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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.
The OA is A.
Easy way:
Statement (1):
(80+4*100+x)/6 > 90 ?
480+x > 540
x>60
There would be more than 90 people per day if on the 5th day there were at least 60 people. We know the least number of people registered at any day was 80. Sufficient.
Easy to check:
(80+400+60)/6 = 90 --> 540 = 540
(80+400+70)/6 > 90 --> 550 > 540
Statement (2):
(3*85+x+y)/6 > 90?
255+x+y > 540?
x+y > 285?
Do you know? Me neither. Insufficient.
Has anyone another strategic approach to solving this DS question? Regards!
(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.
The OA is A.
Easy way:
Statement (1):
(80+4*100+x)/6 > 90 ?
480+x > 540
x>60
There would be more than 90 people per day if on the 5th day there were at least 60 people. We know the least number of people registered at any day was 80. Sufficient.
Easy to check:
(80+400+60)/6 = 90 --> 540 = 540
(80+400+70)/6 > 90 --> 550 > 540
Statement (2):
(3*85+x+y)/6 > 90?
255+x+y > 540?
x+y > 285?
Do you know? Me neither. Insufficient.
Has anyone another strategic approach to solving this DS question? Regards!
