TS 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise

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TS 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise

Question 1.
Can the number 6n, n being a natural number, end with the digit 5 ? Give reason.
Solution:
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.
Solution:
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.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise

Question 3.
Prove that (2\(\sqrt{3}\) + \(\sqrt{5}\) is an irrational number. Also check whether (2\(\sqrt{3}\) + \(\sqrt{5}\))(2\(\sqrt{3}\) – \(\sqrt{5}\)) is rational or irrational.
Solution:
To prove : 2\(\sqrt{3}\) + \(\sqrt{5}\) is an irrational number. On contrary, let us suppose that 2\(\sqrt{3}\) + \(\sqrt{5}\) be a rational number then 2\(\sqrt{3}\) + \(\sqrt{5}\) = \(\frac{\mathrm{p}}{\mathrm{q}}\)
Squaring on both sides, we get
TS 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise 1
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 \(\sqrt{15}\) is an irrational. This is due to our assumption that 2\(\sqrt{3}\) + \(\sqrt{5}\) is a rational. Hence our assumption is wrong and 2\(\sqrt{3}\) + \(\sqrt{5}\) is an irrational number.
Also,
(2\(\sqrt{3}\) + \(\sqrt{5}\)) (2\(\sqrt{3}\) – \(\sqrt{5}\)) = (2\(\sqrt{3}\))2 -(\(\sqrt{5}\))2
[∵ (a + b) (a – b) = a2 – b2]
= 4 × 3 – 5 = 12 – 5 = 7, a rational number.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise

Question 4.
If x2 + y2 = 6xy, prove that 2 log (x +y) = logx + logy + 3 log 2
Solution:
Given : x2 + y22 = 6xy Adding both sides
⇒ x2 + y2 + 2xy = 6xy + 2xy (x + y)2 = 8xy
Taking logarithms on both sides
log (x + y)2 = log 8xy
⇒ 2 log(x + y) = log 8 + log x + log y
[∵ log xm = mlog x]
[∵ log xy = log x + log y]
= log 23 + log x + log y
⇒ 2log (x + y) = log x + log y + 3 log 2

Question 5.
Find the number of digits in 42013, if log10 2 = 0.3010.
Solution:
Given log102 = 0.3010
42013 = (22)2013 = 24026
∴ log10 24026 = 4026 log 2
[∵ log xm = m log x]
= 4026 × 0.3010
= 1211.826 \(\simeq\) 1212
∴ 42013 has 1212 digits in its expansion.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.2

Question 1.
Express each of the following numbers as a product of its prime factors.

(i) 140
Answer:
140
TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 1
∴ 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7

(ii) 156
Answer:
156
TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 2
∴ 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2

(iii) 3825
Answer:
3825
TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 3
∴ 3825 = 3 × 3 × 5 × 5 × 17
= 32 × 52 × 17

(iv) 5005
Answer:
5005
TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 4
∴ 5005 = 5 × 7 × 11 × 13

(v) 7429
Answer:
7429
TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 5
∴ 7429 = 17 × 19 × 23

Question 2.
Find the LCM and HCF of the following integers by the prime factorization method.
(i) 12, 15 and 21
Answer:
12, 15, 21
12 = 2 × 2 × 3, 15 = 3 × 5
21 = 3 × 7
H.C.F of 12, 15, 21 is 3.
L.C.M of 12, 15, 21 = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23, and 29
Answer:
17, 23, 29
17 = 1 × 17, 23 = 1 × 23
29 = 1 × 29
H.C.F of 17, 23 and 29 is 1.
L.C.M of 17, 23, 29 is 17 × 23 × 29 = 11339

(iii) 8, 9 and 25
Answer:
8, 9, 25
8 = 1 × 2 × 2 × 2, 9 = 1 × 3 × 3
25 = 1 × 5 × 5
H.C.F. = 1
L.C.M = 2 × 2 × 2 × 3 × 3 × 5 × 5
= 1800

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2

(iv) 72 and 108
Answer:
72, 108
TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 6
72 = 2 × 36
= 2 × 2 × 18
= 2 × 2 × 2 × 9
= 2 × 2 × 2 × 3 × 3

