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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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Updated on: 10 Feb 2015, 05:17
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5% (low)

Question Stats:

85% (01:43) correct 15% (01:57) wrong based on 364 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.

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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13 Aug 2013, 01:27
7
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. _________________ ##### General Discussion VP Joined: 02 Jun 2006 Posts: 1213 Re: For each customer, a bakery charges p dollars for the first [#permalink] ### Show Tags 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. Retired Moderator Joined: 05 Jul 2006 Posts: 1726 Re: For each customer, a bakery charges p dollars for the first [#permalink] ### Show Tags 20 Sep 2006, 05:44 i cant get the warding of st one...........Help guys VP Joined: 02 Jun 2006 Posts: 1213 Re: For each customer, a bakery charges p dollars for the first [#permalink] ### Show Tags 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 Retired Moderator Joined: 05 Jul 2006 Posts: 1726 Re: For each customer, a bakery charges p dollars for the first [#permalink] ### Show Tags 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 Retired Moderator Joined: 05 Jul 2006 Posts: 1726 Re: For each customer, a bakery charges p dollars for the first [#permalink] ### Show Tags 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 Senior Manager Joined: 13 Sep 2006 Posts: 255 Location: New York Re: For each customer, a bakery charges p dollars for the first [#permalink] ### Show Tags 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. Director Joined: 29 Nov 2012 Posts: 765 Re: For each customer, a bakery charges p dollars for the first [#permalink] ### Show Tags 13 Aug 2013, 00:21 What would be the equations for this question? _________________ Click +1 Kudos if my post helped... Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/ GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html Director Joined: 26 Oct 2016 Posts: 641 Location: United States Concentration: Marketing, International Business Schools: HBS '19 GMAT 1: 770 Q51 V44 GPA: 4 WE: Education (Education) Re: For each customer, a bakery charges p dollars for the first [#permalink] ### Show Tags 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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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..

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, 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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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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16 Sep 2018, 12:32
IMO C.

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

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