Question

Digit in ten's place is four less than twice
The digit in the unit's place. sum of the original number and interchanged number is 55. find the number

Answer 1

Answer:

23

Step-by-step explanation:

Let digit in ten's place be x .

Let digit in unit's place be y .

Number = 10 x + y

Sum of original and interchanged digit = 55

10 x + y + 10 y + x = 55

⇒ 11 x + 11 y = 55

⇒ x + y = 5 ......... ( 1 )

Digit in tens place is 4 less than twice the unit's place .

x + 4 = 2 y

⇒ x = 2 y - 4 ...... ( 2)

Put this in ( 1 ) :

2 y - 4 + y = 5

⇒ 3 y = 9

⇒ y = 9/3

⇒ y = 3

x = 5 - y = 5 - 3 = 2

Hence the number is 10 x + y = 20 + 3 = 23.

The answer is 23 .

Answer 2

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$


$\bf{QUESTION}$
Digit in ten's place is four less than twice
The digit in the unit's place. sum of the original number and interchanged number is 55. find the number


STEP BY STEP EXPLANATION

Let the digit at tens place be x

and digit at unit place be y.

Now

digit=10x+y

interchanged digits

=> 10y+x

GIVEN= the sum of digit and interchanged digits=55

CASE1
________

=> 10x+y+10y+x=55

=> 11x+11y=55

divide both side by 11

$\sf{x + y = 55.........(1)}$


NOW


CASE 2
________



=> x+4=2y

=> x-2y=-4........... (2)

from equation 1 and 2

x+y=5
x-2y=-4
_______

3y=9


$\bf{y = \frac{9}{3} }$


$\bf{y = 3}$

put the value of y in equation 1

x+y=5

x+3=5


x=5-3

$\bf{x=2}$



Hence digit= 10x+y

=> 10(2)+3

$\huge \bf{ \implies2 3.}$


$\huge \bf \pink{ \boxed{ \boxed{THANKS}}}$