NCERT Solutions Class 8 Mathematics Chapter 16 Playing with Numbers

NCERT Solutions Class 8 Mathematics Chapter 16 Playing with Numbers have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 8 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 8 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 8 Mathematics are an important part of exams for Class 8 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 8 Mathematics and also download more latest study material for all subjects. Chapter 16 Playing with Numbers is an important topic in Class 8, please refer to answers provided below to help you score better in exams

Chapter 16 Playing with Numbers Class 8 Mathematics NCERT Solutions

Class 8 Mathematics students should refer to the following NCERT questions with answers for Chapter 16 Playing with Numbers in Class 8. These NCERT Solutions with answers for Class 8 Mathematics will come in exams and help you to score good marks

Chapter 16 Playing with Numbers NCERT Solutions Class 8 Mathematics

Exercise 16.1

Q.1) Find the values of the letters in the following and give reasons for the steps involved.

""NCERT-Solutions-Class-8-Mathematics-Playing-with-Numbers

Sol.1) i) On putting A = 1, 2, 3, 4, 5, 6, 7 and so on
and we get,
7 + 5 = 12 in which ones place is 2.
∴ 𝐴 = 7
And putting 2 and carry over 1, we get B = 6
Hence, A = 7 and B = 6

ii) On putting A = 1, 2, 3, 4, 5, 6, 7 and so on
and we get, 8 + 5 = 13 in which ones place is 3.
∴ A = 5
And putting 3 and carry over 1,
we get B = 4 and C = 1
Hence, A = 5, B = 4 and C = 1

iii) On putting A = 1, 2, 3, 4, 5, 6, 7 and so on
and we get, 𝐴 × 𝐴 = 6 × 6 = 36 in which ones place is 6.
∴ A = 6 Hence, A = 6

iv) Here, we observe that B = 5 so that 7 + 5 = 12.
Putting 2 at ones place and carry over 1 and A = 2,
we get 2 + 3 + 1 = 6
Hence, A = 2 and B = 5

v) Here on putting B = 0,
We get 0 x 3 = 0.
And A = 5,
then 5 × 3 = 15
⇒ A = 5 and C = 1
Hence, A = 5, B = 0 and C = 1

vi) On putting B = 0,
we get 0 x 5 = 0 and A = 5,
then 5 × 5 = 25
⇒ 𝐴 = 5, 𝐶 = 2
Hence, A = 5, B = 0 and C = 2

vii) Here product of B and 6 must be same as ones place digit as B.
6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24
On putting B = 4,
We get the ones digit 4 and remaining two B’s value should be 44.
∴ For 6 × 7 = 42 + 2 = 44
Hence, A = 7 and B = 4

viii) On putting B = 9, we get 9 + 1 = 10
Putting 0 at ones place and carry over 1,
we get For A = 7
⇒ 7 + 1 + 1 = 9
Hence, A = 7 and B = 9

ix) On putting B = 7,
⇒ 7 + 1 = 8
Now A = 4,
then 4 + 7 = 11
Putting 1 at tens place and carry over 1,
we get 2 + 4 + 1 = 7
Hence, A = 4 and B = 7

x) Putting A = 8 and B = 1,
we get 8 + 1 = 9
Now again we add 2 + 8 = 10
Tens place digit is ‘0’ and carry over 1.
Now 1 + 6 + 1 = 8 = A
Hence, A = 8 and B = 1

Exercise 16.2

Q.1) If 21y5 is a multiple of 9, where y is a digit, what is the value of y
Sol.1) Since 21 5𝑦 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.
∴ 2 + 1 + 𝑦 + 5 = 8 + 𝑦
⇒ 8 + 𝑦 = 9
⇒ 𝑦 = 1

Q.2) 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 for the last problem. Why is this so?
Sol.2) Since 31 5𝑧 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.
∴ 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 9
⇒ 𝑧 = 0
⇒ If 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 18
⇒ 𝑧 = 9
Hence, 0 and 9 are two possible answers.

Q.3) 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.)
Sol.3) Since 24x is a multiple of 3.
Therefore, according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
∴ 2 + 4 + x = 6 + x
Since x is a digit.
⇒ 6 + x = 6 ⇒ x = 0
⇒ 6 + x = 9 ⇒ x = 3
⇒ 6 + x = 12 ⇒ x = 6
⇒ 6 + x = 15 ⇒ x = 9
Thus, x can have any of four different values.

Q.4) If 31𝑧5 is a multiple of 3, where z is a digit, what might be the values of 𝑧?
Sol.4) Since 31 5𝑧 is a multiple of 3.
Therefore, according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3. Since 𝑧 is a digit.
∴ 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 9 ⇒ 𝑧 = 0
If 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 12 ⇒ 𝑧 = 3
If 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 15 ⇒ 𝑧 = 6
If 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 18 ⇒ 𝑧 = 9
Hence, 0, 3, 6 and 9 are four possible answers.

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NCERT Solutions Class 8 Mathematics Chapter 16 Playing with Numbers

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