Question

in a two digit number the digit in tens place is three times the digit in the units place . if the digits are reversed,the number decreases by 36. find the number.

Answer 1

Let the digit at the unit place be y and the digit on tens place be z
The digit =10z+y
As per the condition
z=3y
I.e 3y-z=0 (1)
from second condition
10y+z=10z+y-36
10y+z-10z-y = -36
9y-9z=-36
y-z=-4 (2)
equation (1)-(2)
3y-z=0
-
y-z=-4
2y=4
y =2
Put y=2 in (1)
3(2)-z=0
6=z
Therefore digit is 10(6)+2
62




hope it will be useful......

Answer 2

Answer:

The Original Number is 62.

Step-by-step explanation:

Given :

The digit in the tens place of two digit number is = three times that in the unit place.

When the digits are reversed, the new number will be = 36 less than the original number.

To find :

The original number

Solution :

Original Number -

Consider the -

• Units Place as y
• Tens Place as 10(3y)

Number Formed = 10(3y) + y

⇒ 30y + y

⇒ 31y ..... [Original Number]

$\rule{300}{1.5}$

Number with Reversed Digits -

Consider the -

• Units Place = 3y
• Tens Place = 10(y)

Number Formed = 10(y) + 3y

⇒ 10y + 3y

⇒ 13y ..... [Number with Reversed Digits]

$\rule{300}{1.5}$

As given in the Question -

When the digits are reversed, the new number will be = 36 less than the original number.

⇒ 13y = 31y - 36

⇒ 13y - 31y = - 36

⇒ - 18y = - 36

⇒ 18y = 36 .... (Negative signs to be cancelled)

⇒ y = 36/18

⇒ y = 2

$\rule{300}{1.5}$

As we got the value of y, find the original number by placing the y's value.

⇒ 31y

⇒ 31(2)

⇒ 62

Original number = 62

 The Original Number is 62.