let the no. of rs 5 coins be x
& the no. of rs 1 coin be y
& no. of rs 2 coins be 3 x
a.q.t total no. of coins = 160
x + 3x+y = 160
4x + y = 160 eq. 1
again a.q.t total rs = 300
5x + 2(3x) + 1y = 300
11x + y = 300 eq. 2
by elimination method
on subtracting both the equations
4x + y = 160
-11x -y = -300
=-7x =-140
x =20
on putting the the value of x in eq 2
4(20) + y =160
80 + y = 160
y =80
now,
no of rs 2 coins = 3(20)
= 60
no. of rs 5 coins = 20
no. of rs 1 coins=80