sum of a number of two digit and the number format by reversing a digit is 110 and difference of the digit is 6 find the number
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10x+y=(10y+x)=110 and x-y=6. then 11x+11y=110 and x-y=6. x+y=10 and x-y=6. then by solving by elimination method we will get x=8 and y=2
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Here is your solution
Let,
The ten's digit of original number be x
The unit's digit of the original number be y
The original number= 10x + y
The Number obtained on reversing the digits = 10y + x
A/q
Add original numbers and reverse number.
So
=>( 10x + y )+( 10y + x )=110 .
=>11x + 11y = 110
=>11( x + y ) = 110
=>x + y = 110/11
x + y = 10...............(i)
x - y = 6 ..................(ii) (given)
On adding equation (i) and (ii), we get
x + y + x - y= 10 +6
2x = 16
X=8
On putting the value of x in equation (i), we get
x + y = 10
8 + y = 10 .
Y = 10 - 8
y= 2
original number = 10x + y .
=>10 × 8 + 2 .
=>80 + 2 .
=>82
Hence,
The required number is 82 .
Hope it helps you
Let,
The ten's digit of original number be x
The unit's digit of the original number be y
The original number= 10x + y
The Number obtained on reversing the digits = 10y + x
A/q
Add original numbers and reverse number.
So
=>( 10x + y )+( 10y + x )=110 .
=>11x + 11y = 110
=>11( x + y ) = 110
=>x + y = 110/11
x + y = 10...............(i)
x - y = 6 ..................(ii) (given)
On adding equation (i) and (ii), we get
x + y + x - y= 10 +6
2x = 16
X=8
On putting the value of x in equation (i), we get
x + y = 10
8 + y = 10 .
Y = 10 - 8
y= 2
original number = 10x + y .
=>10 × 8 + 2 .
=>80 + 2 .
=>82
Hence,
The required number is 82 .
Hope it helps you
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