⁰ + y¹⁸ + y² and 2u³ + u² + 3 are trinomials as they contain only three terms.

Question 2.
Classify the following polynomials based on their degree.
Solution:
p(x) = 3, p(x) = -7, p(x) = \(\frac{3}{2}\) are constant polynomials
p(x) = x + 3, p(x) = 4x, p(x) = \(\sqrt{3}\)x + 1 are linear polynomials, since the highest degree of the variable x is one.
p(x) = 5x² – 3x + 2, p(y) = \(\frac{5}{2}\) y² + 1, p(x) = 3x² are quadratic polynomials, since the highest degree of the variable is two.
p(x) = 2x³ – x² + 4x + 1, p(x) = x³ + 1, p(y) = y³ + 3y are cubic polynomials, since the highest degree of the variable is three.

Question 3.
Find the product of given polynomials p(x) = 3x³+ 2x – x² + 8 and q (x) = 7x + 2.

Solution:
(7x +2) (3x³ + 2x – x² + 8) = 7x(3x³ + 2x – x² + 8) + 2x
(3x³ + 2x – x² + 8) = 21x⁴ + 14x² – 7x³ + 56x + 6x³ + 4x – 2x²+ 16 = 21x⁴ – x³ + 12x⁴ + 60x + 16

Question 4.
Let P(x) = 4x⁴ – 3x + 2x³ + 5 and q(x) = x² + 2x + 4 find p (x) – q(x).

Solution:

Exercise 3.2

Question 1.
If p(x) = 5x³ – 3x² + 7x – 9, find
(i) p(-1)
(ii) p(2).
Solution:
Given that p(x) = 5x³ – 3x² + 7x – 9
(i) p(-1) = 5(-1)³ – 3(-1)² + 7(-1) – 9 = -5 – 3 – 7 – 9
= -24
(ii) p(2) = 5(2)³ – 3(2)³ + 7(2) – 9 = 40 – 12 + 14 – 9
∴ p(2) = 33

Question 2.
Find the zeros of the following polynomials.
(i) p(x) = 2x – 3
(ii) p(x) = x – 2
Solution:

Exercise 3.3

Question 1.
Find the remainder using remainder theorem, when
(i) 4x³ – 5x² + 6x – 2 is divided by x – 1.
(ii) x³ – 7x² – x + 6 is divided by x + 2.
Solution:
(i) Let p(x) = 4x³ – 5x² + 6x – 2. The zero of (x – 1) is 1.
When p(x) is divided by (x – 1) the remainder is p( 1). Now,
p(1) = 4(1)³ – 5(1)² + 6(1) – 2 = 4 – 5 + 6 – 2 = 3
∴ The remainder is 3.

(ii) Let p(x) = x³ – 7x² – x + 6. The zero of x + 2 is -2.
When p(x) is divided by x + 2, the remainder is p(-2). Now,
p(-2) = (-2)³ – 7(-2)² – (-2) +6 = – 8 – 7(4) + 2 + 6
= – 8 – 28 + 2 + 6 = -28.
∴ The remainder is -28.

Question 2.
Find the value of a if 2x³ – 6x² + 5ax – 9 leaves the remainder 13 when it is divided by x – 2.
Solution:
Let p(x) = 2x³ – 6x² + 5ax – 9
When p(x) is divided by (x – 2) the remainder is p(2).
Given that p(2) = 13
⇒ 2(2)³ – 6(2)² + 5a(2) – 9 = 13
⇒ 2(8) – 6(4) + 10a – 9 = 13
⇒ 16 – 24 + 10a – 9 = 13
⇒ 10a – 17 = 13
⇒ 10a = 30
⇒ ∴ a = 3

Question 3.
If the polynomials f(x) = ax³+ 4x² + 3x – 4 and g (x) = x³ – 4x + a leave the same remainder when divided by x – 3. Find the value of a. Also find the remainder.

Solution:
Let f(x) = ax³ + 4x² + 3x – 4 and g(x) = x³ – 4x + a,When f(x) is divided by (x – 3), the remainder is f(3).
Now f(3) = a(3)³4(3)² + 3(3) – 4 = 27a + 36 + 9 – 4
f (3) = 27a + 41 …(1)
When g(x) is divided by (x – 3), the remainder is g(3).
Now g(3) = 33 – 4(3) + a = 27 – 12 + a = 15 + a …(2)
Since the remainders are same, (1) = (2)
Given that, f(3) = g(3)
That is 27a + 41 = 15 +a
27a – a = 15 – 41
26a = -26

Sustituting a = -1, in f(3), we get
f(3) = 27(-1) + 41 = -27 + 41
f(3) = 14
∴ The remainder is 14.

