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16 Sep 2014, 01:31
Kudos
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There are two schools in the village. The average (arithmetic mean) age of pupils in the first school is 12.2 years, and the average (arithmetic mean) age of pupils in the second school is 13.1 years. What is the average (arithmetic mean) age of all the pupils in the village?
(1) The second school has 40 more pupils than the first school.
(2) The second school has three times as many pupils as the first school.
(1) The second school has 40 more pupils than the first school.
(2) The second school has three times as many pupils as the first school.
16 Sep 2014, 01:31
Kudos
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Official Solution:
There are two schools in the village. The average (arithmetic mean) age of pupils in the first school is 12.2 years, and the average (arithmetic mean) age of pupils in the second school is 13.1 years. What is the average (arithmetic mean) age of all the pupils in the village?
This problem deals with weighted averages. Let \(x\) represent the number of pupils in the first school and \(y\) represent the number of pupils in the second school. Then, the combined average age is given by: \(\text{average} = \frac{12.2x + 13.1y}{x + y}\).
(1) The second school has 40 more pupils than the first school.
From this, we deduce: \(y = x + 40\). This information alone is not sufficient to determine the average age.
(2) The second school has three times as many pupils as the first school.
This implies: \(y = 3x\). Substituting into our average equation, we get: \(\text{average} = \frac{12.2x + 13.1(3x)}{x + 3x} = \frac{12.2 + 13.1*3}{4} \approx 12.9\). Sufficient.
Answer: B
There are two schools in the village. The average (arithmetic mean) age of pupils in the first school is 12.2 years, and the average (arithmetic mean) age of pupils in the second school is 13.1 years. What is the average (arithmetic mean) age of all the pupils in the village?
This problem deals with weighted averages. Let \(x\) represent the number of pupils in the first school and \(y\) represent the number of pupils in the second school. Then, the combined average age is given by: \(\text{average} = \frac{12.2x + 13.1y}{x + y}\).
(1) The second school has 40 more pupils than the first school.
From this, we deduce: \(y = x + 40\). This information alone is not sufficient to determine the average age.
(2) The second school has three times as many pupils as the first school.
This implies: \(y = 3x\). Substituting into our average equation, we get: \(\text{average} = \frac{12.2x + 13.1(3x)}{x + 3x} = \frac{12.2 + 13.1*3}{4} \approx 12.9\). Sufficient.
Answer: B
08 Feb 2018, 02:35
Kudos
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ttaiwo
For (1): if we substitute \(x+40=y\) into \(\text{average}=\frac{12.2x+13.1y}{x+y}\) we'd get \(\text{average}=\frac{25.3 x + 524}{2x+40}\). Without knowing x or y, we cannot get the single numerical value of this expression, so cannot answer the question.
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