A two digit number is three times the sum of its digits. if 45 is added to it, the digits are reversed. the number is
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Let the ten's digit be x. The one's digit is y.
Given that a two digit number is 3 times the sum of its digits.
10x+y = 3(x + y)
10x + y = 3x + 3y
7x = 2y --- (1)
Given that if 45 is added to it, the digits are reversed.
10x + y = 10y - x - 45
9x = 9y - 45
x = y - 5 --- (2)
Substitute (2) in (1), we get
7(y-5) = 2y
7y - 35 = 2y
y = 7.
Substitute y = 7 in (2), we get
x = 7 - 5 =2.
The values of x and y are 72.
The number is 27.
Hope this helps!
Given that a two digit number is 3 times the sum of its digits.
10x+y = 3(x + y)
10x + y = 3x + 3y
7x = 2y --- (1)
Given that if 45 is added to it, the digits are reversed.
10x + y = 10y - x - 45
9x = 9y - 45
x = y - 5 --- (2)
Substitute (2) in (1), we get
7(y-5) = 2y
7y - 35 = 2y
y = 7.
Substitute y = 7 in (2), we get
x = 7 - 5 =2.
The values of x and y are 72.
The number is 27.
Hope this helps!
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67
I hope this will help you
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