Contents
Andhra Pradesh SSC Class 10 Solutions For Maths – Real Numbers (English Medium)
These Solutions are part of AP SSC Class 10 Solutions for Maths. In this article, we have provided Andhra Pradesh SSC Class 10 Solutions For Maths Chapter 1 Real Numbers.
Exercise 1.1:
Question 1:
Use Euclid’s division algorithm to find the HCF of
- 900 and 270
- 196 and 38220
- 1651 and 2032
Solution :
- Let x= be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non negative integers. Then x has a decimal expansion which terminates.
- Let x= be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non negative integers. Then x has a decimal expansion which non terminating repeating (recurring).
- Let x= be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non negative integers. Then x has a decimal expansion which terminates.
- Let x= be a rational number, such that the prime factorization of q is of the form 2n5m,where n, m are non negative integers. Then x has a decimal expansion which terminates.
- Let x= be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non negative integers. Then x has a decimal expansion which non terminating repeating (recurring).
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.
Solution :
Question 3:
Use Euclid’s division lemma to show that the square of any positive integer is of the form 3p, 3p + 1 or 3p + 2.
Solution :
Question 4:
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9 m, 9m + 1 or 9m + 8.
Solution :
Exercise 1.2:
Question 1:
Express each of the following number as a product of its prime factors.
(i) 140 (ii) 156 (iii) 3825 (iv)5005 (v)7429
Solution :
Question 2:
Find the LCM and HCF 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
Solution :
Question 3:
Check whether 6n can end with the digit 0 for any natural number n.
Solution :
For the number 6n to end with digit zero for any natural number n, it should be divisible by 5. This means that the prime factorization of 6n should contain the prime number 5, but it is not possible because 6n = (2)3n so 2 is the only prime in the factorization of 6n.
Since 5 is not present in the prime factorization, so there is no natural number n for which 6n ends with the digit zero.
Question 4:
Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1+5 are composite numbers.
Solution :
7 × 11 × 13 + 13 = 13 (7 × 11 + 1)
= 78 × 13
= 2 × 3 × 13 × 13
= 2 × 3 × 132
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × 1009
Since both the above number is unique and is product of primes. So according to Fundamental Theorem of Arithmetic, every composite number can be factorized as a product of primes.
∴ Given number is composite number.
Question 5:
How will you show that (17 x 11 x 2) + (17 x 11 x 5) is a composite number? Explain.
Solution :
Given,
(17 × 11 × 2) + (17 × 11 × 5) = 17 × 11 (2 + 5)
= 17 × 11 × 7
It is unique and is product of primes. So According to Fundamental Theorem of Arithmetic, every composite number can be factorized as a product of primes
∴ Given number is composite number.
Exercise 1.3:
Question 1:
Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating, repeating decimal.
(i) \(\frac { 3 }{ 8 }\)
(ii) \(\frac { 229 }{ 400 }\)
(iii) \(4\frac { 1 }{ 5 }\)
(iv) \(\frac { 2 }{ 11 }\)
(v) \(\frac { 8 }{ 125 }\)
Solution :
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) \(\frac { 23 }{ { 2 }^{ 3 }.{ 5 }^{ 2 } }\)
(vii) \(\frac { 129 }{ { 2 }^{ 2 }.{ 5 }^{ 7 }.{ 7 }^{ 5 } }\)
(viii) \(\frac { 9 }{ 15 }\)
(ix) \(\frac { 36 }{ 100 }\)
(x) \(\frac { 77 }{ 210 }\)
Solution :
Question 3:
Write the fol lowing rationals in decimal form using Theorem 1.1
(i) \(\frac { 13 }{ 25 }\)
(ii) \(\frac { 15 }{ 16 }\)
(iii) \(\frac { 23 }{ { 2 }^{ 3 }.{ 5 }^{ 2 } }\)
(iv) \(\frac { 7218 }{ { 3 }^{ 2 }.{ 5 }^{ 2 } }\)
(v) \(\frac { 143 }{ 110 }\)
Solution :
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 ?
- 43.123456789
- 0.120120012000120000…
- \(43.\overline { 123456789 }\)
Solution :
- It is terminating, hence it is rational.
- It is non-terminating and proper pattern is not there so, it is not rational.
- It is non-terminating, but recurring so it is rational.
Exercise 1.4:
Question 1:
Prove that the following are irrational.
Question 1(i):
Solution :
Question 1(ii):
\(\sqrt { 3 } +\sqrt { 5 }\)Solution :
Question 1(iii):
\(6+\sqrt { 2 }\)Solution :
Question 1(iv):
\(\sqrt { 5 }\)Solution :
Question 1(v):
\(3+2\sqrt { 5 }\)Solution :
Question 2:
Prove that \(\sqrt { p } +\sqrt { q }\) is irrational, where p, q are primes.
Solution :
Exercise 1.5:
Question 1:
Determine the value of the following
(i) \(\log _{ 25 }{ 5 }\)
(ii) \(\log _{ 81 }{ 3 }\)
(iii) \(\log _{ 2 }{ \left( \frac { 1 }{ 16 } \right) }\)
(iv) \(\log _{ 7 }{ 1 }\)
(v) \(\log _{ x }{ \sqrt { x } }\)
(vi) \(\log _{ 2 }{ 512 }\)
(vii) \(\log _{ 10 }{ 0.01 }\)
(viii) \(\log _{ \frac { 2 }{ 3 } }{ \left( \frac { 8 }{ 27 } \right) }\)
(ix) \({ 2 }^{ 2+\log _{ 2 }{ 3 } }\)
Solution :
Question 2:
Write the following expressions as log N and find their values.
(i) log 2 +log 5 (ii) log2 16 – log2 2 (iii)3 log64 4
(iv) 2 log 3-3 log 2 (v) log 10 + 2 log 3 – log 2
Solution :
Question 3:
Evaluate each of the following in terms of x and y, if it is given x = log2 3 and y = log2 5
(i) log2 15 (ii) log2 7.5 (iii) log2 60 (iv) log2 6750
Solution :
Question 4:
Expand the following.
(i) \(\log { 1000 }\)
(ii) \(\log { \left( \frac { 128 }{ 625 } \right) }\)
(iii) \(\log { { x }^{ 2 }{ y }^{ 3 }{ z }^{ 4 } } \)
(iv) \(\log { \frac { { p }^{ 2 }{ q }^{ 3 } }{ r } }\)
(v) \(\log { \sqrt { \frac { { x }^{ 3 } }{ { y }^{ 2 } } } }\)
Solution :
Question 5:
If x2 +y2 = 25xy, then prove that 2 log(x + y) = 3log3 + logx + logy.
Solution :
Hope given Andhra Pradesh SSC Class 10 Solutions For Maths chapter 1 Real Numbers are helpful to complete your math homework.
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