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In a certain company, the average (arithmetic mean) salary of finance

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In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post Updated on: 07 Oct 2018, 04:26
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E

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Question Stats:

29% (01:19) correct 71% (01:32) wrong based on 75 sessions

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In a certain company, the average (arithmetic mean) salary of finance executives is $x and the average (arithmetic mean) salary of marketing executives is $y. Is the average (arithmetic mean) salary of finance executives and marketing executives combined greater than $\(\frac{(x+y)}{2}\)?

(1) The number of the finance executives is less than the number of the marketing executives.
(2) x is $10,000 less than y.


Weekly Quant Quiz #3 Question No 2


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Originally posted by gmatbusters on 06 Oct 2018, 09:05.
Last edited by gmatbusters on 07 Oct 2018, 04:26, edited 3 times in total.
Renamed the topic and edited the question.
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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 06 Oct 2018, 09:06

Official Explanation



It is very clear that if the number of finance executives = number of marketing executives
Average = \(\frac{(x+y)}{2}\) ( tag me , if it is not clear)
Now the average would be shifted towards x if the number of finance executives is greater than number of marketing executives

Statement 1
: the number of finance executives < number of marketing executives
It means the Average tilts towards y but we don't know whether x is greater or smaller than y
NOT SUFFICIENT

Statement2
: x is $10,000 less than y
or x<y
but we do not know whether the number of finance executives is greater or smaller than number of marketing executives.
NOT SUFFICIENT

Combining Statement 1 & 2, we get
Average tilts towards y and y >x
hence, Average > \(\frac{(x+y)}{2}\)
SUFFICIENT

Answer C

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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 06 Oct 2018, 09:09
IS there something wrong with the question stem?
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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post Updated on: 06 Oct 2018, 09:16
B
statment 1 is not sufficient, the number of pepole in each group crosses out
statement 2 is sufficient because it provides information about which average is bigger

Originally posted by lostboy on 06 Oct 2018, 09:13.
Last edited by lostboy on 06 Oct 2018, 09:16, edited 1 time in total.
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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 06 Oct 2018, 09:16
avg of both = (N *finance avg. + n* market avg)/(N+n)

Option A: N >n but we do not relation between finance avg and market avg
so not sufficient

Option B: says finance avg < market avg
Not sufficient as well

Combining both
N* (market avg - 10000) + n *market avg /(N+n)

we can't conclusive says its more than sum of avg /2 or not..In some scenarios it is greater but rest it is smaller...

Hence E is the answer
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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 06 Oct 2018, 09:21
Total avg = xa+yb/a+b ==> when a=b, avg= x+y/2
1) NS
2) NS
1+2) a(y-10000)+ by / a+b ' even if a is less than a , the avg is less than x+y/2........Ans C
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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post Updated on: 06 Oct 2018, 09:50
(1) We know that finance people are less than marketing but we did not know if x was greater than or less than y.

(2) We know x is less than y but nothing about how many are getting x and how many are getting y.

Together, we know that there are more people getting than the people getting x so.

Average of all executives' salaries is > \(\frac{x+y}{2}\) as it will be more tilted towards y.
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Originally posted by MsInvBanker on 06 Oct 2018, 09:23.
Last edited by MsInvBanker on 06 Oct 2018, 09:50, edited 2 times in total.
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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 06 Oct 2018, 09:25
E ,not sufficient information to tell the values

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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 06 Oct 2018, 09:28
1
Let the total number of finance executives be F and that of marketing be M.

Now, as per question stem the average (arithmetic mean) salary of finance executives is $x and the average (arithmetic mean) salary of marketing executives is $y. Is the average (arithmetic mean)

So average (arithmetic mean) salary of finance executives and marketing executives combined together = (F x $x + M x $y)/(F+M)

We need to find if (F x $x + M x $y)/(F+M) > ( $ (x + y) / 2


From statement 1:

No idea about x or y hence insufficient

From statement 2:
No idea about F and M hence insufficient


Together we know,

F<M and x < y - 10,000

Hence sufficient. Answer C
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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 06 Oct 2018, 10:08
There was an issue with the original question and hence I am revisiting it post finishing the rest of the questions - apologies for the delayed response to Q2 but I feel it is only fair to allow a 10 min extension to it as it had a problem to begin with.

In my opinion Option E is the answer

Please find soln attached in image.

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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 06 Oct 2018, 10:21
Hey,

Thank you for the questions.

It maybe just me, but I couldn't understand the last part of your question
2(x+y)2(x+y)?
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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 06 Oct 2018, 10:39
Thanks for making the changes.

So,

Imagine this question in terms of weights

Imagine one finance person as average of all finance folks with salary x*

same for marketing person salary y*

St 1 tells us about number of people (more finance than marketing) but we know nothing about the relation between x and y

St 2 tells us about the relation bt x and y, but now we don't know relatively number of people

1 and 2 combine to give us info.

Finance X -------------------(x+y)/2-------------(Actual average)-------------Marketing Y

Actual average has to be closer to marketing side because there are more number of people there taking the overall average above (x+y)/2
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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 06 Oct 2018, 13:23
In a certain company, the average (arithmetic mean) salary of finance executives is $x and the average (arithmetic mean) salary of marketing executives is $y. Is the average (arithmetic mean) salary of finance executives and marketing executives combined greater than $(x+y)/2?
1) The number of the finance executives is less than the number of the marketing executives.
2) x is $10,000 less than y.

Weighted average will be closer to the value with the higher number of executives than to strict (x+y)/2

1) the weight average will be closer to the value of $y but $y could be anything larger or smaller than $x. Insufficient
2) Also insufficient because we don't know anything about the number of executives in each group. If there are the same number of both executives for example, the average salary combined will be equal to $(x+y)/2 (not greater than). If there are 20x more marketing executives, then the average will be greater than.
1+2) There are more marketing executives and they make a higher salary, therefore the average will be greater than (x+y)/2

C
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Re: In a certain company, the average (arithmetic mean) salary of finance  [#permalink]

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New post 10 Oct 2018, 06:20
gmatbusters wrote:
In a certain company, the average (arithmetic mean) salary of finance executives is $x and the average (arithmetic mean) salary of marketing executives is $y. Is the average (arithmetic mean) salary of finance executives and marketing executives combined greater than $\(\frac{(x+y)}{2}\)?

(1) The number of the finance executives is less than the number of the marketing executives.
(2) x is $10,000 less than y.


Weekly Quant Quiz #3 Question No 2



Picking numbers works aswell.

Let the number of finance executives be 0 and the number of marketing executives 1. \(x\) is 0$ and \(y\) is 10,000$.

Therefore the average of \(x\) and \(y\) would be \(\frac{(0*0+1*10000)}{1}=10,000$\) therefore bigger than \(\frac{(0+10000)}{2}=5000$\)

Hence, C
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Re: In a certain company, the average (arithmetic mean) salary of finance &nbs [#permalink] 10 Oct 2018, 06:20
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