10th Maths - Book Back Answers - Chapter 2 Excercise 2.2 - English Medium Guides

  

 

SSLC / 10th - Maths - Book Back Answers - Chapter 2 Excercise 2.2 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 2 Excercise 2.2: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 2 Excercise 2.2 – 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 Excercise 2.2 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 2.2

Question 1.
For what values of natural number n, 4th can end with the digit 6?
Answer Key:
4n = (2²)ⁿ = 2²ⁿ
= 2ⁿ × 2ⁿ
2 is a factor of 4ⁿ
∴ 4ⁿ is always even.

 

Question 2.
If m, n are natural numbers, for what values of m, does 2ⁿ × 5ⁿ ends in 5?
Answer Key:
2ⁿ × 5^m
2ⁿ is always even for all values of n.
5^m is always odd and ends with 5 for all values of m.
But 2ⁿ × 5^m is always even and ends in 0.
∴ 2ⁿ × 5^m cannot end with the digit 5 for any values of m. No value of m will satisfy 2ⁿ × 5^m ends in 5.

 

Question 3.
Find the H.C.F. of 252525 and 363636.
Answer Key:
To find the HCF of 252525 and 363636 by using Euclid’s Division algorithm.
363636 = 252525 × 1 + 111111
The remainder 111111 ≠ 0
By division of Euclid’s algorithm
252525 = 111111 × 2 + 30303
The remainder 30303 ≠ 0
Again by division of Euclid’s algorithm
111111 = 30303 × 3 + 20202
The remainder 20202 ≠ 0
Again by division of Euclid’s algorithm.
30303 = 20202 + 10101
The remainder 10101 ≠ 0
Again by division of Euclid’s algorithm.
20202 = 10101 × 2 + 0
The remainder is 0
∴ The H.C.F. is 10101

 

Question 4.
If 13824 = 2^a × 3^b then find a and b?
Answer Key:
Using factor tree method factorise 13824

 

13824 = 2⁹ × 3³

Given 13824 = 2a × 3b

Compare we get a = 9 and b = 3  

 Question 5.
If p₁^x1 × p₂^x2 × p₃^x3 × p₄^x4 = 113400 where p₁ p₂, p₃, p₄ are primes in ascending order and x₁, x₂, x₃, x₄, are integers, find the value of p₁,p₂,p₃,p₄ and x₁,x₂,x₃,x₄.
Answer Key:
Given 113400 = p₁^x1 × p₂^x2 × p₃^x3 × p₄^x4
Using tree method factorize 113400

 

 

 

 

113400 = 23 × 34 × 52 × 7

compare with

113400 = p₁^x1 × p₂^x2 × p₃^x3 × p₄^x4

P₁ = 2, x₁ = 3

P₂ = 3, x₂ = 4

P₃ = 5, x₃ = 2

P₄ = 7, x₄ = 1

Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.
Answer Key:
Factorise 408 and 170 by factor tree method

 

 

408 = 2³ × 3 × 17

170 = 2 × 5 × 17

To find L.C.M. list all prime factors of 408 and 170 of their greatest exponents.

L.C.M. = 2³ × 3 × 5 × 17  = 2040

To find the H.C.F. list all common factors of 408 and 170.

H.C.F. = 2 × 17 = 34

L.C.M. = 2040 ; HCF = 34   

 

Question 7.
Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?
Answer Key:
The greatest number of 6 digits is 999999.
The greatest number must be divisible by L.C.M. of 24, 15 and 36 

24 = 2³ × 3

15 = 3 × 5

36 = 2² × 3²

L.C.M = 2³ × 3² × 5 = 360

To find the greatest number 999999 must be subtracted by the remainder when 999999 is divided by 360

The greatest number in 6 digits = 999999 – 279  = 999720

 

 

Question 8.

What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Answer Key:

35 = 5 × 7

56 = 2 × 2 × 2 × 7

91 = 7 × 13

LCM of 35, 56, 91 = 5 × 7 × 2 × 2 × 2 × 13 = 3640

∴ Required number = 3647 which leaves remainder 7 in each case.

 

Question 9.
Find the least number that is divisible by the first ten natural numbers?
Answer Key:
Find the L.C.M of first 10 natural numbers.

 

 

 

 

 

The least number is 2520