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# Last year the average (arithmetic mean) salary of the 10

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Joined: 12 Jan 2012
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Last year the average (arithmetic mean) salary of the 10 [#permalink]

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05 Feb 2012, 09:58
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68% (01:33) correct 32% (00:30) wrong based on 77 sessions

Last year the average (arithmetic mean) salary of the 10 employees of Company X was $42,800. What is the average salary of the same 10 employees this year? (1) For 8 of the 10 employees, this year’s salary is 15 percent greater than last year’s salary. (2) For 2 of the 10 employees, this year’s salary is the same as last year’s salary. [Reveal] Spoiler: Official answer is E, but I think is C given that using both statements: of 10 employees, 2 kept the same salary but 8 increases by 15%, so the average salary would increase by 1.15*0.8 + 1.0*0.2, then new average salary would be: 42,800*(1.15*0.8 + 1.0*0.2) 42,800*1.12 47,936 Am I right or what am I missing? OPEN DISCUSSION OF THIS QUESTION IS HERE: last-year-the-average-arithmetic-mean-salary-of-the-135830.html [Reveal] Spoiler: OA Senior Manager Status: May The Force Be With Me (D-DAY 15 May 2012) Joined: 06 Jan 2012 Posts: 288 Location: India Concentration: General Management, Entrepreneurship Followers: 3 Kudos [?]: 272 [0], given: 16 Re: Error in Official Guide 12th? [#permalink] ### Show Tags 05 Feb 2012, 10:07 Hi, I think the OA is correct. You have the average which is$ 42,800 from which the sum of total salaries paid can be calculated

Now statement 1 says 8 workers for a hike of 15%. We dont have individual salary's hence increase cannot be accurately applied thus leading to an incorrect average.

=> INSUFFICIENT

Statement 2 says 2 workers did not get a raise. again we don't know anything about the rest of the workers

=> INSUFFICIENT

Combining both we know that 8 got a hike & 2 didn't but again we dont know which ones, hence you cant calculate the mean.

=> INSUFFICIENT Hence E is correct

Hope this helps
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Intern
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Re: Error in Official Guide 12th? [#permalink]

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05 Feb 2012, 10:42
They are asking for new average salary.
If you increase all workers salary by 15%, no matter what is the salary for each one and maintaining the same number of workers, then the new average salary will be exactly 15% higher than past year.
Example:
Worker 1: 10,000
Worker 2: 20,000
Worker 3: 20,000

Avg: 16,667

With 15% Increase

Worker 1: 11,500
Worker 2: 23,000
Worker 3: 23,000

Avg: 19,167 = 16,667*1.15 (increased the 15%)

Given that in the question just 80% of workers increase their salaries by 15% and 2 maintained the same, you don't multiply the average by 15% or 1.15 as I did in the example, you weight this 15% by just 80%, then new average salary would be:

Last Year Salary * (1.15*0.8 + 1.00*0.2)
1.00 weighted by 0.2 because only 20% of the workers maintained the salary, the rest (80%) increased by 15% (1.15)
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Re: Error in Official Guide 12th? [#permalink]

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05 Feb 2012, 11:09
Okay lets assume that 6 out of 10 of the salary were 42800. The last 4 salary will be 2 people with $5 and 2 people with 85600 salary. If the two highest paying employee plus the 6 average paying employee got a 15% raise then the average would be (6*42800*1.15 + 2*85600*1.15 + 2*5)/10 = 49219.85. If the two lowest employee plus the 6 average employee got the 15% raise then (6*42800*1.15 +2*85600+2*5*1.15)/10 = 46652.15. What you're assuming from your equation is that every number will weigh the same, so that's why 8 of the numbers are worth 80%. This is wrong because weight is determine by the magnitude of the number compare to others. 8 number could weigh only 60% of the value or 99% of the total value. Without knowing how much the weight of those 8 numbers represent there is no solution. Hence E Math Expert Joined: 02 Sep 2009 Posts: 37547 Followers: 7390 Kudos [?]: 99176 [0], given: 11008 Re: Error in Official Guide 12th? [#permalink] ### Show Tags 05 Feb 2012, 12:18 Latinapp wrote: Hi guys, maybe I'm wrong but I disagree with the official answer: Last year the average (arithmetic mean) salary of the 10 employees of Company X was$42,800. What is the average salary of the same 10 employees this year?
(1) For 8 of the 10 employees, this year’s salary is 15 percent greater than last year’s salary.
(2) For 2 of the 10 employees, this year’s salary is the same as last year’s salary.

