Exercise 16.1
Find the values of the letters in each of the following and give reasons for the steps involved.
Question 1.
Solution :
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
Question 2.
Solution :
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
Question 3.
Solution :
On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get, A x A = 6 x 6 = 36 in which ones place is 6.
∴ A = 6
Hence A = 6
Question 4.
Solution :
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
Question 5.
Solution :
Here on putting B = 0,
we get 0 × 3 = 0.
And A = 5, then 5 × 3 = 15
⇒ A = 5 and C = 1
Hence A = 5, B = 0 and C = 1
Question 6.
Solution :
On putting B = 0, we get 0 , 5 = 0 and A = 5, then 5×5 = 25
⇒ A = 5, C = 2
Hence A = 5, B = 0 and C = 2
Question 7.
Solution :
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
Question 8.
Solution :
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
Question 9.
Solution :
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
Question 10.
Solution :
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
NCERT Solutions for Class 8 Maths Exercise 16.1
Question 1.
If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Solution :
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
Since 21y5 is a multiple of 9.
Question 2.
If 31z
Solution :
Since 31z5 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.
Question 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.)
Solution :
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 ⇒
⇒ 6 + x = 9 ⇒
⇒ 6 + x = 12 ⇒
⇒ 6 + x = 15 ⇒
Thus, x can have any of four different values.
Question 4.
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution :
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 = 15 ⇒ z = 9
Hence 0, 3, 6 and 9 are four possible answers.