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555-605 Level|   Statistics and Sets Problems|                        
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Bunuel
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The first 6 paychecks are for 750 each and then next 20 paychecks are for 780 each. The average will be closer to 780. Eliminate options A, B and C.

How do we guesstimate the right option between D and E. Upon solving, we get 773 but the calculations are tedious. Any good way to eliminate 775?
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Option D.
A carafe paycheck=Total amt./total no. Of paychecks
Total amt.=750*6+780*20=20100
Total no. Of paychecks=26
Avg.=20100/26=773 approx

Posted from my mobile device
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the solutions posted thus far are good but lots of computation, an area where i tend to make mistakes. here is how i broke it down.

6 checks are at 750. 20 checks are at 780.

difference between these checks is 30.

20 checks * 30 = 600 (easy math)
when removing 600 from the latter checks, all checks are now 'evenly filled' at 750, as if they were glasses filled with 750ml of water, for example.

what's an even way to distribute the 600 among all of the 26 checks? ==> 600/26, but this is also a tedious calculation
however, 600/30 is 20, since our actual denominator is smaller (and 600 numerator the same) we know that the resulting number will be just a little above 770 (750 + 20)

773 is more 'little' above 750 than 775.
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hi,
there are various people in the thread asking for a short cut to estimate between 773 and 775 as answer.
it is easy by seeing the spread of the number of cheques and of the value of cheques..
for eg the two values of cheques are 750 and 780..
the spread away from average if it is 775 is 25:5 or 5:1...
the no of cheques should be in ratio 1:5 or x:5x... here if x=6, 5x=30.. this means you require 30 cheques of 780 if there are 6 cheques of 770 to get an avg of 775..
however we have only 20 cheques, therefore the avg has to be less than 775 .. leaving 773 as answer..
I hope it is slightly clear to people, earlier or now, in doubt of choosing between the two figures
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IMO, the simplest way to deal with this type of questions is by calculating the weighted average.

750..................780 (diff is 30)
6......................20(total=26)
so the distance of 30 will be divided in the respective opposite weights.
So wted avg= 750+ (20/26)*30= 773
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Bunuel
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Last year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $30 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his paychecks for the year?

(A) $752
(B) $755
(C) $765
(D) $773
(E) $775

Manfred received 6 paychecks of $750 each and 20 paychecks of $780 each.

The first 6 paychecks summed to 6(750) = 4,500.

The next 20 paychecks summed to 20(780) = 15,600.

So, the average of all the checks is:

(4,500 + 15,600)/26 = 20,100/26 ≈ 773

Answer: D
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Hi,

This question can be solve in few seconds ( I struggled for minutes at first) :
Notice that the first 6 months the pay is 750$/month and then it's 30$ more for the remaining 20 months which means 30$*20=600$
basically the average for the 26 months is 750$ +600$/26 = 750$+23$= 773$

Answer D)
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Bunuel
SOLUTION

Last year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $30 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his paychecks for the year?

(A) $752
(B) $755
(C) $765
(D) $773
(E) $775

The average amount of the paychecks =

\(\frac{750*6+ (750 +30)*20}{26}=\frac{750*26+ 30*20}{26}=750+\frac{30*20}{26}=750+\frac{30*10}{13}\approx{750+23}=773\).

Answer: D.


Bunuel can you explain how you did the next step as 750*26 + 30*20 ?
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Bunuel
SOLUTION

Last year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $30 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his paychecks for the year?

(A) $752
(B) $755
(C) $765
(D) $773
(E) $775

The average amount of the paychecks =

\(\frac{750*6+ (750 +30)*20}{26}=\frac{750*26+ 30*20}{26}=750+\frac{30*20}{26}=750+\frac{30*10}{13}\approx{750+23}=773\).

Answer: D.


Bunuel can you explain how you did the next step as 750*26 + 30*20 ?

Sure.

\(\frac{750*26+ 30*20}{26}=\frac{750*26}{26}+ \frac{30*20}{26}=750+\frac{30*20}{26}\).
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Bunuel
Last year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $30 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his paychecks for the year?

