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29 May 2024, 20:51
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1: equal to $100,000.
2: less than $90,000.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
48% (02:16) correct
52%
(02:44)
wrong
based on 552
sessions
History
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Companies X, Y and Z are located in the same town and are involved in providing consultancy to firms involved in construction business. The median annual salary of employees at Company X is $10,000 more than that of employees at Company Y but less than the median annual salary of employees at Company Z. The median salary of at least one of the company is $100,000 or more.
Statement: If the median annual salary at Company Z is ___1___, then the median annual salary at Company Y is ___2___.
Select for 1 and for 2 the options that complete the statement so that it most accurately reflects the information provided and would be true under all circumstances. Make only two selections, one in each column.
Statement: If the median annual salary at Company Z is ___1___, then the median annual salary at Company Y is ___2___.
Select for 1 and for 2 the options that complete the statement so that it most accurately reflects the information provided and would be true under all circumstances. Make only two selections, one in each column.
| 1 | 2 | |
| greater than $110,000. | ||
| greater than $100,000. | ||
| equal to $100,000. | ||
| greater than $90,000. | ||
| equal to $90,000 | ||
| less than $90,000. |
ShowHide Answer
Official Answer
1: equal to $100,000.
2: less than $90,000.
30 May 2024, 23:06
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chetan2u
Let the median annual salary of employees of company X, Y and Z be X, Y and Z respectively......
The median annual salary of employees at Company X is $10,000 more than that of employees at Company Y => X = Y+10000
but less than the median annual salary of employees at Company Z. => X<Z
The median salary of at least one of the company is $100,000 or more., so \(Z\geq 100000\)
Now, in the statement we are comparing Z and Y...
X<Z => Y+10000<Z or Z-Y>10000
Look for options that, without any ambiguity, give you the above relation.
A quick observation can save a lot of time: Any option that does not limit the upper limit value of Z, can have two or more values for Y. For example: If Z is greater than 100000, Y could be greater than 100000 too ( Z=150000 and Y = 120000) or Y could be less than 100000 ( Z=150000 and Y = 90000)
Discard options 1, 2 and 4
Next, Z is at least 100000, so discard options 4, 5 and 6.
Only possibility is Z = $100,000, that is option 3.
So, Z-Y > 10,000 or 100,000-Y>10,000 or Y<90,000. Hence option 6.
anish777
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31 May 2024, 07:28
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Hello chetan2u
I have a question to your solution approach. If we fix the lower salary as well, lets say to 90000 for 2, then Z can have salary greater than 100,000, which also logically follows from what is given in the question stem. Just as you are putting a limit to Z's salary, I'm putting the limit to Y's. Can you please let me know what's wrong in my approach?
I have a question to your solution approach. If we fix the lower salary as well, lets say to 90000 for 2, then Z can have salary greater than 100,000, which also logically follows from what is given in the question stem. Just as you are putting a limit to Z's salary, I'm putting the limit to Y's. Can you please let me know what's wrong in my approach?
