Answer:
Bob has 8 nickels and 8 dimes.
Explanation:
• Let the number of dimes bob has = d
,
• Let the number of nickels bob has = n
Bob has a total of 16 coins, therefore:
[tex]\begin{gathered} d+n=16 \\ \implies d=16-n\cdots(1) \end{gathered}[/tex]
Next, it is given that the total worth of the coins is $1.20.
[tex]0.1d+0.05n=1.20\cdots(2)[/tex]
Solve equation (1) and (2) simultaneously:
[tex]\begin{gathered} 0.1d+0.05n=1.20 \\ 0.1(16-n)+0.05n=1.20 \\ 1.6-0.1n+0.05n=1.20 \\ 1.6-1.20=0.1n-0.05n \\ 0.4=0.05n \\ n=\frac{0.4}{0.05} \\ n=8 \end{gathered}[/tex]
Finally solve for d:
[tex]d=16-n=16-8=8[/tex]
Bob has 8 nickels and 8 dimes.
