<p>Can anyone help with this one from the Blue Book. Page 871, #16</p>
<p>To celebrate a colleague's graduation, the m coworkers in an office agreed to contribute equally to a catered lunch that costs a total of y dollars. If p of the coworkers fail to contribute, which of the following represents the additional amount, in dollars, that each of the remaining coworkers must contribute to pay for the lunch?</p>
<p>A: y/m
B: y/(m-p)
C: (py)/(m-p)
D: (y(m-p))/m
E: (py)/(m(m-p))</p>
<p>Correct answer is E</p>
<p>This type of problem becomes really easy if you plug in your own numbers for the variables.</p>
<p>Say that m = 10, y = 20, p = 5. </p>
<p>So if there are 10 co-workers, who agreed to split the cost of a lunch that costs $20, that means each person would have to pay $2. But, if 5 people don't pay, that leaves 5 co-workers who have to pay $4 each to cover the $20 lunch. So originally, the co-workers had to pay $2 but now they have to pay $4. That's a difference of $2. </p>
<p>So plug in the values you chose for m, y, and p and see which one gives you an answer of $2.</p>
<p>The answer is (E) because (5*20)/(10(10-5) = $2</p>
<p>That method takes about 20x as much time as needed!</p>
<p>The total amount of cash to pay for the lunch = y
Each person pays y/m dollars.
But some people 'forgot' to pay, so we only collected y(m-p)/m dollars.</p>
<p>We need to make up some money: y - y(m-p)/ m
Let's simplify first: y - ym/m + yp / m
Which is: y - y + yp / m
Which is: yp / m</p>
<p>Dividing by how many people we need to divide this cost up by:
yp / m(m-p)</p>
<p>I'm not too good with dealing with variables. I usually just plug reasonable numbers in.</p>
<p>My method seems like it's complicated but that's just because I tried to explain the thought process behind it</p>
<p>Ok..</p>
<p>The amount of money each person pays assuming everyone pays is: y/m
The amount of money each person pays when "p" people forgot to pay is: y/(m-p)</p>
<p>Fair enough?</p>
<p>The difference then, must be the extra amount of money that each person that remember to pay must contribute to compensate for the people who didn't pay.</p>
<p>(y/(m-p)) - (y/m)</p>
<p>Common denominator...</p>
<p>(ym/m(m-p)) - (y(m-p)/m(m-p))</p>
<p>Simplify...</p>
<p>(ym/m(m-p)) - (ym-yp/m(m-p))
(ym-ym+yp/m(m-p))</p>
<p>Oh look! Something crosses out...</p>
<p>yp/m(m-p)</p>
<p>Probably same thing as octal's except different steps.</p>
<p>Come on guys, let's get real!<br>
Would you still have enough wind to mill on your algebraic grindstones the last question on the last math section?
And why measure superiority in Nx's, if at all?</p>
<p>Let's use silvermoon's approach with a couple of modifications.
I would change a number of slackers "p" from 5 to some other number, so a number of honest payers would be different (not 10-5=5) to avoid confusion.
Say that m = 10, y = $20, p = 4. </p>
<p>Now comes the METHOD.
If a word problem and the answer choices are in variables:
1. draw a vertical line under your question to split a spare space in two,
2. solve on the left hand side in numbers, and on the right hand side write the same manipulations in corresponding variables and expressions.</p>
<p>Fair share was
$20/10 = $2 $ y/m
Did not pay
4 people p people
Deficit
4<em>($2)=$8 $ p</em>(y/m)
Will have to pick up the slack
10-4= 6 people <a href="m-p">center</a> people[/center]
Each of them will have to compensate
$8/6 $ p*(y/m) / (m-p) = py / m(m-p)
Done.
© gcf101</p>
<p>If you don't know how to solve a certain math problem with given variables, just plug numbers in to satisfy the given information.. It makes it much easier. This is what I do, and I usually average high 700s/800 on practices.</p>
<p>That's a much better explanation, gcf101.</p>
<p>You're right. I hate dealing with variables after about an hour into the test. Then I start losing concentration and making stupid mistakes.</p>
<p>Here's my explanation of the answer to this problem. I am a current Harvard student and have a lot of experience dealing with SAT problems.</p>
<hr>
<p>To celebrate a colleague's graduation, the m coworkers in an office agreed to contribute equally to a catered lunch that costs a total of y dollars. If p of the coworkers fail to contribute, which of the following represents the additional amount, in dollars, that each of the remaining coworkers must contribute to pay for the lunch?</p>
<hr>
<p>We know that the total cost is y dollars. </p>
<p>Originally, each person was expected to contribute y/m dollars. </p>
<p>Since p people don't contribute, the new contribution of the remaining paying members is given by y/(m-p)</p>
<p>The excess is the difference between the original contribution and the second contribution.</p>
<p>y/(m-p) - y/m</p>
<p>Simplifying by common denominators, </p>
<p>[y(m)]/[(m-p)(m)] - [y(m-p)]/[(m-p)(m)]</p>
<p>= (ym - ym + yp) / [(m)(m-p)]</p>
<p>= yp/[m(m-p)]</p>
<p>CORRECT ANSWER : E</p>
<p>This solution looks vaguely familiar...
Yeah, right! Ewardz in post #5 has it. Must be classic. :D</p>
<p>It would be not surprising to learn that xiggi is a present of former student of Harvard, MIT, or Caltech. One can not see his credentials in his posts though.</p>
<p>I remember that xiggi showed how to quickly solve a question, where all the answer choices were expressions - I am not sure if it was this particular question.
I'll try to recreate his reasoning approach (would be nice to hear from xiggi himself whether his disciples are true to the Gospel ;)).
++++++++++++++
Lunch costs $ y.
If p of m coworkers fail to contribute, which of the following represents the additional amount, in dollars, that each of the remaining coworkers must contribute to pay for the lunch?</p>
<p>A: y/m
B: y/(m-p)
C: (py)/(m-p)
D: (y(m-p))/m
E: (py)/(m(m-p))</p>
<p>Let's see, what happens if "p" is really small, and "m" is really big. Say, p=1, and m=1,000.
From a common sense, if 1 out of 1,000 backs out, the rest would hardly have to change their contribution.
Since original contribution was $ y/m,
A, B, C, and D do not work: extra $$, which the rest would have to fork (and spoon) out, can't be that big:
A would double the original contribution,
B and C are almost equal A,
D is almost the total cost of the lunch.
That leaves E.
©xiggi (?) :D</p>
<p>Yes, that is called the "limiting cases" approach to solving algebraic expression problems. It is often taught in university physics and math classes. To check answers, take limits that could prove or disprove your answer. Great way to make sure your answer is right!</p>