In Country X and Country Y, lunch for the vast majority of high...

Question

In Country X and Country Y, lunch for the vast majority of high schoolers is the free lunch provided at their school cafeteria. In Country X, the average high-school cafeteria lunch (AHCL) contains about twice the amount of sugar compared with the AHCL in Country Y. In which of the two countries does the average high schooler consume more sugar per day?

(1) For the average high schooler in Country Y, the average amount of sugar across all the meals of the day is similar to that contained in the AHCL in Country X.

(2) In Country X, the average high schooler has one fewer meal per day compared with Country Y.

HNS · Answer

Shouldnt ans be option C ? My logic is as follows:

Bunuel ,  kingbucky  can you please look into this question and provide with the correct reasoning ?

Let AHCL in country Y = a.
So AHCL in country X = 2a.

Option (i) :- Let total no. of meals in country Y = n.
So Avg. sugar across all meals = ( Total sugar across all meals / n ) = 2a.

But we dont know the number of meals in country X. Hence  insufficent .

Option (ii) :- We dont know about the sugar content of meals except for lunch. Hence  insufficient .

Option (i) + (ii) :

No. of meals in country Y = n.
No. of meals in country X = n - 1. 
Avg. sugar across all meals for country Y = ( Total sugar across all meals / n ) = 2a / n. (As per option (i))

In the worst case we assume that all the meals except for lunch in country X had 0 sugar. Then sugar consumed by avg high schooler in country X = (AHCL + sugar content of other meals ) / (n - 1)
 = (2a + 0) / (n - 1). = 2a / ( n - 1 ) 
For country Y we have average sugar content = 2a / n

Hence in the worst case country X will have more sugar content then country Y. So answer should be C as per the above reasoning.

Dream009 · Answer

This question got me but yes answer is E. Let me explain with few numbers
COuntry X: Breakfast, Lunch in canteen(sugar ratio mentioned as twice of Y), dinner 
Country Y Breakfast, Lunch in canteen(sugar ratio with X mentioned), dinner

Asked who eats more sugar. That will not just depend on lunch but other meals as well. Could be possible 'Y's breakfast and dinner is10 times sugar content of Breakfast and dinner. Also almost lunch is had by everyone - but on avg how much is tendency to skip breakfast or lunch - with this background lets make a formula Sugar content = Sugar in Breakfast*avg probability to have Breakfast 
+ Sugar in Lunch*avg probability to have Lunch + Sugar in Dinner*avg probability to have Dinner.

Now
 (1) For the average high schooler in Country Y, the average amount of sugar across all the meals of the day is similar to that contained in the AHCL in Country X.

We know avg sugar across the meal for 1 day not the probability to have. X could eat more likely in 1 day and Y could eat less. Weighted avgs - its not about what sugar content is present but consumed which is why probability to eat is imp
  No

(2) In Country X, the average high schooler has one fewer meal per day compared with Country Y.
 
Well this answer's the second part I addressed in statement 1 . Reject alone

Now together - I have avg of day and avg of probability but is this correct. Lets check mathematically

Sugar content B=10, L =20, C=30 Avg = 20 
Probability = .1, .7, .4 = Avg probability = .03

IS 20*.03 same as .1*10 + .7*20 + .4*30 ------------NOOOOO 
And that's the trap.... Ans E

Usernamevisible · Answer

One-line intuition
 (1) tells you Y's total. It does not tell you X's total—it only tells you X's lunch.  The sugar in X's remaining meals is unknown, so you can't compare totals.

Given
 
 Lunch sugar in X =  2 ×  lunch sugar in Y. 
 Need to compare  total daily sugar .  
Statement (1)
 
This means:
 
  Total daily sugar in Y = sugar in X's lunch.   
But in X:
 
 Daily sugar =  lunch + other meals .  
So X's total is:
 
 Equal to Y if other meals have  0  sugar. 
 Greater than Y if other meals have  any  sugar.  
Cannot determine  more vs equal .
  Not sufficient. 
 
Statement (2)
X has one fewer meal than Y.
Number of meals alone tells us nothing about total sugar.
 Not sufficient. 
 
Together
(2) tells us X has fewer meals, but still doesn't tell us how much sugar is in X's other meals.
So X's total could still be  equal to  or  greater than  Y's.
 ❌ Not sufficient. 
 Answer: E

nikhilkaps · Answer

This question from the official mocks is confusing. I believe the answer is C. But as per GMAC it is E. Please guide.

Rickooreo · Answer

Let:
* Country X cafeteria lunch = 2S grams of sugar
* Country Y cafeteria lunch = S grams

* Statement (1)
For Country Y, the average sugar across all meals of the day is similar to the sugar in Country X's cafeteria lunch = 2S.
So Country Y's total daily sugar = approximately:
2S × number of meals in Y
But we don't know how many meals Y students eat.
* Insufficient

* Statement (2)
Country X students have one fewer meal per day than Country Y.
But we don't know how much sugar is in the other meals.
For example:
* X: 2 meals × 2S = 4S
* Y: 3 meals × S = 3S → X consumes more

But:
* X: 2 meals × 2S = 4S
* Y: 3 meals × 10S = 30S → Y consumes more
* Insufficient
* Together: (1) + (2)

From (1):
* Country Y's average sugar per meal = 2S
From (2):
* Country X has one fewer meal than Y.
Suppose Y has N meals.

Then:
* Y total = N × 2S = 2NS
* X has N − 1 meals.
But we still don't know the average sugar per meal in X's non-cafeteria meals.
We only know that its cafeteria lunch contains 2S. The other X meals could contain more or less than 2S.
Therefore, even together, we cannot determine which country has higher daily sugar consumption.

Answer: E — Statements (1) and (2) together are insufficient.

egmat · Answer

Hi nikhilkaps,

The C argument floating in this thread (the one that assumes X's non-lunch meals have  0  sugar and then compares per-meal averages) has  two  hidden leaks, and once you plug them the answer falls to  E .

Leak 1 - you compared averages, not totals.  The question asks who consumes  more total sugar per day , not who has a higher average per meal. Dividing by the number of meals answers a different question.

Leak 2 - you can't just set X's other meals to 0.  Nothing in the statements pins down the sugar in X's breakfast/dinner. It could be tiny or huge. That single unknown is what sinks the combination.

Here's what we actually know. Let Y's cafeteria lunch =  10 , so X's lunch =  20 . Statement ( 1 ) says Y averages  20  of sugar per meal across the day. Statement ( 2 ) says X has one fewer meal than Y.

Now build two cases that obey  both  statements:

-  Case A:  Y has  3  meals averaging  20  - Y's daily total =  60 . X has  2  meals: lunch  20  + other meal  10  =  30 . -  Y  eats more.
-  Case B:  Y again totals  60 . X has  2  meals: lunch  20  + other meal  80  =  100 . -  X  eats more.

Both cases satisfy the twice-the-lunch fact, statement ( 1 ), and statement ( 2 ) - yet one says Y, the other says X.

Same two statements - two different answers - not sufficient.  That's the definition of E.

The takeaway: statement ( 1 ) hands you Y's full daily total (average x meals), but for X it only fixes the lunch. The rest of X's day stays unknown, so the totals can't be compared - no matter how you combine the statements.

Answer: E

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