On a hot summer day, a coffee and tea shop sold a container of coffee beans at $15 and a container of tea bags at $25. Did the shop sell more containers of coffee beans than containers of tea bags that day?
(1) The average (arithmetic mean) of all containers of coffee beans and tea bags sold that day was $21.
(2) The average sales of the two items that day was $420.
The OA is A.
Let the number of coffee beans = x
The number of tea bags = y
is y > x?
From (1): (15x+25y)/(x+y) = 21
thus 15x+25y = 21x+21y
3x=2y
since x & y are positive integers
y>x. Sufficient.
From (2): 15x + 25y =420
x can be 28 then y=0 (x>y)
or x = 8 then y= 12 (y>x). Insufficient.
Hence, the correct answer is A.
Has anyone another strategic approach to solving this DS question? Regards!
On a hot summer day, a coffee and tea shop sold a container
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Had the number of containers of coffee beans and the number containers of tea bags were equal, the average (arithmetic mean) of all containers of coffee beans and tea bags sold that day would have been (15 + 25)/2 = 20. However, the actual figure is $21 > $20. Since $21 is closer to $25 than to $15, this means that a greater number of tea bags were sold. Sufficient.AAPL wrote:On a hot summer day, a coffee and tea shop sold a container of coffee beans at $15 and a container of tea bags at $25. Did the shop sell more containers of coffee beans than containers of tea bags that day?
(1) The average (arithmetic mean) of all containers of coffee beans and tea bags sold that day was $21.
(2) The average sales of the two items that day was $420.
The OA is A.
Let the number of coffee beans = x
The number of tea bags = y
is y > x?
From (1): (15x+25y)/(x+y) = 21
thus 15x+25y = 21x+21y
3x=2y
since x & y are positive integers
y>x. Sufficient.
From (2): 15x + 25y =420
x can be 28 then y=0 (x>y)
or x = 8 then y= 12 (y>x). Insufficient.
Hence, the correct answer is A.
Has anyone another strategic approach to solving this DS question? Regards!
Hope this helps!
-Jay
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