Question

guys please answer the 15th question don't spam and help fast......if u wanna be the brainliest.........​

Answer 1

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

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Answer 2

Answer:

Let the number of coin 5 = Rs.x

so the number of coin 2 = 3x

so the number of coin 1 = 160 - 4x

so amount of rs.5 = 5x

so the amount of rs.2 = 6x

a/q

5x + 6x + 160 - 4x = 300

7x = 300 - 160

7x = 140

x = 20

so the number of coin of 5 = 20 coins

so the number of coin of 2 = 60 coins

so the number of coin of 1 = 80 coins

or

Let number of Rs5 coins be x

Number of Rs 2 coins =3× Number of 5 Rs coins =3x

Number of Re1 coins =160−(x+3x)

=160−4x

Amount of Rs5 coins =5×x=5x

Amount of Rs2 coins =2×3x=Rs6x

Amount of Re1 coins =(160−4×x)

=160−4x

As per the question, Total amount =Rs300

160−4x+5x+6x=300

7x=300−160

x=  

7

140  

x=20

Number of Re 1 coins =160−4x

=160−4×20

=80

Number of Rs2 coins =3x=3×20=60

Number of Rs5 coins =x=20

Step-by-step explanation:

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