Question 4.
Show that x + 4 is a factor of x³ + 6x² – 7x – 60.
Solution:

Question 5.
In (5x + 4) a factor of 5x³ + 14x² – 32x – 32
Solution:

Question 6.
Find the value of k, if (x – 3) is a factor of polynomial x³ – 9x² + 26x + k.
Solution:
Let p (x) = x³ – 9x² + 26x + k
By factor theorem, (x – 3) is a factor of p(x), if p(3) = 0
p(3) = 0
3³ – 9(3)² + 26 (3) + k = 0
27 – 81 + 78 + k = 0
k = -24

Question 7.
Show that (x – 3) is a factor of x³ + 9x² – x – 105.
Solution:
Let p(x) = x³ + 9x² – x – 105
By factor theorem, x – 3 is a factor of p(x), if p(3) = 0
p (3) = 3³ + 9(3)³ – 3 – 105
= 27 + 81 – 3 – 105
= 108 – 108
p(3) = 0

Therefore, x – 3 is a factor of x³ + 9x² – x – 105.

Question 8.
Show that (x + 2) is a factor of x³ – 4x² – 2x + 20.
Solution:
Let p(x) = x³ – 4x² – 2x + 20
By factor theorem,
(x + 2) is factor of p(x), if p(-2) = 0
p(-2) = (-2)³ – 4(-2)² – 2(-2) + 20 = -8 – 4(4) + 4 + 20
P(- 2) = 0
Therefore, (x + 2) is a factor of x³ – 4x² – 2x + 20

Exercise 3.4

Question 1.
Expand the following using identities :
(i) (7x + 2y)²
(ii) (4m – 3m)²
(iii) (4a + 3b) (4a – 3b)
(iv) (k + 2)(k – 3)
Solution:
(i) (7x + 2y)² = (7x)² + 2 (7x) (2y) + (2y)² = 49x² + 28xy + 4y²
(ii) (4m – 3m)² = (4m)² – 2(4m) (3m) + (3m)² = 16m² – 24mn + 9n²
(iii) (4a + 3b) (4a – 3b) [We have (a + b ) (a – b) = a² – b² ]
Put [a = 4a, b = 3b]
(4a + 3b) (4a – 3b) = (4a)² – (3b)² = 16a² – 9b²
(iv) (k + 2)(k – 3) [We have (x + a) (x – b) = x² + (a – b)x – ab]
Put [x = k, a = 2, b = 3]
(k + 2) (k – 3) = k² + (2 – 3)x – 2 × 3 = k² – x – 6

Question 2.
Expand : (a + b – c)²
Solution:
Replacing ‘c’ by ‘-c’ in the expansion of
(a + b + c)² = a² + b² + c² + 2 ab+ 2 bc + 2ca
(a + b + (-c))² = a² + b² + (-c)² + 2ab + 2b(-c) + 2(-c)a
= a² + b² + c² + 2ab – 2bc – 2ca

Question 3.
Expand : (x + 2y + 3z)²
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Substituting, a = x, b = 2y and c = 3z
(x + 2y + 3 z)² = x² + (2y)² + (3z)² + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)
= x² + 4y² + 9z² + 4xy + 12y² + 6zx

Question 4.
Find the area of square whose side length is m + n – q.
Solution:
Area of square = side × side = (m + n – q)²
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
[m + n + (-q)]² = m² + n² + (-q)² + 2mn + 2n(-q) + 2(-q)m
= m² + n² + q² + 2mn – 2nq – 2qm
Therefore, Area of square = m² + n² + q² + 2mn – 2nq – 2qm = [m² + n² + q² + 2mn – 2nq – 2qm] sq. units.

