the sum of the digits of a two digit number is 9 . also nine times this number is twice the number obtained by reversing the order of the digits . find the number . { plz explain clearly and fast i need it now !!! }
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Here is your solution
Given :-
The sum of the digits of a two digit number is 9
Let
The digits be x and y, and the number be 10x + y.
I. e
x + y = 9 ------ (1)
Now
A/q
9×(10x + y) = 2×(10y + x)
90x + 9y = 20y + 2x
88x = 11y
y = 8x ---- (2)
putting the value of y in equation (1.) we get,
x + 8x = 9
9x = 9.
x = 1
y = 8×1 = 8.
Hence
The number is 10x +y = 10×1 + 8 = 18.
Hope it helps you
Given :-
The sum of the digits of a two digit number is 9
Let
The digits be x and y, and the number be 10x + y.
I. e
x + y = 9 ------ (1)
Now
A/q
9×(10x + y) = 2×(10y + x)
90x + 9y = 20y + 2x
88x = 11y
y = 8x ---- (2)
putting the value of y in equation (1.) we get,
x + 8x = 9
9x = 9.
x = 1
y = 8×1 = 8.
Hence
The number is 10x +y = 10×1 + 8 = 18.
Hope it helps you
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