Dollars

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Dollars

by Sak32 » Sat Jan 04, 2014 10:54 am
For each customer, a bakery charges p dollars for the first loaf of bread bought by the customer and charges q dollars for each additional loaf of bought by the customer. What is the value of p?

1) A customer who buys 2 loaves is charged 10 percent less per loaf than a customer who buys a single loaf.

2) A customer who buys 6 loaves of bread is charged 10 dollars.
Last edited by Sak32 on Sun Jan 05, 2014 1:47 am, edited 2 times in total.

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by brianlange77 » Sat Jan 04, 2014 2:00 pm
Just for clarity -- are you submitting this question because you've created it and want to see others try it? Or do you have a question?

If it's the first -- feel free to ignore my post.
If it's the second -- can you give us an idea of how you tried approaching this one ... or where your struggle was?

Thanks!

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by Sak32 » Sun Jan 05, 2014 1:43 am
No I didn't create it. I know what the answer is its just I had difficulty with the first statement. So if you could provide me with explanations that would be helpful. Thanks.

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by ceilidh.erickson » Mon Jan 06, 2014 4:34 pm
In this question, we have two unknowns: p and q. The golden rule for solving for 2 variables is that we need 2 equations (although there are some exceptions. See here: https://www.beatthegmat.com/to-find-the- ... tml#706713 )

(1) A customer who buys 2 loaves is charged 10 percent less per loaf than a customer who buys a single loaf.

If we were to set this up algebraically, we'd want to find the average of 2 loaves, and compare it to a single loaf:

(p + q)/2 = 0.9p

Here, we have one equation with two variables. We can quickly test values to see if this works for multiple scenarios. We weren't given any integer constraints (they don't have to be WHOLE dollar amounts), so we can test anything that fits.
If p = $1, then q would equal $0.80
If p = $100, then q would equal $80
Ridiculous for prices of bread, obviously! But the point is - one equation was not enough for us to solve. Insufficient.

2) A customer who buys 6 loaves of bread is charged 10 dollars.

Again, set this up algebraically:

p + 5q = 10

We can easily see that there are multiple possibilities that would add up to 10: p=5 and q=1, p=2.50 and q=1.50, etc. Insufficient.

1&2
If we put the statements together, we have two equations with two variables. We can solve for q in the first equation:
(p + q)/2 = 0.9p
p + q = 1.8p
q = 0.8p
Then we can easily plug it into the second equation and solve. Since it's DS - don't actually do the work! You can see that you'll get a value for p, but you don't have to actually figure out what it is.
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by ceilidh.erickson » Mon Jan 06, 2014 4:36 pm
Brian brings up a good point here, though: please don't just post a problem without commentary. Let us know why you found it confusing, or what kind of help you want. Please also post your sources, so other people can tell where you got the question.

Thanks!
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by vipulgoyal » Thu Jan 09, 2014 3:32 am
from 1&2
let the price of first loaf for $5 for two loves @ 10% less be $ 9
means the second loaf for $4, as of now we dont know the values but ratios
now 5x(1)+4x(5)=10, x = 2/5
price of first loaf = 5(2/5) = 2
price of another 5 loves 4 x 2/5 x 5 =8