The sum of the digits of a two digit number is 12. If the digits are reversed, new number is 12 less than twice the original number. Find the original number
Answers
Step-by-step explanation:
Given :-
The sum of the digits of a two digit number is 12. If the digits are reversed, new number is 12 less than twice the original number.
To find :-
Find the original number ?
Solution :-
Let the digit at tens place of a two digit number be X
The place value of X = 10×X = 10X
Let the digit at ones place be Y
The place value of Y = 1×Y = Y
The original number = 10X+Y
The number formed by reversing the digits
= 10Y+X
Given that
The sum of the digits of a two digit number = 12.
=> X+Y = 12 ------------(1)
or
=> X = 12-Y ------------(2)
And
The digits are reversed, new number is 12 less than twice the original number.
=> 10Y+X = 2(10X+Y)-12
=> 10Y+X = 20X+2Y-12
=> 10Y+X -20X-2Y = -12
=> (10Y-2Y)+(X-20X) = -12
=> 8Y-19X = -12
=> 8Y-19(12-Y) = -12
=> 8Y-228+19Y = -12
=> 27Y = -12+228
=> 27Y = 216
=> Y = 216/27
=> Y = 8
The digit at Ones place = 8
On Substituting the value of Y in (2)
=> X = 12-8
=> X = 4
The digit at tens place = 4
The number = 48
Answer:-
The original number for the given problem is 48
Check:-
The original number = 48
The sum of the digits = 4+8 = 12
The new number formed by reversing the digits
= 84
= 96-12
= 2(48)-12
the digits are reversed, new number is 12 less than twice the original number.
Verified the given relations in the given problem.
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