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Last year, the five employees of company X took an average

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Last year, the five employees of company X took an average [#permalink]

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22 Jan 2010, 12:41
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Last year, the five employees of company X took an average of 16 vacation days each. What was the average number of vacation days taken by the same employees this year?

(1) Three employees had a 50% increase in thier number of vacation days and two employees has a 50% decrease
(2) Three employees had 10 more vacation days each , and two employees has 5 fewer vacation days each.
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Aug 2012, 01:28, edited 2 times in total.
Edited the question.
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22 Jan 2010, 13:08
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5 employees
16 average vacation days
Find: average this year

The easy one first:

2) Average = (e1 + e2 + e3 + e4 + e5) / 5 = (e1 + e2 + e3 + e4 + e5 + 10 + 10 + 10 -5 -5) / 5
= (e1 + e2 + e3 + e4 + e5 + 20) / 5
From the knowns, we know (e1 +... e5) / 5 = 16, so e1+...e5 = 80
Thus, new average = 80 + 20 / 5 = 20
So 2 alone is sufficient.

The hard one:
1) Average = (e1 + e2 + e3 + e4 + e5) / 5
Because the option gives you a percent increase, you can only find an answer IF e1 through e5 are all the same value (I.e. 16). However, as you are not given this information, 1 alone is not sufficient.
For example, take (4 + 8 + 16 + 20 + 32) / 5 = 16.
A 50% increase in e1, e2, and e3 (6+12+24) is different than a 50% increase in e3, e4, and e5 (24+30+48).
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22 Jan 2010, 13:18
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GMAT10 wrote:
Last year, the five employees of company X took an average of 16 vacation days each.what was the average number of vacation days taken by the same employees this year?

1) Three employees had a 50% increase in thier number of vacation days and two employees has a 50% decrease
2) Three employees had 10 more vacation days each , and two employees has 5 fewer vacation days each.

OA is B

(1) $$\frac{x+y}{5}=16$$, where x is the # vacations days taken by the three employees mentioned and y is the the # of days taken by two employees mentioned. --> Question: $$\frac{1.5x+0.5y}{5}=?$$, can not be determined. Not sufficient.

(2) $$\frac{x+y}{5}=16$$ --> $$x+y=80$$, where x is the # vacations days taken by the three employees mentioned and y is the the # of days taken by two employees mentioned. --> Question: $$\frac{(x+3*10)+(y-2*5)}{5}=?$$ --> $$\frac{(x+y)+20}{5}=\frac{80+20}{5}=20$$. Sufficient.

One thing to mention here: stem says that vacation days were "taken" by employees and statements say that employees "had" (more, less) vacation days. What if they were given these vacation days but they didn't take them? But as the credited OA is B, then we should assume that all the vacation days that were given to the employees were used. Though kind of strange to "assume" something in GMAT.
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15 Feb 2010, 06:39
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1) Three employees had a 50% increase in thier number of vacation days and two employees has a 50% decrease
2) Three employees had 10 more vacation days each , and two employees has 5 fewer vacation days each.

given 5 employees 16 days average so 80 days leave last year

stmt 1 doesn't tell for which employess 50% increase and for which employess 50% decrease in vacation days

so not sufficient

stmt 2
if 3 employees had 10 more vacation days each then total vacation days is increased 80+3*10=>110

and if 2 employee vacation days decreased by 5 days each then total vacation days will come down to 110-2*5 =>100 days

so average vac. days for emp. this year will be 100/5 => 20 days

so ans is B
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17 Feb 2010, 08:18
I evaluated statement 1 intuitively:
the 50% increase could be referring to those with the most vacation days, so the average would increase
likewise, it could refer to those with the least vacation days, in which case the average would decrease
no way to tell, so 1 = insufficient

statement 2:
5x16 = 80 (total)
take away 3x10 and add 2x5: 100
100/5 = 20 sufficient
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20 Feb 2010, 03:46
GMAT10 wrote:
Last year, the five employees of company X took an average of 16 vacation days each.what was the average number of vacation days taken by the same employees this year?

1) Three employees had a 50% increase in thier number of vacation days and two employees has a 50% decrease
2) Three employees had 10 more vacation days each , and two employees has 5 fewer vacation days each.

Clearly B...

We have Sum of all vacations taken by 5 employess = 16x5=80
S1. gives % increases of 3 employees and % decrease of 2 employees... This isn't sufficient. INSUFF...
S2: This can gives us the average as we need to add 10 and subtract 5 from 80 and then divide by 5... SUFF
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21 Aug 2012, 12:15
Bunuel wrote:
GMAT10 wrote:
Last year, the five employees of company X took an average of 16 vacation days each.what was the average number of vacation days taken by the same employees this year?

