Question

Carter took 7 coins out of his pocket. He did not have any pennies.He had more than $0.95 and less than $1.13 what could his coins have been ?Prove your anser

Answer 1

Answer:

Carter's coins could be;

3 of 5¢, 2 of 10¢, 1 of 25¢, and 1 of 50¢

Step-by-step explanation:

The available coin denominations are;

1¢ = 1 penny, 5¢, 10¢, 25¢, 50¢, and $1.00

Since the coins are more than $0.95, we can have;

5¢ × 3

10¢× 2  

25¢× 1

50¢

Total = $1.00

That is his coins could have been 3 × 5¢, 2 × 10¢, 1 × 25¢, and 1 × 50¢, to make a total of 7 coins with an amount value of $1.00.