Prasanna

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:

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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

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.14

Solve by any one of the methods
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 two digit number be x y
Its place value = 10x + y
After interchanging the digits the number will be y x
Its place value = 10y + x
Their sum = 10x + y + 10y + x = 110
11x + 11y = 110
x + y = 10 ………….. (1)
If 10 is subtracted from the first number, the new number is 10x + y – 10
The sums of the digits of the first number is x + y
Its 4 more than 5 times is = 5(x + y) + 4
∴ 10x + y – 10 = 5x + 5y + 4
10x + y – 5x – 5y = 4 + 10
5x – 4y = 14 ………. (2)

Substitute y = 4 in (1)
x + 4 = 10
x = 10 – 4
x = 6
The first number is 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
Denominator be y
x + y = 12 ………. (1)
\(\frac{x}{y+3}=\frac{1}{2}\)
2x = y + 3
2x – y = 3 …………. (2)

Substitute y = 7 in (1)
x + 7 = 12
x = 12 – 7
x = 5
∴ The fraction is \(\frac{x}{y}=\frac{5}{7}\)

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:
In a cyclic quadrilateral the sum of the four angles is 360° and
the sum of the opposite angles is 180°.
∠A + ∠C = 180°
4y + 20 + 4x = 180°
4x + 4y = 180 – 20
x + y = \(\frac{160}{4}\)
x + y = 40 ………….. (1)
∠B + ∠D = 180°
3y – 5 + 7x + 5 = 180
7x + 3y = 180 …………. (2)

Substitute y = 25° in (1)
x + 25 = 40
x = 40 – 25
x = 15°
∠A = (4y + 20)° = (4 × 25 + 20) = 100 + 20 = 120°
∠B = (3y – 5)° = (3 × 25° – 5) = 75° – 5° = 70°
∠C = (4x)° = 4 × 15° = 60°
∠D = (7x + 5)° = (7 × 15 + 5) = 105° + 5° = 110°
∴ ∠A = 120°, ∠B = 70°, ∠C = 60°, ∠D = 110°

Question 4.
On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains ₹ 2000. But if he sells the T.V. at 10% gain and the fridge at 5% loss, he gains ₹ 1500 on the transaction. Find the actual price of the T.V. and the fridge.
Solution:
Let the actual price of a T.V. = x
Let the actual price of a Fridge = y
\(\frac{5}{100}\)x + \(\frac{10}{100}\)y = 2000
\(\frac{5}{100}\) (x + 2y) = 2000
x + 2y = 2000 × \(\frac{100}{5}\)
x + 2y = 40000 ………. (1)
\(\frac{10}{100}\) x – \(\frac{5}{100}\)y = 1500
2x – y = 1500 × \(\frac{100}{5}\)
2x – y = 30000 ……….. (2)

Substitute y = 10000 in (1)
x + 2(10000) = 40000
x + 20000 = 40000
x = 40000 – 20000
x = 20000
∴ Actual price of T.V. = ₹ 20000
Actual price of Fridge = ₹ 10000

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, y
\(\frac{x}{y}=\frac{5}{6}\)
⇒ 6x = 5y
6x – 5y = 0 ………… (1)
\(\frac{x-8}{y-8}=\frac{4}{5}\)
⇒ 5(x – 8) = 4(y – 8)
5x – 40 = 4y – 32
5x – 4y = 40 – 32
5x – 4y =8 ………….. (2)

Substitute y = 48 in (1)
6x – 5(48) = 0
6x – 240 = 0
6x = 240
x = \(\frac{240}{6}\) = 40
\(\frac{x}{y}=\frac{40}{48}=\frac{5}{6}\)
∴ The numbers are in the ratio 5 : 6

Question 6.
4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indians 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 for one Indian the rate of working be \(\frac{1}{x}\)
Let for one Chinese the rate of working be \(\frac{1}{y}\)

For cross multiplication method, we write the co-efficients as

∴ 1 Chinese can do the same piece of work = \(\frac{1}{y}\) = b = \(\frac{1}{36}\) = 36 days
1 Indian can do the piece of work = \(\frac{1}{x}\) = a = \(\frac{1}{18}\) = 18 days

