NCERT Solutions for Class 8 Maths Chapter 16: Playing With Numbers
NCERT Solutions for Class 8 Maths Chapter 16, Playing with Numbers, are meticulously crafted by subject-matter experts at Orchid The International School, with a keen understanding of the specific needs of CBSE Class 8 students. These solutions serve as a comprehensive aid for students, offering clear and concise explanations to facilitate a thorough understanding of mathematical concepts.
Access Answers to NCERT Solutions for Class 8 Maths Chapter 16: Playing With Numbers
Exercise 16.1
Find the values of the letters in each of the following and give reasons for the steps involved.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
1. Say, A = 7, and we get
7+5 = 12
In which one’s place is 2.
Therefore, A = 7
And putting 2 and carrying over 1, we get
B = 6
Hence, A = 7 and B = 6.
2. If A = 5, we get
8+5 = 13, in which one’s place is 3.
Therefore, A = 5 and carry over 1, then
B = 4 and C = 1
Hence, A = 5, B = 4 and C = 1.
3. On putting A = 1, 2, 3, 4, 5, 6, 7 and so on, we get
AxA = 6×6 = 36, in which one’s place is 6.
Therefore, A = 6
4. Here, we observe that B = 5, so that 7+5 =12
Putting 2 at one’s place and carrying over 1 and A = 2, we get
2+3+1 =6
Hence, A = 2 and B =5.
5. Here, on putting B = 0, we get 0x3 = 0.
And A = 5, then 5×3 =15
A = 5 and C = 1
Hence A = 5, B = 0 and C = 1.
6. On putting B = 0, we get 0x5 = 0 and A = 5, then 5×5 =25
A = 5, C = 2
Hence A = 5, B = 0 and C =2
7. Here, products of B and 6 must be the same as one’s place digit is B.
6×1 = 6, 6×2 = 12, 6×3 = 18, 6×4 = 24
On putting B = 4, we get the one’s digit 4, and the remaining two B’s value should be 44.
Therefore, for 6×7 = 42+2 =44
Hence, A = 7 and B = 4.
8. On putting B = 9, we get 9+1 = 10
Putting 0 at ones place and carrying over 1, we get A = 7
7+1+1 =9
Hence, A = 7 and B = 9.
9. On putting B = 7, we get 7+1 = 8
Now A = 4, then 4+7 = 11
Putting 1 at tens place and carrying over 1, we get
2+4+1 =7
Hence, A = 4 and B = 7.
10. Putting A = 8 and B = 1, we get
8+1 = 9
Now, again we add 2 + 8 =10
The tens place digit is ‘0’ and carries over 1. Now 1+6+1 = 8 = A
Hence, A = 8 and B =1.
Exercise 16.2
If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Suppose 21y5 is a multiple of 9.
Therefore, according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.
That is, 2+1+y+5 = 8+y
Therefore, 8+y is a factor of 9.
This is possible when 8+y is any one of these numbers 0, 9, 18, 27, and so on
However, since y is a single-digit number, this sum can be only 9.
Therefore, the value of y should be 1 only, i.e. 8+y = 8+1 = 9.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers to the last problem. Why is this so?
Since 31z5 is a multiple of 9,
According to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.
3+1+z+5 = 9+z
Therefore, 9+z is a multiple of 9
This is only possible when 9+z is any one of these numbers: 0, 9, 18, 27, and so on.
This implies, 9+0 = 9 and 9+9 = 18
Hence, 0 and 9 are the two possible answers.
If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3, its sum of digits 6+x is a multiple of 3, so 6+x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … . But since x is a digit, it can only be that 6+x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.)
Let’s say 24x is a multiple of 3.
Then, according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
2+4+x = 6+x
So, 6+x is a multiple of 3, and 6+x is one of the numbers: 0, 3, 6, 9, 12, 15, 18 and so on.
Since x is a digit, the value of x will be either 0 or 3 or 6 or 9, and the sum of the digits can be 6 or 9 or 12 or 15, respectively.
Thus, x can have any of the four different values: 0 or 3 or 6 or 9.
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Since 31z5 is a multiple of 3,
According to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
That is, 3+1+z+5 = 9+z
Therefore, 9+z is a multiple of 3.
This is possible when the value of 9+z is any of the values: 0, 3, 6, 9, 12, 15, and so on.
At z = 0, 9+z = 9+0 = 9
At z = 3, 9+z = 9+3 = 12
At z = 6, 9+z = 9+6 = 15
At z = 9, 9+z = 9+9 = 18
The value of 9+z can be 9 or 12 or 15 or 18.
Hence 0, 3, 6 or 9 are the four possible answers for z.
Frequently Asked Questions
The NCERT solution for class 8 Chapter 16: Playing with Numbers is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tacking more difficult concepts in their further education.
Yes, the NCERT solution for class 8 Chapter 16: Playing with Numbers is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Playing with Numbers ally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for class 8 Maths Chapter 16 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for class 8 Maths Chapter 16 : Playing with Numbers as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for class 8 Maths Chapter 16 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution. Also, you can make Playing with Numbers all notes and jot down the important concepts for your understanding.
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