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29 Jul 2015, 15:44
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
96% (01:22) correct
4%
(02:16)
wrong
based on 171
sessions
History
Date
Time
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Not Attempted Yet
The weight of four dogs is determined to be 25 pounds, 31 pounds, 35 pounds and 41 pounds respectively. The weight of a fifth dog is determined to be y pounds. If the average (arithmetic mean) weight of the first four dogs is the same as that of all five dogs what is the value of y?
A. 31
B. 33
C. 35
D. 37
E. 39
Kudos for a correct solution.
A. 31
B. 33
C. 35
D. 37
E. 39
Kudos for a correct solution.
29 Jul 2015, 16:51
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Bunuel
The problem asks what is the average weight \(y\) without knowing all the dog's weights. Lucky for us, the weights of the dogs were equally distributed, where the difference between the two lesser were the same as that of the two greater, and therefore, the average of all them will be between 31 and 35, which is 33. IMO, option B.
To prove the answer choice, solve for \(y\), where \(y\) is the fifth dog's weight:
\(\frac{25+31+35+41+y}{5}=y\)
\(132=4y\)
\(33=y\)
29 Jul 2015, 17:35
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Total weight of the 4 dogs = (25+31+35+41) = 132
Avg = 132/4 = 33
Total weight of 5 dogs = 132+y or 4(33)+y
Average of 5 dogs as per Question = 33
Equation:
4(33) + y = 5 (33), or y = 33.
Choose B
Avg = 132/4 = 33
Total weight of 5 dogs = 132+y or 4(33)+y
Average of 5 dogs as per Question = 33
Equation:
4(33) + y = 5 (33), or y = 33.
Choose B