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

Question 1.
Solve by cross-multiplication method
(i) 8x – 3y = 12; 5x = 2y + 7
(ii) 6x + 7y – 11 = 0; 5x + 2y = 13
(iii) \(\frac{2}{x}+\frac{3}{y}=5 ; \frac{3}{x}-\frac{1}{y}+9=0\)
Solution:
(i) 8x – 3y = 12 ……….. (1)
5x – 2y = 7 ………… (2)
8x – 3y – 12 = 0
5x – 2y – 7 = 0
For cross multiplication method, we write the co-efficients as

(ii) 6x + 7y – 11 = 0
5x + 2y – 13 = 0
For cross multiplication method, we write the co-efficients as

∴ Solutions: x = 3; y = -1

(iii) \(\frac{2}{x}+\frac{3}{y}\) – 5 = 0 ……….. (1)
\(\frac{3}{x}-\frac{1}{y}\) + 9 = 0 ………… (2)
In (1), (2) Put \(\frac{1}{x}\) = a, \(\frac{1}{y}\) = b
(1) ⇒ 2a + 3b – 5 = 0
(2) ⇒ 3a – b + 9 = 0
For cross multiplication method, we write the co-efficients as

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

Question 2.
Akshaya has 2 rupee coins and 5 rupee coins in her purse. If in all she has 80 coins totalling ₹ 220, how many coins of each kind does she have.
Solution:
Let the number of 2 rupee coins be x
Let the number of 5 rupee coins be y
x + y = 80
2x + 5y = 220
x + y – 80 = 0
2x + 5y – 220 = 0
For cross multiplication method, we write the co-efficients as

∴ No. of 2 rupee coins = 60
No. of 5 rupee coins = 20

Question 3.
It takes 24 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 8 hours and the pipe of the smaller diameter is used for 18 hours. Only half of the pool is filled. How long would each pipe take to fill the swimming pool.
Solution:
Let the time taken by the larger pipe be x hours and Set the time taken by the smaller pipe be y hours.
∴ \(\frac{1}{x}+\frac{1}{y}=\frac{1}{24}\)
In 1 hour the larger pipe can fill it = \(\frac{1}{x}\)
In 1 hour the smaller pipe can fill it \(\frac{1}{y}\)

For cross multiplication method, we write the co-efficients as

∴ To fill the full tank the larger pipe takes 40hrs
To fill the full tank the smaller pipe takes 60hrs

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

Try These (Textbook Page No. 2)

Question 1.
The Successor of 4576 is ____
Solution:
4577

Question 2.
The Predecessor of 8970 is ____
Solution:
8969

Question 3.
999 + 1 equals _____
Solution:
1000

Question 4.
10000 – 1 equals _____
Solution:
9999

Question 5.
The Predecessor of the smallest 5 digit number is ____
Solution:
Greatest 4 digit number 9999

Try These (Textbook Page No. 4)

Question 1.
Give 3 examples where the number of things counted by you would be a 5 digit number or more.
Solution:

1. Number of stars in the sky.
2. The number of people living in Tamilnadu.
3. The number of accidents in India in the year 2017.

Question 2.
There are ten lakh people in a district. What would be the population of 10 such districts?
Solution:
Number of people in the district = 10,00,000
.’. Population of 10 such districts = 10,00,000 × 10 = 1,00,00,000
Total population of 10 districts would be one crore.
10 lakh = 10,000 Hundreds

Question 3.
The Government spends rupees 2 crores for education in a particular district every month. What would be its expenditure for over 10 months?
Solution:
Expenditure for one month = 2 crores,
Expenditure for ten months = 2,00,00,000 × 10 = 20,00,00,000
Expenditure for 10 months = twenty crores.

Try These (Textbook Page No. 5)

Question 1.
Complete the table:

Solution:

Try These (Textbook Page No. 6)

Question 1.
Read and expand the following numbers:
(i) 2304567
(ii) 4509888
(iii) 9553556
Solution:
(i) Number: 23,04,567
Expanded form: 2 × 1000000 + 3 × 100000 + 0 × 10000 + 4 × 1000 + 5 × 100 + 6 × 10 + 7 × 1
Read as: Twenty Three Lakh Four Thousand Five Hundred and Sixty Seven

(ii) Number: 45,09,888
Expanded form: 4 × 1000000 + 5 × 100000 + 0 × 10000 + 9 × 1000 + 8 × 100 + 8 × 10 + 8 × 1
Read as: Forty Five Lakh Nine Thousand Eight Hundred and Eighty Eight

(iii) Number: 95,53,556
Expanded form: 9 × 1000000 + 5 × 100000 + 5 × 10000 + 3 × 1000 + 5 × 100 + 5 × 10 + 6 × 1
Read as: Ninety Five Lakh Fifty Three Thousand Five Hundred and Fifty Six

Question 2.
How many hundreds are there in 10 lakh?
Solution:

There are four places to the left of a Hundred.

