10th Class Abhyasa Deepika Maths Solutions 1 Real Numbers Telangana SSC (English & Telugu Medium)

Are you searching for 10th class abhyasa deepika maths solutions 1 real numbers Telangana SSC? Chapter 1 – Real Numbers is one of the most important chapters in SSC Mathematics. It forms the base for many upcoming concepts and carries direct questions in the public exam.

In this guide, you will find clear explanations, important formulas, key answer highlights, and smart preparation tips for both English and Telugu medium students.

10 Abyasa Deepika Solutions

1. Using Division Algorithm if 25 is expressed as, $25 = (4 \times q) + r$ , then the value of ‘r’ is
a) 0
b) 1
c) 2
d) 3

View Answer

Solution: By Euclid’s Division Lemma, $a = bq + r$

.
Given $25 = 4q + r$

.
Dividing 25 by 4 gives a quotient of 6 and a remainder of 1.
$25 = (4 \times 6) + 1$

.
Comparing the two, r = 1.
Correct option: b

2. The decimal expression of the number $\frac{441}{2^2 \times 5^3 \times 7}$ is
a) a terminating decimal
b) non terminating but repeating decimal
c) non terminating, non repeating decimal
d) terminating after 2 decimal places

View Answer

Solution: First, simplify the fraction:
$\frac{441}{2^2 \times 5^3 \times 7} = \frac{63 \times 7}{2^2 \times 5^3 \times 7} = \frac{63}{2^2 \times 5^3}$

.
Since the denominator is in the form $2^n \times 5^m$

, the decimal expansion is terminating.
Correct option: a

3. Among the following, the number which is not irrational is
a) $(2-\sqrt{3})^2$
b) $(\sqrt{2}+\sqrt{3})^2$
c) $(\sqrt{2}-\sqrt{3}) (\sqrt{2}+\sqrt{3})$
d) $\frac{2\sqrt{7}}{7}$

View Answer

Solution: Let’s check option (c):
Using the identity $(a-b)(a+b) = a^2 - b^2$

.
$(\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$

.
Since -1 is a rational number, option c is not irrational.
Correct option: c

4. HCF of 26 and 91 is
a) 15
b) 13
c) 19
d) 11

View Answer

Solution: Prime factorization:
$26 = 2 \times 13$

$91 = 7 \times 13$

The common factor is 13.
HCF = 13.
Correct option: b

5. The least number that is divisible by all the numbers from 1 to 5 is
a) 15
b) 80
c) 70
d) 60

View Answer

Solution: We need to find the LCM of 1, 2, 3, 4, and 5.
$1 = 1$

, $2 = 2$

, $3 = 3$

, $4 = 2^2$

, $5 = 5$

.
$\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$

.
Correct option: d

6. Find the value of $\log_5 125$ .

View Answer

Solution: Let $\log_5 125 = x$

.
$5^x = 125 \implies 5^x = 5^3$

.
x = 3.

7. Find the value of $\log_{\sqrt{2}} 64$ .

View Answer

Solution: Let $\log_{\sqrt{2}} 64 = x$

.
$(\sqrt{2})^x = 64$

.
$(2^{1/2})^x = 2^6 \implies 2^{x/2} = 2^6$

.
$\frac{x}{2} = 6 \implies x = 12$

8. Is $\log_4 64$ rational or irrational? Justify.

View Answer

Solution: Let $\log_4 64 = x$

.
$4^x = 64 \implies 4^x = 4^3 \implies x = 3$

.
Since 3 is a rational number, the value is rational.

9. Find the HCF of 36 and 48 by using Euclid’s Division Algorithm.

View Answer

Solution:
Step 1: $48 = 36 \times 1 + 12$

Step 2: $36 = 12 \times 3 + 0$

The HCF is the last non-zero divisor, which is 12.

