Mahesh

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 2 Sets Ex 2.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 2nd Lesson Sets Exercise 2.4

10th Class Maths 2nd Lesson Sets Ex 2.4 Textbook Questions and Answers

Question 1.
State which of the following sets are empty and which are not?
i) The set of lines passing through a given point.
ii) Set of odd natural numbers divisible by 2.
iii) {x : x is a natural number, x < 5 and x > 7}
iv) {x: x is a common point to any two parallel lines}
v) Set of even prime numbers.
Answer:
i) Not empty
ii) Empty
iii) Empty
iv) Empty
v) Not empty

Question 2.
Which of the following sets are finite or infinite?
i) The set of months in a year.
ii) {1, 2, 3, …, 99, 100}
iii) The set of prime numbers smaller than 99.
Answer:
i) Finite
ii) Finite
iii) Finite

Question 3.
State whether each of the following sets is finite or infinite.
i) The set of letters in the English alphabet.
ii) The set of lines which are parallel to the X-axis.
iii) The set of numbers which are multiples of 5.
iv) The set of circles passing through the origin (0, 0).
Answer:
i) Finite
ii) Infinite
iii) Infinite
iv) Infinite

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 2 Sets Ex 2.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 2nd Lesson Sets Exercise 2.3

10th Class Maths 2nd Lesson Sets Ex 2.3 Textbook Questions and Answers

Question 1.
Which of the following sets are equal?
A = {x : x is a letter in the word FOLLOW}
ii) B = {x : x is a letter in the word FLOW}
iii) C = {x : x is a letter in the word WOLF}
Answer:
i) Elements in set A are {F, L, O, W}
ii) Elements in set B are {F, L, O, W}
iii) Elements in set C are {F, L, O, W} Sets A, B and C have same elements, Hence, they are equal sets.

Question 2.
Consider the following sets and fill up the blank in the statement given below with = or ≠ so as to make the statement true.
A = {1, 2, 3};
B = {The first three natural numbers};
C = {a, b, c, d};
D = {d, c, a, b};
E = {a, e, i, o, u};
F = {Set of vowels in English Alphabet}
i) A …. B
ii) A …. E
iii) C …. D
iv) D …. F
v) F …. A
vi) D …. E
vii) F …. B
Answer:
i) A = B
ii) A ≠ E
iii) C = D
iv) D ≠ F
v) F ≠ A
vi) D ≠ E
vii) F ≠ B

Question 3.
In each of the following, state whether A = B or not.
i) A = {a, b, c, d} ; B = {d, c, a, b}
ii) A = {4, 8, 12, 16} ; B = {8, 4, 16, 18}
iii) A = {2, 4, 6, 8, 10}; B = {x : x is a positive even integer and x ≤ 10}
iv) A = {x : x is a multiple of 10}; B = {10, 15, 20, 25, 30, …}
Answer:
i) A = B
ii) A ≠ B
iii) A ≠ B
iv) A ≠ B

Question 4.
State the reasons for the following :
i) {1, 2, 3, …., 10} ≠ {x : x ∈ N and 1 < x < 10}
ii) {2, 4, 6, 8, 10} ≠ {x : x = 2n+1 and x ∈ N}
iii) {5, 15, 30, 45} ≠ {x : x is a multiple of 15}
iv) {2, 3, 5, 7, 9} ≠ {x : x is a prime number}
Answer:
i) In R.H.S ‘x’ is greater than 1 and less than 10 but L.H.S is having both 1 and 10.
ii) L.H.S ≠ R.H.S
R.H.S: x = 2n + 1 is definition of odd numbers.
L.H.S: Given set is even numbers set.
iii) x is a multiple of 15.
So 5 does not exist.
iv) x is a prime number but 9 is not a prime number.

Question 5.
List all the subsets of the following sets.
i) B = {p, q}
ii) C = {x, y, z}
iii) D = {a, b, c, d}
iv) E = {1, 4, 9, 16}
v) F = {10, 100, 1000}
Answer:
i) Subsets of ‘B’ are {p}, {q}, {p, q}, φ
ii) Subsets of ‘C’ are {x}, {y} {z}, {x, y}, {y, z}, {z, x}, {x, y, z} and φ (2³ = 8)
iii) Subsets of ‘D’ are {a}, {b}, {c}, {d}, {a,b}, {b,c}, {c, d}, {a, c}, {a, d}, {b, d}, {a, b, c}, {b, c, d}, {a, b, d}, {a, c, d}, {a, b, c, d} and φ
iv) Subsets of ‘E’ are
φ, {1}, {4}, {9}, {16}, {1,4}, {1,9}, {1, 16}, {4, 9}, {4, 16}, (9, 16}, {1, 4, 9}, {1, 9, 16}, {4, 9, 16}, {1, 4, 16}, {1, 4, 9, 16}
v) Subsets of ‘F’ are
φ, {10}, {100}, {1000}, {10, 100}, {100, 1000}, {10, 1000}, {10, 100, 1000}.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 2 Sets Ex 2.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 2nd Lesson Sets Exercise 2.2

10th Class Maths 2nd Lesson Sets Ex 2.2 Textbook Questions and Answers

Question 1.
If A = {1, 2, 3, 4}; B = {1, 2, 3, 5, 6} then find A ∩ B and B ∩ A. Are they equal ?
Answer:
Given sets are A = {1, 2, 3, 4} and B = {1,2,3, 5,6}
A ∩ B = {1,2, 3,4} ∩ {1,2, 3, 5, 6}
= {1,2,3} …… (1)
B ∩ A = {1, 2, 3, 5, 6} ∩ {1, 2, 3, 4}
= {1,2,3} …….(2)
From (1) and (2)
A ∩ B and B ∩ A are same.

Question 2.
A = {0, 2, 4}, find A ∩ φ and A ∩ A. Comment.
Answer:
Given set A = {0, 2, 4} and φ is a null set.
A ∩ φ = {0, 2, 4} ∩ { }
= { } ……. (1)
A ∩ A = {0, 2, 4} ∩ {0, 2, 4}
= {0, 2,4} …….. (2)
From (1) and (2),
We conclude that A ∩ φ = φ and A ∩ A = A

Question 3.
If A = {2, 4, 6, 8, 10} and B = {3, 6, 9, 12, 15}, find A – B and B – A.
Answer:
Given sets are
A {2, 4, 6, 8, 10} and B = {3, 6, 9, 12, 15}
A – B = {2, 4, 6, 8, 10} – {3, 6, 9, 12, 15}
= {2, 4, 8, 10} …… (1)
B – A = {3, 6, 9, 12, 15} – {2, 4, 6, 8, 10}
= {3, 9, 12, 15} …… (2)
From (1) and (2), A – B ≠ B – A

Question 4.
If A and B are two sets such that A ⊂ B then, what is A ∪ B?
Answer:
Let us consider A ⊂ B
Set A = {1, 2, 3} and
Set B = {1, 2, 3, 4, 5}
Now A ∪ B = {1, 2, 3} ∪ {1, 2, 3, 4, 5}
= {1, 2, 3, 4, 5} = B
∴ A ∪ B = B

Question 5.
If A = {x : x is a natural number},
B = {x : x is an even natural number},
C = {x : x is an odd natural number} and
D = {x : x is a prime number}
Find A ∩ B, A ∩ C, A ∩ D, B ∩ C, B ∩ D, C ∩ D.
Answer:
Given sets are
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ……}
B = {2, 4, 6, 8, 10, …….}
C = {1, 3, 5, 7, 9, …….}
D = {2, 3, 5, 7, 11, …….}
A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …….} ∩ {2, 4, 6, 8, 10, ……}
= {2, 4, 6, 8, 10, ……}
A ∩ C = {1, 2, 3,4, 5, 6, 7, 8, 9, 10, …} ∩ {1, 3, 5, 7, 9 }
= {1, 3, 5, 7, 9, ……}
A ∩ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …} ∩ {2, 3, -5, 7, 11,….}
= {2, 3, 5, 7, 11, ……}
B ∩ C = {2, 4, 6, 8, 10, ……} ∩ {1, 3, 5, 7, 9, …….}
= { } = φ
B ∩ D = {2, 4, 6, 8, 10, ……} ∩ {2, 3, 5, 7, 11, ……}
= {2}
C ∩ D = {1, 3, 5, 7, 9, ……} ∩ {2, 3, 5, 7, 11, 13, ……}
= {3, 5, 7, …..}

