Category: Class 10

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.3

10th Class Maths 1st Lesson Real Numbers Ex 1.3 Textbook Questions

Question 1.
Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating repeating decimals.
i) \(\frac{3}{8}\)
ii) \(\frac{229}{400}\)
iii) 4\(\frac{1}{5}\)
iv) \(\frac{2}{11}\)
v) \(\frac{8}{125}\)
Answer:
i) \(\frac{3}{8}\)

[!! Denominator 8 = 2³, consists of only 2’s. Hence a terminating decimal.]
∴ \(\frac{3}{8}\) = 0.375 is a terminating decimal.

ii) \(\frac{229}{400}\)

[!! Denominator 400 = 2⁴ × 5² = 2ⁿ × 5^m. Hence a terminating decimal.]
∴ \(\frac{229}{400}\) = 0.5725 is a terminating decimal.

iii) 4\(\frac{1}{5}\)
4\(\frac{1}{5}\) = 4 + \(\frac{1}{5}\)

[!! Denominator is 5. Hence a terminating decimal.]
∴ 4\(\frac{1}{5}\) = 4.2 is a terminating decimal.

iv) \(\frac{2}{11}\)

[!! Denominator is not of the form 2^m × 5ⁿ. Hence a non-terminating repeating decimal.]
∴ \(\frac{2}{11}\) = \(0 . \overline{18}\) is a non terminating, repeating decimal.

v) \(\frac{8}{125}\)

[!! Denominator 125 = 5³. Hence a terminating decimal.]
∴ \(\frac{8}{125}\) = 0.064 is a terminating decimal.

Question 2.
Without performing division, state whether the following rational numbers will have a terminating decimal form or a non-terminating, repeating decimal form.
i) \(\frac{13}{3125}\)
ii) \(\frac{11}{12}\)
iii) \(\frac{64}{455}\)
iv) \(\frac{15}{1600}\)
v) \(\frac{29}{343}\)
vi) \(\frac{23}{2^{3} \cdot 5^{2}}\)
vii) \(\frac{129}{2^{2} \cdot 5^{7} \cdot 7^{5}}\)
viii) \(\frac{9}{15}\)
iX) \(\frac{36}{100}\)
X) \(\frac{77}{210}\)
Answer:
i) \(\frac{13}{3125}\)
Note: We check whether the denominator is of the form 2ⁿ . 5^m or not? If yes, the rational number can be expressed as a terminating decimal. If not, it can’t be expressed as a terminating decimal.

[!! Denominator is of the form 2^m × 5ⁿ. Hence a terminating decimal.]
3125 = 5⁵
∴ \(\frac{13}{3125}\) is a terminating decimal.

ii) \(\frac{11}{12}\)
The denominator 12 is not a factor of 11. Moreover 12 = 2² × 3.
[!! Denominator is not of the form 2^m × 5ⁿ.]
∴ \(\frac{11}{12}\) is a non terminating, repeating decimal.

iii) \(\frac{64}{455}\)

[!! Denominator is not of the form 2^m × 5ⁿ. Hence a non terminating decimal.]
∴ 455 = 5 × 7 × 13
Hence\(\frac{64}{455}\) is a non terminating, repeating decimal.

iv) \(\frac{15}{1600}\)

∴ 1600 = 2⁶ × 5² [∵ The denominator is of the form 2ⁿ . 5^m]
Hence \(\frac{15}{1600}\) is a terminating decimal.

v) \(\frac{29}{343}\)

343 = 7³ [Not of the form 2ⁿ . 5^m]
∴ \(\frac{29}{343}\) is a non terminating, repeating decimal.

vi) \(\frac{23}{2^{3} \cdot 5^{2}}\)
\(\frac{23}{2^{3} \cdot 5^{2}}\) is a terminating decimal.
[∵ The denominator is of the form 2ⁿ . 5^m]

vii) \(\frac{129}{2^{2} \cdot 5^{7} \cdot 7^{5}}\)
\(\frac{129}{2^{2} \cdot 5^{7} \cdot 7^{5}}\) is a non terminating, repeating decimal.

viii) \(\frac{9}{15}\)
\(\frac{9}{15}\) = \(\frac{3}{5}\)
Denominator is of the form 2ⁿ . 5^m.
∴ \(\frac{9}{15}\) = \(\frac{3}{5}\) is a terminating decimal.

ix) \(\frac{36}{100}\)
100 = 2² × 5² is of the form 2ⁿ . 5^m
Hence \(\frac{36}{100}\) is a terminating decimal.

x) \(\frac{77}{210}\)

210 = 2 × 3 × 5 × 7 is not of the form 2ⁿ . 5^m
Given fraction has a non-terminating, repeating decimal expansion.

