pmal04 wrote:
At the bakery, Lew spent a total of $6.00 for one kind of cupcake and one kind of doughnut. How many doughnuts did he buy?
(1) The price of 2 doughnuts was $0.10 less than the price of 3 cupcakes.
(2) The average (arithmetic mean) price of 1 doughnut and 1 cupcake was $0.35.
Say, you have 2 linear equations of the form:
Ax + By = C, and Px + Qy = R, where A,B,P,Q are the (numerical) coefficients
Case 1: If A/P = B/Q = C/R => Infinite solutions
Case 2: If A/P = B/Q but not equal to C/R => No solution
Case 3: A/P not equal to B/Q => Unique solution
This actually also follows from the standard form of a straight line: y = mx + c
For this question:
Let the price of a cupcake be $c and that of a doughnut be $d
Let the number of cupcakes be x and the number of doughnuts be y
xc + yd = 6 ... (i)
Statement 1: 2d = 3c - 0.1 ... (ii)
Here, (i) and (ii) would NOT result in a unique solution since there are too many variables - Not sufficient
Statement 2: d + c = 0.7 ... (iii)
(ii) and (iii) can be solved to calculate the price of each: c = $0.50 and d = $0.20
But we cannot determine x or y - Not Sufficient
Combining both:
Using the values of c and d, from (i): 5x + 2y = 60
Here, there are 2 unknowns, ideally there should be infinite solutions. However, we know that x and y are positive integers (additional constraint). Hence, there may NOT be infinite solutions - we should check the values:
Starting solution: x = 12 and y = 0
Decrease x by 2 (coefficient of y) and increase y by 5 (coefficient of x) to get the next solutions:
x = 10, y = 5
x = 8, y = 10
x = 6, y = 15
x = 4, y = 20
x = 2. y = 25
x = 0, y = 30 (you don't need to solve all. I was just showing the method)
So, we do NOT have a unique value - Not Sufficient
Answer E
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