108 = 2 × 54
= 2 × 2 × 27
= 2 × 2 × 3 × 9
= 2 × 2 × 3 × 3 × 3
H.C.F. = 2 × 2 × 3 × 3 = 36
L.C.M. = 2 × 2 × 2 × 3 × 3 × 3 = 216

(v) 306 and 657
Answer:
306, 657
TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 7
H.C.F. = 9
L.C.M. = 9 × 34 × 73
= 22338

Question 3.
Check whether 6″ can end with the digit 0 for any natural number n.
Answer:
If the number 6n for any n, were to end with digit ‘0’ then it would be divisible by 5.
The prime factorisation of 6n would contain the prime 5.
6n = (2 × 3)n
Prime factorisation of 6n does not contain 5 as a factor, so 6n can never end with the digit 0 for any natural number.

Question 4.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer:
Given number is 7 × 11 × 13 + 13
= 13 (7 × 11 + 1)
= (7 × 11 + 1) × 13 distributive law
= (77 + 1) × 13
= 78 × 13
= (2 × 3 × 13) × 13
= 2 × 3 × 132
= Product of prime factors
Hence the given number is a composite number.
Given number is
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 (1008 + 1)
= 5 × 1009
= Product of prime numbers
Hence the given number is a composite number.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2

Question 5.
How will you show that (17 × 11 × 2)+ (17 × 11 × 5) is a composite number? Explain.
Answer:
Given number is
(17 × 11 × 2) + (17 × 11 × 5)
= 17 × 11 × (2 + 5)
= 17 × 11 × 7 = Product of primes
We know that every composite number can be expressed as a product of primes.

Question 6.
Which digit would occupy the units place of 6100.
Answer:
6100 = (2 × 3)100
So the last digit of 6100 is 6.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.4

Question 1.
Prove that the following are irrational.

(i) \(\frac{1}{\sqrt{2}}\)
Solution:
\(\frac{1}{\sqrt{2}}\)
Let us assume to the contrary that \(\frac{1}{\sqrt{2}}\) is rational. Then there exist co-prime positive integers ‘a’ and ‘b’ such that 1 a
\(\frac{1}{\sqrt{2}}\) = \(\frac{\mathrm{a}}{\mathrm{b}}\)
\(\sqrt{2}\)a = b
\(\sqrt{2}\) = \(\frac{b}{a}\)
Here ‘a’ and ‘b’ are integers, \(\frac{b}{a}\) is rational.
∴ \(\sqrt{2}\) is rational.
This contradicts the fact that \(\sqrt{2}\) is irrational.
So our assumption is wrong.
Hence \(\frac{1}{\sqrt{2}}\) is irrational.

(ii) \(\sqrt{3}\) + \(\sqrt{5}\)
Solution:
\(\sqrt{3}\) + \(\sqrt{5}\)
Let us assume to the contrary that \(\sqrt{3}\) + \(\sqrt{5}\) is a rational number.
Then there exist co-prime positive integers ‘a’ and ‘b’ such that
\(\sqrt{3}\) + \(\sqrt{5}\) = \(\frac{a}{b}\)
\(\frac{a}{b}\) – \(\sqrt{3}\) = \(\sqrt{5}\)
S.B.S.
TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4 1
This contradicts the fact that \(\sqrt{3}\) is irrational.
∴ Hence, \(\sqrt{3}\) + \(\sqrt{5}\) is irrational.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4

(iii) 6 + \(\sqrt{2}\)
Solution:
6 + \(\sqrt{2}\)
Let us assume on the contrary that 6 + \(\sqrt{2}\) is rational. Then there exist co-prime positive integers ‘a’ and ‘b’ such that
6 + \(\sqrt{2}\) = \(\frac{a}{b}\)
⇒ \(\sqrt{2}\) = \(\frac{a-6 b}{b}\)
\(\sqrt{2}\) is rational.
\(\frac{a-6 b}{b}\) is rational.
This contradicts the fact that \(\sqrt{2}\) is irrational, so our assumption is wrong.
∴ 6 + \(\sqrt{2}\) is irrational.

