Question

can someone please solve this question for me ??

Answer 1

let the digit at tens place of the number be x.
so the digit at ones place is 4x.
so, number=10x+4x=14x.
when the digits are reversed,
digit at the tens place =4x
and digit at the ones place=x.
Then number=40x+x=41x.
by the condition,
41x-14x=54
=>x=2
so the original number =14×2 =28.
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Answer 2

HEY BUDDY HERE IS YOUR ANSWER.....❣️
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Let the tens place digit number be x.
Then,ones place digit number=4x

The 2-digit number=10×x+1×4x
=10x+4x
Number obtained by reversing the digits=4x×10+x×1
=40x+x

As per the condition,
(40x+x)-(10x+4x)=54
=>. 41x-14x=54
=>. 27x=54
=>. x=54/27
=>. x=2

Therefore, tens place digit number=x=2
and ones place digit number=4x
=4×2
=8

Hence,the required number=10×2+1×8
=20+8
=28.
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