Dear guerrero25,
This is a very tricky problem, and I'm happy to help.

First of all, here's a refresher on revenue, profit, and cost, if these ideas are rusty for you:
http://magoosh.com/gmat/2013/profit-and ... -the-gmat/

Part of what's hard is that there are two unknowns:
x = amount of discount
n = number of shirts

Statement #1: The team purchased 200 T-shirts, sold each T-shirt for $12, and made a $900 profit.
So, revenue = 200*12 = $2400.
(profit) = (revenue) - (cost)
$900 = $2400 - cost
cost = $1500
Of that cost, $100 was for set up, so the rest is the cost of n shirts, at a rate of (8-(x/100)n), and we know n = 200
$1400 = n*(8-(x/100)n) = 200*(8-(x/100)*200)
OK, this is DS. At this point, we have a single equation for x, which we could solve. This statement will allow us to solve for x.
This statement, alone and by itself, is sufficient.

Statement #2: In addition to the $100 setup fee, the team paid $7 for each T-shirt.
This tells us 7 = (8-(x/100)n)
We have to be very careful not to import information from Statement #1 here. We have a single equation with two unknowns, so we cannot solve. See:
http://magoosh.com/gmat/2012/gmat-quant ... variables/
This statement, alone and by itself, is insufficient.

It seems to me the answer is (A).

Does all this make sense?
Mike
_________________

Easy one!!!

Profit =selling price-cost

cost=(8-(x/100)*200) here n=200

selling price=200*12

profit=selling-cost price= 200*12-(8-x/2)=900 ,we have a single equation for x, which we could solve
_________________

Fantastic DS Question.

One-time setup fee = $100
Discount per t-shirt = x cents
Price per t-shirt = $ (8 - n(\(\frac{x}{100}\)))
x = ?



1) n = 200
SP = $12
=> Total Selling Price = $2400
Profit = $900
=> Cost = 2400 - 900 - 100
=> Cost = $ 1400

1400 = 8 - (\(\frac{x}{100}\))*200
We can solve for x.
Sufficient.

2) 7 = (8 - n(\(\frac{x}{100}\)))
2 variables and 1 equation.
Insufficient.

A is the answer.
_________________

Dear expert, mikemcgarry, GMATPrepNow

I have a question about statement 2 : I think statement 2 is SUFFICIENT because of n=1. Why we don't plug n=1 in this equation, so we can get X?

Thanks!
_________________

Dear septwibowo,

I'm happy to respond.

My friend, on GMAT DS, it is absolutely crucial to be completely clear on what could be true vs. what has to be true. What could be true tells us bupkis about sufficiency. Only what has to be true tells us about sufficiency.

I am not sure where you got n = 1. The problem tells us explicitly that \(n \geq 50\), so this rules out the possibility of n = 1. Let's say n = 100, which would be possible: yes, knowing n = 100 would make the equation very easy to solve for X, but once again, our convenience and ease, in and of itself, tells us zilch about sufficiency.

The prompt gives us two variables, n & x. As a very good rule of thumb, we need two separate equations to solve for the values of two variables. The GMAT loves to test this fact in word problems on the DS questions. See:
GMAT Quant: How to Solve Two Equations with Two Variables

Does all this make sense?
Mike
_________________

Thank you mikemcgarry for your explanation. I understand 100% with your explanation.

However, this is the reason why I think that n=1.

In addition to the $100 setup fee, the team paid $7 for each T-shirt.

Whatever the n shirts they buy, but if the team paid $7 for EACH means n=1 right?

I buy 100 iPhone and I should pay 1.000 USD each.

This is one that makes me confused.
_________________

Dear septwibowo,

I'm happy to respond.

My friend, this is a very common mistake: people often confuse the language used to express a rate for information about a total amount or number.

For example, suppose a problem says, "On Tuesday, ABC store was selling shirts at the rate of 3 shirts for $50." That "3 shirts for $50" is information about a rate, about a ratio, but this doesn't mean that the store sold exactly three shirts or only three shirts on Tuesday---that store might have sold hundreds of shirts at that price. In fact, we know that if the store sold 150 shirts that day, they would have taken in $2500 in revenue for the day.

A gas station may sell gas at a rate of $3/gallon, but this doesn't mean that a driver can spend only $3 or can get only 1 gallon. This number merely give the ratio: if I put 15 gallons into my tank, that will cost me $45.

Much in the same way, this information:
the team paid $7 for each T-shirt
is rate information: the word "each" does mean 1, and indeed, there's a 1 in the ratio: ($7):(1 shirt). That's the rate or ratio, but that says absolutely nothing about the total number of shirts. By contrast, n is the total number of shirts, so the total cost (not including the flat fee) would be $7n.

Does this distinction make sense?
Mike
_________________

Hello mike , I got the same doubt of septwibowo...Your explanation is very clear thanks a lot!

I think the misunderstanding is generated by wrongly interpreting the equation (8-(x/100)n). At first I thought of it as a Price*Discount*Quantity = total cost paid for the t-shirts (in this case i think septwibowo approach would have been solid). However, this equation has a totally different meaning: it puts in relation discount and total number of tshirts to give the price paid for each tshirt, therefore (8-(x/100)n) = price paid for each t shirt. In this case it becomes clear why we cannot arbitrarily consider n=1 to solve the problem.

Does it make sense? Hope this clarification, if correct, might help others with the same doubt.

In statement 2, we don't know the value of the discount. Based on the statement, we can only conclude that after the t-shirt was discounted, the team paid $7 per shirt.

We can make the following equation provided by the information in statement 2:

\(7 = (8-\frac{x}{100}n)\) dollars

However, without knowing what n is, we can't determine the value of x. If n = 50, then x = 2. If n = 100, then x = 1.

\(7 = (8-\frac{2}{100}50)\) dollars

\(7 = (8-\frac{1}{100}100)\) dollars

We can't determine the value of x cents based off statement 2 alone.
_________________