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On a hot summer day, a coffee and tea shop sold a container of coffee 07 Nov 2019, 03:01
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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 aggregate sales of the two items that day was $420.


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On a hot summer day, a coffee and tea shop sold a container of coffee 07 Nov 2019, 11:29
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Average tells us ratio of tea and coffee ie 6:4
Thus right answer is A
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On a hot summer day, a coffee and tea shop sold a container of coffee 07 Nov 2019, 12:37
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#1
15x+25y=21x+21y
2y=3x
y=3 , x=2
sufficient
#2
15x+25y=420
x=23, y= 3
x=18, y =6
insufficient
IMO A


Bunuel
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 aggregate sales of the two items that day was $420.


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On a hot summer day, a coffee and tea shop sold a container of coffee 07 Nov 2019, 13:13
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Bunuel
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 aggregate sales of the two items that day was $420.
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Fun one. You can do the calculation, but that's a waste of time. Instead, glance at the statements - the word 'average' should jump out at you. If you're combining two different items of different prices to get an average price for everything, then you're doing a weighted average problem.

Another hint that this can be solved with weighted average techniques is the question itself. It only asks whether there were MORE containers of coffee than tea, not the exact numbers (or even the exact ratio.)

Statement 1: The average was $21. Start a weighted average problem by imagining that the numbers for coffee and tea were exactly equal. In that case, the average would be exactly halfway between 15 and 25, or exactly 20.

However, the actual average is higher. Therefore, there must have been a few more of the more expensive tea bags, driving up the average a bit. So, the answer to the question has to be "no." This statement is sufficient. Eliminate B, C, and E.

Statement 2: This is an interesting one. My first instinct is to assume that it's insufficient, because it seems like there should be many different possibilities for the number of coffee containers and tea containers, and some should have more coffee while others should have more tea. But on second thought, I'm not totally sure about that. I've seen other problems where (since you can't buy half of a container, presumably) there was only one way for the total to come out to a specific number. Is that the case here? Let's play with it.

15c + 25t = 420

Divide everything by 5 to make the math easier:

3c + 5t = 84

Try out some numbers for t. Is the remainder a multiple of 3?

3c + 5(0) = 84
3c = 84

Yes, this works! So, one valid possibility is that there were 84/3 = 28 coffees and 0 teas sold, which makes the answer 'yes'.

At this point, if I'm really savvy, I'll remember that the two statements can never directly contradict each other. It's impossible for both statements to be sufficient if one always gives a 'yes' and the other always gives a 'no'. So, because we found a 'yes' here, if this statement was sufficient, that would put us in that impossible position. Therefore, we MUST also be able to find a 'no', assuming this is an official problem.

But just to be safe, I'm going to keep going and look for another case. This time, I'd like to find one with a higher value for t. Here's what I thought of:

3c + 5(12) = 84
3c = 24
c = 8

So, c = 8 and t = 12 is a valid case, and this time, the answer is 'no.' This statement is definitely insufficient.
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On a hot summer day, a coffee and tea shop sold a container of coffee 07 Nov 2019, 17:31
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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 aggregate sales of the two items that day was $420.

Statement 1:\(\frac{15C + 25T}{(C+T)} = 21\) >>> 1 tea and 1 coffee sold, average = $20. 2 tea and 1 coffee sold, average =$21.67. 1 tea and 2 coffee sold, average = $17.33.
Since the difference of average and coffee container price($6) is greater than the difference between average and tea container ($4), so the shop sell more tea than coffee. SUFICIENT.

Statement 2: 15C + 25T = 420. > 3C + 5T = 84. C and T can take more than 1 values. NOT SUFFICIENT.
A is the correct answer.
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On a hot summer day, a coffee and tea shop sold a container of coffee 22 Jun 2020, 19:40
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Bunuel
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 aggregate sales of the two items that day was $420.


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S1: If the shop had sold equal amounts of coffee and tea, the average would be 20 because 20+15/2=20. Since the actual average is greater than 20, more tea must have been sold. SUFFICIENT.

S2: Our equation is 15x+25y=400. We can make things easier by dividing both sides of the equation by 5 and get 3x+5y=80 as our new equation, then look for possible values of y and x. x=0, y=16 and x=12, y=9 both work. NOT SUFFICIENT.

ANSWER: A
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On a hot summer day, a coffee and tea shop sold a container of coffee 10 Jun 2025, 03:32
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