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pratik2018
Joined: 09 Apr 2017
Last visit: 23 Jun 2020
Posts: 35
Given Kudos: 188
Location: Nepal
Concentration: Finance, Entrepreneurship
Schools: Johnson '22 (S)
GMAT 1: 580 Q47 V22
GMAT 2: 640 Q48 V29
GMAT 3: 690 Q48 V36
WE:Information Technology (Computer Software)
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04 Sep 2018, 02:22
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
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Question Stats:
54% (02:59) correct
46%
(02:55)
wrong
based on 170
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In a class of 5 students, average weight of the 4 lightest students is 40 kgs, Average weight of the 4 heaviest students is 45 kgs. What is the difference between the the maximum and minimum possible average weight overall?
a. 2.8 kgs
b. 3.2 kgs
c. 3 kgs
d. 4 kgs
e. 4.2 kgs
a. 2.8 kgs
b. 3.2 kgs
c. 3 kgs
d. 4 kgs
e. 4.2 kgs
04 Sep 2018, 02:57
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pratik2018
Maximum..
Let the weight of the lighted 4 be same so the weight of each is 40.
Now the heaviest 4 weigh 45*4=180
All 5 will weigh 180+40=220
Average = 220/5=44
Minimum
Let the heaviest 4 be same so 45 each
Total weight of 5 will be 40*4+45=205
Average = 205/5=41
Difference=44-41=3
C
General Discussion
07 Sep 2018, 17:01
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pratik2018
Let a, b, c, d and e be the weights of the 5 students such that a ≤ b ≤ c ≤ d ≤ e. Since the total weight of the 4 lightest students is 4 x 40 = 160 kg, we have a + b + c + d = 160. Similarly, since the total weight of the 4 heaviest students is 4 x 45 = 180 kg, we have b + c + d + e = 180. We need to determine the maximum and minimum value of (a + b + c + d + e)/5.
The maximum average weight of all 5 students occurs when a is as large as possible. However, since the average weight of the 4 lightest students is 40 kg, a is at most 40, so the maximum average weight is:
(a + b + c + d + e)/5 = (a + 180)/5 = a/5 + 180/5 = 40/5 + 36 = 8 + 36 = 44 kg
The minimum average weight of all 5 students occurs when e is as small as possible. However, since the average weight of the 4 heaviest students is 45 kg, e is at least 45 kg, so the minimum average weight is:
(a + b + c + d + e)/5 = (160 + e)/5 = 160/5 + e/5 = 32 + 45/5 = 32 + 9 = 41 kg
Therefore, the difference between the maximum and minimum average weight is 44 - 41 = 3 kg.
Answer: C
