Category: Class 10

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

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

10th Class Maths 1st Lesson Real Numbers Optional Exercise Textbook Questions and Answers

Question 1.
Can the number 6ⁿ, n being a natural number, end with the digit 5? Give reason.
Answer:
Given number = 6ⁿ ; n ∈ N
6ⁿ to be end in 5; it should be divisible by 5
6ⁿ = (2 × 3)ⁿ
The prime factors of 6ⁿ are 2 and 3.
It can’t end with the digit 5.

Question 2.
Is 7 × 5 × 3 × 2 + 3 a composite number? Justify your answer.
Answer:
Given:
7 × 5 × 3 × 2 + 3
= 3 (7 × 5 × 2 + 1)
= 3 × (70 + 1)
= 3 × 71
∴ The given number has two factors namely 3 and 71.
Hence it is a composite number.

Question 3.
Prove that (2√3 + √5 ) is an irrational number. Also check whether (2√3 + √5) (2√3 – √5) is rational or irrational.
Answer:
To prove:
2√3 + √5 is an irrational number. On contrary, let us suppose that 2√3 + √5 be a rational number.
Then 2√3 + √5 = \(\frac{p}{q}\)
Squaring on both sides, we get


L.H.S = an irrational number.
R.H.S = p, q being integers, \(\frac{p^{2}-17 q^{2}}{4 q^{2}}\) is a rational number.
This is a contradiction to the fact that √l5 is an irrational. This is due to our assumption that 2√3 + √5 is a rational. Hence our assumption is wrong and 2√3 + √5 is an irrational number. Also,
(2√3 + √5) (2√3 – √5)
= (2√3)² – (√5)²
[∵ (a + b) (a – b) = a² – b²]
= 4 × 3 – 5
= 12 – 5 = 7, a rational number.

Question 4.
If x² + y² = 6xy, prove that 2 log (x + y) = log x + log y + 3 log 2.
Answer:
Given: x² + y² = 6xy
x² + y² + 2xy = 6xy + 2xy
(x + y)² = 8xy
Taking logarithms on both sides log (x + y)² = log8xy
⇒ 2log(x + y)= log8 + logx + logy [∵ logx^m = mlogx]
[∵ logxy = logx + logy]
= log2³ + logx + logy
⇒ 2log(x + y) = logx + logy + 3log2

Question 5.
Find the number of digits in 42013, if log₁₀2 = 0.3010.
Answer:
Given:
log₁₀2 = 0.3010
4²⁰¹³ = (2²)²⁰¹³ = 2⁴⁰²⁶
∴ log₁₀ 2⁴⁰²⁶ = 4026 log₁₀2
[∵ log x^m = m log x]
= 4026 × 0.3010 = 1211.826.
So 1211 + 1 = 1212
∴ 4²⁰¹³ has 1212 digits in its expansion.
(∵ characteristic 1211)

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers InText Questions

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

10th Class Maths 1st Lesson Real Numbers InText Questions and Answers

Do this

Question 1.
Find q and r for the following pairs of positive integers a and b, satisfying a = bq + r. (Page No. 3)
i) a = 13, b = 3
Answer:
13 = 3 × 4 + 1
here q = 4 ; r = 1
ii) a = 8, b = 80
Answer:
Take a = 80, b = 8
80 = 8 × 10 + 0 here q = 10 ; r = 0
iii) a = 125, b = 5
Answer:
125 = 5 × 25 + 0
here q = 25 ; r = 0
iv) a = 132, b = 11
Answer:
132 = 11 × 12 + 0
here q = 12 ; r = 0

Question 2.
Find the HCF of the following by using Euclid division lemma,
i) 50 and 70 (Page No. 4)
Answer:
For given two positive integers a > b;
there exists unique pair of integers q and r satisfying a = bq + r; 0≤r<b.
∴ 70 = 50 × 1 + 20
Here a = 70, b = 50, q = 1, r = 20.
Now consider 50, 20
50 = 20 × 2 + 10
Here a = 50, b = 20, q = 2, r = 10.
Now taking 20 and 10.
20 = 10 × 2 + 0
Here the remainder is zero.
∴ 10 is the HCF of 70 and 50.

