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To celebrate a colleague's retirement, the \(\mathrm{T}\) coworkers in an office agreed to share equally the cost of a catered lunch. If the lunch costs a total of \(\mathrm{x}\) dollars and \(\mathrm{S}\) of the coworkers fail to pay their share, which of the following represents the additional amount, in dollars, that each of the remaining coworkers would have to contribute so that the cost of the lunch is completely paid?
Let's start by understanding what's happening in plain English. We have a group lunch situation where everyone was supposed to split the cost equally, but some people backed out and didn't pay. This means the people who do pay have to cover more than their original share.
Specifically, we need to find: How much EXTRA does each paying person have to contribute beyond what they originally planned to pay?
The key variables are:
Process Skill: TRANSLATE - Converting the word problem into clear mathematical relationships
Let's think about this with a concrete example first. Suppose the lunch costs \(\$120\) and there are 10 coworkers total. Originally, each person was supposed to pay \(\$120 \div 10 = \$12\).
In general terms: When all T coworkers were supposed to pay equally, each person's share would be the total cost divided by the number of people.
Original individual share = \(\frac{\mathrm{x}}{\mathrm{T}}\) dollars per person
Now let's continue our example. Suppose 2 people back out, so only 8 people end up paying for the \(\$120\) lunch. Each of the 8 paying people must now cover \(\$120 \div 8 = \$15\).
In general terms: When only (T-S) coworkers actually pay, they must split the entire cost x among themselves.
New individual payment = \(\frac{\mathrm{x}}{\mathrm{T-S}}\) dollars per person
Back to our example: Each person was supposed to pay \(\$12\), but now must pay \(\$15\). The additional amount is \(\$15 - \$12 = \$3\) per person.
In general terms: The additional amount each remaining coworker must pay is the difference between what they actually pay and what they were originally supposed to pay.
Additional amount = New payment - Original payment
Additional amount = \(\frac{\mathrm{x}}{\mathrm{T-S}} - \frac{\mathrm{x}}{\mathrm{T}}\)
To subtract these fractions, we need a common denominator of T(T-S):
\(\frac{\mathrm{x}}{\mathrm{T-S}} = \frac{\mathrm{xT}}{\mathrm{T(T-S)}}\)
\(\frac{\mathrm{x}}{\mathrm{T}} = \frac{\mathrm{x(T-S)}}{\mathrm{T(T-S)}}\)
Additional amount = \(\frac{\mathrm{xT}}{\mathrm{T(T-S)}} - \frac{\mathrm{x(T-S)}}{\mathrm{T(T-S)}}\)
Additional amount = \(\frac{\mathrm{xT - x(T-S)}}{\mathrm{T(T-S)}}\)
Additional amount = \(\frac{\mathrm{xT - xT + xS}}{\mathrm{T(T-S)}}\)
Additional amount = \(\frac{\mathrm{xS}}{\mathrm{T(T-S)}}\)
Process Skill: MANIPULATE - Carefully working with fractions to find the difference
The additional amount each remaining coworker must contribute is \(\frac{\mathrm{Sx}}{\mathrm{T(T-S)}}\) dollars.
Let's verify with our example: S=2, x=120, T=10, so (T-S)=8
Additional amount = \(\frac{2 \times 120}{10 \times 8} = \frac{240}{80} = \$3\) ✓
This matches answer choice D: \(\frac{\mathrm{Sx}}{\mathrm{T(T-S)}}\)
Students often think the question is asking for the total amount each remaining person pays (\(\frac{\mathrm{x}}{\mathrm{T-S}}\)), rather than the EXTRA amount beyond their original share. This leads them to select answer choice B instead of working toward the correct difference calculation.
Students may incorrectly assume that the S people who don't pay still need to be accounted for in the final payment structure, or conversely, forget that the original agreement involved all T people. This can lead to setting up the wrong fractions from the start.
Students might try to solve this using proportions or ratios instead of recognizing that they need to find: (new individual payment) - (original individual payment). This conceptual error leads to completely different solution approaches.
When calculating \(\frac{\mathrm{x}}{\mathrm{T-S}} - \frac{\mathrm{x}}{\mathrm{T}}\), students frequently make mistakes finding the common denominator T(T-S). They might incorrectly use T or (T-S) as the common denominator, or make errors when converting each fraction.
After getting \(\frac{\mathrm{xT - x(T-S)}}{\mathrm{T(T-S)}}\), students often make errors when distributing and combining like terms. Common mistakes include: xT - x(T-S) = xT - xT - xS (wrong sign) instead of xT - xT + xS, leading to incorrect final expressions.
Students may correctly set up \(\frac{\mathrm{x}}{\mathrm{T-S}} - \frac{\mathrm{x}}{\mathrm{T}}\) but then make calculation errors due to working too quickly with the algebraic manipulation, especially when dealing with multiple variables simultaneously.
Answer choices C (\(\frac{\mathrm{Sx}}{\mathrm{T-S}}\)) and D (\(\frac{\mathrm{Sx}}{\mathrm{T(T-S)}}\)) look very similar. Students who get the right numerator (Sx) but forget about the T in the denominator often select C instead of D, especially if they don't verify their answer with a concrete example.
Students may arrive at the correct algebraic expression but fail to substitute their example values (S=2, x=120, T=10) to confirm their answer matches the expected result of \(\$3\). This verification step could catch errors in their algebraic work.
This problem can be solved efficiently using smart numbers by choosing convenient values for the variables.
Let's select values that will make our calculations clean:
Originally, each of the 10 coworkers was supposed to pay:
Original share per person = \(\$120 \div 10 = \$12\)
After 2 coworkers fail to pay, only (10 - 2) = 8 coworkers remain to cover the full cost.
New share per remaining person = \(\$120 \div 8 = \$15\)
Additional amount each remaining coworker must pay:
Additional amount = New share - Original share = \(\$15 - \$12 = \$3\)
Using our smart numbers in each answer choice:
Answer choice D matches our calculated result of \(\$3\).
The formula \(\frac{\mathrm{Sx}}{\mathrm{T(T-S)}}\) makes intuitive sense: