1- Price of each carton:
Since they both bought different number of cartons of milk from the same store, we can simply find the price of one carton from the difference between what they both paid.
Since Tamara paid 3$ and Toe paid 3.75$, we can simply say that one carton of milk costs 0.75$ and that Toe bought one carton more than Tamra.
2- How many cartons each bought:
Tamra bought : 3 / 0.75 = 4 cartons
Toe bought : 3.75 / 0.75 = 5 cartons
Answer:
the Answer is c I just had the same problem
now, there are 12 months in a year, so 18 months is really 18/12 of a year, thus
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$4000\\ P=\textit{original amount deposited}\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=years\to \frac{18}{12}\dotfill &\frac{3}{2} \end{cases} \\\\\\ 4000=P[1+(0.05)(\frac{3}{2})]\implies 4000=P(1.075) \\\\\\ \cfrac{4000}{1.075}=P\implies 3720.93\approx P](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5Cdotfill%20%26%20%5C%244000%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B18%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%204000%3DP%5B1%2B%280.05%29%28%5Cfrac%7B3%7D%7B2%7D%29%5D%5Cimplies%204000%3DP%281.075%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B4000%7D%7B1.075%7D%3DP%5Cimplies%203720.93%5Capprox%20P)
So you know the whole line is XZ which is 83.Then you find out it is in two parts so that is XY and YZ which are 11 and 4c. You can make the equation:
11+4c=83
4c=72
c=18
c=18 and YZ=72
