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Playing with Numbers — MATHS STD 8 — Question

Gujarat BoardEnglish MediumSTD 8MATHSPlaying with Numbers3 Marks
Question
Solve Cryptarithms. $\ \ \ \ \ \ \ \ \ \ \ 3\ \ \ \ 7\\\ \ \ \ \ \underline{+\text{A}\ \ \ \ \text{B}\ }\\\ \ \ \ \ {\ \ \ \ \ \ 9\ \ \ \ \text{A}\ }$
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Answer

Two possible values of $A$ are:
$i.$ If $7+\text{B}\leq9,3+\text{A}=9$
$\therefore\text{A}=6$
But if $A = 6, 7 + B$ must be larger than $9.$
Hence, it is impossible.
$ii.$ If $7+\text{B}\geq9$
$\therefore1+3+\text{A}=9$
$\Rightarrow\text{A}=5$
If $A = 5$ and $7 + B = 5,$
$B$ must be $8$
$\therefore A = 5, B = 8$

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