Hello friends,
Could you help me with your approach to tackle the following question:
Chris' average (arithmetic mean) daily expenses were $500 per day over a 15-day trip. During this period, if Chris' average daily expenses were $450 for the first 5 days, what were the average daily expenses for the last 10 days?
A)$505
B)$515
C)$525
D)$532
E)$550
Look forward to hearing from you and thanks for looking through this.
Richa
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Average expenses for the last 10 days
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richajoshi11
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Sum of the daily expenses for all 15 days = (number of days)(average per day) = (15)(500) = 7500.richajoshi11 wrote:Hello friends,
Could you help me with your approach to tackle the following question:
Chris' average (arithmetic mean) daily expenses were $500 per day over a 15-day trip. During this period, if Chris' average daily expenses were $450 for the first 5 days, what were the average daily expenses for the last 10 days?
A)$505
B)$515
C)$525
D)$532
E)$550
Sum of the daily expenses for the first 5 days = (number of days)(average per day) = (5)(450) = 2250.
Sum of the daily expenses for the last 10 days = (sum for all 15 days) - (sum for the first 5 days) = 7500-2250 = 5250.
Average for the last 10 days = (sum for the last 10 days)/(number of days) = 5250/10 = 525.
The correct answer is C.
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Jay@ManhattanReview
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Hi Richa,richajoshi11 wrote:Hello friends,
Could you help me with your approach to tackle the following question:
Chris' average (arithmetic mean) daily expenses were $500 per day over a 15-day trip. During this period, if Chris' average daily expenses were $450 for the first 5 days, what were the average daily expenses for the last 10 days?
A)$505
B)$515
C)$525
D)$532
E)$550
Look forward to hearing from you and thanks for looking through this.
Richa
This is a typical beginner-level question based on the concept of weighted average.
We have an average of the two averages = $500; one of the averages is 450. We have to find out the other average.
If two quantities have equal weights, their average would be a simple average.
Coming back to your question...An example: A phone costs $450 and another phone costs $X. If the average cost of the phones is $500, what is the cost of the other phone?
This is a case of a simple average. Both the phones have equal weights--one each.
So, simple average = 500 = (450 + X) / 2 => X = $550.
However, if the number of phones of each type is not equal, we cannot calculate the simple average. We must take into account their weight, here the number of phones of each type, which is the case in your question.
We see that Chris' weight for the first part of the trip was '5' and that for the last part of the trip was '10'--UNEQUAL! So, we would apply weighted average instead of simple average.
Weighted average of the trip = 500 = [(Average for the first 5-days * 5 days) + (Average for the last 10-days * 10 days)] / 15 days
=> 500*15 = 450*5 + Average for the last 10-days * 10
=> 7500 = 2250 + Average for the last 10-days * 10
=> Average for the last 10-days = (7500 - 2250)/10 = 525.
The correct answer: C
Hope this helps!
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Mo2men
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Dear Mitch,GMATGuruNY wrote:Sum of the daily expenses for all 15 days = (number of days)(average per day) = (15)(500) = 7500.richajoshi11 wrote:Hello friends,
Could you help me with your approach to tackle the following question:
Chris' average (arithmetic mean) daily expenses were $500 per day over a 15-day trip. During this period, if Chris' average daily expenses were $450 for the first 5 days, what were the average daily expenses for the last 10 days?
A)$505
B)$515
C)$525
D)$532
E)$550
Sum of the daily expenses for the first 5 days = (number of days)(average per day) = (5)(450) = 2250.
Sum of the daily expenses for the last 10 days = (sum for all 15 days) - (sum for the first 5 days) = 7500-2250 = 5250.
Average for the last 10 days = (sum for the last 10 days)/(number of days) = 5250/10 = 525.
The correct answer is C.
Is this question solved through the Allegation process? or is it un-doable as we deal with averages?
450....50...500.....x..B
A/B= x/50=5/10........x=25
Average =525 Dollars
Thanks
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DavidG@VeritasPrep
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You could really just eye-ball it. Because 10 days is twice as much as 5 days, we know the following: distance of the 10-day average from the overall average = half the distance of 5-day average from overall average. (The relative weight of each group and the distance from the overall average will always have that reciprocal relationship.)Is this question solved through the Allegation process? or is it un-doable as we deal with averages?
Distance of 5-day average from overall average? 500-450 = 50
Distance of 10-day average from overall average? 1/2 of 50 = 25.
