There are 48 coins in total.
Shawna found:
12 Pennies.
12 Nickels.
16 Dimes.
And 8 Quarters.
Step-by-step explanation:
We will let
denote the total amount of coins.
We know that the total worth of the coins were $4.32.
One-fourth of the found coins are pennies. Hence:
![\displaystyle \frac{1}{4}x\text{ are pennies.}](/tpl/images/0589/9316/e4dc5.png)
Since pennies are worth $0.01 each, the worth will be:
![\displaystyle \frac{1}{4}x(0.01)](/tpl/images/0589/9316/fad8a.png)
Similarly, we know that one-sixth of the found coins are quarters. Quarter are worth $0.25. So, the total worth in quarters is:
![\displaystyle \frac{1}{6}x(0.25)](/tpl/images/0589/9316/3fe5d.png)
We have 1.5 times as many nickels as quarters. Therefore, the amount of nickels we have is:
![\displaystyle 1.5(\frac{1}{6}x)=\frac{3}{2}(\frac{1}{6}x)=\frac{1}{4}x](/tpl/images/0589/9316/25e33.png)
Or one-fourth of the total amount.
Since each nickel is worth $0.05, the total worth in nickels is:
![\displaystyle \frac{1}{4}x(0.05)](/tpl/images/0589/9316/4b058.png)
We will now need to determine the amount to dimes. Notice that 1/4 of the total amount of coins were pennies, 1/6 were quarters, and another 1/4 were nickels. Therefore, the amount of coins that were dimes n must be the remaining fraction of the total amount of coins. In other words, the fraction that were dimes is:
![\displaystyle n=100\%-\frac{1}{4}-\frac{1}{6}-\frac{1}{4}](/tpl/images/0589/9316/7dd52.png)
Evaluate for n. Let 100% equal 1. Hence:
![\displaystyle \begin{aligned} n&=1-\frac{1}{4}-\frac{1}{6}-\frac{1}{4} \\ &=\frac{12}{12}-\frac{3}{12}-\frac{2}{12}-\frac{3}{12}\\&=\frac{12-3-2-3}{12}\\&=\frac{4}{12}=\frac{1}{3} \end{aligned}](/tpl/images/0589/9316/fadf6.png)
Therefore, 1/3 of the coins were dimes.
Since dimes are worth $0.10, the total worth in dimes are:
![\displaystyle \frac{1}{3}x(0.1)](/tpl/images/0589/9316/c95fd.png)
The total worth of all the coins found was worth $4.32. Therefore:
![\displaystyle \frac{1}{4}x(0.01)+\frac{1}{6}x(0.25)+\frac{1}{4}x(0.05)+\frac{1}{3}x(0.1)=4.32](/tpl/images/0589/9316/d3f82.png)
Solve for
, the total amount of coins. First, let’s multiply everything by 100 to remove the decimals. Hence:
![\displaystyle 100(\frac{1}{4}x(0.01)+\frac{1}{6}x(0.25)+\frac{1}{4}x(0.05)+\frac{1}{3}x(0.1))=100(4.32)](/tpl/images/0589/9316/a2471.png)
Distribute:
![\displaystyle \frac{1}{4}(1)x+\frac{1}{6}(25)x+\frac{1}{4}(5)x+\frac{1}{3}(10)x=432](/tpl/images/0589/9316/e99b3.png)
Now, let’s multiply everything by 12 to remove the fractions. 12 is the LCM of 4, 6, and 3. Hence:
![\displaystyle 12(\frac{1}{4}(1)x+\frac{1}{6}(25)x+\frac{1}{4}(5)x+\frac{1}{3}(10)x)=12(432)](/tpl/images/0589/9316/54fda.png)
Distribute:
![3(1)x+2(25)x+3(5)x+4(10)x=5184](/tpl/images/0589/9316/9ef4e.png)
Multiply:
![3x+50x+15x+40x=5184](/tpl/images/0589/9316/8c1ce.png)
Combine like terms:
![108x=5184](/tpl/images/0589/9316/332dd.png)
Divide both sides by 108. Hence, the total amount of coins are:
![x=48](/tpl/images/0589/9316/7490a.png)
1/4 of the total coins are pennies. Hence, there are 1/4(48) or 12 pennies.
1/6 of the total coins are quarters. Hence, there are 1/6(48) or 8 quarters.
As we determined, 1/4 of the total coins are also nickels. Hence, there are 1/4(48) or 12 nickels.
Finally, as we determined, 1/3 of the total coins are dimes. Hence, there are 1/3(48) or 16 times.