the digits of a two digit number differ by 5 if the digit are intercha

1.	the digits of a two digit number differ by 5 if the digit are interchanged and the resulting number is added to the orginal number we get 121 find the number
Answer» AnswEr:- Two digits number = 38 $\rule{200}{2}$ Solution:- Let the digit at unit's place be y & digit at ten's place be z where y > z ∴ Two digit number = y + 10z CASE 1:- ⇒ y - z = 5 ⇒ y = z + 5 [Eq.i] Case 2:- ★ Interchanged number:- ⇒ New number = z + 10y According to QUESTION:- ⇒ y + 10z + z + 10y = 121 ⇒ 11y + 11z = 121 ⇒ 11(y + z) = 121 ⇒ y + z = 121/11 ⇒ y + z = 11 ⇒ y = 11 - z [Eq.ii] Comparing both equations - ⇒ z + 5 = 11 - z ⇒ z + z = 11 - 5 ⇒ 2z = 6 ⇒ z = 6/2 ⇒ z = 3 So, digit at ten's place = z = 3 Putting this VALUE in (Eq.i) ⇒ y = 3 + 5 ⇒ y = 8 So, digit at unit's place = y = 8 ★ Two digit number:- ⇒ Number = 8 + 10(3) ⇒ Number = 8 + 30 ⇒ Number = 38 Therefore, $\therefore{\underline{\boxed{\textsf{Original two-digit number = {\textbf{38 }}}}}}$

the digits of a two digit number differ by 5 if the digit are interchanged and the resulting number is added to the orginal number we get 121 find the number

Answer»

AnswEr:-

Two digits number = 38

$\rule{200}{2}$

Solution:-

Let the digit at unit's place be y & digit at ten's place be z where y > z

∴ Two digit number = y + 10z

CASE 1:-

⇒ y - z = 5

⇒ y = z + 5 [Eq.i]

Case 2:-

★ Interchanged number:-

⇒ New number = z + 10y

According to QUESTION:-

⇒ y + 10z + z + 10y = 121

⇒ 11y + 11z = 121

⇒ 11(y + z) = 121

⇒ y + z = 121/11

⇒ y + z = 11

⇒ y = 11 - z [Eq.ii]

Comparing both equations -

⇒ z + 5 = 11 - z

⇒ z + z = 11 - 5

⇒ 2z = 6

⇒ z = 6/2

⇒ z = 3

So, digit at ten's place = z = 3

Putting this VALUE in (Eq.i)

⇒ y = 3 + 5

⇒ y = 8

So, digit at unit's place = y = 8

★ Two digit number:-

⇒ Number = 8 + 10(3)

⇒ Number = 8 + 30

⇒ Number = 38

Therefore,

$\therefore{\underline{\boxed{\textsf{Original two-digit number = {\textbf{38 }}}}}}$