During a 6-day local trade show, the least number of people registered in a single day was 80. Was the average...

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During a 6-day local trade show, the least number of people registered in a single day was 80. Was the average (arithmetic mean) number of people registered per day for the 6 days greater than 90?

1) For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100

2) For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85

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Least no. of people registered in a day = 80
Target question => Was the average (arithmetic mean) number of people registered per day for the 6 days greater than 90?
$$Mean=\frac{total\ no.\ of\ people\ registered\ for\ the\ 6\ days}{6}$$
Total no. of people registered > (90 * 6 = 540)
Statement 1=> For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100.
Therefore, total no. of people registered for 4 days = 4 * 100 = 400.
Least no. of people that can be registered in a day = 80
So, the remaining 2 days will have at least (80*2=160) people and the total people registered for the 6 days = 400 + 160 = 560. This is greater than 540.
Therefore, statement 1 is SUFFICIENT.

Statement 2=> For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85.
Therefore, total no. of people that can be registered in a day = 80
Remaining 3 days = 86 * 3 = 258 and total people registered = 255 + 256 = 513.
This is less than 540 but if the value for the remaining 3 days is greater than the least no. of people that can be registered in a day, then total no of people registered might be greater than 540.
Due to this uncertainty, statement 2 is NOT SUFFICIENT.

Since only statement 1 is SUFFICIENT, the correct answer = option A.