EXERCISE 3.5

Question 1.
Factorise the following
(i) 25m^-2 – 16n²
(ii) x⁴ – 9x²
Solution:
(i) 25m² – 16n² = (5m)² – (4n)²
= (5m – 4n) (5m + 4n) [∵ a² – b² = (a – b)(a + b)]
(ii) x⁴ – 9x² = x²(x² – 9) = x²(x² – 3²) = x²(x – 3)(x + 3)

Question 2.
Factorise the following.
(i) 64m³ + 27n³
(ii) 729x³ – 343y³
Solution:
(i) 64m³ + 27n³ = (4m)³ + (3n)³
= (4m + 3n) ((4m)² – (4m) (3n) + (3n)²) [∵ a² + b² = (a + b) (a² – ab + b²)]
= (4m + 3n) (16m² – 12mn + 9n²)
(ii) 729x³ – 343y³ = (9x)³ – (7y)³ = (9x – 7y) ((9x)² + (9x) (7y) + (7y)²
= (9x – 7y) (81x² + 63xy + 49y²)
[∵ a³ – b³ = (a – b) (a² + ab + b²)]

Question 3.
Factorise 81x² + 90x + 25
Solution:
81x² + 90x + 25 = (9x)² + 2 (9x) (5) + 5²
= (9x + 5)² [ ∵ a² + 2ab + b² = (a + b)²]

Question 4.
Find a³ + b² if a + b = 6, ab = 5.
Solution:
Given, a + b = 6, ab = 5
a² + b² = (a + b)² – 3ab(a + b) = (6)³ – 3(5)(6) = 126

Question 5.
Factorise (a – b)² + 7(a – b) + 10
Solution:
Let a – b = p, we get p² + 7p + 10,

p² + 7p + 10 = p² + 5p + 2p + 10
= p(p + 5) + 2(p + 5) = (p + 5)(p + 2)
Put p = a – b we get,
(a – b)² + 7(a – b) + 10 = (a – b + 5)(a – b + 2)

Question 6.
Factorise 6x² + 17x + 12
Solution:

Exercise 3.6

Question 1.
Factorise x² + 4x – 96
Solution:

Question 2.
Factorise x² + 7xy – 12y²
Solution:

Question 3.
Find the quotient and remainder when 5x³ – 9x² + 10x + 2 is divided by x + 2 using synthetic division.
Solution:

Hence the quotient is 5x² – 19x + 48 and remainder is -94.

Question 4.
Find the quotient and remainder when -7 + 3x – 2x² + 5x³ is divided by -1 + 4x using synthetic division.
Solution:

Question 5.
If the quotient on dividing 5x⁴ + 4x³ + 2x + 1 by x + 3 is 5x³ + 9x² + bx – 97 then find the values of a, b and also remainder.
Solution:

Quotient 5x³ + 11x² + 33x – 97 is compared with given quotient 5x³ + ax² + bx – 97
co-efficient of x² is – 11 = a and co-efficient of x is 33 = b.
Therefore a = -11, b = 33 and remainder = 292.

Exercise 3.7

Question 1.
Find the quotient and the remainder when 10 – 4x + 3x² is divided by x – 2.
Solution:
Let us first write the terms of each polynomial in descending order (or ascending
order). Thus the given problem becomes (3x² – 4x + 10) ÷ (x – 2)

∴ Quotient = 3x + 2 and Remainder = 14, i.e 3x² – 4x + 10 = (x – 2) (3x + 2) + 14 and is in the form
Dividend = (Divisor × Quotient) + Remainder

Question 2.
If 8x³ – 14x² – 19x – 8 is divided by 4x + 3 then find the quotient and the remainder.
Solution:

Exercise 3.8

Question 1.
Factorise 2x³ – x² – 12x – 9 into linear factors.
Solution:
Let p(x) = 2x³ – x² – 12x – 9
Sum of the co-efficients = 2 – 1 – 12 – 9 = -20 ≠ 0
Hence x – 1 is not a factor
Sum of co-efficients of even powers with constant = -1 – 9 = -10
Sum of co-efficients of odd powers = 2 – 12 = -10
Hence x + 1 is a factor of x.
Now we use synthetic division to find the other factors.


Then p (x) = (x + 1)(2x² – 3x – 9)
Now 2x² – 3x – 9 = 2x² – 6x + 3x – 9 = 2x(x – 3) + 3(x – 3)
= (x – 3)(2x + 3)
Hence 2x³ – x² – 12x – 9 = (x + 1) (x – 3) (2x + 3)

Question 2.
Factorize x³ + 3x² – 13x – 15.
Solution:
Let p(x) = x³ + 3x² – 13x – 15
Sum of all the co-efficients = 1 + 3 – 13 – 15 = -24 + 0
Hence x – 1 is not a factor
Sum of co-efficients of even powers with constant = 3 – 15 = – 12
Sum of coefficients of odd powers = 1 – 13 = – 12
Hence x + 1 is a factor of p (x)
Now use synthetic division to find the other factors.

