Question

tens digit of a two-digit number exceeds the units digit by 5. If the digits are reversed, the new
mber is less by 45. If the sum of their digits is 9, find the numbers.
please solve with steps.(50 points)​

Answer 1

AnsWers:-

Let the units digit be y,

and the tens digit be x.

Thus the number is 10x+y

according to the condition,

(10x+y)-(10y+x) =45

9(x - y)=45

x-y=5.......(1)

also x+y=9...(given)

Adding(1) and (2)

2x=14

x=7

Putting in (1)

y=7-5

y=2

Thus the number is 72.

Hope this helps.

Answer 2

$\huge\bold\red{HOLA!!}$

$<font color ="blue">$

Let the digit in the units place be x

Hence the digit in the tens place is (x+5)

The original number =10(x+5)+x=11x+50

Number formed by reversing the digits =10x+(x+5)=11x+5

Given that:- (11x+50)+(11x+5)=99

⇒22x+55=99

⇒22x=99−55=44

⇒x=2

∴ Original number =11x+50=11(2)+50=72

Hence the original number is 72.

Hence the correct answer is 72.