ii) 96 and 72
Answer:
96 = 72 × 1 + 24
72 = 24 × 3 + 0
∴ HCF = 24

iii) 300 and 550
Answer:
550 = 300 × 1 + 250
300 = 250 × 1 + 50
250 = 50 × 5 + 0
∴ HCF = 50

iv) 1860 and 2015
Answer:
2015 = 1860 × 1 + 155
1860 = 155 × 12 + 0
∴ HCF = 155

Think & Discuss

Question 1.
From the above questions in ‘DO THIS’, what is the nature of q and r? (Page No. 3)
Answer:
Given: a = bq + r
q > 0 and r lies in between 0 and b
i.e. q > 0 and 0 ≤ r < b

Question 2.
Can you find the HCF of 1.2 and 0.12? Justify your answer. (Page No. 4)
Answer:
Given: 1.2 and 0.12
we have 1.2 = \(\frac{12}{10}\) = \(\frac{120}{100}\)
0.12 = \(\frac{12}{100}\)
Now considering the numerators 12 and 120, their HCF is 12.
∴ HCF of 1.2 and 0.12 is \(\frac{12}{100}\) = 0.12
i.e., if x is a factor of y then x is the HCF of x and y.

Question 3.
If r = 0, then what is the relationship between a, b and q in a = bq + r of Euclid divison lemma? (Page No. 6)
Answer:
Given: r = 0 in a = bq + r then a = bq
i.e., b divides a completely.
i.e., b is a factor of a.

Do this

Question 1.
Express 2310 as a product of prime factors. Also see how your friends have factorized the number. Have they done it as you ? Verify your final product with your friend’s result. Try this for 3 or 4 more numbers. What do you conclude? (Page No. 7)
Answer:
Given: 2310
2310 = 2 × 1155
= 2 × 3 × 385
= 2 × 3 × 5 × 77
2310 = 2 × 3 × 5 × 7 × 11

We notice that this prime factorization is unique.
And also notice that prime factorization of any number is unique i.e., every composite number can be expressed as a product of primes and this factorization is unique.
E.g: 144 = 2 × 72
= 2 × 2 × 36
= 2 × 2 × 2 × 18
= 2 × 2 × 2 × 2 × 9
= 2 × 2 × 2 × 2 × 3 × 3
= 24 × 32
320 = 2 × 160
= 2 × 2 × 80
= 2 × 2 × 2 × 40
= 2 × 2 × 2 × 2 × 20
= 2 × 2 × 2 × 2 × 2 × 10
= 2 × 2 × 2 × 2 × 2 × 2 × 5
= 26 × 5
125 = 5 × 25
= 5 × 5 × 5
= 5³

Question 2.
Find the HCF and LCM of the following given pairs of numbers by prime factorization, (Page No. 8)
i) 120, 90
Answer:
We have 120 = 2 × 2 × 2 × 3 × 5
= 2³ × 3 × 5
90 = 2 × 3 × 3 × 5
= 2 × 3² × 5

∴ HCF = 2 × 3 × 5 = 30
LCM = 2³ × 3² × 5 = 360

ii) 50, 60
Answer:
We have
50 = 2 × 5 × 5 = 2 × 5²
60 = 2 × 2 × 3 × 5 = 2² × 3 × 5

∴ HCF = 2 × 5 = 10
LCM = 2² × 3 × 5² = 300

iii) 37, 49
Answer:
We have
37 = 1 × 37
49 = 7 × 7 = 7²
∴ HCF = 1
LCM = 37 × 7²
Note: H.C.F. of two relatively prime numbers is 1 and LCM is equal to product of the numbers.