10-day average: 500 + 25 = 525
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Hi Richa,
This prompt is an example of a 'Weighted Average' question, and can be solved in a few different ways. The ratio of days involved is rather convenient though - and it can help us to simplify the math considerably.
There are 15 days total and the first 5 days have an average of $450. Whatever the average is for the other 10 days, we know that the ratio of 5 days to 10 days can be reduced to 1 to 2. In other words, for every 1 day at $450, we have 2 days at the other average. We can now create the following equation:
(450 + 2X)/3 = 500
450 + 2X = 1500
2X = 1050
X = 525
Thus, the average for the other 10 days must be...
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
This prompt is an example of a 'Weighted Average' question, and can be solved in a few different ways. The ratio of days involved is rather convenient though - and it can help us to simplify the math considerably.
There are 15 days total and the first 5 days have an average of $450. Whatever the average is for the other 10 days, we know that the ratio of 5 days to 10 days can be reduced to 1 to 2. In other words, for every 1 day at $450, we have 2 days at the other average. We can now create the following equation:
(450 + 2X)/3 = 500
450 + 2X = 1500
2X = 1050
X = 525
Thus, the average for the other 10 days must be...
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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To learn about alligation, check here:Mo2men wrote:Dear Mitch,richajoshi11 wrote:Chris' average (arithmetic mean) daily expenses were $500 per day over a 15-day trip. During this period, if Chris' average daily expenses were $450 for the first 5 days, what were the average daily expenses for the last 10 days?
A)$505
B)$515
C)$525
D)$532
E)$550
Is this question solved through the Allegation process?
https://www.beatthegmat.com/ratios-fract ... 15365.html
Alligation approach for the problem above:
Let F = the first 5 days and L = the last 10 days.
Since the average for F = 450, and the average score for all 15 days is 500, we get the following number line:
F 450 ------- 500 -------- L
F:L = (5 days):(10 days) = 5:10 = 1:2.
The distances on the number line are equal to the RECIPROCAL of this ratio.
Since the reciprocal of 1:2 is 2x:x, we get:
F 450 ---2x--- 500 ---x--- L
On the resulting number line, 2x is the distance between 450 and 500, while x is the distance between 500 and L.
The number line implies that 2x = 500-450 = 50, with the result that x=25.
Thus, L = 500+x = 500+25 = 525.
The correct answer is C.
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Jay@ManhattanReview
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A lot has so far been discussed. Adding one more dimension...richajoshi11 wrote:Hello friends,
Could you help me with your approach to tackle the following question:
Chris' average (arithmetic mean) daily expenses were $500 per day over a 15-day trip. During this period, if Chris' average daily expenses were $450 for the first 5 days, what were the average daily expenses for the last 10 days?
A)$505
B)$515
C)$525
D)$532
E)$550
Look forward to hearing from you and thanks for looking through this.
Richa
If we take the average of 15-days equals to $500 literally, Chris would have spent $500 on each of the 15 days.
Per the average, his expenses for 15 days: 500-500-500-500-500-500-500-500-500-500-500-500-500-500-500;
However, this has not happened. What happened is given below.
Actual expenses for 15 days: 45-450-450-450-450-X-X-X-X-X-X-X-X-X-X; we have to find out the value of X.
We see that Chris short-spent by 500 - 450 = $50 on each of the first 5 days, thus he short-spent by a total of $50*5=$250.
In order to maintain his per day average spent of $500 for 15 days, he must additionally spend $250 over the 10 days, while equally distributing the amount per day = 250/10 = $25.
Thus, Chris' average per day spent on the last 10 days = $500 + $25 = [spoiler]$525.[/spoiler]
The correct answer: C
Hope this helps!
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Chris' average (arithmetic mean) daily expenses were $500 per day over a 15-day trip. During this period, if Chris' average daily expenses were $450 for the first 5 days, what were the average daily expenses for the last 10 days?
A)$505
B)$515
C)$525
D)$532
E)$550
We are given that Chris's average expenses were $500 per day over the course of 15 days, so his expenses were 500 x 15 = 7,500 dollars.
Since his average expenses were $450 for the first 5-days, he spent 450 x 5 = 2,250 dollars over the course of the first 5 days.
Thus, he spent a total of 7,500 - 2,250 = 5,250 for the last 10 days, which is an average of:
5,250/10 = 525 dollars.
Answer: C
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