In a purse there are 20-rupee notes 10-rupee notes and 5-rupee notes. the number of 5-rupee notes exceeds two times the 10-rupee notes by one. the 20-rupee notes are 5 less than the 10-rupee notes. if the total value of the money in the purse is Rs185 find the number of each variety of notes.
if there are x,y,z notes for 5,10,20 rupees, respectively, then
x = 2y+1
z = y-5
5x+10y+20z = 185
(x,y,z) = (15,7,2)
2,7,15
To solve this problem, we can use a system of equations. Let's denote the number of 20-rupee notes as "x," the number of 10-rupee notes as "y," and the number of 5-rupee notes as "z."
1. The number of 5-rupee notes exceeds two times the 10-rupee notes by one:
z = 2y + 1 -- Equation (1)
2. The number of 20-rupee notes is 5 less than the 10-rupee notes:
x = y - 5 -- Equation (2)
3. The total value of the money in the purse is Rs185:
20x + 10y + 5z = 185 -- Equation (3)
Now, we can solve these equations simultaneously to find the values of x, y, and z.
Substituting equation (2) into equation (1) to eliminate z, we have:
2y + 1 = 5 -- Equation (4)
2y = 5 - 1
2y = 4
y = 4 / 2
y = 2
Substituting y = 2 into equation (2) to find x:
x = 2 - 5
x = -3
Now, let's substitute x = -3 and y = 2 into equation (3) to find z:
20(-3) + 10(2) + 5z = 185
-60 + 20 + 5z = 185
-40 + 5z = 185
5z = 185 + 40
5z = 225
z = 225 / 5
z = 45
Therefore, the number of 20-rupee notes is -3, the number of 10-rupee notes is 2, and the number of 5-rupee notes is 45.
However, since the number of notes cannot be negative, we can conclude that there is no solution to this problem given the constraints.