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

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

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19 Sep 2006, 16:19
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5% (low)

Question Stats:

81% (01:06) correct 19% (00:59) wrong based on 259 sessions

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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.
[Reveal] Spoiler: OA

Last edited by Ergenekon on 10 Feb 2015, 06:17, 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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19 Sep 2006, 17:34
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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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13 Aug 2013, 02:27
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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..

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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20 Sep 2006, 06:44
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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20 Sep 2006, 06:58
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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20 Sep 2006, 08:58
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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20 Sep 2006, 09:04
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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20 Sep 2006, 11:11
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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13 Aug 2013, 01:21
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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03 Jan 2015, 21:11
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Re: For each customer, a bakery charges p dollars for the first [#permalink]

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19 Feb 2016, 00:24
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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26 Jan 2017, 06:51
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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08 Oct 2017, 09:41
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..

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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08 Oct 2017, 13:55
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] 08 Oct 2017, 13:55
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