Try this

Question 1.
Show that 3ⁿ × 4^m cannot end with the digit 0 or 5 for any natural numbers ‘n’ and’m’. (Page No. 8)
Answer:
Given number is 3⁴ × 4^m.
So the prime factors to it are 3 and 2 only.
I: but if a number want to be end with zero it should have 2 and 5 as its prime factors, but the given hasn’t ‘5’ as its prime factor.
So it cannot be end with zero.
II : now if a number went to be end with 5 it should have ‘5’ as its one of prime factors. But given 3ⁿ × 4^m do not have 5 as a factor.
So it cannot be end with 5.
Hence proved.

Do this

Question 1.
Write the following terminating decimals in the form of p/q, q ≠ 0 and p, q are co-primes.
i) 15.265
ii) 0.1255
iii) 0.4
iv) 23.34
v) 1215.8
What can you conclude about the denominators through this process? (Page No. 10)
Answer:
i) 15.265

ii) 0.1255

iii) 0.4
0.4 = \(\frac{4}{10}\) = \(\frac{2}{5}\)
iv) 23.34

v) 1215.8

Two and five are the factors for the denominator.

Question 2.
Write the following rational numbers in the form of p/q, where q is of the form 2ⁿ.5^m where n, m are non-negative integers and then write the numbers in their decimal form. (Page No. 11)
i) \(\frac{3}{4}\)
ii) \(\frac{7}{25}\)
iii) \(\frac{51}{64}\)
iv) \(\frac{14}{25}\)
v) \(\frac{80}{100}\)
Answer:
i) \(\frac{3}{4}\)
\(\frac{3}{4}\) = \(\frac{3}{2 \times 2}\) = \(\frac{3}{2^{2}}\)

Decimal form of \(\frac{3}{4}\) = 0.75

ii) \(\frac{7}{25}\)
\(\frac{7}{25}\) = \(\frac{7}{5 \times 5}\) = \(\frac{7}{5^{2}}\)

Decimal form of \(\frac{7}{25}\) = 0.28

iii) \(\frac{51}{64}\)
\(\frac{51}{64}\) = \(\frac{51}{2^{6}}\)
[∵ 64 = 2 × 32
= 2² × 16
= 2³ × 8
= 2⁴ × 4 = 2⁵ × 2 = 2⁶]

Decimal form of \(\frac{51}{64}\) = 0.796875

iv) \(\frac{14}{25}\)

v) \(\frac{80}{100}\)
\(\frac{80}{100}\) = \(\frac{80}{2^{2} \times 5^{2}}\) = \(\frac{80}{10^{2}}\) = 0.80

Question 3.
Write the following rational numbers as decimal form and find out the block of repeating digits in the quotient. (Page No. 11)
i) \(\frac{1}{3}\)
ii) \(\frac{2}{7}\)
iii) \(\frac{5}{11}\)
iv) \(\frac{10}{13}\)
Answer:
i) \(\frac{1}{3}\)
\(\frac{1}{3}\) = 0.3333…. = \(0 . \overline{3}\)
Block of digits, repeating in the quotient = period = 3.

ii) \(\frac{2}{7}\)
Decimal form of \(\frac{2}{7}\) = 0.285714….
Repeating part/period = 285714
∴ \(\frac{2}{7}\) = \(0 . \overline{285714}\)

iii) \(\frac{5}{11}\)
Period = 45
Decimal form of \(\frac{5}{11}\) = 0.454545.
= \(0 . \overline{45}\)

iv) \(\frac{10}{13}\)
Decimal form of \(\frac{10}{13}\) = 0.769230.
= \(0 . \overline{769230}\)
Period = 769230

Do this

Question 1.
Verify the statement proved above for p = 2, p = 5 and for a² = 1, 4, 9, 25, 36, 49, 64 and 81. (Page No. 14)
Answer:


From the above we can conclude that if a prime number ‘p’ divides a², then it also divides a.

Think and Discuss

Question 1.
Write the nature of y, a and x in y = a^x. Can you determine the value of x for a given y? Justify your answer. (Page No. 17)
Answer:
y = a^x here a ≠ 0
We can determine the value of ‘x’ for a given y.
for example y = 5, a = 2
We cannot express y = a^x for y = 5, a = 2 and for y = 7, a = 3, we cannot express seven (7) as a power of 3.

