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Chapter 16

Playing With Numbers

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You find numbers all around you. The hours in a day or the distance from your home to school, all are expressed as numbers. How do you find the score when watching a cricket or football match? All the scores in the sports are numbers that follow certain rules. When you know how to play with numbers it becomes much easier to perform simple calculations quickly. The Class 8 math Chapter 16 has been developed to help you learn some basic calculations, and how you can quickly perform these calculations. 

The chapter is titled “Playing with Numbers” and contains some fun ways to help you understand the following topics.

  • General form of expression for numbers.

  • Replacement of letters with numbers when calculating.

  • Divisibility tests for numbers.

Here, we take a quick look at these concepts and then see the NCERT solutions for Class 8 Maths Chapter 16 exercises.

Quick Recap of Concepts

Before we proceed with the “Playing with Numbers” Class 8 NCERT solutions, let us have a quick recap of the concepts covered in the chapter.

 

 

The first concept that the Class 8 Maths Chapter 16 discusses is about how to express any number in a general form. This is how any number can be expressed.

  • A 2-digit number “ab” can be expressed as 10a+b.

  • A 3-digit number “abc” can be expressed as 100a+10b+c.

Now, let us see this with a few examples to understand why these general forms are correct.

Example 1: Consider the following 2-digit number

35 = 30 + 5 = (10 x 3) + 5. So, we can see that 35 = (10 x 3) + 5.

Example 2: Consider the following 3- digit number

247 = 200 + 40 + 7 = (100 x 2) + (10 x 4) + 7. In other words, 247 = (100 x 2) + (10 x 4) + 7.

A very interesting concept covered in Chapter 16 Math Class 8 shows how to carry out calculations when digits in numbers are replaced by letters. Using basic mathematical formulae for addition, subtraction, multiplication, and division, you can easily find the values of these letters. This is a very concept to understand when trying to understand NCERT Solutions for Class 8 Maths Chapter 16, so let us clarify with a few examples.

Example 3: Find the value of A in the following

A + A + A = 1 A.

In this example, if we add the unit’s digits, we get A + A + A = A with 1 being carried away.

There are only 2 possible values of A that can give the same value when added 3 times. Let us evaluate both.

  1. A = 0. So, A + A + A => 0 + 0 + 0 = 0. But in this case, we do not get 1 as carry over value. Hence, this cannot be the solution.

  2.  A = 5. When taking the value of A as 5, we get the equation as follows.

A + A + A = 5 + 5 + 5 = 15. This satisfies both conditions. So, A = 5, is the correct answer.

The last concept covered in this chapter before we see the NCERT Solutions for Class 8 Maths Chapter 16 is about divisibility tests for different numbers. Let us recap the tests that show if a given number is divisible by a particular number.

  • Divisibility by 10 – If the unit’s digit of any number is 0, it will be divisible by 10.

  • Divisibility by 5 – If the unit’s digit of any number is either 5 or 0, it will be divisible by 5.

  • Divisibility by 2 – If the unit’s digit of any number is an even number, the number will be divisible by 2.

  • Divisibility by 3 or 9 – If the sum of each digit of any number is divisible by 3 or 9, the number is divisible by 3 or 9, respectively.

Having revised the concepts covered in the Class 8 Maths Chapter 16, let us look at the solutions for the exercise questions.

Exercise 16.1

1. 

Solution: Adding the unit’s digits we see A + 5 = 2 in the unit’s digit. The only possible solution for this is 7 + 5 = 12 since 2 is smaller than 5. Thus, A = 7. Carrying over the 1 from 12, we get the next step, i.e., the addition of ten’s digits as (1) + 3 + 2 = B. So, B = 6.

2. 

Solution: Adding the unit’s digits we see A + 8 = 3 in the unit’s digit. The possible solution for this is 5 + 8 = 13. Thus, A = 5. Carrying over the 1 from 13, we get the next step, i.e., the addition of ten’s digits as (1) + 4 + 9 = B. So, B = 4, with 1 being carried over. Thus, C = 1.

3. 

Solution: Checking the multiplication of the unit’s digits we find that A x A= A in the unit’s digit. The possible values for A, in this case, can be 0, 1, 5, and 6.

