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## AP Board Class 10 Maths Chapter 1 Real Numbers InText Questions Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 1 Real Numbers InText Questions Book Answers AP Board Class 10 Maths Chapter 1 Real Numbers InText Questions Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 1 Real Numbers InText Questions Book Answers

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## AP Board Class 10th Maths Chapter 1 Real Numbers InText Questions Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 10th Maths Chapter 1 Real Numbers InText Questions Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

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
13 = 3 × 4 + 1
here q = 4 ; r = 1

ii) a = 8, b = 80
Take a = 80, b = 8
80 = 8 × 10 + 0 here q = 10 ; r = 0

iii) a = 125, b = 5
125 = 5 × 25 + 0
here q = 25 ; r = 0

iv) a = 132, b = 11
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)
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
96 = 72 × 1 + 24
72 = 24 × 3 + 0
∴ HCF = 24

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

iv) 1860 and 2015
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)
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)
Given: 1.2 and 0.12
we have 1.2 = 1210 = 120100
0.12 = 12100
Now considering the numerators 12 and 120, their HCF is 12.
∴ HCF of 1.2 and 0.12 is 12100 = 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)
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)
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
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 = 2<sup>3</sup> × 3<sup>2</sup> × 5 = 360

ii) 50, 60
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
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 × 4m cannot end with the digit 0 or 5 for any natural numbers ‘n’ and’m’. (Page No. 8)
Given number is 34 × 4m.
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 × 4m 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)
i) 15.265 ii) 0.1255 iii) 0.4
0.4 = 410 = 25
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.5m where n, m are non-negative integers and then write the numbers in their decimal form. (Page No. 11)
i) 34
ii) 725
iii) 5164
iv) 1425
v) 80100
i) 34
34 = 32×2 = 322 Decimal form of 34 = 0.75

ii) 725
725 = 75×5 = 752 Decimal form of 725 = 0.28

iii) 5164
5164 = 5126
[∵ 64 = 2 × 32
= 22 × 16
= 23 × 8
= 24 × 4 = 25 × 2 = 26] Decimal form of 5164 = 0.796875

iv) 1425 v) 80100
80100 = 8022×52 = 80102 = 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) 13
ii) 27
iii) 511
iv) 1013
i) 13
13 = 0.3333…. = 0.3⎯⎯⎯
Block of digits, repeating in the quotient = period = 3.

ii) 27
Decimal form of 27 = 0.285714….
Repeating part/period = 285714
∴ 27 = 0.285714⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

iii) 511
Period = 45
Decimal form of 511 = 0.454545.
= 0.45⎯⎯⎯⎯⎯⎯

iv) 1013
Decimal form of 1013 = 0.769230.
= 0.769230⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
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)  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 = ax. Can you determine the value of x for a given y? Justify your answer. (Page No. 17)
y = ax here a ≠ 0
We can determine the value of ‘x’ for a given y.
for example y = 5, a = 2
We cannot express y = ax 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)
From the graph log2 2 = log4 4 = log8 8 = log10 10 = 1
We conclude that loga a = 1 where a is a natural number.

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

Question 4.
We know that, if 7 = 2x then x = log2 7. Then what is the value of 2log27? Justify your answer. Generalise the above by taking some more examples for 𝐚loga𝐍. (Page No. 21)
We know that if 7 = 2x then x = log2 7
We want to find the value of 2log27;
Now put log2 7 = x in the given
∴ 2log27 = 2x = 7 (given)
∴ 2log27 = 7
Thus 𝐚loga𝐍 = N

a) 3log38
If x = 3log38 then
log3 x = log3 8
⇒ x = 8

b) 5log510
If y = 5log510
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 = 2x
64 = 2x
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 = 5b
Here also 100 cannot be written as any power of 5.
i.e., there exists no integer for b such that 5b = 100

iii) 181 = 3c
We know that 81 = 3 x 27
= 3 × 3 × 9
= 3 × 3 × 3 × 3
= 34
∴ 181 = 3-4  [∵ a-m = 1am]
∴ c = – 4

iv) 100 = 10z
100 = 102
z = 2

v) 1256 = 4a
We know that 256 = 4 × 64
= 4 × 4 × 16
= 4 × 4 × 4 × 4
∴ 1256 = 4-4
∴ a = – 4

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

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

iii) 2437 × 3568
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) 2334
log10 = 2334 = log10 23 – log10 34
[∵ log 𝑥𝑦 = log x – log y]

ii) 373275
log10 = 373275 = log10 373 – log10 275
[∵ log 𝑥𝑦 = log x – log y]

iii) 45253734
log10 = 45253734 = log10 4525 – log10 3734
[∵ log 𝑥𝑦 = log x – log y]

iv) 50553303
log10 = 50553303 = log10 5055 – log10 3303
[∵ log 𝑥𝑦 = log x – log y]

Question 4.
By using the formula logaxn = n loga x, convert the following. (Page No. 21)
i) log2 725
log2 725 = 25 log2 7

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

iii) log 523
log 523 = 23 log 5

iv) log 1024
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
log232 = x
⇒ log225 = x [∵ 32 = 25]
⇒ 5 log22 = x [∵ log am = m log a]
⇒ 5 × 1 = x [∵ loga a = 1]
∴ x = 5

ii) log5625 = y
log5625 = y
⇒ log54 = y [∵ 625 = 54]
⇒ 4 log5 5 = y [∵ log am = m log a]
⇒ 4 × 1 = y [∵ loga a = 1]
∴ y = 4

iii) log1010000 = z
log1010000 = z
=> log10104 = z [∵ 10000 = 10 × 10 × 10 × 10 = 104]
=> 4 log1010 = z [∵ log am = m log a]
=> 4 × 1 = z [∵ loga a = 1]
∴ z = 4

iv) log71343 = -a Question 2.
i) Find the value of log232. (Page No. 21)
log2 32 = log2 25
[∵ 32 = 2 × 2 × 2 × 2 × 2 = 25]
= 5 log2 2 [∵ log am = m log a]
= 5 × 1 [∵ loga a = 1]
= 5

ii) Find the value of logc √c. = 12 × 1 [∵ loga a = 1]
= 12

iii) Find the value of log100.001 iv) Find the value of log23827 ## Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers InText Questions Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers InText Questions Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers InText Questions exam preparation. The AP Board STD 10th Maths Chapter 1 Real Numbers InText Questions Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 1 Real Numbers InText Questions Books State Board syllabus with maximum efficiency.

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