(iv) \(\sqrt{5}\)
Solution:
\(\sqrt{5}\)
Let us assume, to the contrary that \(\sqrt{5}\) is irrational then there exist co-prime positive integers a and b such that
\(\sqrt{5}\) = \(\frac{a}{b}\)
\(\sqrt{5}\) b = a
S.B.S. we get
(\(\sqrt{5}\) b)2 = (a)2
5b2 = a2 ……. (1)
5 divides a2.
Hence 5 divides a.
We can write a = 5c for some integer c.
Substitute a = 5c in (1) we get
5b2 = (5c)2
5b2 = 25c2
b2 = \(\frac{25 c^2}{5}\)
b2 = 5c2
5 divides b2 and 5 divide b.
‘a’ and ‘b’ have atleast as a common factor.
This contradicts the fact that ‘a’ and ‘b’ have no common factor other than 1.
So our assumption is wrong.
∵ \(\sqrt{5}\) is irrational.

(v) 3 + 2\(\sqrt{5}\)
Solution:
3 + 2\(\sqrt{5}\)
Let us assume, to the contrary that 3 + 2\(\sqrt{5}\) is rational. Then there exist co-prime positive integers ‘a’ and ’b’ such that
3 + 2\(\sqrt{5}\) = \(\frac{a}{b}\)
2\(\sqrt{5}\) = \(\frac{a}{b}\) – 3
\(\sqrt{5}\) = \(\frac{a-3 b}{2 b}\)
\(\sqrt{5}\) is rational.
\(\frac{a-3 b}{2 b}\) is rational.
This contradicts the fact that \(\sqrt{5}\) is irrational, so our assumption is wrong.
3 + 2\(\sqrt{5}\) is irrational.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4

Question 2.
Prove that \(\sqrt{p}\) + \(\sqrt{q}\) is an irrational, where p, q are primes.
Solution:
Let us assume to the contrary that \(\sqrt{p}\) + \(\sqrt{q}\) is rational. Then there exist co-prime positive in-tegers ‘a’ and ‘b’.
TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4 2
We know that square root of any prime number is irrational, we get \(\sqrt{q}\) is rational.
This contradicts the fact that \(\sqrt{q}\) is irrational.
So our assumption is wrong.
∵ \(\sqrt{q}\) is irrational.
∵ \(\sqrt{p}\) + \(\sqrt{q}\) is irrational.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.1

Question 1.
Use Euclid’s algorithm to find the HCF of

(i) 900 and 270
Answer:
900 and 270
Euclid Division Lemma
a = bq + r, q > 0 and 0 ≤ r < b
When 900 is divided 270, the remainder is 90 to get 900 = 270 × 3 + 90
Now consider the division of 270 with the remainder 90 in the above and division algorithm to get 270 = 90 × 3 + 0 The remainder is zero, when we cannot proceed further. We conclude that the HCF of (900, 270) = 90

(ii) 196 and 38220
Answer:
196 and 38220
When 38220 is divided 196, the remainder is 0 to get 38220 = 196 × 195 + 0
The remainder is zero, when we cannot proceed further. We conclude that the HCF of (38220, 196) = 196.

TS 6th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1

(iii) 1651 and 2032
Answer:
1651 and 2032
When 2032 is divided by 1651, the remainder is 381 to get 2032 = 1651 × 1 + 381
Now consider the division of 1651 with 381 in the above and division algorithm to get 1651 = 381 × 4 + 127
Now consider the division of 381 with the remainder 127 in the above and division algorithm to get 381 = 127 × 3 + 0
The remainder is zero, when we cannot proceed further we conclude that the HCF (1651, 2032) = 127

Question 2.
Use division algorithm to show that any positive odd integer is ofthe form 6q + 1, or 6 q + 3 or 6 q + 5, where q is some integer.
Answer:
Let ‘a’ be any positive odd integer, we apply the division algorithm with a and b = 6. Since 0 ≤ r ≤ 5, 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 quotient, However, since ‘a’ is odd, a cannot be 6q, 6q+2, 6q+4 (since they both are divisible by 2)
∴ any odd integer in the form of 6q + 1, (or) 6q + 3 (or) 6q + 5.

