Since 31z5 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 ⇒Read more
Since 31z5 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.
Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video
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 and ⇒6+x=9 ⇒ x=3 ⇒ 6+x =12 ⇒x=6 and 6+x=15 ⇒x=9 Thus, x can have any of four different values. Class 8 Maths Chapter 11Read more
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 and ⇒6+x=9 ⇒ x=3
⇒ 6+x =12 ⇒x=6 and 6+x=15 ⇒x=9
Thus, x can have any of four different values.
Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video
Since is a multiple of 9. 315 z 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 Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video for more answers vist to: https://www.tiwariacademyRead more
Since is a multiple of 9. 315 z
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
Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video
Since 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. ∴ 2+1+y+5=8+y ⇒8+y=9 ⇒y=1 Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maRead more
Since 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.
∴ 2+1+y+5=8+y ⇒8+y=9 ⇒y=1
Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video
On putting B = 0, we get 0 x 5 = 0 and A = 5, then 5 x 5 = 25 ⇒ A = 5, C = 2 Hence, A = 5, B = 0 and C = 2 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
On putting B = 0, we get
0 x 5 = 0 and A = 5, then 5 x 5 = 25
⇒ A = 5, C = 2
Hence, A = 5, B = 0 and C = 2
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
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 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/Read more
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
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
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 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
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
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
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 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
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
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
Here product of B and 6 must be same as ones place digit as B. 6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24 On putting B = 4, we get the ones digit 4 and remaining two B’s value should be 44. ⇒ For 6 x 7 = 42 + 2 = 44 Hence, A = 7 and B = 4 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video fRead more
Here product of B and 6 must be same as ones place digit as B.
6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24
On putting B = 4, we get the ones digit 4 and remaining two B’s value should be 44.
⇒ For 6 x 7 = 42 + 2 = 44
Hence, A = 7 and B = 4
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
Here on putting B = 0, we get 0 x 3 = 0. And A = 5, then 5 x 3 = 15 ⇒ A = 5 and C = 1 Hence, A = 5, B = 0 and C = 1 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
Here on putting B = 0, we get 0 x 3 = 0.
And A = 5, then 5 x 3 = 15
⇒ A = 5 and C = 1
Hence, A = 5, B = 0 and C = 1
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
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. 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 ⇒Read more
Since 31z5 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.
Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
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).
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 and ⇒6+x=9 ⇒ x=3 ⇒ 6+x =12 ⇒x=6 and 6+x=15 ⇒x=9 Thus, x can have any of four different values. Class 8 Maths Chapter 11Read more
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 and ⇒6+x=9 ⇒ x=3
⇒ 6+x =12 ⇒x=6 and 6+x=15 ⇒x=9
Thus, x can have any of four different values.
Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
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?
Since is a multiple of 9. 315 z 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 Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video for more answers vist to: https://www.tiwariacademyRead more
Since is a multiple of 9. 315 z
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
Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Since 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. ∴ 2+1+y+5=8+y ⇒8+y=9 ⇒y=1 Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maRead more
Since 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.
∴ 2+1+y+5=8+y ⇒8+y=9 ⇒y=1
Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
Find the values of the letters in the following and give reasons for the steps involved.
On putting B = 0, we get 0 x 5 = 0 and A = 5, then 5 x 5 = 25 ⇒ A = 5, C = 2 Hence, A = 5, B = 0 and C = 2 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
On putting B = 0, we get
0 x 5 = 0 and A = 5, then 5 x 5 = 25
⇒ A = 5, C = 2
Hence, A = 5, B = 0 and C = 2
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
Find the values of the letters in the following and give reasons for the steps involved.
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 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/Read more
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
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
Find the values of the letters in the following and give reasons for the steps involved.
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 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
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
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
Find the values of the letters in the following and give reasons for the steps involved.
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 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
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
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
Find the values of the letters in the following and give reasons for the steps involved.
Here product of B and 6 must be same as ones place digit as B. 6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24 On putting B = 4, we get the ones digit 4 and remaining two B’s value should be 44. ⇒ For 6 x 7 = 42 + 2 = 44 Hence, A = 7 and B = 4 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video fRead more
Here product of B and 6 must be same as ones place digit as B.
6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24
On putting B = 4, we get the ones digit 4 and remaining two B’s value should be 44.
⇒ For 6 x 7 = 42 + 2 = 44
Hence, A = 7 and B = 4
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
Find the values of the letters in the following and give reasons for the steps involved.
Here on putting B = 0, we get 0 x 3 = 0. And A = 5, then 5 x 3 = 15 ⇒ A = 5 and C = 1 Hence, A = 5, B = 0 and C = 1 Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/
Here on putting B = 0, we get 0 x 3 = 0.
And A = 5, then 5 x 3 = 15
⇒ A = 5 and C = 1
Hence, A = 5, B = 0 and C = 1
Class 8 Maths Chapter 11 Exercise 16.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/