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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63% (02:49) correct 