Question 3.
Write the following rationals in decimal form using Theorem 1.4.
i) \(\frac{13}{25}\)
ii) \(\frac{15}{16}\)
iii) \(\frac{23}{2^{3} \cdot 5^{2}}\)
iv) \(\frac{7218}{3^{2} \cdot 5^{2}}\)
v) \(\frac{143}{110}\)
Answer:
i) \(\frac{13}{25}\)

ii) \(\frac{15}{16}\)
\(\frac{15}{16}\) = \(\frac{15}{2 \times 2 \times 2 \times 2}\)

iii) \(\frac{23}{2^{3} \cdot 5^{2}}\)

iv) \(\frac{7218}{3^{2} \cdot 5^{2}}\)

v) \(\frac{143}{110}\)

Question 4.
The decimal form of some real numbers are given below. In each case, decide whether the number is rational or not. If it is rational, and expressed in form p/q, what can you say about the prime factors of q?
i) 43.12345678?
ii) 0.120120012000120000 ……….
iii) \(43 . \overline{123456789}\)
Answer:
i) 43.123456789
The given decimal expansion is terminating. Hence it is a rational number and the denominator q is of the form 2ⁿ . 5^m.

ii) 0.120120012000120000 …………
The given decimal expansion is neither terminating nor repeating.
Hence it is not a rational number. It represents an irrational number.

iii) \(43 . \overline{123456789}\)
The given real number is a repeating decimal with period 123456789. Hence it is a rational number.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.2

10th Class Maths 1st Lesson Real Numbers Ex 1.2 Textbook Questions and Answers

Question 1.
Express each of the following numbers as a product of its prime factors.
i) 140
ii) 156
iii) 3825
iv) 5005
v) 7429
Answer:
i) 140

∴ 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7

ii) 156

∴ 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13

iii) 3825

∴ 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17

iv) 5005

∴ 5005 = 5 × 7 × 11 × 13

v) 7429

∴ 7429 = 17 × 19 × 23

Question 2.
Find the L.C.M and H.C.F of the following integers by the prime factorization method.
i) 12, 15 and 21
ii) 17, 23 and 29
iii) 8, 9 and 25
iv) 72 and 108
v) 306 and 657
Answer:
i) 12, 15 and 21
12 = 2 × 2 × 3 = 22 × 3
15 = 3 × 5
21 = 3 × 7
L.C.M = 2² × 3 × 5 × 7 = 420
H.C.F = 3

ii) 17, 23 and 29
The given numbers 17, 23 and 29 are all primes.
L.C.M = their product
= 17 × 23 × 29 = 11339
∴ H.C.F = 1

iii) 8, 9 and 25
8 = 2 × 2 × 2 = 2³
9 = 3 × 3 = 3²
25 = 5 × 5 = 5²
L.C.M = 2³ × 3² × 5² = 1800
(or)
8, 9 and 25 are relatively prime, therefore L.C.M is equal to their product,
(i.e.,) L.C.M = 8 × 9 × 25 = 1800
H.C.F = 1

iv) 72 and 108

72 = 2³ × 3²
108 = 2² × 3³
L.C.M = 2³ × 3³ = 8 × 27 = 216
H.C.F = 2² × 3² = 4 × 9 = 36

v) 306 and 657

306 = 2 × 3² × 17
657 = 32 × 73
L.C.M = 2 × 3² × 17 × 73 = 22338
H.C.F = 3² = 9

Question 3.
Check whether 6ⁿ can end with the digit ‘0’ for any natural number n.
Answer:
Given number = 6ⁿ = (2 × 3)ⁿ
The prime factors here are 2 and 3 only.
To be end with 0; 6ⁿ should have a prime factor 5 and also 2.
So, 6ⁿ can’t end with zero.

Question 4.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer:
Given numbers are 7 × 11 × 13
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
⇒ 13(7 × 11 + 1) and
5(7 × 6 × 4 × 3 × 2 × 1 + 1)
⇒ 13 K and 5 L, where K = 78 and L = 7 × 6 × 4 × 3 × 2 × 1 + 1 = 1009
As the given numbers can be written as product of two numbers, they are composite.

Question 5.
How will you show that (17 × 11 × 2) + (17 × 11 × 5) is a composite number? Explain.
Answer:
(17 × 11 × 2) + (17 × 11 × 5)
= (17 × 11) (2 + 5)
= (17 × 11) (7)
= 187 × 7
Now the given expression is written as a product of two integers and hence it is a composite number.

Question 6.
What is the last digit of 6100?
Answer: We know that
6¹ = 6
6² = 36
6³ = 216
6⁴ = 1296
6⁵ = 7776
We see that 6ⁿ for any positive integer n ends is 6.
i.e., unit digit is always 6.
∴ Unit digit of 6¹⁰⁰ is 6.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.1

10th Class Maths 1st Lesson Real Numbers Ex 1.1 Textbook Questions and Answers

Question 1.
Use Euclid’s division algorithm to find the HCF of
i) 900 and 270
Answer:
900 = 270 × 3 + 90
270 = 90 × 3 + 0
∴ HCF = 90

ii) 196 and 38220
Answer:
38220 = 196 × 195 + 0
∴ 196 is the HCF of 196 and 38220.

iii) 1651 and 2032
Answer:
2032 = 1651 × 1 + 381
1651 = 381 × 4 + 127
381 = 127 × 3 + 0
∴ HCF = 127

Question 2.
Use Euclid division lemma to show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integers.
Answer:
Let ‘a’ be an odd positive integer.
Let us now apply division algorithm with a and b = 6.
∵ 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5.
i.e., ’a’ can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient.
But ‘a’ is taken as an odd number.
∴ a can’t be 6q or 6q + 2 or 6q + 4.
∴ Any odd integer is of the form 6q + 1, 6q + 3 or 6q + 5.

