Question

Find the values of A and B $$\begin{array}{c}12A\\ +6AB\\ A09\end{array}$$

• A. A = 8, B = 1
• B. A = 1, B = 8
• C. A = 9, B = 3
• D. A = 3, B = 9

Answer 1

The correct option is A

A = 8, B = 1

We know that A + 2 is a two digit number with unit digit as 0. So, A = 8

A + B is a single digit number and is equal to 9.
A + B = 9
B = 9 - A
B = 9 - 8 = 1
Therefore, A = 8 and B = 1

$$\begin{array}{c}128\\ +681\\ 809\end{array}$$