Question

Q(2.) The sum of the digits of a 2-digit number is 7. If the digits are reversed, the new number increased by 3
less than 4 times the original number. Find the original number.
Q(3.) The sum of a number of 2 digits and of the number formed by reversing the digits is 110, and the difference
of the digits is 6. Find the number.
PLEASE FAST.
I WILL MARK THE ANSWER AS BRAINLIST.​

Answer 1

Step-by-step explanation:

Q2:- Let the digits of the number be x and y respectively .

Then we know the number is of the form 10x + y

Reversing the digits of the number makes the number of the form 10y + x

Also x + y = 7

So we can say x = 7 - y

And we know :- 4(10x + y) - 3 = 10y + x as given

So:- 40x + 4y - 3 = 10y + x

=> 40(7 - y) + 4y - 3 = 10y + (7 - y)

=> 280 - 40y + 4y  - 3 = 7 + 9y

=> 270 = 45y

=> y = 6

Hence From x = 7 - y , we get x = 7 - 6 = 1

So the original number is 16 and the new number is 61 .

Q3:- Let the digits of the number be x and y respectively .

Then the number is of the form 10x + y

Reversing the number gives 10y + x .

First we assume if x > y .

Then x - y = 6      

So I can say x = 6 + y

And we know:- (10x + y) + (10y + x) = 110

=> 10(6 + y) + y + 10y + (6 + y) = 110

=> 60 + 10y + y + 10y + 6 + y = 110

=> 22y + 66 = 110

=> 22y = 176

=> y = 8

Hence from x = 6 + y , we get x = 6 + 8 = 14

But that is not possible since x cannot be 2-digit number .

Since we first assumed x>y , our assumption was wrong .

So y>x , and y - x = 6

This gives x = 2

So the original number is 82 and reversing the number gives 28 .