sum of two digit no is 9.no obtained by reversing the order of digit of given no exceed the given no by 9 find the given
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Let the tens place digit and ones place digits be a and b respectively
No. = 10a + b
According to question
a+b = 9 -------(1)
10a +b +9 = 10b +a
=> 9 = 10b+a-10a-b
=> 9b-9a = 9
=> b-a = 1
=> b= a+1 -------(2)
Substituting equation 2 in 1, we get
a+a+1 = 9
=> 2a = 8
=> a = 4
b= 5
Required no. = 45
No. = 10a + b
According to question
a+b = 9 -------(1)
10a +b +9 = 10b +a
=> 9 = 10b+a-10a-b
=> 9b-9a = 9
=> b-a = 1
=> b= a+1 -------(2)
Substituting equation 2 in 1, we get
a+a+1 = 9
=> 2a = 8
=> a = 4
b= 5
Required no. = 45
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