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For each customer, a bakery charges p dollars for the first

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For each customer, a bakery charges p dollars for the first  [#permalink]

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New post Updated on: 10 Feb 2015, 05:17
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Question Stats:

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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 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.

Originally posted by dwag on 19 Sep 2006, 15:19.
Last edited by Ergenekon on 10 Feb 2015, 05:17, edited 2 times in total.
Edited the question and added the OA.
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 13 Aug 2013, 01:27
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5
fozzzy wrote:
What would be the equations for this question?


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 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:

Price of 2 loaves = $(p+q).
Price per loaf = $(p+q)/2

Price of a single loaf = $p.

Given that (p+q)/2=0.9p.

Two unknowns. Not sufficient.

(2) A customer who buys 6 loaves of bread is charged 10 dollars --> p+5q=10. Not sufficient.

(1)+(2) We have two distinct linear equations with two unknowns: (p+q)/2=0.9p and p+5q=10, thus we can solve for both p and q. Sufficient..

Answer: C.

Hope it's clear.
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 19 Sep 2006, 16:34
2
Answer: C

Cost for first loaf = p
Cost for remaining = q

S1: Cost of two loaves = p+q
Price /loaf = (p+q)/2 = 0.9p (10% discount)

p+q = 1.8p

or q = 0.8p

or 4p -5q = 0

Not sufficient.


S2: 10 = p+5q
Not sufficient.

S1 & S2:
p+5q = 10
4p-5q=0

or 5p = 10
p = 2

q = 4x2/5 = 1.6

Sufficient w/ 2 equations.
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 20 Sep 2006, 05:44
i cant get the warding of st one...........Help guys
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 20 Sep 2006, 05:58
Statement 1 says that when you buy 2 loaves of bread instead of 1, you get a discount of 10% per loaf.

If you buy 2, then the cost is (p+q)
Cost per loaf = (p+q)/2 = 0.5p+0.5q



If you buy 1, the cost is p.
Cost per loaf = p/1 = p


You get a 10% discount per loaf..

i.e. 0.5p+0.5q = 0.9p

or 0.5q = 0.4p

or 5q = 4p



yezz wrote:
i cant get the warding of st one...........Help guys
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 20 Sep 2006, 07:58
Statement 1 says that when you buy 2 loaves of bread instead of 1, you get a discount of 10% per loaf.

If you buy 2, then the cost is (p+q)
Cost per loaf = (p+q)/2 = 0.5p+0.5q



If you buy 1, the cost is p.
Cost per loaf = p/1 = p

You get a 10% discount per loaf..

i.e. 0.5p+0.5q = 0.9p

or 0.5q = 0.4p

or 5q = 4p

but dont you think that the part in read is the average cost per loaf not cost per loaf
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 20 Sep 2006, 08:04
For each customer, a bakery charges p dollars for the first loaf of bread bought by the customer and charges q dollars for each add'l loaf bought by the customer. What's the value of p?
(1) A customer who buys 2 loaves is charged 10% less per loaf than a customer who buys a single loaf.
(2) A customer who buys 6 loaves of bread is charged 10 dollars.

general formula to calculate price

x = p+nq where n is number of loafs in excess of one

from one

original price of two loafs is = p+q

fro one i think it means

p+q = 2(0.9)p = 1.8p

thus 0.8p=q

from two
p+5q = 10

both together

p+5(0.8p) = 10

thus 5p = 10 and p = 2 and q = 1.6

what is my mistake here
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 20 Sep 2006, 10:11
I used a combination of math and some guess work.

I have seen these type of problems before and made the mistake of thinking it was E in the past. However this time I knew it was D ...From the two stems it looks like you are going to get two equations with two variables which one can solve. I just made the sure the equations weren't equal to each other when I chose D...I didn't actually work all the way through the problem.
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 13 Aug 2013, 00:21
What would be the equations for this question?
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 26 Jan 2017, 05:51
Question stem tells you that p $ for 1st loaf and q $ for additional loaf.