Question 3.
Use division algorithm to show that the square of any positive integer is of the form 3p or 3p + 1.
Answer:
Consider ‘a’ be the square of an integer
Applying Euclid’s division lemma with ‘a’ and ‘b’ = 3
0 ≤ r < 3 the possible remainders are 0, 1, 2 a = 3p or 3p + 1 (or) 3p + 2
Any square number is of the form 3p, 3p + 1, or 3p + 2, Where p is the quotient.

Question 4.
Use division algorithm to show that the cube of any positive integer is of the form 9 m, 9m + 1 or 9m + 8.
Answer:
Let ‘a’ be the cube of a positive integer
Applying Euclid’s division lemma for ‘a’ and b = 9 a = bm + r, Where’m’ is quotient r is the remainder where 0 ≤ r < 9 ‘a’ can be of the form
9m, 9m + 1, 9m + 2 ………… or 9m + 8
The cube of any positive integer is of the form 9m, 9m + 1, 9m + 2, ….. or 9m + 8.

TS 6th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1

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.
Answer:
Given numbers : n; n + 2 and n + 4

Case (i) : ‘n’ is divisible by 3, then n is of the form 3k. Now, n + 2 = 3k + 2 – leaves a remainder 2 when divided by 3. n + 4 = 3k + 4 = (3k + 3) + 1 = 3(k + 1) + 1 leaves a remainder 1 when divided by 3.

Case (ii) : n + 2 is divisible by 3. then n + 2 is of the form 3k.
Now n = 3k – 2 leaves a remainder 1 when divided by 3 and n + 4 = n + 2 + 2
= 3k + 2 leaves a remainder 2 when divided by 3.

Case (iii) : n + 4 is divisible by 3.
Take n + 4 = 3k
Now : n = 3k – 4 = 3(k – 1) + 3 – 4
= 3(k – 1) – 1 leaves a remainder 2 when divided by 3.
Also n + 2 = 3k-2 = 3(k- 1) + 3-2
= 3(k – 1) + 1 leaves a remainder 1 when divided by 3.
In all the above three cases only one out of n, n + 2 and n + 4 is divisible by 3.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.3

Question 1.
Write the following rational numbers in their decimal form and also state which are terminating and which are non-terminating repeating decimal form.

(i) \(\frac{3}{8}\)
Solution:
\(\frac{3}{8}\) = \(\frac{3}{2.2 .2}\) = \(\frac{3}{2^3}\) (∵ Denominator consists of only 2’s. terminating decimal)
\(\frac{3 \times 5^3}{2^3 \times 5^3}\) = \(\frac{3 \times 125}{10^3}\) = \(\frac{375}{1000}\) = 0.375

(ii) \(\frac{229}{400}\)
Solution:
\(\frac{229}{400}\) = \(\frac{229}{2.2 .2 .2 \times 5 \times 5}\) (∵ Denominator consists of only 2 ‘s.)
= \(\frac{229}{2^4 \cdot 5^2}\) (terminating decimal)
= \(\frac{229 \times 5^2}{2^4 \times 5^2 \times 5^2}\) = \(\frac{229 \times 5^2}{2^4 \times 5^4}\)
= \(\frac{229 \times 25}{(2 \times 5)^4}\) = \(\frac{5725}{10^4}\)
= \(\frac{5725}{10000}\) = 0.5725 (∵ Denominator consists of only 2’s. terminating decimal)

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3

(iii) 4\(\frac{1}{5}\)
Solution:
4\(\frac{1}{5}\) = \(\frac{21}{5}\) (terminating decimal)
\(\frac{21}{5}\) = 4.2 (∵ Denominator consists of only 2’s.)