10. Find HCF and LCM of 80, 120 by Prime Factorization Method.

View Answer

Solution:
$80 = 2^4 \times 5$

$120 = 2^3 \times 3 \times 5$

HCF = $2^3 \times 5 = 40$

.
LCM = $2^4 \times 3 \times 5 = 240$

11. If HCF of 90 and 144 is 18, then find their LCM.

View Answer

Solution: $\text{HCF} \times \text{LCM} = a \times b$

.
$18 \times \text{LCM} = 90 \times 144$

.
$\text{LCM} = \frac{90 \times 144}{18} = 5 \times 144 = 720$

12. Convert the following into logarithmic form: i) $3^y = 25$ ; ii) $\frac{1}{49} = 7^z$ .

View Answer

Solution:
i) $\log_3 25 = y$

ii) $\log_7 (\frac{1}{49}) = z$

13. Write the following in the exponential form: i) $\log_3 27 = 3$ ; ii) $5 \log_2 32 = x$ .

View Answer

Solution:
i) $3^3 = 27$

ii) $\log_2 32^5 = x \implies 2^x = 32^5$

14. Expand the following: i) $\log (200)$ ; ii) $\log (\frac{125}{64})$ .

View Answer

Solution:
i) $\log (2 \times 10^2) = \log 2 + 2 \log 10$

ii) $\log 125 - \log 64 = \log 5^3 - \log 2^6 = 3 \log 5 - 6 \log 2$

15. Express as a single logarithm: i) $2 \log x + 3 \log y - 5 \log z$ ; ii) $5 \log 3 + 7 \log 2 - 3 \log 11 - 4 \log 5$ .

View Answer

Solution:
i) $\log (\frac{x^2 y^3}{z^5})$

ii) $\log (\frac{3^5 \cdot 2^7}{11^3 \cdot 5^4})$

16. Show that $5+3\sqrt{2}$ is an irrational number.

View Answer

Solution: Assume $5+3\sqrt{2}$

is rational.
$5+3\sqrt{2} = \frac{a}{b} \implies 3\sqrt{2} = \frac{a}{b} - 5 = \frac{a-5b}{b}$

.
$\sqrt{2} = \frac{a-5b}{3b}$

.
Since $a, b$

are integers, $\frac{a-5b}{3b}$

is rational, but $\sqrt{2}$

is irrational. This is a contradiction.

17. If $x^2 + y^2 = 27xy$ , then show that $2 \log (x-y) = 2 \log 5 + \log x + \log y$ .

View Answer

Solution: $x^2 + y^2 = 27xy$

.
Subtract $2xy$

from both sides: $x^2 + y^2 - 2xy = 27xy - 2xy$

.
$(x-y)^2 = 25xy$

.
Taking log on both sides: $\log (x-y)^2 = \log (25xy)$

.
$2 \log (x-y) = \log 5^2 + \log x + \log y$

.
$2 \log (x-y) = 2 \log 5 + \log x + \log y$

18. Show that the square of any positive integer is of the form $5m, 5m+1$ or $5m+4$ .

View Answer

Solution: Let the integer be $a = 5q + r$

where $r \in \{0, 1, 2, 3, 4\}$

.
Squaring $a$

:
If $r=0, a^2 = 25q^2 = 5(5q^2) = 5m$

.
If $r=1, a^2 = 25q^2+10q+1 = 5(5q^2+2q)+1 = 5m+1$

.
If $r=2, a^2 = 25q^2+20q+4 = 5(5q^2+4q)+4 = 5m+4$

.
If $r=3, a^2 = 25q^2+30q+9 = 5(5q^2+6q+1)+4 = 5m+4$

.
If $r=4, a^2 = 25q^2+40q+16 = 5(5q^2+8q+3)+1 = 5m+1$

19. Express 225 as product of prime factors.

View Answer

Solution: $225 = 3 \times 75 = 3 \times 3 \times 25 = 3 \times 3 \times 5 \times 5$

.
$225 = 3^2 \times 5^2$ .

20. Show that $\sqrt{2} + \sqrt{3}$ is an irrational number.

View Answer

Solution: Let $x = \sqrt{2} + \sqrt{3}$

be rational.
$x - \sqrt{2} = \sqrt{3}$

.
Squaring: $x^2 + 2 - 2x\sqrt{2} = 3$

.
$x^2 - 1 = 2x\sqrt{2} \implies \sqrt{2} = \frac{x^2 - 1}{2x}$

.
Since x is rational, $\frac{x^2-1}{2x}$

is rational, contradicting that $\sqrt{2}$

is irrational.

10 Abyasa Deepika Solutions

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10th Class Abhyasa Deepika Maths Solutions 1 Real Numbers Telangana SSC (English & Telugu Medium) – Complete Key & Exam Guide

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