(A) $752
(B) $755
(C) $765
(D) $773
(E) $775

w1/w2 = (A2 - Aavg)/(Aavg - A1)

6/20 = (780 - Avg)/(Avg - 750)

3Avg - 2250 = 7800 - 10Avg

Avg = 773

For easier calculation, approximate.
We need to divide 30 in the ratio 3:10. When we double 3:10, we get 6:20 which adds up to 26. We are still 4 short of 30. To keep the ratio same, we can distribute the 4 by giving approximately 1 to 6 and 3 to 20 (because 6:20 is approximately 1:3). This gives us 7:23. Here the total is 30.
So 750 will be 23 away from the average which means average will be 773 (the scale method).

Answer (D)
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Using the method KarishmaB had written about a long time ago.
(Cannot add the article here because of some technical issue but do checkout Bunuel's Ultimate GMAT Quantitative Megathread under "Stats Made Easy" and you can find her article there.)

We will solve this problem using the number line.

So, the question is asking us to find a mean.

The two numbers provided are 750 and 780.
Assume a mean between these, I'm gonna go with 765 because it is equidistant from 750 and 780.

Now, given it is equidistant how far out is 750 from 765 and 780? 15 spaces on either side i.e. $15 on either side in this case.

They also gave information regarding it's weights:
6 pay cheques for $750
20 pay cheques for $780

Side tracking for a moment :
You can intuitively say that the weight towards $780 is greater than towards $750 so the first three options are out of the question right away.

Now,
The left side of $765 we have six pay cheques, hence...
6 pay cheques x $15= $90

On the right side of $765 we have twenty pay cheques, hence...
20 pay cheques x $15 = $300

(While using this method ALWAYS subtract the RHS with the LHS, if it's negative then you need to move down the number line instead of up)
Now $300 - $90 = $210

We need to "evenly spread" this $210 over the 26 pay cheques.

So $210 / 26 = 8.07 ~ 8

We need to move up the number line.
Now move 8 spaces right to $765 to get $773.

Final answer $773.

I suggest you give Karishma's article a read.
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VolatilitySmile
Using the method KarishmaB had written about a long time ago.
(Cannot add the article here because of some technical issue but do checkout Bunuel's Ultimate GMAT Quantitative Megathread under "Stats Made Easy" and you can find her article there.)

We will solve this problem using the number line.

So, the question is asking us to find a mean.

The two numbers provided are 750 and 780.
Assume a mean between these, I'm gonna go with 765 because it is equidistant from 750 and 780.

Now, given it is equidistant how far out is 750 from 765 and 780? 15 spaces on either side i.e. $15 on either side in this case.

They also gave information regarding it's weights:
6 pay cheques for $750
20 pay cheques for $780

Side tracking for a moment :
You can intuitively say that the weight towards $780 is greater than towards $750 so the first three options are out of the question right away.

Now,
The left side of $765 we have six pay cheques, hence...
6 pay cheques x $15= $90

On the right side of $765 we have twenty pay cheques, hence...
20 pay cheques x $15 = $300

(While using this method ALWAYS subtract the RHS with the LHS, if it's negative then you need to move down the number line instead of up)
Now $300 - $90 = $210

We need to "evenly spread" this $210 over the 26 pay cheques.

So $210 / 26 = 8.07 ~ 8

We need to move up the number line.
Now move 8 spaces right to $765 to get $773.

Final answer $773.

I suggest you give Karishma's article a read.


VolatilitySmile

That's great use of the deviations from mean concept.
A couple of further notes:
Since we know that there are more numbers at 780, let's start with the assumed mean as 770. It will lead to smaller calculations. Of course the method followed and the answer obtained would be the same.

Also an interesting way to apply your weighted average formula to reduce calculations would be to shift the scale 750 units to the left.

750 ----- Avg --- 780

Let me move it 750 units to the left: 0 ------- Avg ---30


\(\frac{6}{20} = \frac{(30 - Avg)}{(Avg - 0)} \)

\(\frac{26}{20} = \frac{30}{Avg}\)

Avg = 600/26 = 23

So the actual average = 750 + 23 = 773 (move it back 750 units to the right)

Think about why it works.
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