      1. We know that 0 x 0 = 0. Since the final product is not 0, this is not the correct answer.

      2. Next, if A = 1, the equation becomes 11 x 1 = 91, which is not correct.

      3. For A = 5, we see that 15 x 5 = 75, so our value does not satisfy the equation.

      4. Checking for A = 6, we find that 16 x 6 = 96. Thus, the correct solution is A = 6.

4.

Solution: Adding the unit’s digits we see B + 7 = A in the unit’s digit. Also, adding the ten’s digits gives us the equation A + 3 = 6.

Let us call these equations as

i.                     B + 7 = A

ii.                   A + 3 = 6

  1. Let us see what happens when we take A = 3. We know that 3 + 3 = 6. So, equation ii. becomes correct. But, B + 7 = 3 means that B will be 6, in which case we get a carryover as 1 and equation ii. becomes incorrect. Thus, A cannot be 3.

  2. Taking A = 2, gives us the following solution.

    1. B + 7 = 2 => B = 5. So, 5 + 7 = 2 with 1 carried over.

    2. (1) + 2 + 3 = 6. This equation is also correct. Hence, our solution becomes A = 2, B = 5.

5. 

Solution: Checking the multiplication of the unit’s digits we find that B x 3= B in the unit’s digit. There are 2 possible solutions for this B = 5 or 0. Taking the value of B as 5, we get (A x 3) + 1 = A with C as carryover. No value of A can solve this equation. Now, consider B=0. This gives us the value of C=1 and A=5.

6. 

Solution: Checking the multiplication of the unit’s digits we find that B x 5= B in the unit’s digit. There are 2 possible solutions for this B = 5 or 0. Taking B=0, we get (A x 5) = A with C as carryover. So, we get the following solution A =5, B = 0, C = 2. Now, taking B=5, we get the next equation as (Ax5)+2=A. There are 2 possible values of A that can satisfy this equation, A = 2 or 7. Considering both we get the values as A=2,B=5,C=1 and A=7,B=5,C=3.

7. 

Solution: Checking the multiplication of the unit’s digits we find that B x 6= B in the unit’s digit. There are 2 possible solutions for this B = 4 or 6. Taking B=4, we get (A x6)+2 = 4 with 4 as carryover. A=7 gives us the equation (7x6)+2 = 44. Thus, a possible solution is A=7 and B=4. Now checking with B=6, we get (Ax6)+3=66. There is no value of A that solves this equation.

8.

Solution: Adding the unit’s digits we see 1+B=0 in the unit’s digit. The only possible solution for this is B=9. Thus, addition of tens digits gives us, (1)+A+1=9. So, A=7.

9. 

Solution: Adding the unit’s digits we see B+1=8 in the unit’s digit. The only possible solution for this is B=7. Thus, addition of tens digits gives us, A+7=1. Solving this, we get A=4.

10. 

Solution: Adding the unit’s digits we see A+B=9 in the unit’s digit. Next, on adding tens digits we get 2+A=0. The possible value of A=8. This satisfies the addition of hundreds digits (1)+1+6=8. Taking A=8, we get B=1.

  1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Solution: 2+1+y+5 should be a multiple of 9. Solving this equation, we get the following.

2+1+y+5 = 8+y should be divisible by 9. y being a digit, the only possible solution is y=1.

  1. If 31z5 is a multiple of 9, where z is a digit, what is the value of z?

Solution: 3+1+z+5 should be a multiple of 9. Solving this equation, we get the following.

3+1+z+5 = 9+z should be divisible by 9. z being a digit, the possible solutions are

            1. z=0 => 9+0=9.
            2. z=9 => 9+9=18.

     3. If 24x is a multiple of 3, where x is a digit, what is the value of x?

Solution: 2+4+x should be a multiple of 3. Solving this equation, we get the following.

2+4+x = 6+x should be divisible by 3. x being a digit, the possible solutions are

            1. x=0 => 6+0=6.

            2. x=3 => 6+3=9.

            3. x=6 => 6+6=12.

            4. x=9 => 6+9=15. 

    4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Solution: 3+1+z+5 should be a multiple of 3. Solving this equation, we get the following.

3+1+z+5 = 9+z should be divisible by 3. z being a digit, the possible solutions are

            1. z=0 => 9+0=9.

            2. z=3 => 9+3=12.

            3. z=6 => 9+6=15.

            4. z=9 => 9+9=18.

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