Question 6.
If A = {3, 6, 9, 12, 15, 18, 21}; B = {4, 8, 12, 16, 20}; C = {2, 4, 6, 8, 10, 12, 14, 16}; D = {5, 10, 15, 20} find
(i) A – B
(ii) A – C
(iii) A – D
(iv) B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(x) D – B
Answer:
Given sets are A = {3, 6, 9, 12, 15, 18, 21}
B = {4, 8, 12, 16, 20}
C = {2, 4, 6, 8, 10, 12, 14, 16} and
D = {5, 10, 15, 20}
i) A – B = {3, 6, 9, 12, 15, 18, 21} – {4, 8, 12, 16, 20} = {3, 6, 9, 15, 18, 21}
ii) A – C = {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16} = {3,9,15,18,21}
iii) A – D = {3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20} = {3,6,9,12,18,21}
iv) B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21} = {4, 8, 16, 20}
v) C – A = {2, 4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21} = {2, 4, 8, 10, 14, 16}
vi) D – A = {5, 10, 15, 20} – {3, 6, 9, 12, 15, 18, 21} = {5, 10, 20}
vii) B – C = {4, 8, 12, 16, 20} – {2, 4, 6, 8, 10, 12, 14, 16} = {20}
viii) B – D = {4, 8, 12, 16,20} – {5, 10, 15, 20} = {4, 8, 12, 16}
ix) C – B = {2, 4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20} = {2, 6, 10, 14}
x) D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20} = {5, 10, 15}

Question 7.
State whether each of the following statement is true or false. Justify your answers.
i) {2,3,4,5} and {3,6} are disjoint sets.
ii) {a, e, i, o, u} and {a, b, c, d} are disjoint sets.
iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.
Answer:
i) Rule: If two sets are disjoint their intersection is null set.
= {2, 3, 4, 5} n {3, 6} = { 3 } ≠ φ
∴ Given statement is False.

ii) Given sets are
{a, e, i, o, u} and {a, b, c, d}
= {a, e, i, o, u} ∩ {a, b, c, d}
= { a } ≠ φ
∴ Given statement is False.

iii) Given sets are
{2, 6, 10, 14} and {3, 7, 11, 15}
= {2, 6, 10, 14} ∩ {3, 7, 11, 15}
= { }
∴ Given statement is True.

iv) Given sets are
{2, 6, 10} and {3, 7, 11}
= {2, 6, 10} ∩ {3, 7, 11} = { }
∴ Given statement is True.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 2 Sets Ex 2.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 2nd Lesson Sets Exercise 2.1

10th Class Maths 2nd Lesson Sets Ex 2.1 Textbook Questions and Answers

Question 1.
Which of the following are sets? Justify your answer.
i) The collection of all the months of a year beginning with die letter “J”.
ii) The collection of ten most talented writers of India.
iii) A team of eleven best cricket batsmen of the world.
iv) The collection of all boys in your class.
v) The collection of all even integers.
Answer:
i) There are 3 months as January, June and July beginning with letter ‘J’. Therefore, it is a well defined collection of months and hence it is a set.
ii) The concept of talented writers of India is vague, since there is no rule given for deciding whether a particular writer is talented or not.
Hence, the given collection is not a set.
iii) A team of eleven best cricket batsmen of the world is vague, since there is no rule given for deciding whether a particular batsman is the best.
Hence, the given collection is not a set.
iv) The collection of all boys in my class is well defined. Hence, the given collection is a set.
v) The collection of all even integers i.e., (2, 4, 6, 8, ……) is well defined.
Hence, the given collection is a set.

Question 2.
If A = {0, 2, 4, 6}, B = {3, 5, 7} and C = {p, q, r} then fill the appropriate symbol, ∈ or ∉ in the blanks.
i) 0 ….. A
ii) 3 ….. C
iii) 4 ….. B
iv) 8 ….. A
v) p ….. C
vi) 7 ….. B
Answer:
i) ∈
ii) ∉
iii) ∉
iv) ∉
v) ∈
vi) ∈

Question 3.
Express the following statements using symbols.
i) The elements ‘x’ does not belong to ‘A’.
ii) ‘d’ is an element of the set ‘B’.
iii) ‘1’ belongs to the set of Natural numbers N.
iv) ‘8′ does not belong to the set of prime numbers P.
Answer:
i) x ∉ A
ii) d ∈ B
iii) 1 ∈ N
iv) 8 ∉ P

Question 4.
State whether the following statements are true or false. Justify your answer.
i) 5 ∉ set of prime numbers
ii) S = {5, 6, 7} implies 8 ∈ S.
iii) -5 ∉ W where ‘W’ is the set of whole numbers.
iv) \(\frac{8}{11}\) ∈ Z
where ‘Z’ is the set of integers.
Answer:
i) False
ii) False
iii) True
iv) False

Question 5.
Write the following sets in roster form.
i) B = {x : x is a natural number smaller than 6}.
ii) C = {x : x is a two-digit natural number such that the sum of its digits is 8}.
iii) D = {x : x is a prime number which is a divisor of 60}.
iv) E = {x : x is an alphabet in BETTER}.
Answer:
i) B = {1, 2, 3, 4, 5}
ii) C = {17, 26, 35, 44, 53, 62, 71}
iii) D = {5, 3}
iv) E = {B, E, T, R}

Question 6.
Write the following sets in the set – builder form.
i) {3, 6, 9, 12}
ii) {2, 4, 8, 16, 32}
iii) {5, 25, 125, 625}
iv) {1, 4, 9, 16, 25, …, 100}
Answer:
i) A = {x : x is multiple of 3 and less than 13}
ii) B = {x : x = 2^P, 0 < P < 6, P ∈ N}
iii) C = {x : x = 5^P, 0 < P < 5, P ∈ N}
iv) D = {x : x = P², 0< P < 11, P ∈ W}

Question 7.
Write the following sets in roster form.
i) A = {x: x is a natural number greater than 50 but smaller than 100}
ii) B = {x : x is an integer, x² = 4}
iii) D = {x : x is a letter in the word “LOYAL”}
Answer:
i) A = {51, 52, 53, ……. , 98, 99}
ii) B = {+2, -2}
iii) D = {L, O, Y, A}

Question 8.
Match the roster form with set builder form.
i) {1, 2, 3, 6}                       ( )      a) {x : x is a prime number and a divisor of 6}
ii) {2, 3}                              ( )      b) {x : x is an odd natural number smaller than 10}
iii) {M, A, T, H, E, I, C, S}     ( )      c) {x : x is a natural number and divisor of 6}
iv) {1, 3, 5, 7, 9}                  ( )      d) {x : x is a letter of the word MATHEMATICS}
Answer:
i) c
ii) a
iii) d
iv) b

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.4

10th Class Maths 3rd Lesson Polynomials Ex 3.4 Textbook Questions and Answers

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
ii) p(x) = x⁴ – 3×2 + 4x + 5, g(x) = x² + 1 – x
iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²
Answer:
i) Given polynomials are
p(x) = x³ – 3x² + 5x – 3 and
g(x) = x² – 2
Here, dividend and divisor are both in standard forms.
So, we have

∴ The quotient is x – 3 and the remainder is 7x – 9.

ii) Given polynomials are
p{x) = x⁴ – 3x² + 4x + 5 and
g(x) = x² + 1 – x
Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as x² – x + 1.
We have,

∴ The quotient is x² + x – 3 and the remainder is +8.

iii) Given polynomials are
p(x) = x⁴ – 5x + 6 and
g(x) = 2 – x²
Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as -x² + 2.
So, we have

∴ The quotient is -x² – 2 and the remainder is -5x + 10.