Question 2.
You know that 2¹ = 2, 4¹ = 4, 8¹ = 8 and 10¹ = 10. What do you notice about the values of log₂ 2, log₄ 4, log₈ 8 and log₁₀ 10? What can you generalise from this?  (Page No. 18)
Answer:
From the graph log₂ 2 = log₄ 4 = log₈ 8 = log₁₀ 10 = 1
We conclude that log_a a = 1 where a is a natural number.

Question 3.
Does log₁₀ 0 exist? (Page No. 18)
Answer:
No, log₁₀ 0 doesn’t exist, i.e a^x ≠ 0 ∀ a, x ∈ N.

Question 4.
We know that, if 7 = 2^x then x = log₂ 7. Then what is the value of \(2^{\log _{2} 7}\)? Justify your answer. Generalise the above by taking some more examples for \(\mathbf{a}^{\log _{\mathrm{a}} \mathbf{N}}\). (Page No. 21)
Answer:
We know that if 7 = 2^x then x = log₂ 7
We want to find the value of \(2^{\log _{2} 7}\);
Now put log₂ 7 = x in the given
∴ \(2^{\log _{2} 7}\) = 2^x = 7 (given)
∴ \(2^{\log _{2} 7}\) = 7
Thus \(\mathbf{a}^{\log _{\mathrm{a}} \mathbf{N}}\) = N

a) \(3^{\log _{3} 8}\)
Answer:
If x = \(3^{\log _{3} 8}\) then
log₃ x = log₃ 8
⇒ x = 8

b) \(5^{\log _{5} 10}\)
Answer:
If y = \(5^{\log _{5} 10}\)
then log₅ y = log₅ 10
⇒ y = 10

Do this

Question 1.
Write the powers to which the bases to be raised in the following.  (Page No. 18)
i) 64 = 2^x
Answer:
64 = 2^x
We know that
64 = 2 × 32
= 2 × 2 × 16
= 2 × 2 × 2 × 8
= 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2
64 = 2⁶
⇒ x = 6

ii) 100 = 5^b
Answer:
Here also 100 cannot be written as any power of 5.
i.e., there exists no integer for b such that 5^b = 100

iii) \(\frac{1}{81}\) = 3^c
Answer:
We know that 81 = 3 x 27
= 3 × 3 × 9
= 3 × 3 × 3 × 3
= 34
∴ \(\frac{1}{81}\) = 3^-4   [∵ a^-m = \(\frac{1}{\mathrm{a}^{\mathrm{m}}}\)]
∴ c = – 4

iv) 100 = 10^z
Answer:
100 = 10²
z = 2

v) \(\frac{1}{256}\) = 4^a
Answer:
We know that 256 = 4 × 64
= 4 × 4 × 16
= 4 × 4 × 4 × 4
∴ \(\frac{1}{256}\) = 4^-4
∴ a = – 4

Question 2.
Express the logarithms of the following into sum of the logarithms.   (Page No. 19)
i) 35 × 46
Answer:
log xy = log x + log y
log₁₀35 × 46 = log₁₀35 + log₁₀46

ii) 235 × 437
Answer:
log₁₀235 × 437 = log₁₀235 + log₁₀437   [∵ log xy = log x + log y]

iii) 2437 × 3568
Answer:
log₁₀ 2437 × 3568 = log₁₀2437 + log₁₀3568   [∵ log xy = log x + log y]

Question 3.
Express the logarithms of the follow¬ing into difference of the logarithms.   (Page No. 20)
i) \(\frac{23}{34}\)
Answer:
log₁₀ = \(\frac{23}{34}\) = log₁₀ 23 – log₁₀ 34
[∵ log \(\frac{x}{y}\) = log x – log y]

ii) \(\frac{373}{275}\)
Answer:
log₁₀ = \(\frac{373}{275}\) = log₁₀ 373 – log₁₀ 275
[∵ log \(\frac{x}{y}\) = log x – log y]

iii) \(\frac{4525}{3734}\)
Answer:
log₁₀ = \(\frac{4525}{3734}\) = log₁₀ 4525 – log₁₀ 3734
[∵ log \(\frac{x}{y}\) = log x – log y]

iv) \(\frac{5055}{3303}\)
Answer:
log₁₀ = \(\frac{5055}{3303}\) = log₁₀ 5055 – log₁₀ 3303
[∵ log \(\frac{x}{y}\) = log x – log y]

