In a test, five students of a class scored 39, 37, 40, 34, and 36, respectively. If the sixth student scored n marks, for which of the following values of n does the average (arithmetic mean) score per student for the six students equal the median score?
Indicate all such values.
[Note: Select one or more answer choices]
Indicate all such values.
[Note: Select one or more answer choices]
Show Hint
- 33
- 37
- 42
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
We need to find values of a sixth score, \(n\), such that the mean of the six scores equals their median. The median of six scores is the average of the 3rd and 4th scores when they are arranged in order. The position of \(n\) relative to the other scores will affect the median. Thus, we may have to consider different cases based on the value of \(n\).
Step 2: Key Formula or Approach:
Method 1: Using the definition of average
Let \(n_1\) be the initial number of students and \(n_2\) be the number of new students.
Initial sum of ages = \(n_1 \times 18\).
Sum of ages of new students = \(n_2 \times 16\).
Final total number of students = \(n_1 + n_2\).
Final sum of ages = \(18n_1 + 16n_2\).
Final average age = \(\frac{18n_1 + 16n_2}{n_1 + n_2} = 17\).
We need to solve this equation for the ratio \(n_2 : n_1\).
Method 2: Alligation
Alligation is a rule that enables us to find the ratio in which two or more ingredients at the given prices must be mixed to produce a mixture of a desired price. Here, we use 'age' instead of 'price'.
We set up a diagram:
| Age of Group 1 | Age of Group 2 | |
| 18 | 16 | |
| \(\searrow\) \(\swarrow\) | ||
| Mean Age (17) | ||
| \(\nearrow\) \(\nwarrow\) | ||
| (17 - 16) | (18 - 17) | |
| 1 | 1 |
The ratio of the quantities (number of students) is the inverse of the ratio of the differences.
Ratio \(n_1 : n_2\) = \((17-16) : (18-17)\).
Step 3: Detailed Explanation:
Using Method 1 (Algebraic Equation):
\(\frac{18n_1 + 16n_2}{n_1 + n_2} = 17\)
Multiply both sides by \((n_1 + n_2)\):
\(18n_1 + 16n_2 = 17(n_1 + n_2)\)
\(18n_1 + 16n_2 = 17n_1 + 17n_2\)
Now, group the terms with \(n_1\) on one side and the terms with \(n_2\) on the other.
\(18n_1 - 17n_1 = 17n_2 - 16n_2\)
\(n_1 = n_2\)
This means the number of initial students is equal to the number of new students.
The ratio of the number of students who joined (\(n_2\)) to the number of students who were initially in the class (\(n_1\)) is:
\(\frac{n_2}{n_1} = \frac{n_1}{n_1} = \frac{1}{1}\)
So, the ratio is 1:1.
Using Method 2 (Alligation):
The difference between the initial average age (18) and the final average age (17) is \(18-17=1\).
The difference between the new students' average age (16) and the final average age (17) is \(17-16=1\).
The rule of alligation states that the ratio of the number of initial students (\(n_1\)) to the number of new students (\(n_2\)) is the inverse of the ratio of these differences.
\(\frac{n_1}{n_2} = \frac{17 - 16}{18 - 17} = \frac{1}{1}\)
This gives \(n_1 : n_2 = 1:1\).
The question asks for the ratio of the number of students who joined (\(n_2\)) to the number of students who were initially in the class (\(n_1\)), which is also \(n_2 : n_1 = 1:1\).
Step 4: Final Answer:
The required ratio is 1:1.
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