Statistics -Pls help me to solve this problem

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Q:Matt gets a $1,000 commission on a big sale. This commission alone raises his average
commission by $150. If Matt's new average commission is $400, how many
sales has Matt made?

How to Solve this problem??

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by cans » Sun Sep 18, 2011 9:13 pm
let n sales. then total = 400n
Also previous average = 400-150=250
Thus (n-1)(250) + 1000 = 400n
10n - 10 +40 = 16n -> n=5. thus including this sale, he has made 5 sales in total..
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by Geva@EconomistGMAT » Sun Sep 18, 2011 9:49 pm
dhiren8182 wrote:Q:Matt gets a $1,000 commission on a big sale. This commission alone raises his average
commission by $150. If Matt's new average commission is $400, how many
sales has Matt made?

How to Solve this problem??
If you had included the answer choices, I would probably try to reverse plug in the answer choice back into the question (It's not clear whether the number of sales in the answer choices includes the new sale or not - I'm assuming that it does).

Let's assume that the answer is 5 - meaning that he made 5 sales with an average of $400 per sale. Total commissions is 5*400 = 2000. Subtract the 1000 for the latest sale, and you are left with $1000 for the remaining 4 sales - for an average of $250 per sale. Since everything checks out, this is the right answer choice.

Not that you need to test any other answer choice if you start with 5 as above, but If you were to start with any different answer choice and do the same process, you would find some flaw. For example, if you assumed that the answer was 6, then

6 sales with a $400 commission =$2400 total.
Subtract the $1000 for the latest sale, and you are left with $1400 for 5 sales - the average does not equal $250.
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by bijoyajj » Sun Sep 18, 2011 10:46 pm
n*150 = 600 .. there for n= 4 .. so total sales = 5

New avg is 400... so the last transaction is 600 more than the new avg.. so old transactions sum of avgs would be -600 (then only it balances to 400).. In the question its given that it needs 150 per avg increase.. hence

n*150 = 600 where n is the number of old transactions.. So total transactions is n+1..

** Just draw a line with 400 as avg and put +600 above and then starts thinking for this kinda probs

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by khushbumerchant » Fri Jan 24, 2014 11:39 pm
bijoyajj wrote:n*150 = 600 .. there for n= 4 .. so total sales = 5

New avg is 400... so the last transaction is 600 more than the new avg.. so old transactions sum of avgs would be -600 (then only it balances to 400).. In the question its given that it needs 150 per avg increase.. hence

n*150 = 600 where n is the number of old transactions.. So total transactions is n+1..

** Just draw a line with 400 as avg and put +600 above and then starts thinking for this kinda probs
Can someone explain me what is this above method used? Is this Change of the mean formula?? Also how was this calculated?

thanks in advance!

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by GMATGuruNY » Sat Jan 25, 2014 4:24 am
dhiren8182 wrote:Q:Matt gets a $1,000 commission on a big sale. This commission alone raises his average
commission by $150. If Matt's new average commission is $400, how many
sales has Matt made?

How to Solve this problem??
Since the new average of $400 represents a $150 increase from the previous average, the average commission on the previous sales = 400-150 = 250.

This is a WEIGHTED AVERAGE/MIXTURE problem.
Let P = previous sales and N = new sale.

Average commission for P = 250.
Commission for N = 1000.
Average commission for the MIXTURE of P and N = 400.
To determine the ratio of P to N in the mixture, use ALLIGATION.

Step 1: Plot the 3 averages on a number line, with the averages for P and N on the ends and the average for the mixture in the middle.
P 250-------------400--------------1000 N

Step 2: Calculate the distances between the averages.
P 250-----150-----400-----600-----1000 N

Step 3: Determine the ratio in the mixture.
The required ratio of P to N is equal to the RECIPROCAL of the distances in red:
P:N = 600:150 = 4:1.

Since P:N = 4:1, there are 4 previous sales for the 1 new sale, implying a total of 5 sales.

An alternate approach would be PLUG IN THE ANSWERS, which the GMAT would provide:
Let's say that the answer choices were as follows:
4
5
6
7
8

When we plug in the answers:
The average for the previous sales = $250
The commission on the new sale = $1000.
The average for all of the sales = $400.

Answer choice C: 6
Here, P=5 and N=1, yielding the following average for all 6 sales:
(5*250 + 1*1000)/6 = 2250/6 = less than 400.
Eliminate C.

