GMAT Club Olympics 2024 (Day 1): Last week, Edmon had lunch in two

Let's say that Edmon paid the following amounts at both cafes:
\(A = a(1+x) = a + ax\)
\(B = b(1+y) = b + by\)
where \(a \) and \(b\) are the checks, while \(x \) and \(y \) are the % for the tip.

The question goes: is \(A > B\) ?
Here we know that the tip amount was greater for A - and this equals to the tip percentage by the base value:
\(ax > by\)
This is helpful but not sufficient, given we still need to understand what \(a\) and \(b \) are equal to, or at least how they correspond.
Basically, \(\frac{x}{y} > \frac{3}{1}\)
Which means \(x > 3y\)
Let's take the minimum value for \(x\), which will be \(x = 3y\) (keeping in mind that this is just an assumption and the equality is not correct).
Inserting it in the original equation, we will get:
\(A = a(1+x) = a(1 + 3y)\)
Still, independently it's not sufficient, because it shed no light on the value of \(b\).
 
Combining options 1 + 2:
We insert  \(x = 3y\) into \(ax > by\) and get \(3ya > yb\), and therefore \(3a > b\)
Again, getting the minimal value for \(a \)we can write it out as \(3a = b \) and \(a = \frac{b}{3} \) (keeping in mind it's not actually possible, just an extreme assumption).
As a result:
\(A = a(1+x) = a + ax = \frac{b}{3}  + 3ay =  \frac{b}{3} + 3y( \frac{b}{3}) = [m]b/3\) + yb [/m]
\(B = b(1+y) = b + yb \)

Comparing A and B we obviously get:\(\frac{ b}{3} + yb < b + yb \) and \(A < B.\)
However, we need to remember that for both conditions we took the minimal value for both \(a\) and \(x\), which means that \(A\) can only be increasing. 
And given now we see \(A < B\) with only the possibility of \(A \)growing, we won't be able to solve the task and surely decide on the right sign in this equality.

Hence the right answer is E.
 ­

From the problem, we know that:

\(Paid_A = Cost_A+Cost_A*\frac{x}{100}\)
\(Paid_A = Cost_A(1+\frac{x}{100})\) (from factoring out \(Cost_A\))

\(Paid_B = Cost_B+Cost_B*\frac{y}{100}\)
\(Paid_B = Cost_B(1+\frac{y}{100})\) (from factoring out \(Cost_B\))

­Looking at statement 1, we know that:

\(x>y\)

which means that \((1+\frac{x}{100})>(1+\frac{y}{100})\)

but that gives us no information about \(Cost_A\) or \(Cost_B\), so we can't know how much he paid.

-> Statement 1 alone is not sufficient.

From statement two,

\(\frac{x}{y} > 3\)
\(x>3y\) 

which means that \((1+\frac{x}{100})>(1+\frac{3y}{100})\)

but once again, this is just a ratio and gives us no information about \(Cost_A\) or \(Cost_B\) so we can't find out how much he paid.

-> Statement 2 alone is not sufficient.

Combining both statements:

This does not help as the information contained in \(x>3y\)­ is a subset of the information contained within m]x>y[/m], an no new information is provided. 

Therefore, the answer is E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.­­

­Last week, Edmon had lunch in two cafés, Café A and Café B. In Café A, he left an additional x% tip on the check, while in Café B, he left an additional y% tip on the check. Did Edmon pay more in Café A than in Café B?

(1) The amount of the tip Edmon left in Café A was greater than the tip he left in Café B.
(2) The ratio of x to y is greater than 3 to 1.­

Solution:
Let Ca be the check amount at Cafe A
and Cb be the check amount at Cafe B
Also, % of tip left at Cafe A = x
and % of tip left at Cafe B = y

We need to determine if Ca + \(\frac{(x * Ca)}{100}\) > Cb + \(\frac{(y * Cb)}{100}\)­
or Ca (1 + \(\frac{x}{100}\)) > Cb (1 + \(\frac{y}{100}\))­

Statement 1: The amount of the tip Edmon left in Café A was greater than the tip he left in Café B.
This means that \(\frac{(x * Ca)}{100}\) > \(\frac{(y * Cb)}{100}\)
Here, we have 2 pairs of variables i.e., (x, y) and (Ca, Cb)
We don't know the values of either pair. Hence, we cannot determine if the total amount paid at Cafe A was greater than that paid at Cafe B.

Assume Ca = 1000 and Cb = 100
Assume x = 50% and y = 1%
So Tip at Cafe A = 50% of 1000 = 500
and Tip at Cafe B = 1% of 100 = 1
here the tip amount paid at Cafe A is more than that at Cafe B and the total amount paid at Cafe A is more than that at Cafe B

Now consider another example
assume Ca = 100 and Cb = 1000
assume x = 50% and y = 1%
So Tip at Cafe A = 50% of 100 = 50
and Tip at Cafe B = 1% of 1000 = 10
here the tip amount paid at Cafe A is more than that at Cafe B but the total amount paid at Cafe B is more than that at Cafe A

Hence, INSUFFICIENT

Statement 2: The ratio of x to y is greater than 3 to 1.
Given that \(\frac{x }{ y}\) > 3
or x > 3y
This means the % of tip left at Cafe A is more than 3 times the % tip left at Cafe B.
This information alone is not sufficient to determine if Ca > Cb

Hence, INSUFFICIENT

Combining Statements 1 and 2
­\(\frac{(x * Ca)}{100}\) > \(\frac{(y * Cb)}{100}\)
and
x > 3y

Consider the same example as in Statement 1
Assume Ca = 1000 and Cb = 100
Assume x = 50% and y = 1%
So Tip at Cafe A = 50% of 1000 = 500
and Tip at Cafe B = 1% of 100 = 1
here the tip amount paid at Cafe A is more than that at Cafe B and the total amount paid at Cafe A is more than that at Cafe B

assume Ca = 100 and Cb = 1000
assume x = 50% and y = 1%
So Tip at Cafe A = 50% of 100 = 50
and Tip at Cafe B = 1% of 1000 = 10
here the tip amount paid at Cafe A is more than that at Cafe B but the total amount paid at Cafe B is more than that at Cafe A

Hence, INSUFFICIENT
Option E is the right answer­