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In a class of 40 students, is the average (arithmetic mean)

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In a class of 40 students, is the average (arithmetic mean)

by BrijNath » Thu Oct 04, 2018 10:12 am

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In a class of 40 students, is the average (arithmetic mean) score of the class lesser than the median score of the class?

(1) 60% of the boys and 80% of the girls scored lesser than the average (arithmetic mean) score of the class.

(2) There are 15 boys in the class.


Please explain in the easiest way possible.


Source--------------------> E-Gmat
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Source: — Data Sufficiency |
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by deloitte247 » Sun Oct 21, 2018 4:31 am

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Question = If the average arithemetic mean score of the class is lesser than the median score of the class i.e Average Arithemetic Mean(Score median)

Statement 1 = 60% of the boys and 80% of the girls scored less than the average (arithemetic mean) score of the class.
$$Average\ arithemetic\ mean=\frac{£Fx}{£\text{F}}$$
We don't know the scores and numbers of boys and girls because we cannot determine if average arithemetic mean is lesser than the median score of the class.
$$Therefore\ Statement\ 1\ is\ NOT\ SUFFICIENT$$

Statement 2= There are 15boys in the class hence class consist of 15 boys and 35 girls. There is no information about their scores hence we cannot determine if average arithemetic mean score is lesser than the median scores of the class.
$$Statement\ 2\ is\ NOT\ SUFFICIENT$$
Combine Statement 1 and 2 together
Total Student in class=40
Boys =15
Girls =35
$$\frac{60}{100}\cdot15=no\ of\ \ boysthat\ scored\ less\ than\ the\ average\ arithemetic\ mean\ =9boys$$
$$\frac{80}{100}\cdot35=no\ of\ \ girlsthat\ scored\ less\ than\ the\ average\ arithemetic\ mean\ =28girls$$
$$28+9=37\ students\ scored\ less\ than\ the\ average\ arithemetic\ mean$$
$$if\ arithemetic\ mean\ is\ =x$$
$$37\ students\ didn't\ score\ up\ to\ x$$
$$3students\ scored\ more\ than\ x$$
All these doesn't give specific information about the scores of all the students so we cannot determine if the average arithemetic mean score (median score of the class)
$$Hence,\ the\ statement\ combined\ is\ NOT\ SUFFICIENT$$
$$Answer\ is\ option\ E$$
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