Question

The sum of the digits of a two digit number is 15. The number obtained by interchanging its digits exceeds the given number by 9. Find the original number.

Answer 1

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Let the number be xy.

Given,the sum of the digits of a two digit number is 15.

x+y = 15 ...(1)

Also, if the digits are interchanged the result exceeds the original number by 9.

10y+x=10x+y+9

10y+x-10x-y=9

9y-9x=9

y-x=1 ...(2)

(1)+(2) => 2y = 16

y = 16/2 = 8

(1) => x+8=15

x=15-8=7

Therefore the given number is 78

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Answer 2

Answer:

78

Step-by-step explanation:

let the digit at tens place be x and the digit at once place be y

sum of the digits(x+y) = 15

x=15-y (statement 1)

let the original number be = 10x + y (for example 23 = 2*10 + 3)

let the number after reversing the digits be = 10y + x (for example 32 = 3*10 + 2)

difference between new number and old number is 9

new number-9 = old number

(10y+x)-9 = 10x + y

10y+x-9= 10x+y

-9= 10x+y-10y-x

-9= 9x-9y

-9=9(x-y)

-9/9=x-y

x-y = (-1)

x = y-1 (statement 2)

from statement 1and 2 we get

y-1 = 15-y

y+y = 15+1

2y= 16

y= 16/2

y= 8

from statement 1

x + y= 15

x + 8 = 15

x = 15-8

x =7

the number is = 10x+y = 10*7+8 = 70+8 = 78