Find out the values of the letters and give reasons. 3 1 Q + 1 Q 3 5 0 1

Given, 3 1 Q + 1 Q 3 5 0 1 Here, Q + 3 > 3 , since Q is added to 3 . Also, Q + 3 is a two-digit number whose units digit must be 1 . Q + 3 ≠ 21 and Q + 3 ≠ 31 because in this case, Q will be a one-digit number. ∴ Q + 3 = 11 ⇒ Q = 8 Substituting the value of Q , we get carry → 1 1 3 1 8 + 1 8 3 5 0 1 Therefore, the value of Q is 8 .

Find out the values of the letters and give reasons. 2 A B + A B 1 B 1 8

Given, 2 A B + A B 1 B 1 8 Here, B + 1 = 8 , since B is added to 1 . B + 1 is one-digit number whose units digit must be 8 . Since, B + 1 = 8 ⇒ B = 8 - 1 = 7 Substituting the value of B , we get 2 A 7 + A 7 1 7 1 8 ⇒ A + 7 is two-digit number whose units digit must be 1 . Since, A + 7 = 11 ⇒ A = 11 - 7 = 4 Substituting the value of A and B , we get carry → 1 2 4 7 + 4 7 1 7 1 8 Hence, the value of A = 4 and B = 7 .

Find out the values of the letters and give reasons. 8 A 5 + 9 4 A 1 A 3 3

Given, 8 A 5 + 9 4 A 1 A 3 3 Here, 5 + A > 3 , Since A is added to 5 . Also, 5 + A is two-digit number whose units digit must be 3 . 5 + A ≠ 23 and 5 + A ≠ 33 . Because in this case, A is one-digit number. ∴ 5 + A = 13 ⇒ A = 13 - 5 = 8 Substituting the value of A , we get carry → 1 1 1 8 8 5 + 9 4 8 1 8 3 3 Hence, the value of A is 8 .

Find out the values of the letters and give reasons. A B × 5 C A B

Given, A B × 5 C A B The multiplication of 5 and B gives a number whose ones digit is B again. Hence, B must be 0 or 5 . On putting B = 5 , we get 5 × 5 = 25 , Here 2 + A × 5 does not gives A . So, B = 5 is not possible. On putting B = 0 , we get 0 × 5 = 0 , The multiplication of 5 and A gives a number whose ten's digit is A again. i.e., A × 5 = A So, A × 5 = A ⇒ 5 × 5 = 25 and 2 will be a carry for the next step. ∴ A = 5 Substituting the value of B and A , we get. 5 0 × 5 2 5 0 Therefore, the value of A = 5 , B = 0 , C = 2 .

Find out the values of the letters and give reasons. A B 3 + 2 B 1 B A A

Given, A B 3 + 2 B 1 B A A Add the ones column: 3 + 1 = 4 Therefore, A = 4 On putting B = 7 , then we get 7 + 7 = 14 Putting 4 at tens place and carry over 1 in the hundred place. Now, 1 + 4 + 2 = 7 Substituting the value of A and B , we get carry → 1 4 7 3 + 2 7 1 7 4 4 Hence, A = 4 , B = 7 .