Question 2.
Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
Answer:
i) Given first polynomial is t² – 3.
Second polynomial is
2t⁴ + 3t³ – 2t² – 9t – 12.
Let us divide 2t⁴ + 3t³ – 2t² – 9t – 12 by t² – 3, we get

Since the remainder is 0, therefore, t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

ii) Given first polynomial is x² + 3x + 1
Second polynomial is 3x⁴ + 5x³ – 7x² + 2x + 2
Let us divide 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1, we get

Since the remainder is 0, therefore x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

iii) Given first polynomial = x³ – 3x + 1
Second polynomial = x⁵ – 4x³ + x² + 3x + 1
Let us divide x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1, we get

Here, remainder is 2(≠ 0).
Therefore, x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\).
Answer:
Let the other two zeroes are α and β.
Now compare the given polynomial 3x⁴ + 6x³ – 2x² – 10x – 5 with the standard form ax⁴ + bx³ + cx² + dx + e we get a = 3, b = 6, c = -2, d = -10, e = -5

–\(\frac{5}{3}\)αβ = \(\frac{-5}{3}\) ⇒ αβ = 1
now (α – β)² = (α + β)² – 4αβ
= (-2)² – 4(1)
= 4 – 4 = 0
α – β = 0 …. (2)
Now solving (1) and (2) we get

⇒ α = -1, β = -1
Then the remaining the zeroes are -1 and -1.
Hence all zeroes of it = –\(\sqrt{\frac{5}{3}}\), \(\sqrt{\frac{5}{3}}\), -1, -1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).
Answer:
Given, p(x) = x³ – 3x² + x + 2
q(x) = x – 2 and
r(x) = -2x + 4
By division algorithm, we know that Dividend = Divisor × Quotient + Remainder
p(x) = q(x) × g(x) + r(x)
Therefore, x³ – 3x² + x + 2
= (x – 2) × g(x) + (- 2x + 4)
⇒ x³ – 3x² + x + 2 + 2x – 4 = (x – 2) × g(x)
g(x) = \(\frac{x^{3}-3 x^{2}+3 x-2}{x-2}\)
On dividing x³ – 3x² + x + 2, by x – 2, we get

First term of g(x) = \(\frac{\mathrm{x}^{3}}{\mathrm{x}}\) = x²
Second term of g(x) = \(\frac{-x^{2}}{x}\) = -x
Third term of g(x) = \(\frac{x}{x}\) = 1
Hence, g(x) = x² – x + 1.

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
i) deg p(x) = deg q(x)
ii) deg q(x) = deg r(x)
iii) deg r(x) = 0
Answer:
Let q(x) = 3x² + 2x + 6, degree of q(x) = 2
p(x) = 12x² + 8x + 24, degree of p(x) = 2
Given degree p(x) = degree q(x)
i) Using division algorithm,
We gave, p(x) = q(x) × g(x) + r(x)
On dividing 12x² + 8x + 24 by 3x² + 2x + 6, we get

Since, the remainder is zero, therefore 3x² + 2x + 6 is a factor of 12x² + 8x + 24.
∴ g(x) = 4 and r(x)= 0

ii) Let p(x) = x⁵ + 2x⁴ + 3x³ + 5x² + 2
q(x) = x² + x + 1, degree q(x) = 2
Given degree q(x) = degree r(x)
On dividing x⁵ + 2x⁴ + 3x³ + 5x² + 2 by x² + x + 1, we get

Here, g(x) = x³ + x² + x + 1 and r(x) = 2x² – 2x + 1
degree of r(x) = 2.
∴ deg g(x) = deg r(x).

iii) Let p(x) = 2x⁴ + 8x³ + 6x² + 4x + 12, r(x) = 2
Here, degree r(x) = 0
On dividing 2x⁴ + 8x³ + 6x² + 4x + 12 by 2, we get

Here, g(x) = x⁴ + 4x³ + 3x² + 2x + 1 and r(x) = 10
so degree of r(x) = 0

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.3

10th Class Maths 3rd Lesson Polynomials Ex 3.3 Textbook Questions and Answers

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
i) x² – 2x – 8
ii) 4s² – 4s + 1
iii) 6x² – 3 – 7x
iv) 4u² + 8u
v) t² – 15
vi) 3x² – x – 4
Answer:
i) Given polynomial is x² – 2x – 8
We have x² – 2x – 8 = x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4) (x + 2)
So, the value of x² – 2x – 8 is zero
when x – 4 = 0 or x + 2 = 0 i.e.,
when x = 4 or x = -2
So, the zeroes of x² – 2x – 8 are 4 and -2.
Sum of the zeroes = 4 – 2 = 2 Coefficient of ,x -(-2)
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-2)}{1}\) = 2
And product of the zeroes = 4 × (-2) = -8
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-8}{1}\) = -8

ii) Given polynomial is 4s² – 4s + 1
We have, 4s² – 4s + 1
= 4s² – 2s – 2s + 1
= 2s (2s – 1) – 1(2s – 1)
= (2s – 1) (2s – 1)
= (2s – 1)²
So, the value of 4s2 – 4s + 1 is zero
when 2s-1 = 0 or s = \(\frac{1}{2}\)
∴ Zeroes of the polynomial are \(\frac{1}{2}\) and \(\frac{1}{2}\)
∴ Sum of the zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1.
= – \(\frac{\text { Coefficient of } s}{\text { Coefficient of } s^{2}}\) = –\(\frac{-4}{4}\) = 1
And product of the zeroes = \(\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)\) = \(\frac{1}{4}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{1}{4}\)

iii) Given polynomial is 6x² – 3 – 7x
We have, 6x² – 3 – 7x = 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
The value of 6x² – 3 – 7x is zero, when the value of (3x +1) (2x – 3) is 0
i.e., when 3x + 1 = 0 and 2x – 3 = 0
3x = -1 and 2x = 3
x = \(\frac{-1}{3}\) and x = \(\frac{3}{2}\)
∴ The zeroes of 6x² – 3 – 7x = \(\frac{-1}{3}\) and \(\frac{3}{2}\)
∴ Sum of the zeroes = \(\frac{1}{3}\) + \(\frac{3}{2}\) = \(\frac{7}{6}\).
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-7)}{6}\) = \(\frac{7}{6}\)
And product of the zeroes = \(\left(\frac{-1}{3}\right) \times\left(\frac{3}{2}\right)\) = \(\frac{-1}{2}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\)

iv) Given polynomial is 4u² + 8u
We have, 4u² + 8u = 4u (u + 2)
The value of 4u² + 8u is 0,
when the value of 4u(u + 2) = 0, i.e.,
when u = 0 or u + 2 = 0, i.e.,
when u = 0 (or) u = – 2
∴ The zeroes of 4u² + 8u are 0 and – 2.
Therefore, sum of the zeroes = 0 + (-2) = -2
= – \(\frac{\text { Coefficient of } u}{\text { Coefficient of } u^{2}}\) = \(\frac{-8}{4}\) = -2
And product of the zeroes 0 . (-2) = 0
= \(\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}\) = \(\frac{0}{4}\) = 0

v) Given polynomial is t² – 15.
We have, t² – 15 = (t – √15 ) (t + √l5)
The value of t² – 15 is 0,
when the value of (t – √15 ) (t + √l5) = 0, i.e.,
when t – √15 = 0 or t + √15 = 0, i.e.,
when t = √15 (or) t = -√15
∴ The zeroes of t² – 15 are √15 and -√15.
Therefore, sum of the zeroes = √15 + (-√15) = 0
= – \(\frac{\text { Coefficient of } t}{\text { Coefficient of } t^{2}}\) = –\(\frac{0}{1}\) = 0
And product of the zeroes √15 × (-√15) = -15
= \(\frac{\text { Constant term }}{\text { Coefficient of } t^{2}}\) = \(\frac{-15}{1}\) = -15

vi) Given polynomial is 3x² – x – 4
we have, 3x² – x – 4 = 3x² + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1) (3x – 4)
The value of 3x² – x – 4 is 0 when the value of (x + 1) (3x – 4) is 0.
i.e., when x + 1 = 0 or 3x – 4 = 0
i.e., when x = -1 or x = \(\frac{4}{3}\)
∴ The zeroes of 3x² – x – 4 are -1 and \(\frac{4}{3}\)
Therefore, sum of the zeroes = -1 + \(\frac{4}{3}\) = \(\frac{-3+4}{3}\) = \(\frac{1}{3}\)
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-1)}{3}\) = \(\frac{1}{3}\)
And product of the zeroes -1 × \(\frac{4}{3}\) = \(\frac{-4}{3}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-4}{3}\)

Question 2.
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.
i) \(\frac{1}{4}\), -1
ii) √2, \(\frac{1}{3}\)
iii) 0, √5
iv) 1, 1
v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
vi) 4, 1
Answer:
Let the polynomial be ax² + bx + c
and its zeroes be α and β.
i) Here, α + β = \(\frac{1}{4}\) and αβ = -1
Thus, the polynomial formed = x² – (sum of the zeroes)x + product of the zeroes
= x² – (\(\frac{1}{4}\))x – 1
= x² – \(\frac{x}{4}\) – 1
The other polynomials are (x² – \(\frac{x}{4}\) – 1)
then the polynomial is 4x² – x – 4.

ii) Here, α + β = √2 and αβ = \(\frac{1}{3}\)
Thus, the polynomial formed = x² – (sum of the zeroes)x + product of the zeroes
= x² – (√2)x + \(\frac{1}{3}\)
= x² – √2x + \(\frac{1}{3}\)
The other polynomials are (x² – √2x + \(\frac{1}{3}\))
then the polynomial is 3x² – 3√2x + 1.

iii) Here, α + β = 0 and αβ = √5
Thus, the polynomial formed = x² – (sum of the zeroes)x + product of the zeroes
= x² – (0)x + √5
= x² + √5

iv) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = 1 = \(\frac{-(-1)}{1}\) = \(\frac{-b}{a}\) and
αβ = 1 = latex]\frac{1}{1}[/latex] = \(\frac{c}{a}\)
If a = 4, then b = 1 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is 4x² + x + 1.

v) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = \(\frac{-1}{4}\) = \(\frac{-b}{a}\) and
αβ = \(\frac{1}{4}\) = \(\frac{c}{a}\)
If a = 4, then b = 1 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is 4x² + x + 1.

vi) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = 4 = \(\frac{-(-4)}{1}\) = \(\frac{-b}{a}\) and
αβ = 1 = \(\frac{1}{1}\) = \(\frac{c}{a}\)
If a = 1, then b = -4 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is x² – 4x + 1.

