SOLUTION: Stopping at the apple stand for his weekly purchase of fifty cents’ worth of apples, Professor Digit was pleased to find that he was given five more apples than usual. “Hmm!” he

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Stopping at the apple stand for his weekly purchase of fifty cents’ worth of apples, Professor Digit was pleased to find that he was given five more apples than usual. “Hmm!” he      Log On


   

Question 629642: Stopping at the apple stand for his weekly purchase of fifty cents’ worth of apples, Professor Digit was pleased to find that he was given five more apples than usual.
“Hmm!” he said. “I see the price has gone down ten cents per dozen.”
What was the new price per dozen?


This will be a quadratic equation. I can solve it if you can figure out what the equation is. Thank you!!
Found 2 solutions by MathTherapy, KMST:
Answer by MathTherapy(10549) About Me  (Show Source):
You can put this solution on YOUR website!
Stopping at the apple stand for his weekly purchase of fifty cents’ worth of apples, Professor Digit was pleased to find that he was given five more apples than usual.
“Hmm!” he said. “I see the price has gone down ten cents per dozen.”
What was the new price per dozen?

This will be a quadratic equation. I can solve it if you can figure out what the equation is. Thank you!!

Let amount bought before be A
Then cost of each = .5%2FA, and cost of 1 dozen = 12+%2A+.5%2FA, or 6%2FA
Since price of a dozen was reduced by .10, then new price of 1 dozen = 6%2FA+-+.1, or %286+-+.1A%29%2FA
New price of 1 apple = %28%286+-+.1A%29%2FA%29%2F12, or %286+-+.1A%29%2F12A
Amount he got after discount = (A + 5)
Therefore, new price of 1 apple, times amount he now receives, equals 50c, OR

%28%286+-+.1A%29%2F12A%29%28A+%2B+5%29+=+.5
%28%286+-+.1A%29%28A+%2B+5%29%29%2F12A+=+.5

(6 - .1A)(A + 5) = 6A ----- Cross-multiplying
6A+%2B+30+-+.1A%5E2+-+.5A+=+6A
.1A%5E2+%2B+6A+-+6A+%2B+.5A+-+30+=+0
.1A%5E2+%2B+.5A+-+30+=+0
highlight_green%28A%5E2+%2B+5A+-+300+=+0%29 ----- Multiplying by 10

There is your quadratic equation, which can be factored to determine A, or amount he used to get for 50c.

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Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Another of many ways to solve it:
Let the old price per dozen (in cents) be p.
The number of apples he used to get was n=%2850%2Fp%29%2A12=600%2Fp, so
n=600%2Fp --> np=600
This time he got 5 more apples, or n%2B5 apples,
at a price of p-10 cents per dozen, so
%28%28n%2B5%29%2F12%29%2A%28p-10%29=50 --> %28n%2B5%29%28+p-10%29=50%2A12=600 --> np-10n%2B5p-50=600 --> 600-10n%2B5p-50=600 --> -10n%2B5p-50=0
Here I would divide both side of the equal sign by 5
-10n%2B5p-50=0 --> -2n%2Bp-10=0 --> -2np%2Bp%5E2-10p=0 --> -2%2A600%2Bp%5E2%2A10p=0 --> highlight%28+p%5E2-10p-1200=0%29
I would solve that equation by factoring; discard the negative value for p,
and then I would subtract 10 to find the new price per dozen (in cents).