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
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/