Question 4.
By using the formula log_axⁿ = n log_a x, convert the following.   (Page No. 21)
i) log₂ 7²⁵
Answer:
log₂ 7²⁵ = 25 log₂ 7

ii) log₅ 8⁵⁰
Answer:
log₅ 8⁵⁰ = 50 log₅ 8 = 50 log₅ 23
= 3 × 50 log₅2 = 150 log₅2

iii) log 5²³
Answer:
log 5²³ = 23 log 5

iv) log 1024
Answer:
log 1024 = log 2¹⁰ [∵ 1024 = 2¹⁰]
= 10 log 2

Try this

Question 1.
Write the following relation in exponential form and find the values of respective variables.   (Page No. 18)
i) log₂32 = x
Answer:
log₂32 = x
⇒ log₂2⁵ = x     [∵ 32 = 2⁵]
⇒ 5 log₂2 = x     [∵ log a^m = m log a]
⇒ 5 × 1 = x      [∵ log_a a = 1]
∴ x = 5

ii) log₅625 = y
Answer:
log₅625 = y
⇒ log₅4 = y    [∵ 625 = 5⁴]
⇒ 4 log₅ 5 = y     [∵ log a^m = m log a]
⇒ 4 × 1 = y     [∵ log_a a = 1]
∴ y = 4

iii) log₁₀10000 = z
Answer:
log₁₀10000 = z
=> log₁₀10⁴ = z     [∵ 10000 = 10 × 10 × 10 × 10 = 10⁴]
=> 4 log₁₀10 = z     [∵ log a^m = m log a]
=> 4 × 1 = z     [∵ log_a a = 1]
∴ z = 4

iv) \(\log _{7} \frac{1}{343}\) = -a
Answer:

Question 2.
i) Find the value of log₂32. (Page No. 21)
Answer:
log₂ 32 = log₂ 2⁵
[∵ 32 = 2 × 2 × 2 × 2 × 2 = 2⁵]
= 5 log₂ 2 [∵ log a^m = m log a]
= 5 × 1 [∵ log_a a = 1]
= 5

ii) Find the value of log_c √c.
Answer:

= \(\frac{1}{2}\) × 1 [∵ log_a a = 1]
= \(\frac{1}{2}\)

iii) Find the value of log₁₀0.001
Answer:

iv) Find the value of \(\log _{\frac{2}{3}} \frac{8}{27}\)
Answer:

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.5 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.5

10th Class Maths 1st Lesson Real Numbers Ex 1.5 Textbook Questions and Answers

Question 1.
Determine the values of the following,
i) log₂₅5
Answer:

ii) log₈₁3
Answer:

iii) log₂(\(\frac{1}{16}\))
Answer:

iv) log₇1
Answer:
log₇1 = log₇7⁰ = 0 log₇7 = 0

v) log_x√x
Answer:

vi) log₂512
Answer:
log₂512 = log₂2⁹    [∵ 512 = 2⁹]
= 9log₂2   [∵ log x^m = m log x]
= 9 × 1    [∵ log_aa = 1]
= 9

vii) log₁₀0.01
Answer:

viii) \(\log _{\frac{2}{3}}\left(\frac{8}{27}\right)\)
Answer:

ix) \(2^{2+\log _{2} 3}\)
Answer:
\(2^{2+\log _{2} 3}\) = 2² . \(2^{\log _{2} 3}\)   [∵ a^m . aⁿ = a^m+n]
= 4 × 3 [∵ \(\log _{\mathrm{a}} \mathrm{N}\) = N]
= 12

