[Guest]

I went to the Bank to cash a cheque. The teller, by mistake gave me Dollars instead of Cents and Cents instead of Dollars. From the amount I spent 20 Cents for the Tram drive back. On reaching Home, I found the balance amount with me was exactly double the amount on the cheque. What was the amount on the cheque?. I had no money other than the one from the Bank.

Solve by calculation and not by mere trial and error.

[Guest]

The amount on the cheque is originally 26.53!

let assum originally is x.y, (x = dollar, y = cent),

and the teller give you y.x

based on the story, it become y.x - 0.20 = 2 (x.y)

so it become (y).(x-20) = 2x.2y or

y.(x-20) = (2x+1).(2y-100), due to 2y is greather than 1 dollar, so minus 100 cent, bring it to 2x+1, or

(y-1).(x+80) = 2x.2y, due to x < 20, so need to take 100 from y, or

(y-1).(x+80) = (2x+1).(2y-100)

Pattern 1:

y.(x-20)=2x.2y, so y =2x, x-20 = 2y, solving this will give you 3y= -20, not possible.

Pattern 2:

y=2x+1, x-20 = 2y-100, then solving this will give you x = 26, y =53. so it is 26.53

Pattern 3:

y -1 = 2x, x+80 = 2y, this will give you also x =26, y = 53

Pattern 4:

will give you 3x = 176, not able to be divide fully.

[bonanova]

Similar to this one, but different numbers.

[Guest]

(x*100+y)*2 = y*100+x-20

200x + 2y = 100 y + x - 20

98 y = 199 x + 20

y = (199 x + 20)/ 98

x & y are whole numbers below 100.

Remainder of 199 is 3 when divided by 98.

so 3*x+20 has to be a multiple of 98. Try multiple of 1. Multiplier cannot be more than 2 due to the limit of 100.

3*x+20 = 98

x = 26 $

Y = 53 cents

Other multiples of 98 do not yield answers.