Download RS Aggarwal Solutions Class 8 Solutions Chapter 5 in PDF
FAQs on Easy Learning with RS Aggarwal Class 8 Mathematics Solutions for Chapter-5 Playing with Numbers
1. How do you solve problems involving the general form of a 2-digit number in RS Aggarwal Class 8 Chapter 5?
To solve problems using the general form of a 2-digit number, you represent the number as 10a + b, where 'a' is the digit in the tens place and 'b' is the digit in the ones place. For example, if you are asked to work with a number and the number formed by reversing its digits, you would set up the original number as (10a + b) and the reversed number as (10b + a). You can then form equations based on the problem's conditions (e.g., their sum or difference) to find the values of 'a' and 'b'.
2. What is the step-by-step method to test the divisibility of a large number by 9, as per RS Aggarwal solutions?
The method to check if a number is divisible by 9 involves a simple two-step process:
- Step 1: Find the sum of all the digits of the given number.
- Step 2: Check if this sum is divisible by 9. If the sum is a multiple of 9 (e.g., 9, 18, 27), then the original number is also divisible by 9. Otherwise, it is not. For example, for the number 4581, the sum is 4+5+8+1 = 18. Since 18 is divisible by 9, the number 4581 is also divisible by 9.
3. How do you find the value of a missing digit in a number if the entire number is given to be a multiple of 3?
To find a missing digit in a number that is a multiple of 3, you use the divisibility rule of 3. Let's say the number is 5X42. First, sum the known digits: 5 + 4 + 2 = 11. The full sum is 11 + X. For the number to be divisible by 3, (11 + X) must be a multiple of 3 (e.g., 12, 15, 18, etc.). By testing single-digit values for X, we find that if X=1, the sum is 12; if X=4, the sum is 15; and if X=7, the sum is 18. Therefore, the possible values for X are 1, 4, or 7.
4. What is the correct method for solving letter-for-digit addition puzzles, also known as cryptarithmetic, in Exercise 5C?
The correct method for solving cryptarithmetic addition puzzles follows these key rules:
- Each letter represents a unique single digit from 0 to 9.
- The first digit of any number cannot be zero.
- You must solve the puzzle column by column, starting from the rightmost (ones) column, and account for any carry-over to the next column on the left.
By logically deducing the values based on these rules and simple arithmetic, you can find the value for each letter.
5. Why does the divisibility test for 11 involve finding the difference between the sums of alternating digits?
The divisibility rule for 11 works because of the properties of place value in powers of 10. Any power of 10 is either one more or one less than a multiple of 11. For example, 10 = 11-1, 100 = 99+1, 1000 = 1001-1, and so on. When you expand a number in its generalised form (e.g., 1000a + 100b + 10c + d) and group the terms based on their relation to 11, you are effectively isolating a part that is divisible by 11 and a remaining part which is the alternating sum and difference of the digits (d - c + b - a). For the whole number to be divisible by 11, this remaining part must also be divisible by 11 (or be zero).
6. How is the concept of representing numbers in generalised form fundamental to proving the divisibility tests for 3 and 9?
The generalised form is the key to proving these rules. For a 3-digit number 'abc', its value is 100a + 10b + c. This can be rewritten as (99a + 9b) + (a + b + c). Since (99a + 9b) is always divisible by both 3 and 9, the divisibility of the entire number depends solely on the remaining part, which is (a + b + c), the sum of its digits. This proves that a number is divisible by 3 or 9 if and only if the sum of its digits is divisible by 3 or 9, respectively.
7. In cryptarithmetic puzzles, what is the most common mistake students make, and how can it be avoided?
The most common mistake is forgetting the carry-over from one column to the next. Students often solve each column in isolation without considering that a sum greater than 9 in the ones column will generate a '1' to be added to the tens column. To avoid this, always solve the puzzle systematically from right to left. After solving for a column, immediately check if the sum is 10 or more. If it is, make a note of the carry-over '1' above the next column before you begin solving for it.
8. What is the correct method to check for divisibility by 6, and why is checking for divisibility by 3 alone not enough?
To check if a number is divisible by 6, you must verify that it is divisible by both 2 and 3. A number is divisible by 2 if it is an even number (ends in 0, 2, 4, 6, or 8). It is divisible by 3 if the sum of its digits is a multiple of 3. Checking for divisibility by 3 alone is insufficient because 6 is a composite number (2 × 3). For example, the number 21 is divisible by 3, but it is not an even number, so it is not divisible by 6.
