Question

I have a total of 300 in coins of denomination 1, 2 and 5.The number of 2 coins is three times the number of 5 coins. The total number of coins is 160. How many coins of each denomination are with meh?

Answer 1

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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

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Answer 2

Let the number of ₹5 coins be x.

Then,

number ₹2 coins = 3x

and, number of ₹1 coins = (160 – 4x) Now,

Value of ₹5 coins = x × 5 = 5x

Value of ₹2 coins = 3x × 2 = 6x

Value of ₹1 coins = (160 – 4x) × 1 = (160 – 4x)

According to the question,

5x + 6x + (160 – 4x) = 300

⇒ 11x + 160 – 4x = 300

⇒ 7x = 140

⇒ x = 140/7

⇒ x = 20

Number of ₹5 coins = x = 20

Number of ₹2 coins = 3x = 60

Number of ₹1 coins = (160 – 4x) = 160 – 80 = 80