Question

A digit in tens place of a two digit number
is twice the digit in place. One
fourth of the sum of the original number
obtained by reversing the order of the
digits is 33 Find the original number​

Answer 1

Answer:

Let the digit in the unit’s place be  x.  

The digit in the ten’s place is twice the digit in the unit’s place.

⇒  The digit in the ten’s place is  2x.  

⇒  The number is  2x×10+x=21x.  

⇒  The number formed by reversing the digits is  x×10+2x=12x.  

⇒  The sum of the number and the number formed by reversing the digits is  21x+12x=33x.  

It is given that the sum of the number and the number formed by reversing the digits is  66.  

⇒33x=66⇒x=2⇒2x=4.  

⇒  The number is  42.  

if u find it helpful pls mark as brainliest

Answer 2

Answer:

Let x be the number at the unit's place and y be the number at the ten's place.

Two digit number = 10y + x

 

According to the conditions,

y = 2x  

2x - y = 0 ...(i)

 

 

10x + y + 10y + x = 66

x + y =11 ....(ii)

 

Solve (i) and  (ii), we get

x = 2 and y = 4

Two digit number = 10y + x = 42

Hope it helps u...........

Step-by-step explanation: