Question

the sum of the digits of a two digit number is 10, when the number is reversed,the number increases by 72.find the number​

Answer 1

Answer:

Here is your answer

Step-by-step explanation:

Let the number be xy

So 10x + y = 10 ..... (i)

When the number is reversed the new number is yx

So (10y + 10 ) - ( 10x +y ) = 72 ....(ii)

From Eqs. (i) & (ii)

x = 1 and y =9

∴number = 19

Answer 2

SOLUTION :

• Let's Ten's digit = a

• One's digit = b

Sum of the digit

$\sf a + b = 10 \\ \\ \sf a = 10 - b \: .....(eq \: 1)$

After reversed the number increased by 72

$\sf 10b + a = 10a + b + 72 \\ \\ \sf 10b - b + a - 10a = 72 \\ \\ - \sf9a + 9b = 72 \\ \\ \boxed{ \sf9a - 9b = - 72 \: ............( eq \: 2)}$

Put the value of a in ( eq 2 )

$\sf 9a - 9b = - 72 \\ \\ \sf9(10 - b) - 9b = - 72 \\ \\ \sf 90 - 9b - 9b = - 72 \\ \\ \sf - 18b = - 162 \\ \\ \sf b = \frac{ - 162}{ - 18} \\ \\ \boxed{ \sf b = 9}$

Put value of b in ( eq 1 )

$\sf a = 10 - b \\ \\ \sf a = 10 - 9 \\ \\ \boxed{ \sf a = 1}$

Number formed by these digit is 19