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28 Jan 2019, 06:05
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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:
75%
(hard)
Question Stats:
65% (03:09) correct
35%
(02:48)
wrong
based on 241
sessions
History
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Three math classes: X, Y, and Z, take an algebra test.
The average score in class X is 83.
The average score in class Y is 76.
The average score in class Z is 85.
The average score of all students in classes X and Y together is 79.
The average score of all students in classes Y and Z together is 81.
What is the average score for all the three classes, taken together?
A. 81
B. 81.5
C. 82
D. 82.5
E. 84.5
The average score in class X is 83.
The average score in class Y is 76.
The average score in class Z is 85.
The average score of all students in classes X and Y together is 79.
The average score of all students in classes Y and Z together is 81.
What is the average score for all the three classes, taken together?
A. 81
B. 81.5
C. 82
D. 82.5
E. 84.5
28 Jan 2019, 06:43
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By using alligation, we can find that the ratio of students in each class x:y:z is 3:4:5
and the average = \(\frac{(3*83)+(4*76)+(5*85)}{3+4+5} = \frac{978}{12} = 81.5\)
B
and the average = \(\frac{(3*83)+(4*76)+(5*85)}{3+4+5} = \frac{978}{12} = 81.5\)
B
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General Discussion
29 Jan 2019, 04:50
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Solution
Given:
- • The average score of class X = 83
• The average score of class Y = 76
• The average score of class Z = 85
• The average score of all students in X and Y together = 79
• The average score of all students in Y and Z together = 81
To find:
- • The average score of all the three classes
Approach and Working:
- • \(\frac{S_1}{X} = 83\)
o Implies, \(S_1 = 83X\)
• \(\frac{S_2}{Y} = 76\)
- o Implies, \(S_2 = 76Y\)
• \(\frac{S_3}{Z} = 85\)
- o Implies, \(S_3 = 85Z\)
\(\frac{(S_1 + S_2)}{(X + Y)} = 79\)
• Implies, \(S_1 + S_2 = 79X + 79Y\)
• Thus, 83X + 76Y = 79X + 79Y
• 4X = 3Y
\(\frac{(S_2 + S_3)}{(Y + Z)} = 81\)
• Implies, \(S_2 + S_3 = 81Y + 81Z\)
• Thus, 85Z + 76Y = 81Y + 81Z
• 4Z = 5Y
The average score of all the three classes = \(S_1 +S_2 + S_3/X + Y + Z = \frac{83X + 76Y + 85Z}{X + Y + Z}\)
- • Therefore, total average score = (\(83 * \frac{3}{4} + 76 + 85 * \frac{5}{4})/ (\frac{3}{4}+ 1 + \frac{5}{4}) = \frac{163}{2} = 81.5\)
Hence, the correct answer is Option B
Answer: B
