a number consisting of two digits is equal to 7 times the sum of its digits.When 27 is subtracted from the number, the digita interchange their places. find the number
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Here is your answer,
As the number is a two digit , then let x be tens's place and y be one's place.
Therefore the two digit number would form as 10 x +y
according to the question,
a number consisting of two digits is 7 times the sum of its digits.
=> 10 x + y = 7( x+y)
=> 3x-6y=0
=> x = 2y ............(1)
when 27 is subtracted from the number the digits are reversed.
After reversing the digits, the number = 10 y +x
=> (10 x+ y )- 27 = 10y + x
=> 9 x- 9y = 27
=>x -y = 3 ...............(2)
Substituting the value of x from eq (1) to eq(2) , we get
=> 2y-y = 3
=> y = 3
x= 2y
=> x = 2(3) = 6
Therefore the number is = 10 (6) + 3 = 63
Answer : the required two digit number is 63
Hope it helps you!
Here is your answer,
As the number is a two digit , then let x be tens's place and y be one's place.
Therefore the two digit number would form as 10 x +y
according to the question,
a number consisting of two digits is 7 times the sum of its digits.
=> 10 x + y = 7( x+y)
=> 3x-6y=0
=> x = 2y ............(1)
when 27 is subtracted from the number the digits are reversed.
After reversing the digits, the number = 10 y +x
=> (10 x+ y )- 27 = 10y + x
=> 9 x- 9y = 27
=>x -y = 3 ...............(2)
Substituting the value of x from eq (1) to eq(2) , we get
=> 2y-y = 3
=> y = 3
x= 2y
=> x = 2(3) = 6
Therefore the number is = 10 (6) + 3 = 63
Answer : the required two digit number is 63
Hope it helps you!
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