1) Three employees had a 50% increase in thier number of vacation days and two employees has a 50% decrease
2) Three employees had 10 more vacation days each , and two employees has 5 fewer vacation days each.

OA is B

(1) $$\frac{x+y}{5}=16$$, where x is the # vacations days taken by the three employees mentioned and y is the the # of days taken by two employees mentioned. --> Question: $$\frac{1.5x+0.5y}{5}=?$$, can not be determined. Not sufficient.

(2) $$\frac{x+y}{5}=16$$ --> $$x+y=80$$, where x is the # vacations days taken by the three employees mentioned and y is the the # of days taken by two employees mentioned. --> Question: $$\frac{(x+3*10)+(y-2*5)}{5}=?$$ --> $$\frac{(x+y)+20}{5}=\frac{80+20}{5}=20$$. Sufficient.

One thing to mention here: stem says that vacation days were "taken" by employees and statements say that employees "had" (more, less) vacation days. What if they were given these vacation days but they didn't take them? But as the credited OA is B, then we should assume that all the vacation days that were given to the employees were used. Though kind of strange to "assume" something in GMAT.

Hi . .Can you pls explain the variables x and y and how you arrived at the equations ?
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22 Aug 2012, 01:38
Bunuel wrote:
GMAT10 wrote:
Last year, the five employees of company X took an average of 16 vacation days each.what was the average number of vacation days taken by the same employees this year?

(1) Three employees had a 50% increase in thier number of vacation days and two employees has a 50% decrease
(2) Three employees had 10 more vacation days each , and two employees has 5 fewer vacation days each.

OA is B

(1) $$\frac{x+y}{5}=16$$, where x is the # vacations days taken by the three employees mentioned and y is the the # of days taken by two employees mentioned. --> Question: $$\frac{1.5x+0.5y}{5}=?$$, can not be determined. Not sufficient.

(2) $$\frac{x+y}{5}=16$$ --> $$x+y=80$$, where x is the # vacations days taken by the three employees mentioned and y is the the # of days taken by two employees mentioned. --> Question: $$\frac{(x+3*10)+(y-2*5)}{5}=?$$ --> $$\frac{(x+y)+20}{5}=\frac{80+20}{5}=20$$. Sufficient.

One thing to mention here: stem says that vacation days were "taken" by employees and statements say that employees "had" (more, less) vacation days. What if they were given these vacation days but they didn't take them? But as the credited OA is B, then we should assume that all the vacation days that were given to the employees were used. Though kind of strange to "assume" something in GMAT.

Hi . .Can you pls explain the variables x and y and how you arrived at the equations ?

(1) Three employees had a 50% increase in their number of vacation days and two employees has a 50% decrease.

Say x is the total # vacations days taken last year by the three employees mentioned and y is the total # of days taken last year by two employees mentioned.

Now, since we are told that "last year the five employees of company X took an average of 16 vacation days each", then (total # of vacation days)/(# of employees)=(x+y)/5=16.

Next, the first statement says that "the three employees had a 50% increase in their number of vacation days", so those three had 1.5x vacation days this year, and the other two had 0.5y vacation days this year. We need the new average for this year, so the value of (1.5x+0.5y)/5.

The same way for the second statement.

Hope it's clear.
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Re: Last year, the five employees of company X took an average [#permalink]

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28 May 2014, 09:19
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Re: Last year, the five employees of company X took an average [#permalink]

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A non algebraic approach:

Total number of vacationing days => $$16 * 5 = 80 days$$

A) Insufficient:
Let the employees be segregated as $$a--b--16--x--y$$ (where x & y have taken more than 16 days of vacationing last year and a & b have taken less than 16 days of vacationing last year). Nowhere does it mention the three/two are < 16 days guy (ie a/b) or > 16 days guy (ie x/y). It only mentions that 3 employees had a 50% incre. and 2 employees 50% decrease.

B) Sufficient:
Three guys have taken more than 10 days of last years' average, each (each is important here!). Therefore, these three have taken $$16*3 (avg.) + 10*3 (this years' incre.) =$$ $$78 days$$ of vacationing. Similarly, the remaining two have taken $$22 days$$ of vacationing $$[ie 2*(16-5)=11].$$

Avg. days of vacationing this year = $$\frac{78+22}{5}$$=$$20 days.$$

Ans: B

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Last year, the five employees of company X took an average [#permalink]

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08 Nov 2015, 17:25
Bunuel wrote:
GMAT10 wrote:
Last year, the five employees of company X took an average of 16 vacation days each.what was the average number of vacation days taken by the same employees this year?

