The sum of digits of a two digit number is 8. If the digits are reversed, the new number increases by 18. Find the number. Solve with statement.
Answers
Let the digits in tens place be x and the digits in units place be y.
Therefore, the original two digit number is 10x + y.
Given that Sum of digits of a two digit number is 8.
= > x + y = 8 ------- (1)
Given that The new number increases by 18, when the digits are reversed.
= > 10y + x = 10x + y + 18
= > 10y - y = 10x - x + 18
= > 9y = 9x + 18
= > 9y - 9x = 18
= > y - x = 2
= > x - y = -2 ----- (2)
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On solving (1) & (2), we get
= > x + y = 8
= > x - y = -2
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2x = 6
x = 3.
Substitute x = 3 in (1), we get
= > x + y = 8
= > 3 + y = 8
= > y = 8 - 3
= > y = 5.
So,
= > 10x + y
= > 10(3) + 5
= > 35.
Therefore, the original number = 35.
Hope this helps!
It is given that ,
Sum of digits of a two digit number = 8
Let unit place digit = x
ten's place digit = 8 - x
The number = 10( 8 - x ) + x
= 80 - 10x + x
= 80 - 9x ---( 1 )
If the digits are reversed the new so
formed = 10x + 8 - x
= 9x + 8 ----( 2 )
According to the problem given ,
( 2 ) = ( 1 ) + 18
9x + 8 = 80 - 9x + 18
9x + 9x = 98 - 8
18x = 90
x = 90/18
x = 5
THEREFORE ,
required number = 80 - 9x
= 80 - 9 × 5
= 80 - 45
= 35
I hope this helps you.
: )