Question 3.
10 lakh candidates write the Public Exam this year. If each exam centre is allotted with 1000 candidates. How many exam centres would be needed?
Solution:
Candidate for one centre = 1000
For 10 lakh people
10,00,000
Ten lakh contains = \(\frac{10,00,000}{1000}\) = 1000 Thousands
For 10 lakh people 1000 centres are needed

Try These (Textbook Page No. 7)

Question 1.
Find the place value of underlined digits
(i) 3841567
(ii) 94,43,810
Solution:
(i) Place value of 8 is 8 × 1,00,000 = 8,00,000 (Eight Lakh)
(ii) Place value of 4 is 4 × 10,000 = 40,000 (Forty Thousand)

Question 2.
Write down the numerals and place value of 5 in the numbers represented by the following number names.
(i) Forty-Seven Lakh Thirty Fight Thousand Five Hundred Sixty One.
(ii) Nine Crore Eighty-Two lakh Fifty Thousand Two Hundred Forty-One
(iii) Nineteen Crore Fifty-Seven Lakh Sixty Thousand Three Hundred Seventy
Solution:
(i) 47,38,561
Place value of 5 is 5 × 100 = 500 (Five Hundred)

(iii) 9,82,50,241
Place value of 5 is 5 × 10000 = 50,000 (Fifty Thousand)

(iv) 19,57,60,370
Place value of 5 is 5 × 10,00,000 = 50,00,000 (Fifty Lakhs)

Try These (Textbook Page No. 9)

Question 1.
Identify the incorrect places of comma and rewrite correctly. Indian System: 56, 12, 34, 0, 1, 5 ; 9,90,03,2245
International System: 7,5613,4534; 30,30,304,040
Indian System: 56, 12, 34, 015; 99,00,32,245
Solution:
Indian System: 56,12,34,015; 99,00,32,245
International System: 756,134,534 ; 3,030,304,040

Try These (Textbook Page No. 13)

Question 1.
Write the numbers in ascending order: 688, 9, 23005, 50, 7500.
Solution:
Ascending order: 9, 50, 688,7500, 23005
9 < 50 < 688 < 7500 < 23005

Question 2.
Find the least and greatest among the numbers: 478, 98, 6348,3, 6007, 50935
Solution:
The lease number is 3.
The greatest number is 50935

Try These (Textbook Page No. 14)

Question 1.
Compare the two numbers and put <, > and = using a place value chart.

Solution:
(i) 15475, 3214
Comparing the place value using a place value chart.

comparing the place values from left we have 15475 > 3214
(ii) 73204, 973561
Place value chart

comparing the digits or two numbers 73204 < 973561
(iii) 8975430, 8975430

From the place value chart comparing tne digits trom left 8 = 8, 9 = 9, 7 = 7, 5 = 5, 4 = 4, 3 = 3, 0 = 0
8975430 = 8975430
(iv) 1899799, 1899799.

From the place value chart comparing the digits of the two numbers from the highest place value we have 1 = 1, 8 = 8, 9 = 9, 9 = 9, 7 = 7, 9 = 9, 9 = 9
1899799 = 1899799

Try These (Textbook Page No. 16)

Question 1.
The area in sq.km of 4 Indian states are given below:

List the areas of the above 4 Indian States in the ascending and the descending order.
Solution:
We can prepare place value chart

5 digit number 38,863 is the least value.
Comparing digits of other 6 digit numbers from left.
1 = 1 = 1, 3 < 6 < 9
Ascending order = 38,863 < 1,30,058 < 1,62,968 < 1,91,791
Kerala < Tamilnadu < Andhra Pradesh < Karnataka
Descending order = 1,91,791 > 1,62,968 > 1,30,058 > 38,863
Karnataka > Andhra Pradesh > Tamilnadu > Kerala

Try These (Textbook Page No. 17)