Hence p(x) = (x + 1) (x² + 2x – 15)
Now x³ + 3x² – 13x – 15 = (x + 1)(x² + 2x – 15)
Now x² + 2x – 15 = x² + 5x – 3x – 15 = x(x + 5) – 3 (x + 5) = (x + 5)(x – 3)
Hence x³ + 3x² – 13x – 15 = (x + 1)(x + 5)(x – 3)

Question 3.
Factorize x³ + 13x² + 32x + 20 into linear factors.

Solution:
Let, p(x) = x³ + 13x² + 32x + 20
Sum of all the coefficients = 1 + 13 + 32 + 20 = 66 ≠ 0
Hence, (x – 1) is not a factor.
Sum of coefficients of even powers with constant = 13 + 20 = 33
Sum of coefficients of odd powers = 1 + 32 = 33
Hence, (x + 1) is a factor of p(x)
Now we use synthetic division to find the other factors

Exercise 3.9

Question 1.
Find GCD of 25x³y² z, 45x²y⁴z³b
Solution:

Question 2.
Find the GCD of (y³ – 1) and (y – 1).
Solution:
y³ – 1 = (y – 1)(y³ + y + 1)
y – 1 = y – 1
Therefore, GCD = y – 1

Question 3.
Find the GCD of 3x² – 48 and x² – 7x + 12.
Solution:
3x² – 48 = 3(x² – 16) = 3(x² – 4³) = 3(x + 4)(x – 4)
x² – 7x + 12 = x² – 3x – 4x + 12 = x(x – 3) – 4 (x – 3) = (x – 3) (x – 4)
Therefore, GCD = x – 4

Question 4.
Find the GCD of a^x , a^{x + y}, a^{x + y + z}.
Solution:

Exercise 3.10

Question 5.
Solve graphically. x – y = 3 ; 2x – y = 11
Solution:

Exercise 3.11

Question 1.
Solve using the method of substitution. 5x – y = 5, 3x + y = 11
Solution:

Exercise 3.12

Question 2.
Solve by the method of elimination. 2x + y = 10; 5x – y = 11
Solution:

Exercise 3.13

Question 3.
Solve 5x – 2y = 10 ; 3x + y = 17 by the method of cross multiplication.
Solution:

Exercise 3.14

Question 1.
The age of Arjun is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
Solution:
Let the age of the father be x
Let the sum of the age of his sons be y
At present x = 2y ⇒ x – 2y = 0 …. (1)
After 20 years x + 20 = y + 40
x – y = 40 – 20

Exercise 3.15

Multiple Choice Questions :

Question 1.
The polynomial 3x – 2 is a
(1) linear polynomial
(2) quadratic polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is linear if its degree is 1
Solution:
(1) linear polynomial

Question 2.
The polynomial 4x² + 2x – 2 is a
(1) linear polynomial
(2) quadratic polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is quadratic of its highest power of x is 2
Solution:
(2) quadratic polynomial]

Question 3.
The zero of the polynomial 2x – 5 is

Hint:

Solution:
(1) \(\frac{5}{2}\)

Question 4.
The root of the polynomial equation 3x – 1 = 0 is

Hint:

Solution:
(2) x = \(\frac{1}{3}\)

Question 5.
Zero of (7 + 4x) is ___

Solution:
(2) \(\frac{-7}{4}\)

Question 6.
Which of the following has as a factor?
(1) x² + 2x
(2) (x – 1)²
(3) (x + 1)²
(4) (x² – 2²)
Solution:
(1) x² + 2x

Question 7.
If x – 2 is a factor of q (x), then the remainder is _____
(1) q (-2)
(2) x – 2
(3) 0
(4) -2
Solution:
(3) 0

Question 8.
(a – b) (a² + ab + b²) =
Solution:
(1) a³ + b³ + c³ – 3abc
(2) a² – b²
(3) a³ + b³
(4) a³ – b³
Solution:
(4) a³ – b³

Question 9.
The polynomial whose factors are (x + 2) (x + 3) is
(1) x² + 5x + 6
(2) x² – 4
(3) x² – 9
(4) x² + 6x + 5
Solution:
(1) x² + 5x + 6

Question 10.
(-a – b – c)² is equal to
(1) (a – b + c)²
(2) (a + b – c)²
(3) (-a + b + c)²
(4) (a + b + c)²
Solution:
(4) (a + b + c)²

Text Book Activities

Activity – 1
Write the Variable, Coefficient and Constant in the given algebraic expression,
Solution:

Activity – 2
Write the following polynomials in standard form.