Question 3.
Use Euclid’s division lemma to show that the square of any positive integer is of the form 3p, 3p + 1.
Answer:
Let ‘a’ be the square of an integer.
Applying Euclid’s division lemma with a and b = 3
Since 0 ≤ r < 3, the possible remainders are 0, 1, and 2.
∴ a = 3q (or) 3q + 1 (or) 3q + 2
∴ Any square number is of the form 3q, 3q + 1 or 3q + 2, where q is the quotient.
(or)
Let ‘a’ be a positive integer
So it can be expressed as a = bq + r (from Euclideans lemma)
now consider b = 3 then possible values of ‘r’ are ‘0’ or ‘1’ or 2.
then a = 3q + 0 = 3q (or) 3q + 1 or 3q + 2 now square of given positive integer (a²) will be
Case – I: a² – (3q)² = 9q²=3(3q²) = 3p (p = 3q²)
Case-II: a² = (3q + l)² = 9q² + 6q+ 1
= 3[3q² + 2q] + 1 = 3p+l (Where p = 3q² + 2q) or
Case – III: a² = (3q + 2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1
= 3[3q² + 4q + 1] + 1
= 3p + 1 (where ‘p’ = 3q² + 4q + 1)
So from above cases 1, 2, 3 it is clear that square of a positive integer (a) is of the form 3p or 3p + 1
Hence proved.

Question 4.
Use Euclid’s division lemma to show that the cube of a positive integer is of the form 9m, 9m + 1 or 9m + 8.
(OR)
Show that the cube of any positive integer is of form 9m or 9m + 1 or 9m + 8, where m is an integer.
Answer:
Let ‘a’ be positive integer. Then from Euclidean lemma a = bq + r;
now consider b = 9 then 0 ≤ r < 9, it means remainder will be 0, or 1, 2, 3, 4, 5, 6, 7, or 8
So a = bq + r
⇒ a = 9q + r (for b = 9)
now cube of a = a³ + (9q + r)³
= (9q)³ + 3.(9q)³r + 3. 9q.r + r³
= 9³q³ + 3.9²(q²r) + 3.9(q.r) + r³
= 9[9².q³ + 3.9.q²r + 3.q.r] + r³
a³ = 9m + r³ (where ‘m’ = 9²q³ + 3.9.q2r + 3.q.r)
if r = 0 ⇒ r³ = 0 then a³ = 9m + 0 = 9m
and for r = 1 ⇒ r³ = l³ then a³ = 9m + 1
and for r = 2 ⇒ r³ = 2³ then a³ = 9m + 8
for r = 3 ⇒ r³, = 3³ ⇒ a³ = 9m + 27 = 9(m) where m = (9m +3)
for r = 4 ⇒ r³ = 4³ ⇒ a³ = 9m + 64 = (9m + 63) + 1 = 9m + 1
for r = 5 ⇒ r³ = 125 ⇒ a³ = 9m + 125 = (9m + 117) + 8 = 9m + 8
for r = 6 ⇒ r³ — 216 ⇒ a³ = 9m + 216 = 9m + 9(24) = 9m
for r = 7 ⇒ r³ = 243
⇒ a³ = 9m + 9(27) = 9m
for r = 8 ⇒ r³ = 512
⇒ a³ = 9m + 9(56) + 8 = 9m + 8
So from the above it is clear that a³ is either in the form of 9m or 9m + 1 or 9m + 8.
Hence proved.

Question 5.
Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.
(Or)
Show that one and only one out of a, a + 2 and a + 4 is divisible by 3 where ‘a’ is any positive integer.
Answer:
Let ‘n’ be any positive integer.
Then from Euclidean’s lemma n = bq + r (now consider b = 3)
⇒ n = 3q + r (here 0 ≤ r < 3) which means the possible values of ‘r’ = 0 or 1 or 2
Now consider r = 0 then ‘n’ = 3q (divisible by 3)
and n + 2 = 3q + 2 (not divisible by 3)
n + 4 = 3q + 4 (not divisible by 3)
Case – II: For r = 1
n = 3q + 1 (not divisible by 3)
n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + l) divisible by 3
n + 4 = 3q + 1 + 4 = 3q + 5 not divisible by 3
Case – III: For r = 2,
n = 3q + 2 not divisible by 3
n + 2 = 3q + 2 + 2 = 3q + 4, not divisible by 3
n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) divisible by 3
So in all above three cases we observe, only one of either (n) or (n + 1) or (n + 4) is divisible by 3.
Hence proved.

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