Question 3.
Find the quadratic polynomial, for the zeroes α, β given in each case.
i) 2, -1
ii) √3, -√3
iii) \(\frac{1}{4}\), -1
iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Answer:
i) Let the polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α and β.
Here α = 2 and β = – 1
Sum of the zeroes = α + β = 2 + (-l) = 1
Product of the zeroes = αβ = 2 × (-1) = -2
Therefore the quadratic polynomial ax² + bx + c is x² – (α + β)x + αβ = [x² – x – 2]
the quadratic polynomial will be x² – x – 2.

ii) Let the zeroes be α = √3 and β = -√3
Sum of the zeroes = α + β
= √3 + (-√3) = 0
Product of the zeroes = αβ
= √3 × (-√3) = -3
∴ The quadratic polynomial
ax² + bx + c is [x² – (α + β)x + αβ]
= [x² – 0.x + (-3)] = [x² – 3]
the quadratic polynomial will be x² – 3.

iii) Let the zeroes be α = \(\frac{1}{4}\) and β = -1
Sum of the zeroes = α + β
= \(\frac{1}{4}\) + (-1) = \(\frac{1+(-4)}{4}\) = \(\frac{-3}{4}\)
Product of the zeroes = αβ
= \(\frac{1}{4}\) × (-1) = \(\frac{-1}{4}\)
∴ The quadratic polynomial
ax² + bx + c is [x² – (α + β)x + αβ]
= [x² – \(\left(\frac{-3}{4}\right)\).x + (\(\frac{-1}{4}\))]
the quadratic polynomial will be 4x² + 3x – 1.

iv) Let the zeroes be α = \(\frac{1}{2}\) and β = \(\frac{3}{2}\)
Sum of the zeroes = α + β
= \(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{1+3}{2}\) = \(\frac{4}{2}\) = 2
Product of the zeroes = αβ
= \(\frac{1}{2}\) × \(\frac{3}{2}\) = \(\frac{3}{4}\)
∴ The quadratic polynomial
ax² + bx + c is [x² – (α + β)x + αβ]
= [x² – 2x + (\(\frac{3}{4}\))]
the quadratic polynomial will be 4x² – 8x + 3.

Question 4.
Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³ + 3x² – x – 3 and check the relationship between zeroes and the coefficients.
Answer:
Given cubic polynomial
p(x) = x³ + 3x² – x – 3
Comparing the given polynomial with ax³ + bx² + cx + d, we get a = 1, b = 3, c = -1, d = -3
Futher given zeroes are 1,-1 and – 3
p(1) = (1)³ + 3(1)² – 1 – 3
= 1 + 3 – 1 – 3 = 0
p(-1) = (-1)³ + 3(-1)² – 1 – 3
= -1 + 3 + 1 – 3 = 0
p(-3) = (-3)³ + 3(-3)² – (-3) – 3
= -27 + 27 + 3 – 3 = 0
Therefore, 1, -1 and -3 are the zeroes of x³ + 3x² – x – 3.
So, we take α = 1, β = -1 and γ = -3 Now,
α + β + γ = 1 + (-1) + (-3) = -3
αβ + βγ + γα = 1(-l) + (-1) (-3) + (-3)1
= -1 + 3 – 3 = -1
= \(\frac{c}{a}\) = \(\frac{-1}{1}\) = -1
αβγ = 1 (-1) (-3) = 3 = \(\frac{-d}{a}\) = \(\frac{-(-3)}{1}\) = 3

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.2

10th Class Maths 3rd Lesson Polynomials Ex 3.2 Textbook Questions and Answers

Question 1.
The graphs of y = p(x) are given in the figure below, for some polynomials p(x). In each case, find the number of zeroes of p(x).

Answer:
i) There are no zeroes as the graph does not intersect the X – axis.
ii) The number of zeroes is one as the graph intersects the X – axis at one point only.
iii) The number of zeroes is three as the graph intersects the X – axis at three points.
iv) The number of zeroes is two as the graph intersects the X – axis at two points.
v) The number of zeroes is four as the graph intersects the X – axis at four points.
vi) The number of zeroes is three as the graph intersects the X – axis at three points.

Question 2.
Find the zeroes of the given polynomials,
(i) p(x) = 3x
(ii) p(x) = x² + 5x + 6
(iii) p(x) = (x + 2) (x + 3)
(iv) p(x) = x⁴ – 16
Answer:
i) Given p(x) = 3x
Let p(x) = 0
So, 3x = 0
x = \(\frac{0}{3}\) = 0,
Zeroes of p(x) = 3x is zero.
∴ No. of zeroes is one.

ii) Given p(x) = x² + 5x + 6 is a quadratic polynomial.
It has atmost two zeroes.
To find zeroes, let p(x) = 0
⇒ x² + 5x + 6 = 0
⇒ x² + 3x + 2x + 6 = 0
⇒ x(x + 3) + 2 (x + 3) = 0
⇒ (x + 3) (x + 2) = 0
⇒ x + 3 = 0 or x + 2 = 0
⇒ x = -3 or x = -2
Therefore the zeroes of the polynomial are -3 and -2.

iii) Given p(x) = (x + 2) (x + 3)
It is a quadratic polynomial.
It has atmost two zeroes.
Let p(x) = 0
⇒ (x + 2) (x + 3) = 0
⇒ (x + 2) = 0 or (x + 3) = 0
⇒ x = -2 or x = -3
Therefore the zeroes of the polynomial are -2 and – 3.

iv) Given p(x) = x⁴ – 16 is a biquadratic polynomial. It has atmost two zeroes.
Let p(x) = 0
⇒ x⁴ – 16 = 0
⇒ (x²)² – 4² = 0
⇒ (x² – 4) (x² + 4) = 0
⇒ (x + 2) (x – 2) (x² + 4) = 0
⇒ (x + 2) = 0 or (x – 2) = 0 or (x² + 4) = 0
⇒ x = -2 (or) x = 2 (or) x² = -4
Therefore the zeroes of the polynomial are 2, – 2, we do not consider √-4 since it is not real.

Question 3.
Draw the graphs of the given polynomial and find the zeroes. Justify the answers,
i) p(x) = x² – x – 12
ii) p(x) = x² – 6x + 9
iii) p(x) = x² – 4x + 5
iv) p(x) = x² + 3x – 4
v) p(x) = x² – 1
Answer:
i) Given polynomial p(x) = x² – x – 12.
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-3, 0) and (4, 0).
So, the zeroes of the polynomial are -3 and 4.
Justification:
Given p(x) = x² – x – 12 = 0
⇒ x² – 4x + 3x – 12 = 0
⇒ x(x – 4) + 3(x – 4) = 0
⇒ (x – 4) (x + 3) = 0
⇒ x – 4 = 0 and x + 3 = 0
x = 4 and x = – 3

ii) Given polynomial p(x) = x² – 6x + 9
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (3, 0).
So, the zeroes of the given polynomial are same i.e., 3.
Justification:
Given p(x) = x² – 6x + 9
⇒ x² – 3x – 3x + 9 = 0
⇒ x(x – 3) – 3(x – 3) = 0
⇒ (x – 3) (x – 3) = 0
⇒ x – 3 = 0 and x – 3 = 0
x = 3 and x = 3

iii) Given polynomial p(x) = x² – 4x + 5
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph does not cut the X – axis at any point.
So, the quadratic polynomial p(x) has no zeroes.
Justification: For the given p(x) = x² – 4x + 5 not possible to split in factors.

iv) Given polynomial p(x) = x² + 3x – 4.
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-4, 0) and (1, 0).
So, the zeroes of the polynomial are -4 and 1.
Justification:
Given p(x) = x² + 3x – 4 = 0
⇒ x² + 4x – x – 4 = 0
⇒ x(x + 4)- 1(x + 4) = 0
⇒ (x + 4) (x – 1) = 0
⇒ x + 4 = 0 and x – 1 = 0
x = – 4 and x = 1

v) Given polynomial p(x) = x² – 1
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-1, 0) and (1,0).
So, the zeroes of the polynomial are – 1 and 1.
Justification:
Given p(x) = x² – 1 = 0
⇒ p(x) = (x + 1) (x – 1) = 0 [∵ a² – b² = (a + b) (a – b)]
⇒ x + 1 = 0 and x – 1 = 0
x = -1 and x = 1

Question 4.
Why are \(\frac{1}{4}\) and -1 zeroes of the polynomial p(x) = 4x² + 3x – 1 ?
Answer:
Given polynomial p(x) = 4x² + 3x – 1
Given zeroes are \(\frac{1}{4}\) and -1
Let x = \(\frac{1}{4}\)

Let x = -1
⇒ p(-1) = 4(-1)² + 3(-1)-1 = 4 – 3 – 1 = 4 – 4 = 0
∴ P(\(\frac{1}{4}\)) = 0 and p(-1) = 0
So these values are zeroes of the polynomial p(x).