1. if someone buys 2 loaf , he will be charged p+q $.
price per laof = (p+q) / 2 $

if one buys only 1 loaf then one pays only p $

St1 tells that

(p+q)/2 = (1-10%) p
p/2+q/2 = 9/10p
q=0.8 p

Cant figure out p insufficient.

2. 6 loaves price is p+5q given as 10$
p+5q=10

Insufficient.

Together 1 & 2
p+5*0.8p = 10
p =2$
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 08 Oct 2017, 08:41
Hi

In this why cant the answer be B.

We have the equation, p+5q=10

The only possible value that q can take is 1. Thus, we can get p as 5.



Bunuel wrote:
fozzzy wrote:
What would be the equations for this question?


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 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:

Price of 2 loaves = $(p+q).
Price per loaf = $(p+q)/2

Price of a single loaf = $p.

Given that (p+q)/2=0.9p.

Two unknowns. Not sufficient.

(2) A customer who buys 6 loaves of bread is charged 10 dollars --> p+5q=10. Not sufficient.

(1)+(2) We have two distinct linear equations with two unknowns: (p+q)/2=0.9p and p+5q=10, thus we can solve for both p and q. Sufficient..

Answer: C.

Hope it's clear.
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 08 Oct 2017, 12:55
sinhap07 wrote:
Hi

In this why cant the answer be B.

We have the equation, p+5q=10

The only possible value that q can take is 1. Thus, we can get p as 5.


You don't necessarily know that q is an integer. Dollar amounts can be non-integers.

For instance, the first loaf could cost 3.2 dollars, and the remaining 5 loaves could cost 1.36 dollars each. 3.2 + 5(1.36) = 10, which fits statement 2.
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 16 Sep 2018, 11:46
dwag wrote:
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 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.

\({\text{\$ }}p\,\,\,:\,\,{\text{first}}\,\,{\text{loaf}}\)

\({\text{\$ }}q\,\,\,{\text{:}}\,\,\,{\text{any}}\,\,{\text{additional}}\,\,{\text{loaf}}\)

\(? = q\)

\(\left( 1 \right)\,\,{\left( {\frac{{\,{\text{charge}}\,}}{{{\text{loaf}}}}} \right)_{\,2\,\,{\text{loaves}}}} = \frac{{p + q}}{2}\,\,\,\,\mathop = \limits^{{\text{given}}} \,\,\frac{9}{{10}}p\,\,\,\,\, \Rightarrow \,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\frac{q}{p} = \frac{4}{5}\)

\(\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {p,q} \right) = \left( {0.5,0.4} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0.4 \hfill \\
\,{\text{Take}}\,\,\left( {p,q} \right) = \left( {1,0.8} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0.8 \hfill \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,6\,\,{\text{loaves}}\,\,{\text{for}}\,\,\$ 10\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {p\,;q} \right) = \left( {2\,;\frac{8}{5}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = \frac{8}{5} \hfill \\
\,{\text{Take}}\,\,\left( {p\,;q} \right) = \left( {3\,;\frac{7}{5}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = \frac{7}{5} \hfill \\
\end{gathered} \right.\)

\(\left( {1 + 2} \right)\,\,\,\left\{ \begin{gathered}
p + 5q = 10 \hfill \\
\frac{q}{p} = \frac{4}{5} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{k}}\,\,{\text{technique}}} \,\,\,\,\left( {5k} \right) + 5\left( {4k} \right) = 10\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 4k\,\,{\text{unique}}\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: For each customer, a bakery charges p dollars for the first  [#permalink]

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New post 16 Sep 2018, 12:32
IMO C.

Both will give equations of 2 variables, and then we can solve to get p.
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Re: For each customer, a bakery charges p dollars for the first &nbs [#permalink] 16 Sep 2018, 12:32
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