Question

free points!!!
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L was just kidding so please solve the problem above .

Answer 1

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

Answer 2

Let the no. of rs 5 coins be x& the no. of rs 1 coin be y& no. of rs 2 coins be 3 xa.q.t total no. of coins = 160x + 3x+y = 1604x + y = 160 eq. 1again a.q.t total rs = 3005x + 2(3x) + 1y = 30011x + y = 300 eq. 2by elimination methodon subtracting both the equations4x + y = 160-11x -y = -300=-7x =-140x =20on putting the the value of x in eq 24(20) + y =16080 + y = 160y =80now,no of rs 2 coins = 3(20)= 60no. of rs 5 coins = 20no. of rs 1 coins=80