RS Aggarwal Solutions for Class 10 Maths Chapter 1 Exercise 1.2: RS Aggarwal's solutions for Class 10 Maths, Chapter 1, Exercise 1.2, focus on real numbers and their properties. This exercise helps students understand the fundamental theorem of arithmetic, which states that every composite number can be expressed as a product of prime numbers uniquely, except for the order of the primes.
The problems in this exercise involve finding the prime factorization of given numbers, determining the highest common factor (HCF) and least common multiple (LCM) using prime factorization, and solving word problems based on these concepts. By working through these solutions, students can strengthen their grasp of prime numbers and their applications, laying a strong foundation for more advanced topics in mathematics.RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2 PDF
Question 16.
Solution:
We can represent any integer number in the form of: pq + r, where ‘p is divisor, ‘q is quotient’, r is remainder*. So, we can write given numbers from given in formation As : 43 = pq1 + r ...(i) 91 = pq2 + r ...(ii) And 183 = pq3 + r ...(iii) Here, we want to find greatest value of ‘p’ were r is same. So, we subtract eq. (i) from eq. (ii), we get Pq2 – Pq1 = 48 Also, subtract eq. (ii) from eq. (iii), we get pq3 – pq2 = 92 Also, subtract eq. (i) from eq. (iii), we get Pq3 – Pq1 = 140 Now, to find greatest value of ‘p’ we find HCF of 48, 92 and 140 as, 48 = 2 x 2 x 2 x 2 x 3 92 = 2 x 2 x 2 x 2 x 2 x 3 and 140 = 2 x 2 x 5 x 7 So, HCF (48, 92 and 140) = 2 x 2 = 4 Greatest number that will divide 43, 91 and 183 as to leave the same remainder in each case = 4.Question 17.
Solution:
Remainder in all the cases is 6, i.e., 20 – 14 = 6 25 – 19 = 6 35 – 29 = 6 40 – 34 = 6 The difference between divisor and the corresponding remainder is 6. Required number = (LCM of 20, 25, 35, 46) – 6 = 1400 – 6 = 1394