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20 Jan 2021, 01:15
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B
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Difficulty:
95%
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Question Stats:
42% (02:31) correct
58%
(02:56)
wrong
based on 78
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Alberto, Barbara, and Chris each earned money working on a certain job. Is the average (arithmetic mean) amount earned by the three workers more than 1.5 times the median amount earned?
(1) The difference in earnings between the two lowest paid workers is less than 1/3 of the difference in earnings between the highest paid worker and lowest paid worker.
(2) Alberto, who earned the most money, made more than four times as much as the next highest paid worker.
Are You Up For the Challenge: 700 Level Questions
(1) The difference in earnings between the two lowest paid workers is less than 1/3 of the difference in earnings between the highest paid worker and lowest paid worker.
(2) Alberto, who earned the most money, made more than four times as much as the next highest paid worker.
Are You Up For the Challenge: 700 Level Questions
20 Jan 2021, 02:37
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Alberto, Barbara, and Chris each earned money working on a certain job. Is the average (arithmetic mean) amount earned by the three workers more than 1.5 times the median amount earned?
Let the amounts earned in decreasing order be x, y and z.
So we are looking for \(\frac{x+y+z}{3}>1.5y.........x+y+z>4.5y.......x+z>3.5y\)
(1) The difference in earnings between the two lowest paid workers is less than 1/3 of the difference in earnings between the highest paid worker and lowest paid worker.
So \(y-z<\frac{1}{3}(x-z).........3y-3z<x-z........3y-x<2z\)
Insufficient
(2) Alberto, who earned the most money, made more than four times as much as the next highest paid worker.
x>4y
the least possible value of mean will be slightly greater than case when we take the least value of x and z as 4y and 0.
So mean =( 4y+y+0)/3=5y/3
Now is 5y/3>1.5y or 5y>4.5y => YES as y is positive.
So least possible value of mean is greater than 1.5 times median.
Sufficient
B
Let the amounts earned in decreasing order be x, y and z.
So we are looking for \(\frac{x+y+z}{3}>1.5y.........x+y+z>4.5y.......x+z>3.5y\)
(1) The difference in earnings between the two lowest paid workers is less than 1/3 of the difference in earnings between the highest paid worker and lowest paid worker.
So \(y-z<\frac{1}{3}(x-z).........3y-3z<x-z........3y-x<2z\)
Insufficient
(2) Alberto, who earned the most money, made more than four times as much as the next highest paid worker.
x>4y
the least possible value of mean will be slightly greater than case when we take the least value of x and z as 4y and 0.
So mean =( 4y+y+0)/3=5y/3
Now is 5y/3>1.5y or 5y>4.5y => YES as y is positive.
So least possible value of mean is greater than 1.5 times median.
Sufficient
B
20 Jan 2021, 10:26
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Alberto, Barbara, and Chris each earned money working on a certain job. Is the average (arithmetic mean) amount earned by the three workers more than 1.5 times the median amount earned?
Say, A>B>C, just for statements verification. So, Given, (A+B+C)/3 > 1.5*B ?
Stat1: (B-C) < (A-C)/3, or, B-C+B/3+2C/3 < (A-C)/3 +B/3 +2C/3 or, (4B-C)/3 < (A+B+C)/3, so, need value of C also to get the expression verified. Not sufficient.
Stat2: Alberto, who earned the most money, made more than four times as much as the next highest paid worker.
A= 4B, So, (A+B+C)/3 > 1.5*B or, (5B+C)/3 > 1.5*B or, 1.6B +C/3 >1.5B; which is true. Sufficient
So, I think B.
Say, A>B>C, just for statements verification. So, Given, (A+B+C)/3 > 1.5*B ?
Stat1: (B-C) < (A-C)/3, or, B-C+B/3+2C/3 < (A-C)/3 +B/3 +2C/3 or, (4B-C)/3 < (A+B+C)/3, so, need value of C also to get the expression verified. Not sufficient.
Stat2: Alberto, who earned the most money, made more than four times as much as the next highest paid worker.
A= 4B, So, (A+B+C)/3 > 1.5*B or, (5B+C)/3 > 1.5*B or, 1.6B +C/3 >1.5B; which is true. Sufficient
So, I think B.