Question 1.
In the same way, try placing the digit 4 in thousandth place and get six different 4-digit numbers. Also, make different 4-digit numbers by fixing 8 and 5 in the thousandth place.
Solution:

Question 2.
In the same way, make different 4 digit numbers by exchanging the digits and check every time whether the number made is small or big.
1432 < 4321 4321 > 3214
3214 > 2143
Solution:

Question 3.
Pedometer used in walking practice contains 5 digit number. What could be the largest measure?
Solution:
99,999

Try These (Textbook Page No. 25)

Question 1.
Round off the following numbers to the nearest ten
(i) 57
(ii) 189
(iii) 3,956
(iv) 57,312
Solution:
(i) 57
Given number = 57
Place value to be rounded off is ten.
Digit in tens place is 5.
Digit to’the right is 7 > 5
Adding 1 to 5 = 1 + 5 = 6
Changing the digits to the right of 6 to zero = 60
rounded off number is 60.

(ii) 189
Place value to be rounded off is ten
Digit is ten places is 8
Digit to the right is 9 > 5
Adding 1 to 8 = 1 + 8 = 9.
changing the digits to the right of 19 to zero = 190
Required rounded off number is 190

(iii) 3956
Place value to be rounded off is ten.
Digit in tens place is 5
Digit to the right is 6 > 5
Adding 1 to 5 = 1 + 5 = 6
Changing the right digits of 396 to zero = 3960
Required rounded off number is 3960.

(iv) 57312
Place value to be rounded off is ten.
Digit in tens place is 1
Digit to the right is 2 < 5
Leaving the number 2 as it is changing the digits to the right of 5731 to zero = 57310.
The rounded number is 57310

Question 2.
Round off the following numbers to the nearest ten, hundred and thousand.
(i) 9,34,678
(ii) 73,43,489
(iii) 17,98,45,673
Solution:
(i) 9,34,678
Nearest Tens: 9,34,680
Nearest Hundreds: 9,34,700
Nearest Thousands: 9,35,000

(ii) 73,43,489
Nearest Tens: 73,43,490
Nearest Hundreds: 73,43,500
Nearest Thousands: 73,43,000

(iii) 17,98,45,673
Nearest Tens: 17,98,45,670
Nearest Hundreds: 17,98,45,700
Nearest Thousands: 17,98,46,000

Question 3.
The tallest mountain in the world Mount Everest, located in Nepal is 8,848 m high. Its height can be rounded off to the nearest thousands as ______
Solution:
9000 m

Try These (Textbook Page No. 27)

Question 1.
Estimate the sum and difference:
8457 and 4573
Solution:
(a) Sum
8457 ⇒ 8000
4573 ⇒ 5000
Sum = 13,000

(b) Difference
8457 ⇒ 8000
4573 ⇒ 5000
Difference = 3,000

Question 2.
Estimate the product 39 × 53
Solution:
39 ⇒ 40
53 ⇒ 50
Product 40 × 50 = 2000

Question 3.
Estimate the quotient 5546 ÷ 524
Solution:
5546 ⇒ 5500
524 ⇒ 500
Quotient is 11

Try These (Textbook Page No. 29)

Question 1.
Find the value of 6 + 3 + 8 and 3 + 6 + 8
(i) Are the same?
(ii) Is there any other way of arranging these three numbers?
Solution:
6 + 3 + 8 = 3 + 6 + 8 = 17
(i) Yes, 6 + 3 + 8 = 3 + 6 + 8 = 17, Both are same
(ii) Yes, we can arrange these numbers as
3 + 8 + 6 = 8 + 6 + 3 = 8 + 3 + 6 = 6 + 8 + 3

Question 2.
Find the value of 5 × 2 × 6 and 2 × 5 × 6
(i) Are the same?
(ii) Is there any other way of arranging these three numbers?
Solution:
5 × 2 × 6 = 2 × 5 × 6 = 60
(i) Yes, they are the same
(ii) They can be arranged as
2 × 6 × 5 = 6 × 5 × 2 = 5 × 6 × 2 = 6 × 2 × 5.