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.1

10th Class Maths 3rd Lesson Polynomials Ex 3.1 Textbook Questions and Answers

Question 1.
a) If p(x) = 5x⁷ – 6x⁵ + 7x – 6, find
i) coefficient of x⁵
ii) degree of p(x)
iii) constant term.
Answer:
Given p(x) = 5x⁷ – 6x⁵ + 7x – 6
i) coefficient of x⁵ is -6
ii) degree of p(x) is 7
iii) constant term is -6

b) Write three more polynomials and create three questions for each of them.
Answer:
Polynomial – 1: P(x) = x + 5
Questions:
1) What is the order of given polynomial?
2) What are maximum possible zeroes to the above polynomial?
3) What is the zero value of given polynomial?

Polynomial – 2: P(x) = x² – 5x + 6
Questions:
1) What is the sum of zeroes of given polynomial?
2) What is the product of zeroes of it?
3) At how many points, do the polynomial crosses x-axis?

Polynomial – 3: P(x) = ax^p + bx² + cx + d
Questions:
1) What will be the value of ‘p’, if the given is cubic polynomial?
2) What is the product of zeroes of it?
3) What can you say about the value of ‘a’ if the given is a cubic polynomial?

Question 3.
If p(t) = t³ – 1, find the values of p(1), p(-1), p(0), p(2), p(-2).
Answer:
Given polynomial p(t) = t³ – 1
p(1) = 1³ – 1 = 1 – 1 = 0
p(-1) = (-1)³ – 1 = – 1 – 1 = – 2
p(0) = 0³ – 1 = 0 – 1 = – 1
p(2) = 2³ – 1 = 8 – 1 = 7
p(-2) = (-2)³ – 1 = – 8 – 1 = – 9

Question 4.
Check whether – 2 and 2 are the zeroes of the polynomial x⁴ – 16.
Answer:
Given polynomial is x⁴ – 16
Let p(x) = x⁴ – 16
We have p(-2) = (-2)⁴ – 16
= 16 – 16 = 0 and
p(2) = (2)⁴ – 16
= 16 – 16 = 0
p(-2) = 0 and p(2) = 0.
So these are zeroes of the polynomial.

Question 5.
Check whether 3 and -2 are the zeroes of the polynomial p(x) when p(x) = x² – x – 6.
Answer:
Given polynomial p(x) = x² – x – 6
We have, p(3) = 3² – 3 – 6
= 9 – 3 – 6
= 9 – 9
= 0 and
p(-2) = (-2)² – (-2) – 6
= 4 + 2 – 6
= 6 – 6
= 0
We see that p(3) = 0 and p(-2) = 0
∴ 3 and – 2 are the zeroes of the polynomial p(x).

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.3

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers

Question 1.
Solve each of the following pairs of equations by reducing them to a pair of linear equations.
i) \(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}\) + \(\frac{3}{y-2}\) = 1
Answer:
Given
\(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}\) + \(\frac{3}{y-2}\) = 1
Put \(\frac{1}{x-1}\) = a and \(\frac{1}{y-2}\) = b,
then the given equations reduce to
5a + b = 2 ……… (1)
6a – 3b = 1 ………. (2)

⇒ b = \(\frac{7}{21}\) = \(\frac{1}{3}\)
Substituting b = \(\frac{1}{3}\) in equation (1) we get

⇒ (x – 1) . 1 = 3 × 1
⇒ x – 1 = 3
⇒ x = 3 + 1 = 4
b = \(\frac{1}{y-2}\) ⇒ \(\frac{1}{3}\) = \(\frac{1}{y-2}\)
⇒ (y – 2) . 1 = 3 × 1
⇒ y – 2 = 3
⇒ y = 3 + 2 = 5
∴ Solution (x, y) = (4, 5)

ii) \(\frac{x+y}{xy}\) = 2;
\(\frac{x-y}{xy}\) = 6
Answer:
Given
\(\frac{x+y}{xy}\) = 2
⇒ \(\frac{x}{xy}\) + \(\frac{y}{xy}\) = 2
⇒ \(\frac{1}{y}\) + \(\frac{1}{x}\) = 2
\(\frac{x-y}{xy}\) = 6
⇒ \(\frac{x}{xy}\) – \(\frac{y}{xy}\) = 6
⇒ \(\frac{1}{y}\) – \(\frac{1}{x}\) = 6
Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,
then the given equations reduces to

⇒ b = \(\frac{8}{2}\) = 4
Substituting b = 4 in equation (1) we get
a + 4 = 2 ⇒ a = 2 – 4 = -2
but a = \(\frac{1}{x}\) = -2 ⇒ x = \(\frac{-1}{2}\)
b = \(\frac{1}{y}\) = 4 ⇒ y = \(\frac{1}{4}\)
∴ Solution (x, y) = \(\left(\frac{-1}{2}, \frac{1}{4}\right)\)

iii) \(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2;
\(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1
Answer:
Given
\(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2 and \(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1
Take \(\frac{1}{\sqrt{x}}\) = a and \(\frac{1}{\sqrt{y}}\) = b,
then the given equations reduces to
2a + 3b = 2 …….. (1)
4a – 9b = – 1 …….. (2)

⇒ b = \(\frac{5}{15}\) = \(\frac{1}{3}\)
Substituting b = \(\frac{1}{3}\) in equation (1) we get
2a + 3\(\left(\frac{1}{3}\right)\) = 2
⇒ 2a + 1 = 2 ⇒ 2a = 2 – 1 ⇒ a = \(\frac{1}{2}\)

∴ Solution (x, y) = (4, 9)

iv) 6x + 3y = 6xy
2x + 4y = 5xy
Answer:
Given
6x + 3y = 6xy
⇒ \(\frac{6x+3y}{xy}\) = 6
⇒ \(\frac{6x}{xy}\) + \(\frac{3y}{xy}\) = 6
⇒ \(\frac{6}{y}\) + \(\frac{3}{x}\) = 6
2x + 4y = 5xy
⇒ \(\frac{2x+4y}{xy}\) = 5
⇒ \(\frac{2x}{xy}\) + \(\frac{4y}{xy}\) = 6
⇒ \(\frac{2}{y}\) + \(\frac{4}{x}\) = 6
Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,
then the given equations reduces to
3a + 6b = 6 ……. (1)
4a + 2b = 5 ……. (2)

⇒ b = \(\frac{9}{18}\) = \(\frac{1}{2}\)
Substituting b = \(\frac{1}{2}\) in equation (1) we get
3a +6\(\left(\frac{1}{2}\right)\) = 6
⇒ 3a = 6 – 3
⇒ a = \(\frac{3}{3}\) = 1
but a = \(\frac{1}{x}\) = 1 ⇒ x = 1
b = \(\frac{1}{y}\) = \(\frac{1}{2}\) ⇒ y = 2
∴ Solution (x, y) = (1, 2)

v) \(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1
\(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10
where x ≠ 0, y ≠ 0
Answer:
Given
\(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1 and
\(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10
Take \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b, then
the given equations reduce to
5a – 2b = – 1 ……… (1)
15a + 7b = 10 ……… (2)

