I have a total of? 300 in c ins,of denominationき1,? 2 and 5. Thenumb

1.	I have a total of? 300 in c ins,of denominationき1,? 2 and 5. Thenumber of 32coins is 3 times thcoins is 160, How many coins of each denomination are with me?e number off5 coins. The total number oftition degide that a winner in the
Answer» 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 it should not be done with two variables

I have a total of? 300 in c ins,of denominationき1,? 2 and 5. Thenumber of 32coins is 3 times thcoins is 160, How many coins of each denomination are with me?e number off5 coins. The total number oftition degide that a winner in the

Answer»

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

it should not be done with two variables