Word Prob (Percentages): student enrolment; candy bars

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mcheytan

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Jan 25, 2008
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Problem #1: College admission officer predicts that 20% of students admited will not attend college. According to the prediction, how many students need to be accepted to achieve planned enrolment by x students?

I was thinking that they need to accept 20% more=1/2x, but the answer is 1.25x...I don't understand why?

Problem #2: The similar problem is: A candy manufacturer reduced the weight of a candy bar by 20%, but left the price the same. WHat is the resulting % increase in the price per ounce of candy bar??

Again I was expecting to be 20% and it is 25%.....
Can somebody explain...
Is it because it is actually 20% of the old price (higher price) and 25% of the new price/ounce??? because the more money for less weight will be a bigger percentage??? But what about the students?
 
Re: Word PRoblem PErcentages

Kind of tricky, huh?.

Let's do the candy bar. Try setting up the college student problam the same.

Let x=weight and y=price

The we have old price per oz.= \(\displaystyle \frac{y}{x}\)

New price per oz. = \(\displaystyle \frac{y}{x-.20x}\)

Subtract the two and divide by the old price to find the ratio of the increase per oz.

\(\displaystyle \frac{\frac{y}{x-.20x}-\frac{y}{x}}{\frac{y}{x}}\)

Do the algebra. You may be surprised at what you get.

For the college problem, think about x-.20x being the number who actually attend.

Then \(\displaystyle \frac{x}{x-.20x}\)
 
Ok I got it, but I have a question.

WHy in the candy problem we use the formula: (new-old)/old
and in the students problem we use only: selected/attended
we are looking at percentages in both cases....
 
mcheytan said:
Problem #1: College admission officer predicts that 20% of students admited will not attend college. According to the prediction, how many students need to be accepted to achieve planned enrolment by x students?

20% of those admitted not attending translates to 80% of those admitted will attend ...
(.80)(number admitted) = (number enrolled) = x
number admitted = x/(.80) = 1.25x

Problem #2: The similar problem is: A candy manufacturer reduced the weight of a candy bar by 20%, but left the price the same. WHat is the resulting % increase in the price per ounce of candy bar??

reducing the weight by 20% translates to the new weight = (.80)(old weight)
let W = old weight, P = total price
for the old bar, price per ounce is P/W.
for the new bar, price per ounce is P/(.8W) = 1.25(P/W) ... a 25% increase.
Click to expand...
 
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