Question

Sum of the digit of a two digit is 9 . when we interchange the digits , it is found that the resulting new number is greater than the original number by 63. find the two digit number.​

Answer 1

Let two digit number be '10x + y'

★  Sum of the digit of a two digit is 9 .

➠  x + y = 9 ... (1)

★  When we interchange the digits , it is found that the resulting new number is greater than the original number by 63 .

• Original number = 10x + y
• Reversed number = 10y + x = Number whose digits got interchanged

➠ 10y + x = ( 10x + y ) + 63

➠ 10y + x - 10x - y = 63

➠ 9y - 9x = 63

➠ 9 ( y - x ) = 9 ( 7 )

➠ y - x = 7

➠ y = 7 + x

Sub. this 'y' in (1) ,

➳ x + ( 7 + x ) = 9

➳ 2x = 9 - 7

➳ 2x = 2

➳ x = 1  $\orange{\bigstar}$

Sub. x value in (1) ,

➤ (1) + y = 9

➤ y = 9 - 1

➤ y = 8  $\pink{\bigstar}$

• Two digit number = 10x + y

➠ 10(1) + (8)

➠ 10 + 8

➠ 18  $\green{\bigstar}$

Answer 2

Answer:

• The two digit number is 18.

Step-by-step explanation:

Let one digit number be "R" and tenth digit number be "S".

→ So, two digit number be 10R + S.

According to first statement,

Sum of the digit of a two digit is 9.

→ R + S = 9 ...(i)

According to second statement,

When we interchange the digits , it is found that the resulting new number is greater than the original number by 63.

• Original number = 10R + S
• Reversed number = 10S + R

→ 10S + R = 10R + S + 63

→ 10S + R - 10R - S = 63

→ 9R - 9S = 63

→ R - S = 7 ....(ii)

From eqn.(1) & (2),

$\sf R + S = 9 \\ \sf R - S = 7 \\ - - - - - - - - \\ \sf 2R = 2$

→ So, R = 1 $\</strong><strong>p</strong><strong>i</strong><strong>n</strong><strong>k</strong><strong>\bigstar$

Put R = 1 in eqn. (1), We get ;

→ R + S = 9

→ 1 + S = 9

→ S = 9 - 1

→ S = 8 $\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>\bigstar$

Now,

Two digit number = 10R + S

→ 10(1) + 8

→ 10 + 8

→ 18 $\red\bigstar$