(iv) \(\frac{2}{11}\)
Solution:
\(\frac{2}{11}\) (∵ Denominator doesn’t contain 2s or 5’s or both. Hence it is an non-terminating, repeating decimal) (non-terminating, re-peating decimal because the denomination does not contain power of 2 or power of 5 of both 2 and 5)
\(\frac{2}{11}\) = 0.181818……..
= \(0 . \overline{18}\)

(v) \(\frac{8}{125}\)
Solution:
\(\frac{8}{125}\) = \(\frac{8}{5^3}\) (∵ Denominator does not consists of only 2’s.) (terminating decimal)
= \(\frac{8}{5^3}\) × \(\frac{2^3}{2^3}\) = \(\frac{8 \times 8}{(5 \times 2)^3}\) = \(\frac{64}{1000}\) = 0.064

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) \(\frac{13}{3125}\)
Solution:
\(\frac{13}{3125}\). It is of the form \(\frac{p}{q}\)
\(\frac{13}{3125}\) = \(\frac{13}{5 \times 5 \times 5 \times 5 \times 5}\) = \(\frac{13}{5^5}\)
∵ q = 55 which is of the form 2n5m (n = 0; m = 5)
Given rational number has a terminating decimal expansion.

(ii) \(\frac{11}{12}\)
Solution:
\(\frac{11}{12}\) it is of the form \(\frac{p}{q}\).
\(\frac{11}{12}\) = \(\frac{11}{2 \times 2 \times 3}\) = \(\frac{11}{2^2 \times 3}\) = 0.916666.
∵ Here q = 22 × 3 which is not of the form 2n × 5m.
∴ Given rational number has a non-termi-nating repeating decimal expansion.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3

(iii) \(\frac{64}{455}\)
Solution:
\(\frac{64}{455}\) It is of the form \(\frac{p}{q}\).
\(\frac{64}{455}\) = \(\frac{64}{5 \times 7 \times 13}\)
∵ q = 5 × 7 × 13 which is not of the form 2n.5m.
∵ Given rational number has a non-terminating, repeating decimal expansion.

(iv) \(\frac{15}{1600}\)
Solution:
\(\frac{15}{1600}\) It is of the form \(\frac{p}{q}\).
\(\frac{15}{1600}\) = \(\frac{3}{320}\) = \(\frac{3}{2^6 \times 5}\)
∵ Here q = 26 × 51 which is of the form 2n.5m (m = 1, n = 6)
∴ Given rational number has a terminating decimal expansion.

(v) \(\frac{29}{343}\)
Solution:
\(\frac{29}{343}\) It is of the form \(\frac{p}{q}\).
\(\frac{29}{343}\) = \(\frac{29}{7 \times 7 \times 7}\) = \(\frac{29}{7^3}\)
∵ Here q = 73 which is not of the form 2n5m.
∴ Given rational number has a non-terminating, repeating decimal expansion.

(vi) \(\frac{23}{2^3 \cdot 5^2}\)
Solution:
\(\frac{23}{2^3 \times 5^2}\) It is of the form \(\frac{p}{q}\).
∵ Here q = 23 × 52 which is of the form of 2n5m (n = 3, m = 2).
∴ Given rational number has a terminating decimal expansion.

(vii) \(\frac{129}{2^2 \cdot 5^7 \cdot 7^5}\)
Solution:
\(\frac{129}{2^2 \times 5^7 \times 7^5}\) It is of the form \(\frac{p}{q}\).
∵ Here q = 22 × 57 × 75 which is not of the form 2n5m.
∴ Given rational number has a non-terminating, repeating decimal expansion.

(viii) \(\frac{9}{15}\) = \(\frac{3}{5}\) It is of the form \(\frac{p}{q}\).
Solution:
\(\frac{9}{15}\) = \(\frac{3}{5}\) It is of the form \(\frac{p}{q}\).
Here q = 51 which is of the form 2n5m (n = 0; m = 1). .
∴ Given rational number has a terminating decimal expansion.

(ix) \(\frac{36}{100}\)
Solution:
\(\frac{36}{100}\) It is of the form \(\frac{p}{q}\).
\(\frac{36}{100}\) = \(\frac{36}{2 \times 2 \times 5 \times 5}\) = \(\frac{36}{2^2 5^2}\)
∵ Here q = 22. 52 which is of the form 2n5m (n = 2, m = 2).
∴ Given rational number has a terminating decimal expansion.

