Question

Consider a two digit number such that when its digits are reversed the new number obtained is 6 more than three times the original number also one digit is five times that of the other find the original number​

Answer 1

Answer:

the anwer will be like this

Step-by-step explanation:

let the no. be 'x'

so ,

x*3+6 = 3x+6

as one digit is five times the orignal number , it will be like

6(3x+6)

=18x + 36

pls mark as brainliest i wrote it by myself

thank you so much kamya ;-)

Answer 2

Answer:

Let the digit in units place be x.

Then, digit in ten's place =3x

So, original number =10(3x)+x=31x

On interchanging the digits, new number =10x+3x=13x

According to the given condition,

31x+13x=88

⇒44x=88

⇒x=2

So, the original number is 31x or 31×2=62.

On the other hand, if we consider the digit in ten's place as x, then the digit in unit's place will be 3x.

So, the resulting number we get is 26.

Here, both answers are correct, as 26+62=88.