Question

Solve please question 9

Answer 1

let the number of coins of (1) be x and the number of coins of (2) be y and the number of coins of (5) be z.

⇒ x + 2y + 5z = 740

given the number of coins of (2) are 3 times of (1)

⇒ y = 3x

gievn the number of coins of (5) are twice the number of coins of (2)

⇒ z = 2y

substituting the above values in x + 2y + 5z = 740

⇒ y/3 + 2y + 10y = 740

⇒ y =60

⇒ z = 120

⇒ x = 20

⇒ the coins of (1) are 20 and (2) are 60 and (5) are 120