Official answer is E, but I think is C given that using both statements: of 10 employees, 2 kept the same salary but 8 increases by 15%, so the average salary would increase by 1.15*0.8 + 1.0*0.2, then new average salary would be:

42,800*(1.15*0.8 + 1.0*0.2)
42,800*1.12
47,936

Am I right or what am I missing?

Last year the average (arithmetic mean) salary of the 10 employees of Company X was $42,800. What is the average salary of the same 10 employees this year? (1) For 8 of the 10 employees, this year’s salary is 15 percent greater than last year’s salary. (2) For 2 of the 10 employees, this year’s salary is the same as last year’s salary. Given: total salary is$42,800*10. Question: new total salary=?

Each statement together is not sufficient.

When taken together. Consider two possible cases:
2 lowest salaries didn't change and 8 highest salaries increased by 15%;
2 highest salaries didn't change and 8 lowest salaries increased by 15%;

Ask yourself, would new total salary be the same for both cases? No, because increase in amount for the first case will be more than increase in amount for the second case. Hence even taken together statements are not sufficient.

Hope it's clear.
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Re: Last year the average (arithmetic mean) salary of the 10 [#permalink]

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29 Sep 2013, 02:37
I don't think your reasoning is right.. Lets say the two highest salaries are h1 and h2 and the two lowest salaries are l1,l2

previous sum = h1+h2+l1+l2

new sum = h1 +(h1*.15) + h2+(h2*.15) + l1+(l1*.15) + l2+(l2*.15)

= h1+h2+l1+l2 +(h1*.15) + (h2*.15) + (l1*.15) + (l2*.15)

= h1+h2+l1+l2 +.15(h1+h2+l1+l2)

Replacing "h1+h2+l1+l2" as previous sum it becomes,

new sum = previous sum +.15 *previous sum

Therefore,

new average = new sum/4

Which proves that answer should be "C"
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Re: Last year the average (arithmetic mean) salary of the 10 [#permalink]

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29 Sep 2013, 12:52
bkhan wrote:
I don't think your reasoning is right.. Lets say the two highest salaries are h1 and h2 and the two lowest salaries are l1,l2

previous sum = h1+h2+l1+l2

new sum = h1 +(h1*.15) + h2+(h2*.15) + l1+(l1*.15) + l2+(l2*.15)

= h1+h2+l1+l2 +(h1*.15) + (h2*.15) + (l1*.15) + (l2*.15)

= h1+h2+l1+l2 +.15(h1+h2+l1+l2)

Replacing "h1+h2+l1+l2" as previous sum it becomes,

new sum = previous sum +.15 *previous sum

Therefore,

new average = new sum/4

Which proves that answer should be "C"

That's not correct. This is OG13 question ans the official answer is E, not C.

Notice that we don't know which 8 salaries increased and which 2 salaries remained the same.

If the salaries were {0, 0, 0, 0, 0, 0, 0, 0, 0, $42,800*10} and among the 8 salaries that increased there was a salary of$42,800*10 you get one answer (15% increase in total salary) and if among the 8 salaries that increased there was NOT a salary of \$42,800*10 you get the different answer (0% increase in total salary).

In case of any further questions please post here: last-year-the-average-arithmetic-mean-salary-of-the-135830.html
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Re: Last year the average (arithmetic mean) salary of the 10   [#permalink] 29 Sep 2013, 12:52
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