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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
60% (02:39) correct
40%
(02:20)
wrong
based on 1280
sessions
History
Date
Time
Result
Not Attempted Yet
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.
(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.
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.
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.
Smita04
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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General Discussion
Posts: 148
Smita04
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..!!
Posts: 6,727
pmal04
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
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
BrentGMATPrepNow VeritasKarishma
Aditya110
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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sujoykrdatta
Expert
Tutor
Joined: 26 Jun 2014
Last visit: 23 Aug 2026
Posts: 598
Status:Mentor & Coach | GMAT Q51 | CAT 99.98
GMAT 1: 750 Q51 V39
Expert reply
GMAT 1: 750 Q51 V39
Posts: 598
pmal04
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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Posts: 43
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
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
sujoykrdatta
how did you get c=0.5 and d = 20?
deep1624
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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samarpan.g28
Joined: 08 Dec 2023
Last visit: 18 Feb 2026
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Concentration: General Management, Human Resources
GPA: 8.88
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pmal04
You will find 2 equations from statements 1 and 2 where the variables will be individual prices of cupcakes and doughnuts. However, by solving the equations, you will find the prices only, not their quantities. Choose (E).
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