tens digit of a no. of two digits is greater by 7 than thane units digit. sum of the digits is 1/9 of that no. find the no.
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6
let no is xy and ans is 81 is right ans
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neha606:
verify the sum of digit is 1/9
see solution
but the sum of digit is only9
x+y=9
the sum of digit is 1/9 th of no not equal to the no
Answered by
19
let the number yx
given tens digit is greater by 7
y= x+7
a two digit no can be expressed as : x+10y = x +10(x+7) = x+10x+70 =11x
given sum of the digits is 1/9 of that no
so 1/9of 11x+70 = x+x+7
1/9 X (11X+70)=2x+7
11x+70 = 9(2x+7)
18x+63 = 11x+70
7x=7
x = 1
y = x+7= 8
therefore no = 81
given tens digit is greater by 7
y= x+7
a two digit no can be expressed as : x+10y = x +10(x+7) = x+10x+70 =11x
given sum of the digits is 1/9 of that no
so 1/9of 11x+70 = x+x+7
1/9 X (11X+70)=2x+7
11x+70 = 9(2x+7)
18x+63 = 11x+70
7x=7
x = 1
y = x+7= 8
therefore no = 81
now verify the sum of digit is 1/9 of that no.
please
1/9 of 81=9
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