Activity – 3

Question 1.
Find the value of k for the given system of linear equations satisfying the condition below:
(i) 2x + ky = 1; 3x – 5y = 7 has a unique solution
(ii) kx + 3y = 3; 12x + ky = 6 has no solution
(iii) (k – 3)x + 3y = k; kx + ky = 12 has infinite number of solution
Solution:

Question 2.
Find the value of a and b for which the given system of linear equation has infinite number of solutions 3x – (a + 1)y = 2b – 1, 5x + (1 – 2a)y = 3b
Solution:

Activity – 4
For the given linear equations, find another linear equation satisfying each of the given condition
Solution:

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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 = ?
(1) -6
(2) -7
(3) -8
(4) 11
Solution:
(4) 11
Hint: P(-2) = (-2)³ + 6 (-2)² + k (-2) + 6 = 0
⇒ – 8 + 24 – 2k +6 = 0
⇒ 22 = 2k
⇒ k = 11

Question 2.
The root of the polynomial equation 2x + 3 = 0 is
(1) \(\frac{1}{3}\)
(2) \(-\frac{1}{3}\)
(3) \(-\frac{3}{2}\)
(4) \(-\frac{2}{3}\)
Solution:
(3) \(-\frac{3}{2}\)

Question 3.
The type of the polynomial 4 – 3x³ is
(1) constant polynomial
(2) linear polynomial
(3) quadratic polynomial
(4) cubic polynomial.
Solution:
(4) cubic polynomial.
Hint: Polynomial of degree 3 is called cubic.

Question 4.
If x⁵¹ + 51 is divided by x + 1, then the remainder is
(1) 0
(2) 1
(3) 49
(4) 50
Solution:
(4) 50
Hint: P(-1 = (-1)⁵¹ + 51 = -1 + 51 = 50

Question 5.
The zero of the polynomial 2x + 5 is

Solution:
(2) \(-\frac{5}{2}\)

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

Question 7.
Degree of the polynomial (y³ – 2) (y³ + 1) is
(1) 9
(2) 2
(3) 3
(4) 6
Solution:
(4) 6
Hint: (y³ – 2)(y³ + 1) = (y³ – 2)(y³ – 2) × 1 = y⁶ – 2y³ + y³ – 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
(1) A, B, D, C
(2) A, B, C, D
(3) B, C, D, A
(4) B, A, C, D
Solution:
(4) B, A, C, D
Hint: Degree of (A), (B) (C) & (D) are respectively be 5, 4, 8, 12

Question 9.
If p (a) = 0 then (x – a) is a ________ of p(x)
(1) divisor
(2) quotient
(3) remainder
(4) factor
Solution:
(4) factor

Question 10.
Zeros of (2 – 3x) is _____
(1) 3
(2) 2
(3) \(\frac{2}{3}\)
(4) \(\frac{3}{2}\)
Solution:
(3) \(\frac{2}{3}\)
Hint: 2 – 3x =0
-3x = – 2
x = \(\frac{2}{3}\)

Question 11.
Which of the following has x – 1 as a factor?
(1) 2x – 1
(2) 3x – 3
(3) 4x – 3
(4) 3x – 4
Solution:
(2) 3x – 3
Hint: p(x) = 3x – 3
P( 1) = 3(1) – 3 = 0
∴ (x – 1) is a factor of p(x)

Question 12.
If x – 3 is a factor of p (x), then the remainder is
(1) 3
(2) -3
(3) p(3)
(4) p(-3)
Solution:
(3) p(3)

Question 13.
(x + y)(x² – xy + y²) is equal to
(1) (x + y)³
(2) (x – y)³
(3) x³ + y³
(4) x³ – y³
Solution:
(3) x³ + y³

Question 14.
(a + b – c)² is equal to
(1) (a – b + c)²
(2) (-a – b + c)²
(3) (a + b + c)²
(4) (a – b – c)²
Solution:
(2) (-a – b + c)²
Hint: (a + b – c)² = [- (- a – b + c)]² = (-a – b + c)²

Question 15.
In an expression ax² + bx + c the sum and product of the factors respectively,
(1) a, bc
(2) b, ac
(3) ac, b
(4) bc, a
Solution:
(2) 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
(1) 1, 2, 3
(2) 1, 2, 15
(3) 1, 2, -15
(4) 1, -2, 15
Solution:
(3) 1, 2, -15
Hint: p(-5) = a (-5²) + b (-5) + c = 25a – 5b + c = 0 ……… (1)
p( 3) = a (3²) + bc + 3 + c = 9 + 3b + c = 0 …….. (2)
25a – 5b = 9a + 3b
25a – 9a = 3b + 5b
16a = 8 b
\(\frac{a}{b}=\frac{8}{16}=\frac{1}{2}\)
Substitute a = 1,b = 2 in (1)
25(1) – 5 (2) = -c
25 – 10 = 15 = – c
c = -15

