A certain number between 10 and 100 is 8 times the sum of its digit , and if 45 be subtracted from it the digit will be reversed . find the number
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let the ten's digit be x and unit digit be y.
ATQ,
=> 8(x + y) = 10x + y
=> 8x + 8y = 10x + y
=> 2x - 7y = 0 ------(1)
Also
=> 10x + y - 45 = 10y + x
=> 9x - 9y = 45
=> (x - y = 5)------(2)
Subtract (2) from (1)
=> 2x - 7y - (2x - 2y) = -10
=> -5y = -10
=> y = 2
Thus x - 2 = 5
=> x = 7
Thus original number = 70 + 2 = 72
Hope this helps you again
I answered you in inbox yet I'll answer again
let the ten's digit be x and unit digit be y.
ATQ,
=> 8(x + y) = 10x + y
=> 8x + 8y = 10x + y
=> 2x - 7y = 0 ------(1)
Also
=> 10x + y - 45 = 10y + x
=> 9x - 9y = 45
=> (x - y = 5)------(2)
Subtract (2) from (1)
=> 2x - 7y - (2x - 2y) = -10
=> -5y = -10
=> y = 2
Thus x - 2 = 5
=> x = 7
Thus original number = 70 + 2 = 72
Hope this helps you again
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