Q1) What is the median number of employees assigned per project for the projects at
Company Z?
(1) 25 percent of the projects at Company Z have 4 or more employees assigned to
each project.
(2) 35 percent of the projects at Company Z have 2 or fewer employees assigned to
each project
Why the OA is C. Please explain
Q2)One kilogram of a certain coffee blend consists of x kilogram of type I coffee and y
kilogram of type II coffee. The cost of the blend is C dollars per kilogram, where C =
6.5x + 8.5y. Is x < 0.8?
(1) y > 0.15
(2) C >=7.30
OA is b how??
DS some querries
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- codesnooker
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Q2)One kilogram of a certain coffee blend consists of x kilogram of type I coffee and y
kilogram of type II coffee. The cost of the blend is C dollars per kilogram, where C =
6.5x + 8.5y. Is x < 0.8?
(1) y > 0.15
(2) C >=7.30
OA is (B) because...
Taking 1st statement, if y=0.16 then x>0.8 but if y=0.3 then x<0.8
Hence insufficient.
Taking 2nd statement,
C = 6.5x + 8.5y
According to second statement, the minimum cost of blended coffee would be 7.30
Let's check the value of C when x = 0.8
C = 6.5*0.8 + 8.5*0.2 = 6.9
Now, let's suppose that x>0.8, then by placing x = 0.81, the value of C is,
C = 6.5*0.81 + 8.5*0.19 = 6.88
Since by increasing the value of x reduces is the value of C and minimum value of C should be equal to 7.3, therefore, x should be always less than 0.8
Hence II statement is alone sufficient. Choose (B).
Hope this clears your doubt.
kilogram of type II coffee. The cost of the blend is C dollars per kilogram, where C =
6.5x + 8.5y. Is x < 0.8?
(1) y > 0.15
(2) C >=7.30
OA is (B) because...
Taking 1st statement, if y=0.16 then x>0.8 but if y=0.3 then x<0.8
Hence insufficient.
Taking 2nd statement,
C = 6.5x + 8.5y
According to second statement, the minimum cost of blended coffee would be 7.30
Let's check the value of C when x = 0.8
C = 6.5*0.8 + 8.5*0.2 = 6.9
Now, let's suppose that x>0.8, then by placing x = 0.81, the value of C is,
C = 6.5*0.81 + 8.5*0.19 = 6.88
Since by increasing the value of x reduces is the value of C and minimum value of C should be equal to 7.3, therefore, x should be always less than 0.8
Hence II statement is alone sufficient. Choose (B).
Hope this clears your doubt.
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IMO q1) Ans must be E.
As we dont know how the rest of the 40% employees are allocated.
Please correct me if I am wrong.
As we dont know how the rest of the 40% employees are allocated.
Please correct me if I am wrong.
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We do know, in fact. If 25% have four or more employees, and 35% have two or fewer employees, then the remaining 40% must have exactly three employees. That ensures that the median will be 3.Tryingmybest wrote:IMO q1) Ans must be E.
As we dont know how the rest of the 40% employees are allocated.
Please correct me if I am wrong.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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The median is the number that stands in the middle, if you order the number from lowest to greatest. If there are two numbers in the middle, its their average.
For the problem above lets assume you have 100 projects. Then you get
number of projects number of employees
25 >=4
40 3
35 <=2
So, from the statements you can be sure that the "number in the middle" is 3.
(Note that the table above is a frequency distribution, so you have 35*(2 or less), 25*(4 or more) and 40*3)
For the problem above lets assume you have 100 projects. Then you get
number of projects number of employees
25 >=4
40 3
35 <=2
So, from the statements you can be sure that the "number in the middle" is 3.
(Note that the table above is a frequency distribution, so you have 35*(2 or less), 25*(4 or more) and 40*3)
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