The commission on the new sale (1000) is GREATER than the previous average (250).
Thus, to increase the average for all of the sales to $400, the $1000 new sale must be given MORE WEIGHT in the mixture.
Implication:
There must be FEWER previous sales, implying that the total number of sales must be LESS than 6.
Eliminate D and E.

Answer choice B: 5
Here, P=5 and N=1, yielding the following average for all 5 sales:
(4*250 + 1*1000)/5 = 2000/5 = 400.
Success!

The correct answer is B.

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https://www.beatthegmat.com/ratios-fract ... 15365.html
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by khushbumerchant » Sun Jan 26, 2014 1:00 pm
GMATGuruNY wrote:
Since the new average of $400 represents a $150 increase from the previous average, the average commission on the previous sales = 400-150 = 250.

This is a WEIGHTED AVERAGE/MIXTURE problem.
Let P = previous sales and N = new sale.

Average commission for P = 250.
Commission for N = 1000.
Average commission for the MIXTURE of P and N = 400.
To determine the ratio of P to N in the mixture, use ALLIGATION.

Step 1: Plot the 3 averages on a number line, with the averages for P and N on the ends and the average for the mixture in the middle.
P 250-------------400--------------1000 N

Step 2: Calculate the distances between the averages.
P 250-----150-----400-----600-----1000 N

Step 3: Determine the ratio in the mixture.
The required ratio of P to N is equal to the RECIPROCAL of the distances in red:
P:N = 600:150 = 4:1.

Since P:N = 4:1, there are 4 previous sales for the 1 new sale, implying a total of 5 sales.
Ok I understood above calculation. However, I have further query.

Out here, how can one compare an average commission (P = 250) to actual commission amount (N = $1000) on a number line, as that seems not-so-conceptually correct to me. I mean, to me out here, one would always compare average commission i.e. 250 (old average) & 400 (new average).

Also, just for more clear understanding of this concept could you please share another example, wherein one could similarly make comparisons between average & other element like actual commission amount here, to arrive at an answer with this easy & fast calculations as mentioned above.

thanks in advance!

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by GMATGuruNY » Sun Jan 26, 2014 2:32 pm
khushbumerchant wrote: Ok I understood above calculation. However, I have further query.

Out here, how can one compare an average commission (P = 250) to actual commission amount (N = $1000) on a number line, as that seems not-so-conceptually correct to me. I mean, to me out here, one would always compare average commission i.e. 250 (old average) & 400 (new average).
In the problem above, the average for the P commissions is being combined with the average for the N commissions to yield a weighted average for the MIXTURE of P and N.
It just so happens that there is ONLY ONE N commission.

Average = sum/number.
Since there is only one N commission, the number of N commissions = 1.
Since the value of this one commission is $1000, the sum for the N commissions = $1000.
Thus, the average for the N commissions = sum/number = 1000/1 = 1000.

In short:
Because there is only one N commission, the AVERAGE for the N commissions is the equal to the actual VALUE of the sole N commission:
$1000.
Also, just for more clear understanding of this concept could you please share another example, wherein one could similarly make comparisons between average & other element like actual commission amount here, to arrive at an answer with this easy & fast calculations as mentioned above.

thanks in advance!
PS207 in the OG12 is a very similar problem:
For the past n days, the average (arithmetic mean) daily production at a company was 50 units If today's production of 90 units raises the average to 55 units per day, what is the value of n?

a) 30
b) 18
c) 10
d) 9
e) 7
This is a WEIGHTED AVERAGE/MIXTURE problem.
Let N = average daily production for the past n days and T = average daily production for today.

N = 50.
T = 90
Average daily production for the MIXTURE of N and T = 55.
To determine the ratio of N and T in the mixture, use ALLIGATION.

Step 1: Plot the 3 averages on a number line, with the averages for N and T on the ends and the average for the mixture in the middle.
N 50------------55-------------90 T

Step 2: Calculate the distances between the averages.
N 50-----5-----55------35-----90 T

Step 3: Determine the ratio in the mixture.
The required ratio of N to T is equal to the RECIPROCAL of the distances in red:
N:T = 35:5 = 7:1.

Since N:T = 7:1, the mixture of N and T is composted of 7 days with an average production of 50 units and 1 day with an average production of 90 units.
Thus, n=7.

The correct answer is E.
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