the digit at ones place of a 2-digit number is four times the digit at tens place. The number obtained by reversing digits exceeds the given number by 54. Find the given number.
Answers
answer:
28
Step-by-step explanation:
let ten's digit = x
one's digit = 4x
according to question:-
10x+ 4x = 10× 4x+x -54
14x = 41x-54
27x = 54
x = 2
number is - 10× x+4x
= 10×2+4×2
= 28
Step-by-step explanation:
Given :-
- The digit at ones place of a 2-digit number is four times the digit at tens place. The the given number obtained by reversing the digits exceeds the given number by 54.
To find :-
- Required numbers
Solution :-
Let the tens digit be x then ones digit be y
Original number = 10x + y
The digit at ones place of a 2-digit number is four times the digit at tens place.
→ y = 4x ---(i)
The the given number obtained by reversing the digits exceeds the given number by 54.
Reversed number = 10y + x
→ 10x + y + 54 = 10y + x
→ 10x - x + y - 10y = - 54
→ 9x - 9y = - 54
→ 9(x - y) = - 54
→ x - y = - 6 --(ii)
Put the value of y in equation (ii)
→ x - 4x = - 6
→ - 3x = - 6
→ x = 2
Substitute the value of x in equation (i)
→ y = 4x
→ y = 4 × 2
→ y = 8
Hence,
Tens digit = x = 2
Ones digit = y = 8
Therefore,
Original number = 10x + y = 28
Reversed number = 10y + x = 82