RS Aggarwal Class 8 Maths Chapter 5 Ex 5.2 Solutions: In this exercise, the students will learn the methods which can be followed while answering questions from Playing with Numbers. A problem can be solved in different ways & these solutions will be talking about those ways in the form of explanations. Practicing the RS Aggarwal textbook questions enables the students to analyze their exam preparation level & knowledge of each topic.
Frequent practice of RS Aggarwal Class 8 Maths Chapter 5 Ex 5.2 Solutions also increases the speed of writing the exam. These solutions assist the students in gaining knowledge and strong command over the Maths subject. The Mathematics expert faculty team has created these solutions in order to assist the students in their exam preparation to attain high marks in the Maths exam. The students get familiarized with the format of the class 8th Maths question paper so as to secure excellent ranks in the exam.
Access RS Aggarwal Class 8 Maths Chapter 5 Solutions PDF
Download RS Aggarwal Class 8 Maths Chapter 5 Ex 5.2 Solutions
RS Aggarwal Class 8 Chapter 5 Ex 5.2 Solutions
Important Definition for RS Aggarwal Class 8 Maths Chapter 5 Ex 5.2 Solutions
- Playing with 2 – Digit and 3 – Digit Numbers
- Numbers in General form
A two digit number (ab) in its general form is represented as ab = (10 x a) + (1 x b)
A three digit number (abc) is represented as abc = (100 x a) + (10 x b) + c
- Reversing the 2 digit numbers and adding them
When a 2-digit number is reversed & added with the number, the resulting number is divided by 11 & the quotient is equal to the sum of the digits
For instance: The reverse of 29 is 92.
The sum of 29 & 92 = 29 +92 = 121
On dividing the sum by 11, the students will get 121/11 = 11 = 9 + 2.
Therefore, the sum is divisible by 11 & the quotient is equal to the sum of the digits of the number.
- Reversing the 2 digit numbers and Subtracting them
When a two-digit number is reversed & the larger number is subtracted from the smaller number, the resulting number is perfectly divisible by 9 and the quotient is equal to the difference of the digits of the number.
For instance, the reverse of the number 39 is 93.
Then, 93 > 39
Therefore, 93 – 39 = 54
On dividing the difference of the two number by 9, the students will get, 54/9 = 6 = 9 − 3
Thus, the difference is divisible by 9 & the quotient is equal to the difference of the digits.
- Here we have puzzles in which letters take the place of digits in arithmetic sum & the problem is to find out which letter represents which digit.
For instance if 4 Q 1 3 8 Q——— (+) 8 0 3———
When 1 is added to it, the value of Q will be 2 as 2 is the only digit that results in 3. Also, 2 + 8 = 10, 1 as carry & then 1 + 4 + 3 = 8. So, the value of Q is 2.
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