⇒ b = \(\frac{-13}{-13}\) = 1
Substituting b = 1 in equation (1) we get
5a – 2(1) = -1
⇒ 5a = -1 + 2
⇒ 5a = 1
⇒ a = \(\frac{1}{5}\)
but a = \(\frac{1}{x+y}\) = \(\frac{1}{5}\) ⇒ x + y = 5
b = \(\frac{1}{x-y}\) = 1 ⇒ x – y = 1
⇒ x = \(\frac{6}{2}\) = 3
Solving the above equations
Substituting x = 3 in x + y = 5 we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ Solution (x, y) = (3, 2)

vi) \(\frac{2}{x}\) + \(\frac{3}{y}\) = 13
\(\frac{5}{x}\) – \(\frac{4}{y}\) = -2
where x ≠ 0, y ≠ 0
Answer:
Given
\(\frac{2}{x}\) + \(\frac{3}{y}\) = 13 and
\(\frac{5}{x}\) – \(\frac{4}{y}\) = -2
Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b, then
the given equations reduce to
2a + 3b = 13 ……… (1)
5a – 4b = -2 ……… (2)

⇒ b = \(\frac{69}{23}\) = 3
Substituting b = 3 in equation (1) we get
2a + 3 (3) = 13
⇒ 2a = 13 – 9
⇒ a = \(\frac{4}{2}\) = 2
but a = \(\frac{1}{x}\) = 2 ⇒ x = \(\frac{1}{2}\)
b = \(\frac{1}{y}\) = 3 ⇒ y = \(\frac{1}{3}\)
∴ Solution (x, y) = (\(\frac{1}{2}\), \(\frac{1}{3}\))

vii) \(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4
\(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2
Answer:
Given
\(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4 and
\(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2
Take \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b, then
the given equations reduce to
10a + 2b = 4 ……… (1)
15a – 5b = – 2 ……… (2)

⇒ b = \(\frac{16}{16}\) = 1
Substituting b = 1 in equation (1) we get
10a + 2(1) = 4
⇒ 10a = 4 – 2
⇒ a = \(\frac{2}{10}\) = \(\frac{1}{5}\)
but a = \(\frac{1}{x+y}\) = \(\frac{1}{5}\) ⇒ x + y = 5 ……. (3)
b = \(\frac{1}{x-y}\) = 1 ⇒ x – y = 1 …….. (4)
Adding (3) and (4)

⇒ x = \(\frac{6}{2}\) = 3
Substituting x = 3 in x + y = 5 we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ Solution (x, y) = (3, 2)

viii) \(\frac{1}{3x+y}\) + \(\frac{1}{3x-y}\) = \(\frac{3}{4}\)
\(\frac{1}{2(3x+y)}\) – \(\frac{1}{2(3x-y)}\) = \(\frac{-1}{8}\)
Answer:
Given
\(\frac{1}{3x+y}\) + \(\frac{1}{3x-y}\) = \(\frac{3}{4}\) and
\(\frac{1}{2(3x+y)}\) – \(\frac{1}{2(3x-y)}\) = \(\frac{-1}{8}\)
Take \(\frac{1}{3x+y}\) = a and \(\frac{1}{3x-y}\) = b, then
the given equations reduce to

⇒ a = \(\frac{2}{8}\) = \(\frac{1}{4}\)
Substituting a = \(\frac{1}{4}\) in equation (1) we get

Solving (3) and (4)

⇒ x = \(\frac{6}{6}\) = 1
Substituting x = 1 in 3x + y = 4
⇒ 3(1) + y = 4
⇒ y = 4 – 3 = 1
∴ The solution (x, y) = (1, 1)

Question 2.
Formulate the following problems as a pair of equations and then find their solutions.
i) A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Answer:
Let the speed of the boat in still water = x kmph
and the speed of the stream = y kmph
then speed in downstream = x + y
Speed in upstream = x – y
and time = \(\frac{\text { distance }}{\text { speed }}\)
By problem,
\(\frac{30}{x-y}\) + \(\frac{44}{x+y}\) = 10
\(\frac{40}{x-y}\) + \(\frac{55}{x+y}\) = 13
Take \(\frac{1}{x-y}\) = a and \(\frac{1}{x+y}\) = b, then
the given equations reduce to
30a + 44b = 10 ……… (1)
40a + 55b = 13 ……… (2)

⇒ b = \(\frac{1}{11}\)
Substituting b = \(\frac{1}{11}\) in equation (1) we get

⇒ x = 8
Substituting x = 8 in x – y = 5 we get
8 – y = 5
⇒ y = 8 – 5 = 3
∴ The solution (x, y) = (8, 3)
Speed of the boat in still water = 8 kmph
Speed of the stream = 3 kmph.

ii) Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and rest by car. He takes 20 minutes more if he travels 200 km by train and rest by car. Find the speed of the train and the car.
Answer:
Let the speed of the train be x kmph
and the speed of the car = y kmph
By problem,

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b, then
the given equations reduce to
15a + 60b = 1 ……… (1)
8a + 16b = \(\frac{1}{3}\) ⇒ 24a + 48b = 1 ……… (2)

⇒ a = \(\frac{-1}{-60}\) = \(\frac{1}{60}\)
Substituting a = \(\frac{1}{60}\) in equation (1) we get

but a = \(\frac{1}{x}\) = \(\frac{1}{60}\) ⇒ x = 60 kmph
b = \(\frac{1}{y}\) = \(\frac{1}{80}\) ⇒ y = 80 kmph
Speed of the train = 60 kmph and
speed of the car = 80 kmph

iii) 2 women and 5 men can together finish an embroidery work in 4 days while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone and 1 man alone to finish the work.
Answer:
Let the time taken by 1 woman to complete the work = x days
and time taken by 1 man to complete the work = y days
∴ Work done by 1 woman in 1 day = \(\frac{1}{x}\)
Work done by 1 man in 1 day = \(\frac{1}{y}\)
By problem,

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,
then the above equations reduce to
2a + 5b = \(\frac{1}{4}\) and 3a + 6b = \(\frac{1}{3}\)
⇒ 8a + 20b = 1 …….. (1) and
9a + 18b = 1 ……… (2)

⇒ b = \(\frac{1}{36}\)
Substituting b = \(\frac{1}{36}\) in equation (1) we get

but a = \(\frac{1}{x}\) = \(\frac{1}{18}\) ⇒ x = 18 and
b = \(\frac{1}{y}\) = \(\frac{1}{36}\) ⇒ y = 36
∴ Time taken by 1 woman = 18 days
1 man = 36 days

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.2

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers

Form a pair of linear equations for each of the following problems and find their solution.
Question 1.
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly income.
Answer:
Given ratio of incomes of two persons = 9 : 7
So let the incomes of each = Rs. 9x and Rs. 7x
and ratio of expenditures = 4 : 3
So let the expenditures of each = 4y and 3y
then earnings of each = (income – expenditure) of each
⇒ 9x – 4y = Rs. 2000 and 7x – 3y = 200
∴ 9x – 4y = 7x – 3y = 2000
⇒ 9x – 7x = 4y – 3y
⇒ y = 2x
now putting y = 2x in 9x – 4y = 2000 we get
9x – 4(2x) = 2000 ⇒ x = 2000
∴ Income of each = 9x = 9(2000) = 18000
and 7x = 7(2000) = 14,000

Question 2.
The sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Answer:
Let the digit in units place be x
and the digit in tens place be y
then the value of the number = 10y + x
Number obtained by reversing the digits = 10x + y
By problem,
(10y + x) + (10x + y) = 66
and x – y = 2
⇒ 11x – 11y = 66 and x – y = 2
⇒ x + y = 6 and x – y = 2
Solving these two equations
x + y = 6
x – y = 2
(+) 2x = 8
x = \(\frac{8}{2}\) = 4
Substituting x = 4 in x + y = 6
we get 4 + y = 6 ⇒ y = 2
Substituting x, y values in equations (10y + x) & (10x + y),
We get 10y + x
= 10(2) + 4 = 20 + 4 = 24
and 10x + y = 10(4) + 2
= 40 + 2 = 42
∴ The number is 42 or 24
Thus we have two such numbers.

Question 3.
The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Answer:
Let the pair of supplementary angles be x and y [and x > y]
then we have x + y = 180° …… (1)
By problem, x = y + 18°
⇒ x – y = 18° …… (2)
Solving the equations (1) and (2) we get

and x = \(\frac{198}{2}\) = 99°
Substituting x = 99° in equation (2) we get
99° – y° = 18°
⇒ y° = 99° – 18 = 81°
∴ The angles are 99° and 81°.