Question 3.
Is 7 – 5 the same as 5 – 7? Why?
Solution:
7 – 5 ≠ 5 – 7.
Because subtraction is not commutative [∵ 7 – 5 = 2; 5 – 7 = -2]

Question 4.
What is the value of (15 – 8) – 6? Is it the same as 15 – (8 – 6)? Why?
Solution:
(15 – 8) – 6 = 7 – 6 = 1
(15 – 8) – 6 = 1
It is not same as 15 – (8 – 6).
15 – (8 – 6) = 15 – 2 = 13
(15 – 8) – 6 ≠ 15 – (8 – 6)

Question 5.
What is 15 ÷ 5? Is it the same as 5 ÷ 15? Why?
Solution:
(i) 15 ÷ 5 = 3
(ii) 15 ÷ 5 ≠ 5 ÷ 15
(iii) Division is not commutative for whole numbers.

Question 6.
What is the value of (100 ÷ 10) ÷ 5? Is it the same as 100 ÷ (10 ÷ 5)? Why?
Solution:
(i) (100 ÷ 10) ÷ 5 = 10 ÷ 5 = 2
(ii) 100 ÷ (10 ÷ 5) ≠ (100 ÷ 10) ÷ 5
(iii) Because division of whole numbers are not associative.
Also 100 ÷ (10 ÷ 5) = 100 ÷ 2 = 50
But (100 ÷ 10) ÷ 5 = 10 ÷ 5 = 2 = 50 ≠ 2
(i. e) (100 ÷ 10) ÷ 5 ≠ 100 ÷ (10 ÷ 5)

Try These (Textbook Page No. 30, 32, 33)

Question 1.
Use at least three different pairs of whole numbers to verify that subtraction is not commutative
Solution:
(a) 7 and 20
20 – 7 ≠ 7 – 20
(b) 300 and 100
300 – 100 ≠ 100 – 300
(c) 60 and 5
60 – 5 ≠ 5 – 60

Question 2.
Is 10 ÷ 5 the same as 5 ÷ 10? Justify it by taking two more combinations of numbers
Solution:
10 ÷ 5 ≠ 5 ÷ 10
Example:
(a) 20 ÷ 10 ≠ 10 ÷ 20 i.e. 2 ≠ \(\frac { 1 }{ 2 }\)
(b) 100 ÷ 50 ≠ 50 ÷ 100 i.e. 2 ≠ \(\frac { 1 }{ 2 }\)

Question 3.
Complete the following tables.

Solution:

Question 4.
Complete the Table.

Solution:

Question 5.
How will you read the large number given below?
731, 687, 303, 715, 884, 105, 727
Solution:
This is read as 731 quintillions, 687 quadrillions, 303 trillion, 715 billion, 884 million, 105 thousand, 727.

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 Additional Questions

Exercise 4.1

Question 1.
Find the complement of each of the following angles
(i) 63°
(ii) 24°
(iii) 48°
Solution:
(i) The complement of 63° = 90° – 63° = 27°
(ii) The complement of 24° = 90° – 24° = 66°
(iii) The complement of 48° = 90° – 48° = 42°

Question 2.
Find the supplement of each of the following angles
(i) 58°
(ii) 148°
(iii) 120°
Solution:
(i) The supplement of 58° = 180° – 58° = 122°
(ii) The supplement of 148° = 180° – 148° = 32°
(iii) The supplement of 120° = 180° – 120° = 60°

Question 3.
Find the value of x

Solution:

Question 4.
Find the values of x, y in the following figures

Solution:

Question 5.
In the given figure at right, side BC of ∆ABC is produced to D. Find ∠A and ∠C.

Solution:
From the figure
Exterior angle = 120°
⇒ ∠C = 180° – 120° = 60° (linear pair)
∴ ∠A = 180° – (40° + 60°) = 80°

Exercise 4.2

Question 1.
If the measures of three angles of a quadrilateral are 100°, 84° and 76° then, find the measure of fourth angle.
Solution:
Let the measure of the fourth angle be x°.
The sum of the angles of a quadrilateral is 360°
So, 100° + 84° + 76° + x° = 360°
260° + x° = 360°
x = 360° – 260° = 100°
Hence, the measure of the fourth angle is 100°.

