I go to a bank with a cheque written x rs and y paise.the teller by mistake gives paise column as rs and rs column as paise.i spend 5 paise out of the money recd and while checking iam surprised to have double the amount written in the check.though i found the ans by trial and error metod how to find using algebrical method
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Let x and y be positive integers.
Amount on the cheque A = Rs x and y paise. = 100 x + y paise
as paise are less than 100, 1<= y <= 99
Actually given amount B = Rs y and x paise. = 100 y + x paise
1 <= x <= 99
Amount Remaining after 5 paise are spent : B - 5 = 100 y + x - 5 paise
B - 5 = 2 A
100 y + x - 5 = 2 (100 x + y)
98 y - 5 = 199 x
=> 98 y = 199 x + 5
98 y = 196 x + (3 x + 5 )
98 (y - 2x) = 3 x + 5
Since x and y are integers, y-2x = integer and is > 0 as RHS is positive.
3 x + 5 is a multiple of 98.
=> x = (98-5)/3 = 31 one solution.
=> x = (2*98-5)/3 ≠ integer
=> x = (3*98-5)/3 ≠ integer
=> x = (4*98-5)/3 > 100 so not a solution.
Hence, x = Rs 31 and y - 2x = 1 => y = 63 paise.
Amount on the cheque A = Rs x and y paise. = 100 x + y paise
as paise are less than 100, 1<= y <= 99
Actually given amount B = Rs y and x paise. = 100 y + x paise
1 <= x <= 99
Amount Remaining after 5 paise are spent : B - 5 = 100 y + x - 5 paise
B - 5 = 2 A
100 y + x - 5 = 2 (100 x + y)
98 y - 5 = 199 x
=> 98 y = 199 x + 5
98 y = 196 x + (3 x + 5 )
98 (y - 2x) = 3 x + 5
Since x and y are integers, y-2x = integer and is > 0 as RHS is positive.
3 x + 5 is a multiple of 98.
=> x = (98-5)/3 = 31 one solution.
=> x = (2*98-5)/3 ≠ integer
=> x = (3*98-5)/3 ≠ integer
=> x = (4*98-5)/3 > 100 so not a solution.
Hence, x = Rs 31 and y - 2x = 1 => y = 63 paise.
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