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Answer :
Let unit place digit be x.
And the ten’s place digit be y= 3x
Original number = x + 3x × 10 = x + 30x = 31x
Condition I: 10x + 3x = 13x [interchanging the digits]
Condition II: New number + original number = 88
13x + 31x = 88
\(\Rightarrow\) 44x = 88
\(\Rightarrow\) x = 88 ÷ 44 [Transposing 44 to RHS]
\(\Rightarrow\) x = 2
Thus, we have,
The original number = 31x = 31 × 2 = 62
Hence the required number = 62