Question 2.
In the parallelogram ABCD if ∠A = 65°, find ∠B, ∠C and ∠D.
Solution:
Let ABCD be a parallelogram in which ∠A = 65°
Since AD || BC we can treat AB as a transversal. So
∠A+∠B = 180°

65° +∠B = 180°
∠B = 180°-65°
∠B = 115°
Since the opposite angles of a parallelogram are equal, we have
∠C = ∠A = 65° and ∠D = ∠B = 115°
Hence, ∠B = 115°, ∠C = 65° and ∠D = 115°

Question 3.
If ABCD is a rhombus and if ∠A = 76°, find ∠CDB.
Solution:
∠A = ∠C = 76° (Opposite angles of a rhombus)
Let ∠CDB = x°. In ∆CDB, CD = CB
∠CDB + ∠CBD + ∠DCB = 180° (Angles of a triangle)

2x° + 76° = 180° ⇒ 2x° = 104°
x° = 52°
∴ ∠CDB = 52°

Question 4.
In a parallelogram, opposite sides are equal
Solution:
Given ABCD is a parallelogram
To Prove ABCD and DA = BC
Construction Join AC
Proof
Since ABCD is a parallelogram

AD || BC and AC is the transversal
∠DAC = ∠BCA ➝ (1) (alternate angles are equal)
AB || DC and AC is the transversal
∠BAC = ∠DCA ➝ (2) (alternate angles are equal)
In ∆ADC and ∆CBA
∠DAC = ∠BCA from (1)
AC is common
∠DCA = ∠BAC from (2)
∆ADC ≅ ∆CBA (By ASA)
Hence AD = CB and DC = BA (Corresponding sides are equal)

Question 5.
The angles of a quadrilateral are ¡n the ratio 1 : 2 : 3 : 4. Find all the angles. Let each ratio be x.

Solution:
Then the angles are x°, 2x°, 3x°, 4x°
x° + 2x° + 3x° + 4x° = 360°

Exercise 4.3

Question 1.
The radius of a circle 15 cm and the length of one of its chord is 24 cm. Find the distance of the chord from the centre.
Solution:
Distance of the chord from the centre.

Question 2.
The chord of length 32 cm is drawn at the distance of 12 cm from the centre of the circle. Find the radius of the circle.
Solution:

Question 3.
In a circle, AB and CD are two parallel chords with centre O and radius 5 cm such that AB = 8 cm and CD = 6 cm determine the distance between the chords?
Solution:

Question 4.
Find the value of x°

Solution:

Question 5.
Find the value of x°

Solution:

Exercise 4.4

Question 1.
Find the value of x in the figure.
Solution:

In the cyclic quadrilateral ABCD
∠ABC – 180°- 140° = 40°
∠BCA = 90°
∴ x = ∠BAC = 180°- (90° + 40°) = 50°

Question 2.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.

Solution:

Question 3.
AB and CD are two parallel sides of a cyclic quadrilateral ABCD in the figure. such that AB = 12 cm, CD = 16 cm and the radius of the circle is 10cm. Find the shortest distance between the two sides AB and CD.
Solution:
In this figure,

The shortest distance between the two sides = 8 + 6 = 14 cm

Question 4.
In the given figure, AB and CD are the parallel chords of a circle with centre O, such that AB = 30 cm and CD = 40 cm. If OM ⊥ AB and OL ⊥ CD distance between LM is 35 cm. Find the radius of the circle?

Solution:

Exercise 4.5

Question 1.
Construct an equilateral triangle of sides 6 cm and locate its orthocentre.
Solution:

Construction:
(1) Draw the ∆ABC with the given measurements.
(2) Construct the altitudes from any two vertices (A and B) to their opposite sides BC and AC respectively.
(3) The point of intersection of the altitudes H is the orthocentre of the given ∆ABC.

Question 2.
Draw and locate the orthocentre of a right triangle PQR right angled at Q, with PQ = 4.5 cm and QR = 6 cm.
Solution:

Construction:
(1) Draw the ∆PQR with the given measurements.
(2) Construct altitudes from any two vertices (Q and R) to their opposite sides PR and PQ respectively.
(3) The point of intersection of the altitudes H is the orthocentre of the given ∆PQR.

Question 3.
Construct the circumcentre of the ∆ABC with AB = 5 cm, ∠A = 60° and ∠B = 80°, also draw two circumcircle and find the circum radius of the ∆ABC.
Solution:



Solution:
Step 1: Draw the ∆ABC with the given measurements.
Step 2 : Construct the perpendicular bisector of any two sides (AC and BC) and let them meet at S which is the circumcentre.
Step 3 : S as centre and SA = SB = SC as radius, draw the Circumcircle to passes through A,B and C. Circumradius = 3.9 cm.

