I have a total of rs.300 in coins of denomination rs1,rs2,rs5. The no of rs2 coins is 3 times the no of 5rs coins the total no of coins is rs 160 . how many coins of each denomination are with me
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Let, No. of rs.5 =x coins.
.
. . No.of rs.2 =3x coins
The total no. of coins= 160
So, the no.of rs.1 =(160-4x)coins
Total cash with me = rs.300
The amount of rs.1 = 1×(160-4x) = 160-4x
The amount of rs.2 = 2×3x =6x
The amount of rs.5=5×x= 5x
Since, the equation and solving=
160-4x+6x+5x = 300
160+(7x)=300
7x= 300-160= 140
x=140/7=20
X=20
The no.of rs.1 = (160-4×20)=rs.80
The no. of rs.2 =3x (3×20)=rs.60
The no.of rs.5 =x =20
.
. . No.of rs.2 =3x coins
The total no. of coins= 160
So, the no.of rs.1 =(160-4x)coins
Total cash with me = rs.300
The amount of rs.1 = 1×(160-4x) = 160-4x
The amount of rs.2 = 2×3x =6x
The amount of rs.5=5×x= 5x
Since, the equation and solving=
160-4x+6x+5x = 300
160+(7x)=300
7x= 300-160= 140
x=140/7=20
X=20
The no.of rs.1 = (160-4×20)=rs.80
The no. of rs.2 =3x (3×20)=rs.60
The no.of rs.5 =x =20
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Answer:
Answer : x=80, y=60 & z=20
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