10th Maths - Book Back Answers - Chapter 2 Exercise 2.3 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 2 Exercise 2.3 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 2 Exercise 2.3: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 2 Exercise 2.3 – Maths from the Tamil Nadu State Board 10th Standard Maths textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 2 Exercise 2.3 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 2 Numbers and Sequences Ex 2.3

1. Find the least positive value of x such that

(i) 71 = x (mod 8) 
(ii) 78 + x = 3 (mod 5)

(iii) 89 = (x + 3) (mod 4) 

(iv) 96 = x7 (mod 5)

(v) 5x = 4 (mod 6)  

Answer Key:

(i) 71 = x (mod 8)
71 = 7 (mod 8)
∴ The value of x = 7

 

(ii) 78 + x = 3 (mod 5)

78 + x – 3 = 5n (n is any integer)

75 + x = 5n

(Let us take x = 5)

75 + 5 = 80 (80 is a multiple of 5)

∴ The least value of x is 5

 

(iii) 89 = (x + 3) (mod 4)

89 – (x + 3) = 4n
(n may be any integer)
89 – x – 3 = 4n
89 – x = 4n
86 – x is a multiple of 4
(84 is a multiple of 4)
86 – 2 = 4n
84 = 4n
The value of x is 2

 

(iv) 96 = $\frac{x}{7}$ (mod 5)
96 – $\frac{x}{7}$ = 5n (n may be any integer)
672 – x = 35n (multiple of 35 is 665)
672 – 7 = 665
∴ The value of x = 7

 

(v) 5x = 4 (mod 6)
5x – 4 = 6n (n may be any integer)
5x = 6n + 4

x=$\frac{6n+4}{5}$
x = 6n+45
Substitute the value of n as 1, 6, 11, 16 …. as n values in x =$\frac{6n+4}{5}$ 6n+45 which is divisible by 5.
2, 8, 14, 20,…………
The least positive value is 2.

 

2. If x is congruent to 13 modulo 17 then 7x -3 is congruent to which number modulo 17?Answer Key:
x ≡ 13 (mod 17)
Let p be the required number …………. (1)
7x – 3 ≡ p (mod 17) ………….. (2)
From (1),
x – 13 = 17n for some integer M.
x – 13 is a multiple of 17.
x must be 30.
∴ 30 – 13 = 17
which is a multiple of 17.
From (2),
7 × 30 – 3 ≡ p (mod 17)
210 – 3 ≡ p (mod 17)
207 ≡ p (mod 17)
207 ≡ 3 (mod 17)
∴ P ≡ 3

 

3. Solve 5x ≡ 4 (mod 6)

Answer Key:

5x – 4 = 6n (n may be any integer)

5x = 6n + 4

x =$\frac{6\mathrm{n}+4}{5}$

∴ The value of x is 2, 8, 14, 20 …………..

4. Solve 3x – 2 = 0 (mod 11)
Answer Key:
Given 3x – 2 = 0(mod 11)
3x – 2 = 11n (n may be any integer)
3x = 2 + 11n

 x =$x=\frac{11n+2}{3}$

Substitute the value of n = 2, 5, 8, 11 ….
When n ≡ 2 ⇒ x=$\frac{22+2}{3}=\frac{24}{3}=8$

 When n = 5 ⇒ x=$\frac{55+2}{3}=\frac{57}{3}=19$

 When n = 8 ⇒ x=$\frac{88+2}{3}=\frac{90}{3}=30$

 When n = 11 ⇒x=$\frac{121+2}{3}=\frac{123}{3}=41$
∴ The value of x is 8, 19, 30,41

 

5. What is the time 100 hours after 7 a.m.?
Answer Key:

100 ≡ x (mod 12) Note: In a clock every 12 hours
100 ≡ 4 (mod 12) the numbers repeats.

 ∴The time repeat after 7 am is 7 + 4 = 11 o’ clock (or) 11 am.

6. What is time 15 hours before 11 p.m.?
Answer Key:
15 ≡ x (mod 12)
15 – x = 12n

15 – x is a multiple of 12 x must be 3

∴ The time 15 hrs before 11 O’clock is 11 – 3 = 8 O’ clock i.e. 8 p.m

 

7. Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?
Answer Key:
Number of days in a week = 7
45 ≡ x (mod 7)
45 ≡ 3 (mod 7) 

The value of x must be 3.

Three days after tuesday is friday uncle will come on friday.  

 

8. Prove that 2ⁿ + 6 × 9ⁿ is always divisible by 7 for any positive integer n.

Answer Key:

2¹ + 6 × 9¹ = 2 + 54 = 56 is divisible by 7
When n = k,
2^k + 6 × 9^k = 7 m [where m is a scalar]
⇒ 6 × 9^k = 7 m – 2^k …………. (1)
Let us prove for n = k + 1
Consider 2^k+1 + 6 × 9^k+1 = 2^k+1 + 6 × 9^k × 9
= 2^k+1 + (7m – 2^k)9 (using (1))
= 2^k+1 + 63m – 9.2^k = 63m + 2^k.2¹ – 9.2^k
= 63m – 2^k (9 – 2) = 63m – 7.2^k
= 7 (9m – 2^k) which is divisible by 7
∴ 2ⁿ + 6 × 9ⁿ is divisible by 7 for any positive integer n

9. Find the remainder when 2⁸¹ is divided by 17?

Answer Key:

2⁸¹ ≡ x(mod 17)
2⁴⁰ × 2⁴⁰ × 2¹ ≡ x(mod 17)
(2⁴)¹⁰ × (2⁴)¹⁰ × 2¹ ≡ x(mod 17)
(16)¹⁰ × (16)¹⁰ × 2¹ ≡ x(mod 17)
(16²)⁵ × (16²)⁵ × 2¹ ≡ x(mod 17)
= 1 × 1 × 2 (mod 17)
[(16)² = 256 = 1 (mod 17)]
= 2 (mod 17)
2⁸¹ = 2(mod 17)
∴ x = 2
The remainder is 2

 

10. The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is four and half hours ahead to that of London’s time, then find the time at London, when will the flight lands at London Airport?
Answer Key:

The duration of flight from Chennai to London is 11 hours.

Starting time at Chennai is 23.30 hrs. = 11.30 p.m.

Travelling time = 11.00 hrs. = 22.30 hrs = 10.30 a.m.

1 Chennai is 4-$\frac{1}{2}$ hrs ahead to London. 2

10.30-4.30 = 6.00

At 6 a.m. on Monday the flight will reach at London Airport.