Question 4.
The taxi charges in Hyderabad are fixed, along with the charge for the distance covered. For a distance of 10 km., the charge paid is Rs. 220. For a journey of 15 km. the charge paid is Rs. 310.
i) What are the fixed charges and charge per km?
ii) How much does a person have to pay for travelling a distance of 25km?
Answer:
Let the fixed charge be = Rs. x.
and the charge per one km = Rs. y.
By problem, x + 10y = 220 x + 15y = 310
Solving (1) and (2) we get

∴ y = \(\frac{-90}{-5}\) = 18
i.e., charge per one km = Rs. 18
Substituting y = 18 in equation (1) we get
x + 10 × 18 = 220
⇒ x = 220 – 180
⇒ x = Rs. 40
∴ Fixed charge = Rs. 40;
Charge per km = Rs. 18.

ii) Now, the charge for travelling a distance of 25 km = 25 × 18
= Rs. 450 + 40
= Rs. 490

Question 5.
A fraction becomes equal to \(\frac{4}{5}\) if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes equal to \(\frac{1}{2}\). What is the fraction?
Answer:
Let the numerator of the fraction = x
and the denominator of the fraction = y
By problem,
\(\frac{x+1}{y+1}\) = \(\frac{4}{5}\) and \(\frac{x-5}{y-5}\) = \(\frac{1}{2}\)
⇒ 5(x + 1) = 4(y + 1) and 2(x – 5) = 1(y – 5)
5x + 5 = 4y + 4 and 2x – 10 = y – 5
⇒ 5x – 4y = 4 – 5 and 2x – y = – 5 + 10
⇒ 5x – 4y = – 1 …… (1)
and 2x – y = 5 …… (2)

∴ y = \(\frac{-27}{-3}\) = 9
Substituting y = 9 in equation (2) we get
2x – 9 = 5
⇒ 2x = 5 + 9
⇒ 2x = 14 and
x = \(\frac{14}{2}\) = 7
Thus the fraction is \(\frac{x}{y}\) = \(\frac{7}{9}\)

Question 6.
Places A and B are 100 km apart on a highway One car starts from A and another from B at the same time at different speeds. If the cars travel in the same direction, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Answer:

Let the speed of the car which started from the place A = x kmph
and B = y kmph
Distance travelled by first car in 5h = 5x and in 1h = x
The distance covered by second car in 5h = 5y and in 1h = y
By problem when travelled in same direction,
5x – 5y = 100 ⇒ x – y = 20 …… (1)
and when travelled towards each other
x + y = 100 ……. (2)
Solving (1) and (2),

∴ x = \(\frac{120}{2}\) = 60
Substituting x = 60 in equation (1) we get
60 – y = 20
⇒ y = 60 – 20 = 40 kmph
Thus the speed of the cars are 60 kmph and 40 kmph.

Question 7.
Two angles are complementary. The larger angle is 3° less than twice the measure of the smaller angle. Find the measure of each angle.
Answer:
Let the pair of complementary angles be x° and y° with x° > y°
then x° + y° = 90° and
By problem
x = 2y – 3° ⇒ x – 2y = – 3°
Solving these two equations we get,

∴ y = \(\frac{93}{3}\) = 31°
Substituting y = 31°in x + y = 90° we get
x + 31° = 90°
⇒ x = 90° – 31° = 59°
The angles are 59° and 31°.

Question 8.
An algebra textbook has a total of 1382 pages. It is broken up into two parts. The second part of the book has 64 pages more than the first part. How many pages are in each part of the book?
Answer:
Let the first part of the book contains x pages
and the second part of the book contains y pages By problem,
x + y = 1382 ….. (1)
y = x + 64 ⇒ x – y = -64 …… (2)
Solving equations (1) and (2) we get

∴ x = \(\frac{1318}{2}\) = 659
Substituting x = 659 in equation (1) we get
659 + y = 1382
⇒ y = 1382 – 659 = 723
∴ The number of pages in the first part = 659
Second part = 723

Question 9.
A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the other is 80% solution. How much of each should be used to obtain 100 ml of a 68% solution?
Answer:
Let the first solution contains 50% acid.
Second solution contains 80% acid.
Let x ml of 1st solution and y ml of second solution are added.
Then x + y = 100
Acid content in the ‘mix’ is 50% of x + 80% of y = 68%

∴ y = \(\frac{180}{3}\) = 60
Substituting y = 60 in equation (1) we get
x + 60 = 100
⇒ x = 100 – 60 = 40
∴ Quantity of first solution = 40 ml
Quantity of second solution = 60 ml

Question 10.
Suppose you have Rs. 12000 to invest. You have to invest some amount at 10% and the rest at 15%. How much should be invested at each rate to yield 12% on the total amount invested ?
Answer:
Let the amount to be invested @ 10% be Rs. x
and the amount to be invested @ 15% be Rs. y
By problem x + y = 12000 ……. (1)
Also 10% of x + 15% of y = 12% of 12000

⇒ y = \(\frac{-24000}{-5}\) = Rs. 4800
Substituting y = 4800 in equation (1) we get
x + 4800 = 12000
⇒ x = 12000 – 4800 = 7200
The invested @ 10% = Rs. 7200
@ 15% = Rs. 4800

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.1

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers

Question 1.
By comparing the ratios \(\frac{a_{1}}{a_{2}}\), \(\frac{b_{1}}{b_{2}}\), \(\frac{c_{1}}{c_{2}}\) K find out whether the lines represented by the following pairs of linear equations intersect at a point, are parallel or are coincident.
a) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
Answer:
Given: 5x – 4y + 8 = 0
7x + 6y – 9 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{5}{7}\); \(\frac{b_{1}}{b_{2}}\) = \(\frac{-4}{6}\); \(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{-9}\)
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
Hence the given pair of linear equations represents a pair of intersecting lines.

b) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Answer:
Given : 9x + 3y + 12 = 0
18x + 6y + 24= 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{9}{18}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{12}{24}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
The lines are coincident.

c) 6x – 3y + 10 = 0
2x – y + 9 = 0
Answer:
Given: 6x – 3y + 10 = 0
2x – y + 9 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{6}{2}\) = \(\frac{3}{1}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-1}\) = \(\frac{3}{1}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{10}{9}\)
Here \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
∴ The lines are parallel.

Question 2.
Check whether the following equations are consistent or inconsistent. Solve them graphically. (AS2, AS5)
a) 3x + 2y = 8
2x – 3y = 1
Answer:
Given equaions are 3x + 2y = 8 and 2x – 3y = 1
\(\frac{a_{1}}{a_{2}}\) = \(\frac{3}{2}\);
\(\frac{b_{2}}{b_{-3}}\) = \(\frac{-4}{6}\);
\(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
Hence the linear equations are consistent.


The lines intersect at (2, 1), so the solution is (2, 1).

b) 2x – 3y = 8
4x – 6y = 9
Answer:
Given: 2x – 3y = 8 and 4x – 6y = 9
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-6}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{9}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Lines are inconsistent and have no solution.

Lines are parallel.

The lines are parallel and no solution exists.

c) \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7
9x – 10y = 12
Answer:
Given pair of equations \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7 and 9x – 10y = 12
Now take \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7 ⇒ \(\frac{9x+10y}{6}\) = 7 ⇒ 9x + 10y = 42
and 9x – 10y =12
\(\frac{a_{1}}{a_{2}}\) = \(\frac{9}{9}\) = \(\frac{1}{1}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{10}{-10}\) = \(\frac{1}{-1}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-42}{-12}\) = \(\frac{7}{2}\)
Since \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\) they are intersecting lines and hence consistent pair of linear equations.


Solution: The unique solution of given pair of equations is (3.1, 1.4)

d) 5x – 3y = 11
-10x + 6y = -22
Answer:
Given pair of equations 5x – 3y = 11 and -10x + 6y = -22
\(\frac{a_{1}}{a_{2}}\) = \(\frac{5}{-10}\) = \(\frac{-1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{11}{-22}\) = \(\frac{-1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
∴ The lines are consistent.
∴ The given linear equations represent coincident lines.
Thus they have infinitely many solutions.

e) \(\frac{4}{3}\)x + 2y = 8
2x + 3y = 12
Answer:
Given pair of equations \(\frac{4}{3}\)x + 2y = 8 ⇒ \(\frac{4x+6y}{3}\) = 8 ⇒ 4x + 6y = 24 ⇒ 2x + 3y = 12
\(\frac{a_{1}}{a_{2}}\) = \(\frac{4}{2}\) = 2;
\(\frac{b_{1}}{b_{2}}\) = \(\frac{6}{3}\) = 2;
\(\frac{c_{1}}{c_{2}}\) = \(\frac{24}{12}\) = 2
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
Thus the equations are consistent.
∴ The given equations have infinitely many solutions.

f) x + y = 5
2x + 2y = 10
Answer:
Given pair of equations x + y = 5 and 2x + 2y = 10
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
Thus the equations are consistent and have infinitely many solutions.

g) x – y = 8
3x – 3y = 16
Answer:
Given pair of equations x – y = 8 and 3x – 3y = 16
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{3}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-1}{-3}\) = \(\frac{1}{3}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{16}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Thus the equations are inconsistent.
∴ They represent parallel lines and have no solution.

h) 2x + y – 6 = 0 and 4x – 2y – 4 = 0
Answer:
Given pair of equations 2x + y – 6 = 0 and 4x – 2y – 4 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{1}{-2}\) = \(\frac{-1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-6}{-4}\) = \(\frac{3}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
The equations are consistent.
∴ They intersect at one point giving only one solution.


