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The average (arithmetic mean) cost of three computer models is $900.

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The average (arithmetic mean) cost of three computer models is $900.  [#permalink]

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New post 05 Mar 2016, 12:08
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The average (arithmetic mean) cost of three computer models is $900. If no two computers cost the same amount, does the most expensive model cost more than $1,000?

(1) The most expensive model costs 25% more than the model with the median cost.
(2) The most expensive model costs $210 more than the model with the median cost.


The OA uses the concept of assigning variables a, b, c, and plugging in "LT" or "less than" to set up multiple equations. Is there an easier method to accomplish this?
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Re: The average (arithmetic mean) cost of three computer models is $900.  [#permalink]

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New post 05 Mar 2016, 12:39
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GinGMAT wrote:
The average (arithmetic mean) cost of three computer models is $900. If no two computers cost the same amount, does the most expensive model cost more than $1,000?

(1) The most expensive model costs 25% more than the model with the median cost.
(2) The most expensive model costs $210 more than the model with the median cost.


The OA uses the concept of assigning variables a, b, c, and plugging in "LT" or "less than" to set up multiple equations. Is there an easier method to accomplish this?


A good question to understand the limits of the information given.

Try to learn the "best" method rather than the "easiest" method.

Average of 3 unequally priced computers = 900 --->a+b+c =2700 such that a<b<c. The limiting value asked in the question is 1000$

Per statement 1, c=1.25b ---> now let us look at the case when c=1000, b = 1000/1.25 = 800 ----> a= 2700-800-1000=900 and this makes a>b which is NOT allowed as b is the median value. So you can clearly see that the value of c MUST be >1000 for the values to make sense with this statement. Hence the answer to the question asked is a definite yes. Sufficient.

Per statement 2, c=210+b . Again use the same logic as above to use the limiting value of 1000$. This gives us c=1000, b=790 and a=910 , again a>b and as such this isnt allowed as the median value of 3 data points must be the centermost value. Hence c > 1000 in order for this statement to be true. Sufficient.

Hence D is the correct answer.

Sometimes in a DS question, you have to manipulate the given question to your advantage as is done above.

Hope this helps.
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Re: The average (arithmetic mean) cost of three computer models is $900.  [#permalink]

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New post 25 Jan 2019, 11:38
GinGMAT wrote:
The average (arithmetic mean) cost of three computer models is $900. If no two computers cost the same amount, does the most expensive model cost more than $1,000?

(1) The most expensive model costs 25% more than the model with the median cost.
(2) The most expensive model costs $210 more than the model with the median cost.

\({\rm{costs}}\,\,:\,\,a < b < c\,\,\,\,\,\,\left[ \$ \right]\)

\(a + b + c = 3 \cdot 900 = 2700\,\,\,\left( * \right)\)

\(c\,\,\mathop > \limits^? \,\,1000\)


\(\left( 1 \right)\,\,c = {5 \over 4}b\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,a + {9 \over 5}c = 2700\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\)

\(\left( {**} \right)\,\,\,\,c \le 1000\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,{9 \over 5}c\,\, \le \,\,1800\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 2700 - {9 \over 5}c\,\, \ge \,\,2700 - 1800 = 900 \hfill \cr
\,b = {4 \over 5}c\,\, \le \,\,800 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,b \le 800 < 900 \le a\,\,\,{\rm{impossible}}\)


\(\left( 2 \right)\,\,c - b = 210\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,a + \left( {c - 210} \right) + c = 2700\,\,\,\,\, \Rightarrow \,\,\,\,\,a + 2c = 2910\,\,\,\,\mathop \Rightarrow \limits^{\left( {***} \right)} \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\)

\(\left( {***} \right)\,\,\,\,c \le 1000\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,a = 2910 - 2c\,\, \ge \,\,2910 - 2000 = 910 \hfill \cr
\,b = c - 210\,\, \le \,\,790 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,b \le 790 < 910 \le a\,\,\,{\rm{impossible}}\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: The average (arithmetic mean) cost of three computer models is $900.  [#permalink]

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New post 26 Jan 2019, 15:32
GinGMAT wrote:
The average (arithmetic mean) cost of three computer models is $900. If no two computers cost the same amount, does the most expensive model cost more than $1,000?

(1) The most expensive model costs 25% more than the model with the median cost.
(2) The most expensive model costs $210 more than the model with the median cost.


The OA uses the concept of assigning variables a, b, c, and plugging in "LT" or "less than" to set up multiple equations. Is there an easier method to accomplish this?


The alternative would be to try testing cases! You'd have to be careful, though, to make sure that your cases fit all of the constraints. The constraint in the problem itself is that the average is 900. Then there will be extra constraints from the statements. Let's try it:

Statement 1: Suppose that the median computer costs $1000, and the most expensive one is 25% more than that, or $1250. Make sure that's a valid case: for the average price to be $900, the sum of the three prices would have to be $2700. So, the cheapest computer would be $2700-$1000-$1250, or $450. Everything checks out! $450/$1000/$1250 is a valid case that gives us a 'yes' answer.

Now, when we test cases, the goal is always to try to get both a 'yes' and a 'no' answer. We have our 'yes' answer, so let's try to come up with a 'no'. We want the most expensive model to cost no more than $1000. I know that 25% more than 800 is 1000, so let's say that the median model costs $800. Then, the most expensive model is $1000, which is not more than 1000. (These are the biggest numbers we could test, though - we can't go bigger, or we'd get another 'yes' instead of a 'no'.)

Does this case make sense? Well, the cheapest computer would have to cost $2700-$800-$1000 = $900. That's a problem! The cheapest computer can't be more expensive than the median computer! So this case doesn't make sense. We tried to get a 'no' case and failed (and since we were using the biggest possible numbers, we know that any smaller case will have the same problem.) That means this statement is sufficient - only 'yes' cases will work.

Statement 2: Suppose that the median computer costs $1000, and the most expensive one is $1210. Then the cheapest would be $2700-$1000-$1210 = $490. Everything checks out and we have our 'yes'.

Let's try the same approach from the other statement to find a 'no'. The biggest values we can try here, while still getting a 'no', would be a median price of $790 and a high price of $1000. But that doesn't work correctly, because once again, the cheapest computer would have to cost at least $910! The cheapest computer can't cost more than the median computer, so our 'no' case doesn't make sense. Only 'yes' cases work here.

Since both statements only allow prices over $1000, they're each sufficient, and the answer is (D).
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Re: The average (arithmetic mean) cost of three computer models is $900.   [#permalink] 26 Jan 2019, 15:32
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