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Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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25 Jun 2018, 04:02
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Over the past 7 weeks, the Smith family had weekly grocery bills of $74, $69, $64, $79, $64, $84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7week period? A. $64 B. $70 C. $73 D. $74 E. $85 NEW question from GMAT® Official Guide 2019 (PS07369)
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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29 Jun 2018, 05:58
First, choose a nice, round number, such as 70, within the range of values.
Then calculate the average with the following formula:
Average = Nice number + Average of differences from the nice number
Average = 70 + (4  1  6 + 9  6 + 14 + 7)/7 = 70 + 21/7 =73
Answer: C




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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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25 Jun 2018, 04:10
Solution Given:• Over the past 7 weeks, the Smith family had weekly grocery bills of $74, $69, $64, $79, $64, $84, and $77 To find:• Smith’s average weekly grocery bill over the 7week period Approach and Working:• Smith’s total bill amount over the 7week period = (74 + 69 + 64 + 79 + 64 + 84 + 77) = 511 • Therefore, Smith’s average bill amount over the same period = \(\frac{511}{7}\) = 73 Hence, the correct answer is option C. Answer: C
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Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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25 Jun 2018, 04:28
The first way to find the average weekly grossery bill is to find the sum of the bills and to divide it with seven. In this case we have that: Average=[64+64+69+74+77+79+84][/7]=[511][/7]=73.
An another quickest way is to arrange the seven values in ascending order. The arrengement is: 64 64 69 74 77 79 84 The median value is 74 but the three smaller values have a total difference of 25 from the median (7464=10, 7464=10, 7469=5 so 10+10+5=25). The three bigger values have a total difference of 18 from the median (7774=3, 7974=5, 8474=10 so 3+5+10=18). So clearly, the average weekly grossery bill has to be a little smaller from the median. The next smaller value from the median is 73.
In my opinion, if it is possible to make simple calculations without wasting a lot of time (as is the case in this problem), the first way is preferable. But if we have many different values and especially if the given numbers are difficult to handle, the second way can be really helpful.
In both cases, the correct answer is C.



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Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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27 Jun 2018, 17:28
Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of $74, $69, $64, $79, $64, $84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7week period? A. $64 B. $70 C. $73 D. $74 E. $85 NEW question from GMAT® Official Guide 2019 (PS07369) I took the approach described by Iakovos1995 (second method) +1. I do not want to attempt crunching these numbers. Finding the mean as a balance point with numbers this large is faster. The arithmetic mean can be viewed as a balancing point of a scale or seesaw. The mean is the point at which the total distance of numbers below the mean is equal to the total distance above the mean VeritasPrepKarishma explains the theory and method in The Meaning of Arithmetic Mean, here, here, and here (important!) This "balancing" method makes finding the average of a lot of large numbers much easier. We have 64, 64, 69, 74, 77, 79, 84 74 is the median. The lowest and highest data values (64 and 84) are both 10 away from the median, so the list is not heavily skewed. We make an educated guess about the mean: 74 is the middle number, and the center of the range. Now measure distance from (difference from) the middle value. At the mean, the totals on each side will be equal. [74] is NOT the mean  it is a number that is probably very close to the mean. 64, 64, 69, [74] 77, 79, 84 64 is 10 less than 74 64 is 10 less than 74 69 is 5 less than 74 Total: 25 below the middle VeritasPrepKarishma calls this total the "overall shortfall." 77 is 3 more than 74 79 is 5 more than 74 84 is 10 more than 74 Total: 18 above the middle This is called "overall excess." Shortfall must balance out excess. Not true here, so the scale is not balanced. The overall shortfall is greater than the overall excess by 7: (25  18) = 7 This shortfall of 7 needs to be distributed evenly. There are a total of 7 numbers. So the mean must decrease (because we have a shortfall) by \(\frac{7}{7}=1\). (74  1) = 73 Answer C



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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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28 Jun 2018, 17:07
Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of $74, $69, $64, $79, $64, $84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7week period? A. $64 B. $70 C. $73 D. $74 E. $85 We can determine the average using the formula: average = sum / number: average = (74 + 69 + 64 + 79 + 64 + 84 + 77)/7 = 511/7 = 73 Answer: C
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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28 Aug 2018, 02:42
Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of $74, $69, $64, $79, $64, $84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7week period? A. $64 B. $70 C. $73 D. $74 E. $85 NEW question from GMAT® Official Guide 2019 (PS07369) Easy way out is consider avg 70  so diff for all terms is 4, 1 , 6, 9 , 6,14, 7 = sum is 21 = divided by 7 = equal to 3 so 70 +3 = 73 avg



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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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10 Sep 2018, 00:08
https://youtu.be/laNkexjIN2Y
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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10 Sep 2018, 06:34
Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of $74, $69, $64, $79, $64, $84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7week period? A. $64 B. $70 C. $73 D. $74 E. $85 NEW question from GMAT® Official Guide 2019 (PS07369) \(\frac{( 70 + 4 ) + ( 70  1 ) + ( 60 + 4 ) + ( 80  1 ) + ( 60 + 4 ) + ( 80 + 4 ) + ( 70 + 7 )}{7}\) = \(\frac{( 70 + 70 + 60 + 80 + 60 + 80 + 70 ) + ( 4  1 + 4  1 + 4 + 4 + 7 )}{7}\) = \(\frac{490 + 21}{7}\) = \(\frac{490}{7} + \frac{21}{7}\) = \(70 + 3\) = \(73\), Answer must be (C)
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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10 Sep 2018, 06:46
We can rule out 64, 74 and 85 as they are the extreme numbers. That leaves just two choices to choose from 70 and 73.
Only 3 values in the 60s and therefore less likely for 70.
73 is the most likely choice. Confirm 73 by taking difference and summing it up



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Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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01 Oct 2018, 23:09
This solution might be too blunt for some people. Sum the unit digits and you will have 41. Now there are 7 numbers, which mean if x multiply by 7 must will result a number with unit digit of 1, so the x is 3. The only answer with 3 as its unit digit is C.



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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7
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19 Nov 2018, 15:13
Admittedly I made a stupid mistake and I organised the numbers in ascending order then selected the median (74) instead of solving Sum/# Terms
The method I used is only for consecutive integers and the punishment answer D is there for suckers like me.




Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 &nbs
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19 Nov 2018, 15:13