Question 2.
Write the following expressions as log N and find their values.
i) log 2 + log 5
Answer:
log 2 + log 5
= log 2 × 5   [∵ log m + log n = log mn]
= log 10
= 1

ii) log₂ 16 – log₂ 2
Answer:

iii) 3 log₆₄4
Answer:

iv) 2 log 3 – 3 log 2
Answer:
2 log 3 – 3 log 2
= log 3² – log 2³
= log 9 – log 8
= log \(\frac{9}{8}\)

v) log 10 + 2 log 3 – log 2
Answer:
log 10 + 2 log 3 – log 2
= log 10 + log 3² – log 2
= log 10 + log 9 – log 2    [∵ m log a = log a^m]
= log \(\frac{10 \times 9}{2}\)    [∵ log a + log b = log ab; log a – log b = log \(\frac{a}{b}\)]
= log 45

Question 3.
Evaluate each of the following in terms of x and y, if it is given x = log₂3 and y = log₂ 5.
i) log₂15
Answer:
log₂15 = log₂ 3 × 5
= log₂3 + log₂5   [∵ log mn = log m + log n]
= x + y

ii) log₂7.5
Answer:

iii) log₂60
Answer:
log₂60 = log₂2² × 3 × 5
= log₂2² + log₂3 + log₂5
= 2 log₂2 + x + y
= 2 + x + y

iv) log₂6750
Answer:

log₂6750
= log₂2 × 3³ × 5³
= log₂2 + log₂33 + log₂5³
= 1 + 3 log₂3 + 3 log₂5
= 1 + 3x + 3y

Question 4.
Expand the following,
i) log 1000
Answer:
log 1000 = log 10³
= 3 log 10
= 3 × 1
= 3

ii) \(\log \left[\frac{128}{625}\right]\)
Answer:

iii) log x²y³z⁴
Answer:
log x²y³z⁴ = logx² + logy³ + logz⁴ [∵ log ab = log a + log b]
= 2 log x + 3 log y + 4 log z
[∵ log a^m = m log a]

iv) \(\log \frac{\mathbf{p}^{2} \mathbf{q}^{3}}{\mathbf{r}}\)
Answer:

iv) \(\log \sqrt{\frac{x^{3}}{y^{2}}}\)
Answer:

Question 5.
If x² + y² = 25xy, then prove that 2 log (x + y) = 3log3 + logx + logy.
Answer:
Given: x² + y² = 25xy
We know that (x + y)² = x² + y² + 2xy
= 25xy + 2xy    [∵ x² + y² = 25xy given]
(x + y)² = 27xy
Taking ‘log’ on both sides
log (x + y)² = log 27xy
2 log (x + y) = log 27 + log x + log y
= log 3³ + log x + log y
⇒ 2 log (x + y) = 3log3 + log x + log y

Question 6.
If \(\log \left(\frac{\mathbf{x}+\mathbf{y}}{3}\right)\) = \(\frac{1}{2}\) (log x + log y), then find the value of \(\frac{x}{y}+\frac{y}{x}\).
Answer:

(squaring on both sides)
⇒ (x + y)² = (3√xy)²
⇒ x² + y² + 2xy = 9xy
⇒ x² + y² = 9xy – 2xy = 7xy

Question 7.
If (2.3)^x = (0.23)^y = 1000 then find the value of \(\frac{1}{x}-\frac{1}{y}\).
Answer:
Given (2.3)^x = (0.23)^y = 1000 = 10³

Question 8.
If 2^x+1 = 3^1-x then find the value of x.
Answer:
Given: 2^x+1 = 3^1-x
log 2^x+1 = log 3^1-x
(x + 1) log 2 = (1 – x) log 3
x log 2 + log 2 = log 3 – x log 3
x log 2 + x log 3 = log 3 – log 2
x (log 3 + log 2) = log 3 – log 2

Question 9.
Is
i) log 2 is rational or irrational? Justify your answer.
Answer:
Let log₁₀2 = x
Then 10^x = 2
But 2 can’t be written as 10^x for any value of x
∴ log 2 is irrational.

