Question

jared has 12 coin 4th 75 cents. 3 of the coins are worth twice as much as tge rest. construct a math argument to justify the conjecture thqt jared has 9 nickels and 3 dimes

Answer 1

To solve this question, we proceed as follows:

Step 1: Let x be the worth of one of the type of coins Jared has, and let y be the worth of the other type of coin

Thus:

Since 3 of the coins are of a different type, we have that:

`\begin{gathered} 3x+(12-3)y=75 \\ \Rightarrow3x+9y=75 \end{gathered}`

Also, since 3 of the coins are worth twice as much as the rest, we have that:

`x=2y`

Now, substitute for x in the first equation:

`\begin{gathered} 3x+9y=75 \\ \Rightarrow3(2y)+9y=75 \\ \Rightarrow6y+9y=75 \\ \Rightarrow15y=75 \\ \Rightarrow y=(75)/(15) \\ \Rightarrow y=5cents \end{gathered}`

Since y = 5 cents, we have that:

`\begin{gathered} x=2y \\ \Rightarrow x=2(5) \\ \Rightarrow x=10cents \end{gathered}`

Now, since x = 10 cents (the equivalent worth of a dime), and y = 5 cents (the equivalent worth of a nickel), we have from the first equation that:

`3x+9y=75\text{cents}`

From the above equation, therefore, we can conclude that Jared has nine 10 cents coins (dimes), and three 5 cents coins (nickels)