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# A store sold 6 bicycles with an average sale price of \$1,000. What was

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Math Expert
Joined: 02 Sep 2009
Posts: 47015
A store sold 6 bicycles with an average sale price of \$1,000. What was [#permalink]

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06 Oct 2015, 05:59
00:00

Difficulty:

65% (hard)

Question Stats:

62% (01:31) correct 38% (01:22) wrong based on 119 sessions

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A store sold 6 bicycles with an average sale price of \$1,000. What was the price of the most expensive bicycle?

(1) The median price was \$1,000.
(2) The range of prices was \$600.

Kudos for a correct solution.

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Re: A store sold 6 bicycles with an average sale price of \$1,000. What was [#permalink]

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06 Oct 2015, 07:18
1
Total cost of 6 bicycles = 6x1000= 6000 \$
B1+B2+B3+B4+B5+B6=6000

1. Median price was 1000 \$
B3+B4=2000
B1+B2+B5+B6=4000
Not sufficient

2. Range of prices was 600 \$
Difference between B1 and B6 is 600 , but the most expensive still can take any different values.
Suppose B1= 900 and B6 =1500
B1+B6=2100
B2+B3+B4+B5=3900

Or B1= 800 and B6= 1400
B1+B6=2200
B2+B3+B4+B5=3900

Combining 1 and 2,
B3+B4=2000
=>B1+B2+B5+B6=4000
If ,B1=B2 =700 , then B5=B6=1300

If , B1=B2 = 800 , then B5=1000 and B6=1400

Still ,Not sufficient .
hence E.
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Re: A store sold 6 bicycles with an average sale price of \$1,000. What was [#permalink]

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06 Oct 2015, 07:27
Bunuel wrote:
A store sold 6 bicycles with an average sale price of \$1,000. What was the price of the most expensive bicycle?

(1) The median price was \$1,000.
(2) The range of prices was \$600.

Kudos for a correct solution.

Let, a, b, c, d, e, f refer to prices of 6 bicycles in ascending (increasing) order of their prices

Given: a + b+ c + d + e + f = 100*6 = 6000

Question: f = ?

Statement 1: The median price was \$1,000.
i.e. (c+d)/2 = 1000
i.e. c+d = 2000
but it doesn't give us any information to conclude about f. hence
NOT SUFFICIENT

Statement 2: The range of prices was \$600.
i.e. f - a = 600
but it doesn't give us any information about a to conclude about f. hence
NOT SUFFICIENT

Combining the two statements
a + b+ c + d + e + f = 6000
Since, c+d = 2000
therefore, a + b+ e + f = 4000
Since, f - a = 600
therefore, (f-600) + b+ e + f = 4000
i.e. b + e + 2f = 4600
Since we still lack information about b and e therefore
NOT SUFFICIENT

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Re: A store sold 6 bicycles with an average sale price of \$1,000. What was [#permalink]

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06 Oct 2015, 08:18
Bunuel wrote:
A store sold 6 bicycles with an average sale price of \$1,000. What was the price of the most expensive bicycle?

(1) The median price was \$1,000.
(2) The range of prices was \$600.

Kudos for a correct solution.

Target question: What was the price of the most expensive bicycle?

Given: The store sold 6 bicycles with an average sale price of \$1,000.
This means the SUM of the 6 bikes = \$6000 (since \$6000/6 bikes = \$1000 average)

Statement 1: The median price was \$1,000.
This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several scenarios that satisfy statement 1. Here are two:
Case a: the prices are {1000, 1000, 1000, 1000, 1000, 1000} in which case the most expensive bike is \$1000
Case b: the prices are {900, 1000, 1000, 1000, 1000, 1100} in which case the most expensive bike is \$1100
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: http://www.gmatprepnow.com/articles/dat ... lug-values

Statement 2: The range of prices was \$600
This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several scenarios that satisfy statement 2. Here are two:
Case a: the prices are {700, 1000, 1000, 1000, 1000, 1300} in which case the most expensive bike is \$1300
Case b: the prices are {600, 1000, 1000, 1000, 1200, 1200} in which case the most expensive bike is \$1200
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
There are STILL several scenarios that satisfy BOTH statements. Here are two:
Case a: the prices are {700, 1000, 1000, 1000, 1000, 1300} in which case the most expensive bike is \$1300
Case b: the prices are {600, 1000, 1000, 1000, 1200, 1200} in which case the most expensive bike is \$1200

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Cheers,
Brent
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Re: A store sold 6 bicycles with an average sale price of \$1,000. What was [#permalink]

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03 Mar 2016, 08:05
Hi,

The statement in Q and statement 1 tell that mean=median. Thus the set of 6 prices should be equally space and hence an AP. So by using statement 1 and 2,we can get the highest price.

Can someone please clarify the discrepancy here

Regards
P
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Joined: 16 Oct 2010
Posts: 8121
Location: Pune, India
Re: A store sold 6 bicycles with an average sale price of \$1,000. What was [#permalink]

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03 Mar 2016, 23:02
1
prakharkaushik wrote:
Hi,

The statement in Q and statement 1 tell that mean=median. Thus the set of 6 prices should be equally space and hence an AP. So by using statement 1 and 2,we can get the highest price.

Can someone please clarify the discrepancy here

Regards
P

Your premise is incorrect. Mean = Median does not tell you that the set is equally spaced.

e.g.

1, 2, 4, 5, 7, 7, 9

Median = 5
Mean = 5

The reverse is true - An equally spaced set does have mean = median.
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Re: A store sold 6 bicycles with an average sale price of \$1,000. What was [#permalink]

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15 Sep 2017, 20:30
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Re: A store sold 6 bicycles with an average sale price of \$1,000. What was   [#permalink] 15 Sep 2017, 20:30
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