(x) \(\frac{77}{210}\)
Solution:
\(\frac{77}{210}\) it is of the form \(\frac{p}{q}\).
\(\frac{77}{210}\) = \(\frac{11}{30}\) = \(\frac{11}{2 \times 3 \times 5}\)
∵ q = 2 × 3 × 5 which is not of the form 2n5m
∴ Given rational number has non-terminating, repeating decimal expansion.

Question 3.
Write the following rational numbers in decimal form using Theorem 1.4.

(i) \(\frac{13}{25}\)
Solution:
\(\frac{13}{25}\) = \(\frac{13}{5 \times 5}\) = \(\frac{13 \times 2^2}{5^2 \times 2^2}\)
= \(\frac{13 \quad 4}{10^2}\) = \(\frac{52}{100}\) = 0.52

(ii) \(\frac{15}{16}\)
Solution:
\(\frac{15}{16}\) = \(\frac{15}{2 \times 2 \times 2 \times 2}\) = \(\frac{15 \times 5^4}{2^4 \times 5^4}\)
= \(\frac{15 \quad 5^4}{10^4}\) = \(\frac{9375}{10000}\) = 0.9375

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3

(iii) \(\frac{23}{2^3 \cdot 5^2}\)
Solution:
\(\frac{23}{2^3 \times 5^2}\) = \(\frac{23 \times 5}{2^3 \times 5^2 \times 5}\)
= \(\frac{115}{10^3}\) = \(\frac{115}{1000}\) = 0.115

(iv) \(\frac{7218}{3^2 .5^2}\)
Solution:
\(\frac{7218}{3^2 \times 5^2}\) = \(\frac{3 \times 3 \times 802}{3^2 \times 5^2}\)
= \(\frac{802 \times 2^2}{5^2 \times 2^2}\) = \(\frac{3208}{100}\) = 32.08

(v) \(\frac{143}{110}\)
Solution:
\(\frac{143}{110}\) = \(\frac{11 \times 13}{11 \times 10}\) = \(\frac{13}{10}\) = 1.3

Question 4.
Express the following decimal numbers in the form of q and write the prime factors of q. What do you observe?
(i) 43.123
Solution:
43.123456789
Given decimal expansion terminates. Hence given real number is a rational.
It is in the form of \(\frac{p}{q}\)
TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 1
∵ Here q = 29.59 ; q is of the form 2n5m (n = 9; m = 9)

TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3

(ii) 0.120112001120001…
Solution:
0.120120012000120000 …………
Given decimal expansion is not either ter-minating or non-terminating repeating.
∵ Hence given real number is not rational.

(iii) \(43 . \overline{12}\)
Solution:
\(43 . \overline{123456789}\)
Given decimal expansion is non-terminating, repeating.
Given real number is rational and so of the \(\frac{p}{q}\)
Let x = \(43 . \overline{123456789}\)
x = \(43 . \overline{123456789}\) …… (2)
Multiplying both sides of (1) by 1000000000, we get
x = 43123456789.123456789 ……. (2)
Substracting (1) from (2) we get
999999999 x = 43123456746
x = \(\frac{43123456746}{999999999}\)
x = \(\frac{14374485582}{333333333}\)
q = 333333333 which is not of the form 2n5m.

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TS 10th Class Study Material

TS 10th Class Maths Bits with Answers Pdf in English & Telugu Medium | TS Maths Bits for 10th Class

TS 10th Class Maths Important Bits | TS 10th Class Maths Bit Bank Pdf

  1. Real Numbers Bits for 10th Class
  2. Sets Bits for 10th Class
  3. Polynomials Bits for 10th Class
  4. Pair of Linear Equations in Two Variables Bits for 10th Class
  5. Quadratic Equations Bits for 10th Class
  6. Progressions Bits for 10th Class
  7. Coordinate Geometry Bits for 10th Class
  8. Similar Triangles Bits for 10th Class
  9. Tangents and Secants to a Circle Bits for 10th Class
  10. Mensuration Bits for 10th Class
  11. Trigonometry Bits for 10th Class
  12. Applications of Trigonometry Bits for 10th Class
  13. Probability Bits for 10th Class
  14. Statistics Bits for 10th Class

TS 10th Class Study Material