Exercise 4.6

Question 1.
Draw the circumcircle for an equilateral triangle of side 6 cm.
Solution:

Construction:
(1) Draw the ∆ABC with the given measurements.
(2) Construct the perpendicular bisectors of AC and BC and let them meet at S which is the circumcentre.
(3) With S as centre and SA = SB = SC as radius, draw the circumcircle to pass through A, B and C.

Question 2.
Construct the centroid of ∆PQR such that PQ = 9 cm, PQ = 7cm, RP = 8 cm.
Solution:
In ∆PQR,
PQ = 5 cm,
PR = 6 cm
∠QPR = 60°

Construction :
Step 1 : Draw ∆PQR using the given measurements PQ = 9 cm, QR = 7 cm and RP = 8 cm and construct the perpendicular bisector of any two sides (PQ and QR) to find the mid-points M and N of PQ and QR respectively.
Step 2 : Draw the medians PN and RM and let them meet at G. The point G is the centroid of the given ∆PQR.

Question 3.
Draw and locate the centroid of the triangle ABC where right angle at A, AB = 8 cm and AC = 6 cm.
Solution:

Step 1 : Draw ∆ABC with the given measurements AB = 8 cm, ∠A = 90° and AC = 6 cm and construct the perpendicular bisector of any two sides (AB and AC) to find the mid points M and N of AB and BC respectively.
Step 2 : Draw the medians (C and BN and let them meet at G. The point G is the centroid of the given ∆ABC.

Question 4.
Construct the centroid of Ø PQR whose sides are PQ = 8 cm, QR = 6 cm, RP = 7 cm.

Solution:
Side = 6.5 cm

Construction:
Step 1 : Draw ∆ABC with AB = BC = CA = 6.5 cm
Step 2 : Construct angle bisectors of any two angles (A and B) and let them meet at 1.1 is the incentre of ∆ABC.
Step 3 : Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4 : With I as centre, ID as radius draw the circle. This circle touches all the sides of triangle internally.
Step 5 : Measure in radius. In radius = 1.9 cm.

Exercise 4.7

Multiple Choice Questions :

Question 1.
If an angle is equal to one third of its supplement, its measure is equal to
(1) 40°
(2) 50°
(3) 45°
(4) 55°
Hint:

Solution:
(3) 45°

Question 2.
In the given figure, OP bisect ∠BOC and OQ bisect ∠AOC. Then ∠POQ is equal to
(1) 90°
(2) 120°
(3) 60°
(4) 100°
Hint:


Solution:
(1) 90°

Question 3.
The complement of an angle exceeds the angle by 60°. Then the angle is equal to
(1) 25°
(2) 30°
(3) 15°
(4) 35°
Hint:

Solution:
(3) 15°

Question 4.
ABCD is a parallelogram, E is the mid-point of AB and CE bisects ∠BCD. Then ∠DEC is
(1) 60°
(2) 90°
(3) 100°
(4) 120°
Solution:
(2) 90°

Question 5.
If the length of a chord decreases, then its distance from the centre.
(1) increases
(2) decreases
(3) same
(4) cannot say
Solution:
(1) increases

Question 6.
In the figure, O is the centre of the circle and ∠ACB = 60° then ∠AOB =
(1) 60°
(2) 90°
(3) 120°
(4) 180°

Solution:
(3) 120°

Question 7.
The angle subtend by a semicircle at the centre is.
(1) 60°
(2) 90°
(3) 120°
(4) 180°
Solution:
(4) 180°

Question 8.
The angle subtend by a semicircle at the remaining part of the circumference is ___.
(1) 60°
(2) 90°
(3) 120°
(4) 180°

Text Book Activities

Activity – 3

Angle sum for a polygon.
Draw any quadrilateral ABCD.
Mark a point P in its interior. Join the segments PA, PB, PC and PD.
You have 4 triangles now.

How much is the sum of all the angles of the 4 triangles?
How much is the sum of the angles at their vertex, now P?
Can you now find the ‘angle sum’ of the quadrilateral ABCD?
Can you extend this idea to any polygon?
Solution:
Sum of the angles of the 4 triangle = 180° × 4 = 720°
Sum of the angles at their vertex, now p = 360°
Angle sum of the quadrilateral ABCD = 720°- 360° = 360°
Yes we can extend this idea to any polygon.