ii) log 100 is rational or irrational? Justify your answer.
Answer:
Let log₁₀100 = x
⇒ log₁₀10² = x
⇒ 2 log₁₀10 = x = 2
∴ log 100 is rational.
∴ log 100 = 2
Hence rational.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.4

10th Class Maths 1st Lesson Real Numbers Ex 1.4 Textbook Questions and Answers

Question 1.
Prove that the following are irrational,
i) \(\frac{1}{\sqrt{2}}\)
ii) √3 + √5
iii) 6 + √2
iv) √5
v) 3 + 2√5
Answer:
i) \(\frac{1}{\sqrt{2}}\)
On the contrary, suppose that is a \(\frac{1}{\sqrt{2}}\) rational number;
then \(\frac{1}{\sqrt{2}}\) is of the form where \(\frac{p}{q}\) and q are integers.
∴ \(\frac{1}{\sqrt{2}}\) = \(\frac{p}{q}\)
⇒ \(\frac{\sqrt{2}}{1}\) = \(\frac{q}{p}\)
(i.e.,) √2 is a rational number and it is a contradiction. This contradiction arised due to our supposition that \(\frac{1}{\sqrt{2}}\) is a rational number.
Hence \(\frac{1}{\sqrt{2}}\) is an irrational number.

ii) Suppose √3 + √5 is not an irrational number.
Then √3 + √5 must be a rational number.
√3 + √5 = \(\frac{p}{q}\), q ≠ 0 and p, q ∈ Z
Squaring on both sides

but √15 is an irrational number.
\(\frac{p^{2}-8 q^{2}}{2 q^{2}}\) is a rational number
(p² – 8q², 2q² ∈ Z, 2q² ≠ 0)
but an irrational number can’t be equal to a rational number, so our supposition that √3 + √5 is not an irrational number is false.
∴ √3 + √5 is an irrational number.

iii) 6 + √2
To prove: 6 + √2 is an irrational number.
Let us suppose that 6 + √2 is a rational number.
∴ 6 + √2 = \(\frac{p}{q}\), q ≠ 0
⇒ √2 = \(\frac{p}{q}\) – 6
⇒ √2 = Difference of two rational numbers
⇒ √2 = rational number But this contradicts the fact that √2 is an irrational number.
∴ Our supposition is wrong.
Hence the given statement is true.
6 + √2 is an irrational number.

iv) √5
To prove: √5 is an irrational number.
On the contrary, let us assume that √5 is a rational number.
∴ √5 = \(\frac{p}{q}\), q ≠ 0
If p, q have a common factor, on cancelling the common factor let it be
reduces to \(\frac{a}{b}\) where a, b are co-primes.
Now √5 = \(\frac{a}{b}\), where HCF (a, b) = 1
Squaring on both sides we get
⇒ (√5)² = \(\left(\frac{a}{b}\right)^{2}\)
⇒ 5 = \(\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}\)
⇒ 5b² = a²
⇒ 5 divides a² and thereby 5 divides 5
Now, take a = 5c
then, a² = 25c²
i.e., 5b² = 25c²
⇒ b² = 5c²
⇒ 5 divides b² and thereby b.
⇒ 5 divides both b and a.
This contradicts that a and b are co-primes.
This contradiction arised due to our assumption that √5 is a rational number.
Hence our assumption is wrong and the given statement is true, i.e., √5 is an irrational number,

v) 3 + 2√5
To Prove: 3 + 2√5 is an irrational.
On the contrary, let us assume that 3 + 2√5 is a rational number.

Here p, q being integers we can say that \(\frac{p-3q}{2q}\) is a rational number.
This contradicts the fact that √5 is an irrational number. This is due to our assumption “3 + 2√5 is a rational number”.
Hence our assumption is wrong.
∴ 3 + 2√5 is an irrational number.

Question 2.
Prove that √p + √q is an irrational, where p, q are primes.
Answer:
Given that p, q are primes.
Hence fp and fq are irrationals.
[∵ p, q have no factors other than 1 ∵ they are primes.]
Now √p + √q = sum of two irrational numbers = an irrational number
Hence proved.

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