The solution is x = 2 and y = 2

i) 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
Answer:
Given pair of equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-2}{-4}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-2}{-5}\) = \(\frac{2}{5}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Thus the equations are inconsistent.
∴ They represent parallel lines and have no solution.

Question 3.
Neha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts are two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased.”
Help her friend to find how many pants and skirts Neha bought.
Answer:
Let the number of pants = x and the number of skirts = y
By problem y = 2x – 2 ⇒ 2x – y = 2
y = 4x – 4 ⇒ 4x – y = 4


The two lines are intersecting at the point (1,0)
∴ x = 1; y = 0 is the required solution of the pair of linear equations.
i.e., pants =1
She did not buy any skirt.

Question 4.
10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys then, find the number of boys and the number of girls who took part in the quiz.
Answer:
Let the number of boys be x.
Then the number of girls = x + 4
By problem, x + x + 4 = 10
∴ 2x + 4 = 10
2x = 10-4
x = \(\frac{6}{2}\) = 3
∴ Boys = 3 Girls = 3 + 4 = 7 (or)
Boys = x, Girls = y
By problem x + y = 10 (total)
and y = x + 4 (girls)
⇒ x + y = 10 and x – y = – 4


∴ Number of boys = 3 and the number of girls = 7

Question 5.
5 pencils and 7 pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Answer:
Let the cost of each pencil be Rs. x
and the cost of each pen be Rs. y.
By problem 5x + 7y = 50
7x + 5y = 46


The lines are intersecting at the point (3, 5).
x = 3 and y = 5 is the solution of given equations.
∴ Cost of one pencil = Rs. 3 and pen = Rs. 5

Question 6.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden.
Answer:
Let the width of the garden = x cm
then its length = x + 4 cm
Half the perimeter = \(\frac{1}{2}\) × 2(7+ b) = l + b
By problem, x + x + 4 = 36
2x + 4 = 36
2x = 36 – 4 = 32
∴ x = 16 and x + 4 = 16 + 4 = 20
i.e., length = 20 cm and breadth = 16 cm.
(or)
Let the breadth be x and length = y
then x + y = 36 ⇒ x + y = 36
y = x + 4 ⇒ x – y = -4


The two lines intersect at the point (16, 20)
i.e., length = 20 cm and the breadth = 16 cm.

Question 7.
We have a linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is intersect¬ing lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.
Answer:
i) Given: 2x + 3y – 8 = 0
The lines are intersecting lines.
Let the other linear equation be ax + by + c = 0
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\); we have to choose appropriate values satisfying the condition above.
Thus the other equation may be 3x + 5y – 6 =0

ii) Parallel line \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
⇒ 2x + 3y – 8 = 0
4x + 6y – 10 = 0

iii) Coincident lines \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
⇒ 2x + 3y – 8 = 0 ⇒ 8x + 12y – 32 = 0

Question 8.
The area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will increase by 50 sq. units. Find the length and breadth of the rectangle.
Answer:
Let the length of the rectangle = x units
breadth = y units Area = l . b = xy sq. units
By problem, (x – 5) (y + 2) = xy – 80 and          (x + 10) (y – 5) = xy + 50
⇒ xy + 2x – 5y – 10 = xy – 80 and                    xy – 5x + 10y – 50 = xy + 50
⇒ 2x – 5y = xy – 80 – xy + 10 and                   -5x + 10y = xy + 50 – xy + 50
⇒ 2x – 5y = – 70 and                                       -5x + 10y = 100


The two lines intersect at the point (40, 30)
∴ The solution is x = 40 and y = 30
i.e., length = 40 units; breadth = 30 units.

Question 9.
In X class, if three students sit on each bench, one student will be left. If four students sit on each bench, one bench will be left. Find the number of students and the number of benches in that class.
Answer:
Let the number of benches = x say and the number of students = y
By problem
y = 3x + 1 ⇒ 3x – y + 1 = 0
and y = 4(x – 1) ⇒ 4x – y – 4 = 0


The two lines intersect at (5, 16)
∴ The solution of the equation is x = 5 and y = 16
i.e., Number of benches = 5 and the number of students = 16

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.4

10th Class Maths 5th Lesson Quadratic Equations Ex 5.4 Textbook Questions and Answers

Question 1.
Find the nature of the roots of the following quadratic equations. If real roots exist, find them.
i) 2x² – 3x + 5 = 0
Answer:
Given: 2x² – 3x + 5 = 0
a = 2; b = -3; c = 5
Discriminant = b² – 4ac
b² – 4ac = (-3)² – 4(2)(5)
= 9 – 40
= -31 < 0
∴ Roots are imaginary.

ii) 3x² – 4√3x + 4 = 0
Answer:
Given: 3x² – 4√3x + 4 = 0
a = 3; b = -4√3; c = 4
b² – 4ac = (-4√3)² – 4(3)(4)
= 48 – 48 = 0
∴ Roots are real and equal and they
\(\frac{-b}{2a}\), \(\frac{-b}{2a}\)

iii) 2x² – 6x + 3 = 0
Answer:
Given: 2x² – 6x + 3 = 0
a = 2; b = -6; c = 3
b² – 4ac = (-6)² – 4(2)(3)
= 36 – 24
= 12 > 0
∴ The roots are real and distinct. They are

Question 2.
Find the values of k for each of the fol-lowing quadratic equations so that they have two equal roots.
i) 2x² + kx + 3 = 0
Answer:
Given : 2x² + kx + 3 = 0 has equal roots
∴ b² – 4ac = 0
Here a = 2; b = k; c = 3
b² – 4ac = (k)² – 4(2)(3) = 0
⇒ k² – 24 = 0
⇒ k² = 24
⇒ k = √24 = ± 2√6

ii) kx(x – 2) + 6 = 0
Answer:
Given: kx(x – 2) + 6 = 0
kx² – 2kx + 6 = 0
As this Q.E. has equal roots,
b² – 4ac = 0
Here
a = k; b = -2k; c = 6
∴ b² – 4ac = (-2k)² – 4(k)(6) = 0
⇒ 4k² – 24k = 0
⇒ 4k(k – 6) = 0
⇒ 4k = 0 (or) k – 6 = 0
⇒ k = 0 (or) 6
But k = 0 is trivial
∴ k = 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Answer:
Let the breadth = x m
Then length = 2x m
Area = length x breadth = x.(2x)
= 2x² m²
By problem 2x² = 800 ⇒ x² = 400
and x = √400 = ± 20
∴ Breadth x = 20 m and
length 2x = 2 × 20 = 40 m.

Question 4.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is the above situation possible? If so, deter¬mine their present ages.
Answer:
Let the age one of the two friends be x years.
Then the age of the other = 20 – x
Then, 4 years ago their ages would be (x – 4) and (20 – x – 4) = 16 – x
∴ Product of their ages 4 years ago = (x – 4) (16 – x)
By problem (x – 4) (16 – x) = 48
⇒ x(16 – x) – 4(16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ x² – 20x + 112 = 0
Here a = 1; b = -20; c = 112
b² – 4ac = (-20)² – 4(1) (112)
= 400 – 448
= -48 < 0
Thus the roots are not real.
∴ The situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Answer:
Given: Perimeter of a rectangle 2(1 + b) = 80
⇒ 6 + b = \(\frac{80}{2}\) = 40
Area of the rectangle, l × b = 400
If possible, let us suppose that length of the rectangle = x m say
Then its breadth by equation (1) = 40 – x
By problem area = x . (40 – x) = 400
⇒ 40x – x² = 400
⇒ x² – 40x + 400 = 0
Here a = 1; b = -40; c = +400
b² – 4ac = (-40)² – 4(1)(+400)
= 1600 – 1600 = 0
∴ The roots are real and equal.
They are \(\frac{-b}{2a}\), \(\frac{-b}{2a}\)
i.e., \(\frac{-(-40)}{2 \times 1}\) = \(\frac{40}{2}\) = 20
∴ The dimensions are 20 m, 20 m.
(∴ The park is in square shape)