Question

In a two digit number, the unit’s digit is 1 more than twice the ten’s digit. The number formed by reversing the digits is 36 more than the original number. Find the numbers.​

Answer 1

Answer:

@sukumarsawant

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Step-by-step explanation:

Let the tens place digit be a

Unit place digit = 2a+1

No. = 10(a) + (2a+1)

= 10a+2a+1

= 12a + 1

Reversed no. = 10(2a+1)+ a

= 20a+10 + a

= 21a + 10

According to question

Original no. +36 = Reversed no.

=> 12a+1 + 36 = 21a + 10

=> 21a-12a = 37 - 10

=> 9a = 27

=> a = 3

Unit place digit = 3

Tens place digit = 7

Required no. = 37

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Answer 2

Solution :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

In a two digit number, the unit's digit is 1 more than twice the ten's digit. The number formed by reversing the digits is 36 more than the original number.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The number.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Let the unit's digit be 2r + 1

Let the ten's digit be r

So;

$\bf{\underline{\sf{The\:Original\:number\:=\pink{10(r)+(2r+1)}}}}\\\bf{\underline{\sf{The\:Reversed\:number\:=\pink{10(2r+1)+(r)}}}}$

A/q

$\mapsto\sf{10(r)+(2r+1)+36=10(2r+1)+(r)}\\\\\mapsto\sf{10r+2r+1+36=20r+10+r}\\\\\mapsto\sf{12r+1+36=21r+10}\\\\\mapsto\sf{12r+37=21r+10}\\\\\mapsto\sf{12r-21r=10-37}\\\\\mapsto\sf{-9r=-27}\\\\\mapsto\sf{r=\cancel{\dfrac{-27}{-9} }}\\\\\mapsto\sf{\orange{r=3}}$

Thus;

The original number = 10(r) + (2r+1)

The original number = 10(3) + [2(3)+1]

The original number = 30 + (6+1)

The original number = 30 + 7

The original number = 37