At the bakery, Lew spent a total of $6.00 for one kind of

There is an error in this question.
"Average price of 1 doughnut and 1 cupcake was $0.35"
It doesn't make sense if the average price is 3.5 cents. The numbers don't work. It must be 35 cents. Convert everything to cents.

Using both statements together, you get
3C - 2D = 10 (C is the price of cupcakes and D is the price of doughnuts)
C + D = 35*2

Solving them simultaneously, you get C = 30, D = 40.

If Nc is number of cupcakes and Nd is number of doughnuts,
30*Nc + 40*Nd = 600
3*Nc + 4*Nd = 60
The first solution I get is Nd = 3, Nc = 16
We will get many more solutions such as Nd = 6, Nc = 12. Also, Nd = 9, Nc = 8 etc
Hence both statements together are not sufficient.
Answer (E)

Note that in this equation: 3*Nc + 4*Nd = 60, if instead of 60, the total price were say 18, there would have been only one solution.
3*Nc + 4*Nd = 18
Nc = 2, Nd = 3
Other possible solutions will be Nc = 6, Nd = 0; Or Nc = -2, Nd = 6 etc
Since we cannot buy negative or 0 number of items (she spends money on doughnuts AND cupcakes so she must have bought at least one of each), no other solution works.
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D = price of one donut
C = price of one cupcake
x = number of donuts bought
y = number of cupcakes bought

xD + yC = $6.00

1) 2D = 3C - 0.10
We do not know x or y, NS

2) (D + C) / 2 = 0.35
Same problem, NS

1+2) We can solve for D and C, but we can't solve for the quantity of each that has been bought.

E.

let
cupcakes purchased= x
doughnut purchased= y

price of one cupcake and doughnut be c and d respectively,
then

cx+dy = 6

we need to find y.

statement 1) 2d= 3c - 0.1
INsufficient

statement 2) (c+d)/2 = 0.035
c+d = 0.035*2

Insufficient

1) and 2)

we can have c and d but we have no info about x and y.

Insufficient.

hence E

Hope this helps..!!

Target question: How many doughnuts did Lew buy?

Let D = the NUMBER of donuts purchased.
Let C = the NUMBER of cupcakes purchased.
Let X = the PRICE per donut (in CENTS)
Let Y = the PRICE per cupcake (in CENTS)

ASIDE: Given that we have 4 different variables, we will likely need 4 equations to answer the target question.

Given: Lew spent a total of $6.00 for one kind of cupcake and one kind of doughnut.
In other words, Lew spent 600 CENTS
We can write: DX + CY = 600

Okay that's 1 equation. When I SCAN the two statements, I can see that I will be able to create one equation for each statement.
This means we will have a total of 3 equations, which likely means the combined statements are insufficient.
Given this let's jump to ......

Statements 1 and 2 combined
From statement 1, we can write: 2X = 3Y - 10
From statement 2, we can write: 1X + 1Y = 70 (CENTS)

We can solve this system to get, X = 40 and Y = 30
When we can plug these values into our first equation, DX + CY = 600, we get: D(40) + C(30) = 600
Rewrite as: 40D + 30C = 600
Divide both sides by 10 to get: 4D + 3C = 60

There are several solutions to this equation. Here are two:
Case a: D = 3 and C = 16. In this case, the answer to the target question is Lew bought 3 donuts
Case b: D = 6 and C = 12. In this case, the answer to the target question is Lew bought 6 donuts

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent
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Hi. Is there any quick method to know if there are only one or several solutions in a linear equation with two variables? In this case, there are several solutions, but does one have to use only trial and error method?

BrentGMATPrepNow VeritasKarishma

Check this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2011/0 ... -of-thumb/

It discusses integer solution in detail.
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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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Combine 1 and 2, we can solve out price for C and D, C=$0.3, D=$0.4
To fulfill the total cost $6.00, number of C and D have more than one combination, for
example: 4C and 12D, 8C and 9D…
Answer is E

how did you get c=0.5 and d = 20?

In the solution you quote, c and d denote prices of cupcakes and doughnuts in dollars.

(1) The price of 2 doughnuts was $0.10 less than the price of 3 cupcakes:
2d = 3c - 0.1

(2) The average (arithmetic mean) price of 1 doughnut and 1 cupcake was $0.35:
d + c = 0.7

So, we have a system of two linear equation: 2d = 3c - 0.1 and d + c = 0.7. Multiply the second equation by 2 and subtract the result from the first equqation:

2d - 2(d + c) = (3c - 0.1) - 2*0.7
-2c = 3c - 1.5
c = 0.3.

Substitute c = 0.3 into d + c = 0.7 to get d = 0.4.

Hope it's clear.
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