Money!

Form equation : 199 X - 98 Y= - 20 [First convert rupees to paise]

Now Find integral Solutions to this equation, i.e. X=26 , Y=53.

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

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

98y-199x= 20

x & y should be integers.

x=26 & y=53 is the only answer to satisfy the above two conditions

Therefore, X+Y = 53+26

=79

This can be solved by Diophantine equations. But there is a much simpler way---

Note that after spending 20 paise, the person has left 2x+2y. The change that he has left, from the x paise given him by the cashier, will be x-5.

We know that y is less than 100, but we do not know yet whether it is less than 50 paise. If it is less than 50 paise, we can write the following equations:

2x=y

2y=x-5

If y is 50 paise or more, then the person will be left with an amount of paise (2y) that is a rupee or more. We therefore have to modify the above equations by taking 100 from 2y and adding 1 to 2x. The equations become:

2x+1=y

2y-100=x-5

Solving the set of simultaneous equations gives us the solution.The first set gives x a minus value, which is ruled out. The second set gives the correct values.

x= 26 y= 53

He wanted X rupee and Y paise. Converting this to one unit, he wanted 100 X+Y paise.
He actually received Y rupee and X paise. Converting this to one unit, he received 100
Y + X paise. After spending money, he was left with 100 Y+X-20 paise, which is twice what he wanted. So: 100 Y+X-20 = 2(100 X+Y) 100Y+X-20 = 200X+2Y 98Y=199X+20 This looks like 1 equation with 2 variables, so we need to consider additional constraints. X and Y must be integers, since it wouldn't make sense to have a fractional paise. So let's look at an example and go from there.
IF X=1, RHS = 199+20 --> Y = (199+20)/98, which is not an integer. But looking at this example, we may deduce that Y is an integer iff 199X+20 = 0mod98. Rather than think about mods, let's use a couple points.
Consider X = 1, 199

1 + 20 = 23mod98.
Consider X = 2, 199 2 + 20 = 26mod98.
Consider X = 3, 199
3 + 20 = 29mod98. It looks like an arithmetic sequence with initial term 23 and common difference 3.
Since it's purely increasing, to find 0mod98, we find 98mod98.
So 98 = 20+3*X --> X = 26. So we have X, we just plug back into our 98Y = 199X + 20 to find Y = 53 26 + 53 = 79.

Step 1: Convert rupees to paise

=> Want: 100X + Y

Get: 100Y + X

After spend: 100Y + X -20 = 200X + 2Y (After spend = 2 want)

=> Equation: 98Y -199X = 20 (1)

Step 2: This part is a bit tricky and i might not explain it well (since i ain't study math in English)

We see that 98 x 2 - 199 = -3 (which mean Y =2 and X = 1)

=> (98 x 2 - 199) x N = (-3) x N = A

98 = 20 +78

Now what we want is to eliminate 78

=> 78 + A = 0

=> (-3) x N = -78

=> N = 26

Step 3 Now back to equation (1)

=> A + 98 = 20 (as said above 78 + A = 0)

=> (-3) x N + 98 = 20

=> (98 x 2 - 199) x N + 98 = 20

and with N = 26 as prove above

=> 98 x 2 x 26 - 199 x 26 + 98 = 20

=> X= 2 x 26 + 1 = 53

Y = 26

=> X + Y = 26 + 53 = 79

There are many integer solutions to this: (26,53),(124,252), (222,451),(320,650)....and so forth.