Question 309030: there is a two-digit figure in which one digit is exactly half the other. if the position of both fonts or digits is interchanged, the resultant figure when added with the original figure sum to 99. what is the original figure?
Found 2 solutions by mananth, mollukutti:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! let x be in the ten"s place
Let y be in the units place
x=2y
the number will be 10x+y
10x+y + 10y+X =99
11X+11Y= 99
plug the value of x in the above equation
22y +11 y = 99
33y=99
y= 3
x= 6
The number is 63
Answer by mollukutti(30)  (Show Source):
You can put this solution on YOUR website! Basic facts:
1. 2 digit figure say xy can also be written as 10x + y where y is the digit in the units place and x is the digit in the tens place
Eg., 56 = 5x10 + 6
As per the question, in the original number let the digit in the tens place be x
Therefore the other digit in units place is x/2
Therefore the original number : 10x + x/2
When the digits are reversed the digit in the tens place becomes x/2
and the digit in the units place becomes x
Hence the new reversed number is 10x/2 + x
As per the question when both the numbers are added the sum is 99.
Therefore the equation is:
(10x + x/2) + (10x/2 + x)=99
or, 10x + x/2 + 5x + x = 99
or, (10x x 2 + x + 5x x 2 + x x 2) / 2 = 99
or, (20x + x + 10x + 2x)= 99 x 2
or, 33x = 198
or, x = 198/33
or x = 6
hence one digit is 6 and the other digit is 3
Therefore the original number is 63
NOTE:
As this is a tricky qs please note that if the digit in tens place is taken as x/2 and that in the units place is taken as x, the answer will be 36.
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