For each customer, a yogurt shop charges a cents for the

Hi,
if we look at the problem, we have three unknowns, two values/prices and one qty, and one number 5..
lets see what does each eq tells us..

(1) A customer who buys 10oz of yogurt is charged 10 percent less per ounce than a customer who buys just 5oz of yogurt.
we are given a qty value and a percentage relation to 5..
since they are no numeric values about the two values/price, this statement will give us a ratio of the two values.. insuff

(2) A customer who buys 20oz of yogurt is charged $2.50
this eq gives us the QTY and sum of two prices..
so, what we get from this statement is a linear equation between the two prices..... insuff as we still do not know indl prices..

combined we have a linear equation of two variables and the ratio of these two variables...
hence SUFF to find the value of two variables...

Lets put this Algebrically too..

(1) A customer who buys 10oz of yogurt is charged 10 percent less per ounce than a customer who buys just 5oz of yogurt.
(5a+5b)/10=0.9 * a... 5a+5b=9a... or 4a=5b..
a=5b/4... insuff..

(2) A customer who buys 20oz of yogurt is charged $2.50
5a + 15 b =2.5... insuff..

combined substitute a as 5b/4 in eq formed from statement II...
5(5b/4) + 15b = 2.5..
25b +60b =10..
b= 10/85.... and a can be found.... suff

Merging topics. Please refer to the discussion above.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

For each customer, a yogurt shop charges a cents for the first 5oz of purchased yogurt. Each additional ounce of yogurt costs the customer b cents. What is the value of a ?

(1) A customer who buys 10oz of yogurt is charged 10 percent less per ounce than a customer who buys just 5oz of yogurt.

(2) A customer who buys 20oz of yogurt is charged $2.50.

In the original condition, there are 3 variables(a,b,n:n ounces), which should match with the number of equations. So you need 3 equations. For 1) 1 equation, for 2) 1 equation, and you need 1 more equation, which is likely to make E the answer.
When 1) & 2), it becomes (a+5b)/10=a/5*(0.9), and you can calculate a and b from a+15b=2.5 in a unique way. Therefore, the answer is C.


 For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

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