What is the average height of n people in the group?
1. The average height of n/3 tallest people in the group is 6 feet and the rest average height is 5 feet.
2. The sum of n height is 190 feet.
I really doing understand the wording of statment 1. Anyone care to explain. The correct answer is A
Tricky Wording
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Statement 1 is sufficient. Ans will be 16/3 any number you substituteGHong14 wrote:What is the average height of n people in the group?
1. The average height of n/3 tallest people in the group is 6 feet and the rest average height is 5 feet.
2. The sum of n height is 190 feet.
I really doing understand the wording of statment 1. Anyone care to explain. The correct answer is A
Statement 2 is insufficient. Average of n height is 190/n. n is still an unknown.
So answer is A
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????ksundar wrote:Statement 1 is sufficient. Ans will be 16/3 any number you substituteGHong14 wrote:What is the average height of n people in the group?
1. The average height of n/3 tallest people in the group is 6 feet and the rest average height is 5 feet.
2. The sum of n height is 190 feet.
I really doing understand the wording of statment 1. Anyone care to explain. The correct answer is A
Statement 2 is insufficient. Average of n height is 190/n. n is still an unknown.
So answer is A
GHong14, read the solution below
Statement (1)
'The average height of n/3 tallest people in the group is 6 feet' - means, Average (s1) for n/3 is equal to 6 feet; the weighted average coefficient of this sample (n/3) within our general pool (n) is 1/3 - just remember for a moment
'... the rest average height is 5 feet' - means, Average (s2) for n-n/3 or 2n/3 is equal to 5 feet; the weighted average coefficient of this sample (2n/3) within our general pool (n) is 2/3 - remember this too
Now bring together Averages s1 and s2 and find the average height of pool/people (1/3) * 6 + (2/3) * 5 = 6/3 + 10/3 = 16/3
Statement (1) is sufficient since we find the average height of n people as 16/3 or 5 1/3
Statement (2)
'The sum of n height is 190 feet' - means, the sum of all (n) people heights is equal to 190 feet.
We need to find the average of n people height; average formula, S(average) = 190/n. We are missing n here.
Statement (2) is insufficient.
Select A.
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One of the things I always do on a DS question, is to identity if its a value question or a yes/no question. This question is a value question-hence we need to find a specific value.
In order to find the average height, we would need the sum of all the heights & the number of people. Answer choice which gives us the 2 pieces of information( or a relationship between the 2 variables) is sufficient. if not then it is insufficient.
stmt 1 gives us the avg height of n/3 people & avg height of remaining(2n/3) people. Hence, for finding the average, we can create the equation: (6n/3+5*2n/3)/n. We can cancel n from numerator and denominator & find out the avg weight. hence sufficient
stmt2 gives us the total weight but not the number of people n. Hence insufficient
In order to find the average height, we would need the sum of all the heights & the number of people. Answer choice which gives us the 2 pieces of information( or a relationship between the 2 variables) is sufficient. if not then it is insufficient.
stmt 1 gives us the avg height of n/3 people & avg height of remaining(2n/3) people. Hence, for finding the average, we can create the equation: (6n/3+5*2n/3)/n. We can cancel n from numerator and denominator & find out the avg weight. hence sufficient
stmt2 gives us the total weight but not the number of people n. Hence insufficient
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Hi,
if we know the concept of balanced average, we can do far less math to see that (1) is sufficient:
(1) tells us that a third of the people are 6 ft while two-thirds of the people are 5 ft. Because there are twice as many 5-footers, the grand average is twice as close to 5 than it is to 6...5 ft 4 in.
Another example:
For 3/4 of a math class, the average score on the midterm was 60 while the other 1/4's average was 40. What is the average of the entire class?
Well, since there are three times as many 60-scorers than 40-scorers, the average should be three times closer to 60 than it is to 40...55.
The less math you do in a DS question, the quicker you can solve it; NOT doing math is an important skill in DS.
if we know the concept of balanced average, we can do far less math to see that (1) is sufficient:
(1) tells us that a third of the people are 6 ft while two-thirds of the people are 5 ft. Because there are twice as many 5-footers, the grand average is twice as close to 5 than it is to 6...5 ft 4 in.
Another example:
For 3/4 of a math class, the average score on the midterm was 60 while the other 1/4's average was 40. What is the average of the entire class?
Well, since there are three times as many 60-scorers than 40-scorers, the average should be three times closer to 60 than it is to 40...55.
The less math you do in a DS question, the quicker you can solve it; NOT doing math is an important skill in DS.
Kaplan Teacher in Toronto