Activity – 4

Procedure.

1. Draw a circle with centre O and with suitable radius.
2. Make it a semi-circle through folding. Consider the point A, B on it.

3. Make crease along AB in the semi circles and open it.
4. We get one more crease line on the another part of semi circle, name it as CD (observe AB = CD)

5. Join the radius to get the ∆OAB and ∆OCD.

6. Using trace paper, take the replicas of triangle ∆OAB and ∆OCD.
7. Place these triangles ∆OAB and ∆OCD one on the other.

Activity – 6
Procedure:
1. Draw a circle of any radius with centre O.
2. Mark any four points A, B, C and D on the boundary. Make a cyclic quadrilateral ABCD and name the angles as in figure.

3. Figure Make a replica of the cyclic quadrilateral ABCD with the help of tracing paper.
4. Make the cutout of the angles A, B, C and D
5. Paste the angle cutout ∠1, ∠2, ∠3 and ∠4 adjacent to the angles opposite to A, B, C and D as in Figure.
6. Measure the angles ∠1 + ∠3, and ∠2 + ∠4.
Solution:
∠1 + ∠3 = 180°
∠2 + ∠4 = 180°

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 Intext Questions

Try These (Text hook Page No. 41)

Question 1.
Observe the following patterns and complete them.
(i) 5, 8, 11, 14, ___, ____, ____
(ii) If 15873 × 7 = 111111 and 15873 × 14 = 222222 then, what is 15873 × 21 = ? and 15873 × 28 = ?
Solution:
(i) 17, 20, 23
Hint: 5 + 3 = 8, 8 + 3 = 11, 11+3 = 14
(ii) 15873 × 21 = 333333; 15873 × 28 = 444444
Hint: 15873 × 14 = 15873 × 7 × 2 = 111111 × 2 = 222222

Question 2.
Draw the next two patterns and complete the table.

Solution:
The next two patterns:

Question 3.
Create your own patterns of shapes and prepare a table.
Solution:
(i) Match stick pattern of triangles.

(ii) Pattern of squares and circles.

Try These (Text hook Page No. 46 to 48)

Question 4.

Solution:

Question 5.

Solution:

Question 6.
Find the unknown.
(i) 37 + 43 = 43 + ____
(ii) (22 + 10) + 15 = ____ + (10 + 15)
(iii) If 7 × 46 = 322, then 46 × 7 = _____
Solution:
(i) 37 + 43 = 43 + 37
(ii) (22 + 10) + 15 = 22 + (10 + 15)
(iii) If 7 × 46 = 322, then 46 × 7 = 322

Question 7.
Find the suitable value of ‘m’, to get a sum of 9?

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 5 Coordinate Geometry Ex 5.5

Question 1.
Find the centroid of the triangle whose vertices are
(i) (2, -4 ), (-3, -7) and (7, 2)
(ii) (-5, -5), (1, -4) and (-4, -2)
Solution:


Question 2.
If the centroid of a triangle is at (4,-2) and two of its vertices are (3, -2) and (5, 2) then find the third vertex of the triangle.
Solution:
Centroid G (x, y) = (4, -2)

Question 3.
Find the length of median through Aof a triangle whose vertices are A(-1, 3), B(1, -1) and C(5, 1).
Solution:

Question 4.
The vertices of a triangle are (1, 2), (h, -3) and (-4, k). If the centroid of the triangle is at the point (5, -1) then find the value of \(\sqrt{(h+k)^{2}+(h+3 k)^{2}}\)
Solution:

Question 5.
Orthocentre and centroid of a triangle are A(-3, 5) and B(3, 3) respectively. If C is the circumcentre and AC is the diameter of this circle, then find the radius of the circle.
Solution:

Question 6.
ABC is a triangle whose vertices are A(3, 4), B(-2, -1) and C(5, 3). If G is the centroid and BDCG is a parallelogram then find the coordinates of the vertex D.
Solution:

∴ The co-ordinates of the vertex D(x, y) = (1, 0)

Question 7.
If \(\) \(\) and \(\) are mid points of the sides of a triangle, then find the centroid of the triangle.
Solution:
“The centroid of the triangle obtained by joining the mid points of the sides of a triangle is the same as the centroid of the original triangle.”
∴ The mid points of the sides of the triangle are given as