Question 17.
Cubic polynomial may have a maximum of
(1) 1
(2) 2
(3) 3
(4) 4
Solution:
(3) 3

Question 18.
Degree of the constant polynomial is
(1) 3
(2) 2
(3) 1
(4) 0
Solution:
(4) 0

Question 19.
Find the value of m from the equation 2x + 3y = m. If its one solution is x = 2 and y = – 2.
(1) 2
(2) -2
(3) 10
(4) 0
Solution:
(2) -2
Hint: 2x + 3y = m, x = 2, y = – 2
m = 2(2) + 3(-2)
= 4 – 6 = -2

Question 20.
Which of the following is a linear equation?
(1) x + \(\frac{1}{x}\) = 2 x
(2) x(x – 1) = 2
(3) 3x + 5 = \(\frac{2}{3}\)
(4) x³ – x = 5
Solution:
(3) 3x + 5 = \(\frac{2}{3}\)
Hint: x + \(\frac{1}{x}\) = 2 ⇒ x² – 2x + 1 = 0; x(x – 1) = 2 ⇒ x² -x – 2 = 0

Question 21.
Which of the following is a solution of the equation 2x – y = 6?
(1) (2, 4)
(2) (4, 2)
(3) (3, -1)
(4) (0, 6)
Solution:
(2) (4, 2)
Hint: 2x – y = 6
2(4) – 2 = 8 – 2 = 6 = RHS

Question 22.
If (2, 3) is a solution of linear equation 2x + 3y = k then, the value of k is
(1) 12
(2) 6
(3) 0
(4) 13
Solution:
(4) 13
Hint: 2x + 3y = k
2(2) + 3(3) = 4 + 9 = 13

Question 23.
Which condition does not satisfy the linear equation ax + by + c = 0
(1) a ≠ 0,b = 0
(2) a = 0, b ≠ 0
(3) a = 0, b = 0, c ≠ 0
(4) a ≠ 0, b ≠ 0
Solution:
(3) a = 0, b = 0, c ≠ 0
Hint: a = 0, b =0, c ≠ 0 ⇒ (0)x + (0) y + c = 0 False

Question 24.
Which of the following is not a linear equation in two variable
(1) ax + by + c = 0
(2) 0x + 0y + c = 0
(3) 0x + by + c = 0
(4) ax + 0y + c = 0
Solution:
(2) 0x + 0y + c = 0
Hint: a and b both can not be zero

Question 25.
The value of k for which the pair of linear equations 4x + 6y – 1 = 0 and 2x + ky – 7 = 0 represents parallel lines is
(1) k = 3
(2) k = 2
(3) k = 4
(4) k = -3
Solution:
(1) k = 3
Hint: 4x + 6y = 1
6y = -4x + 1
y = \(\frac{-4}{6} x+\frac{1}{6}\) ……….. (1)
2x + ky – 7 = 0
ky = -2x + 7
y = \(\frac{-2}{k} x+\frac{7}{k}\) ……….. (2)
Since the lines (1) and (2) parallel m₁ = m₂
\(\frac{-4}{6}=\frac{-2}{k}\)
k = -2 × \(\frac{-6}{4}\) = 3

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

Solution:
(2)

Hint: Parallel lines have no solution

Question 27.
If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) where a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 then the given pair of linear equation has ______ solution(s)
(1) no solution
(2) two solutions
(3) unique
(4) infinite
Solution:
(3) unique
Hint: \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\); unique solution

Question 28.
If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) where a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 then the given pair of linear equation has
(1) no solution
(2) two solutions
(3) infinite
(4) unique
Solution:
(1) no solution
Hint: \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\): parallel

Question 29.
GCD of any two prime numbers is _______
(1) -1
(2) 0
(3) 1
(4) 2
Solution:
(3) 1

Question 30.
The GCD of x⁴ – y⁴ and x² – y² is
(1) x⁴ – y⁴
(2) x² – y²
(3) (x + y)²
(4) (x + y)⁴
Solution:
(2) x² – y²
Hint:
x⁴ – y⁴ = (x² )² – (y²)² = (x² + y²) (x² – y²)
x² – y² = x² – y²
G.C.D. is = x² – y²