1) Three employees had a 50% increase in thier number of vacation days and two employees has a 50% decrease
2) Three employees had 10 more vacation days each , and two employees has 5 fewer vacation days each.

OA is B

(1) $$\frac{x+y}{5}=16$$, where x is the # vacations days taken by the three employees mentioned and y is the the # of days taken by two employees mentioned. --> Question: $$\frac{1.5x+0.5y}{5}=?$$, can not be determined. Not sufficient.

(2) $$\frac{x+y}{5}=16$$ --> $$x+y=80$$, where x is the # vacations days taken by the three employees mentioned and y is the the # of days taken by two employees mentioned. --> Question: $$\frac{(x+3*10)+(y-2*5)}{5}=?$$ --> $$\frac{(x+y)+20}{5}=\frac{80+20}{5}=20$$. Sufficient.

One thing to mention here: stem says that vacation days were "taken" by employees and statements say that employees "had" (more, less) vacation days. What if they were given these vacation days but they didn't take them? But as the credited OA is B, then we should assume that all the vacation days that were given to the employees were used. Though kind of strange to "assume" something in GMAT.

Hello Sir,

Option B never mentions that the information is for this year, this might as well be of the last year. In the case the we dont have any information for the this year and so B is not the answer. thoughts?
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Re: Last year, the five employees of company X took an average [#permalink]

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14 Jan 2016, 10:57
Bunuel wrote:
GMAT10 wrote:
Last year, the five employees of company X took an average of 16 vacation days each.what was the average number of vacation days taken by the same employees this year?

1) Three employees had a 50% increase in thier number of vacation days and two employees has a 50% decrease
2) Three employees had 10 more vacation days each , and two employees has 5 fewer vacation days each.

OA is B

(1) $$\frac{x+y}{5}=16$$, where x is the # vacations days taken by the three employees mentioned and y is the the # of days taken by two employees mentioned. --> Question: $$\frac{1.5x+0.5y}{5}=?$$, can not be determined. Not sufficient.

(2) $$\frac{x+y}{5}=16$$ --> $$x+y=80$$, where x is the # vacations days taken by the three employees mentioned and y is the the # of days taken by two employees mentioned. --> Question: $$\frac{(x+3*10)+(y-2*5)}{5}=?$$ --> $$\frac{(x+y)+20}{5}=\frac{80+20}{5}=20$$. Sufficient.

One thing to mention here: stem says that vacation days were "taken" by employees and statements say that employees "had" (more, less) vacation days. What if they were given these vacation days but they didn't take them? But as the credited OA is B, then we should assume that all the vacation days that were given to the employees were used. Though kind of strange to "assume" something in GMAT.

Hi Bunuel, this very reasoning made me select option E. The GMAT uses such vague language sometimes it becomes really hard to try to understand what the question is truly asking. So, sometimes you get the right answer and sometimes you don't. What to do in such situations?
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Last year, the five employees of company X took an average [#permalink]

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15 Jul 2016, 02:55
GMAT10 wrote:
Last year, the five employees of company X took an average of 16 vacation days each. What was the average number of vacation days taken by the same employees this year?

(1) Three employees had a 50% increase in thier number of vacation days and two employees has a 50% decrease
(2) Three employees had 10 more vacation days each , and two employees has 5 fewer vacation days each.

Since we are talking about averages; it is quite safe to assume :-
There can be a great range in the elements of the set
or
There can be no difference in the elements of set.

Statement 1)Three employees had a 50% increase in thier number of vacation days and two employees has a 50% decrease
Insufficient :- We don't know which employees increased or decreased their vacations percent.
Case 1) The employees that took more holidays last year increased their holidays by 50%
Case 2) The employee who took less holidays last year increased their holidays by 50 %

Cannot reach a definite value

(2) Three employees had 10 more vacation days each , and two employees has 5 fewer vacation days each.
Sufficient
Last year (a)+(b)+(c)+(d)+(e)=80

this year (a+10)+(b+10)+(c+10)+(d-5)+(e-5)?

==> a+b+c+d+e+30-10
==> a+b+c+d+e+20
==>80+20
==> 100
average number of vacation per employee=100/5= 20 holidays per employee

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Last year, the five employees of company X took an average [#permalink]

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15 Jul 2016, 03:59
simple B

no calculation also required.

1) its clearly infsuff we don't know each employee vacation days .

2)avg=16 for each employee

10+10+10-5-5=20

20/5=4 increase avg by 4 from 16.Easy
Last year, the five employees of company X took an average   [#permalink] 15 Jul 2016, 03:59
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