Last visit was: 03 Aug 2024, 18:59 It is currently 03 Aug 2024, 18:59
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Intern
Joined: 05 Jan 2009
Posts: 48
Own Kudos [?]: 1005 [51]
Given Kudos: 2
Current Student
Joined: 08 Jan 2009
Posts: 245
Own Kudos [?]: 449 [10]
Given Kudos: 7
GMAT 1: 770 Q50 V46
Tutor
Joined: 16 Oct 2010
Posts: 15181
Own Kudos [?]: 67094 [8]
Given Kudos: 436
Location: Pune, India
General Discussion
Manager
Joined: 13 Mar 2012
Posts: 154
Own Kudos [?]: 477 [4]
Given Kudos: 48
Concentration: Operations, Strategy
2
Kudos
2
Bookmarks
Smita04 wrote:
At the bakery lew spent a total of 6\$ for one kind of cupcake and one kind of doughnut. How many donuts did he buy?
1) price of 2 doughty was \$.10 less than 3 cupcakes
2) average price of 1 doughnut and 1 cupcake was \$.035

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..!!
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6804
Own Kudos [?]: 30936 [3]
Given Kudos: 799
Re: At the bakery, Lew spent a total of \$6.00 for one kind of [#permalink]
2
Kudos
1
Bookmarks
Top Contributor
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.

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.

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

Cheers,
Brent
Intern
Joined: 02 Nov 2019
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 36
Re: At the bakery, Lew spent a total of \$6.00 for one kind of [#permalink]
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?

Tutor
Joined: 16 Oct 2010
Posts: 15181
Own Kudos [?]: 67094 [0]
Given Kudos: 436
Location: Pune, India
Re: At the bakery, Lew spent a total of \$6.00 for one kind of [#permalink]
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?

Check this post:

It discusses integer solution in detail.
Tutor
Joined: 26 Jun 2014
Status:Mentor & Coach | GMAT Q51 | CAT 99.98
Posts: 450
Own Kudos [?]: 814 [1]
Given Kudos: 8
Re: At the bakery, Lew spent a total of \$6.00 for one kind of [#permalink]
1
Bookmarks
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
Intern
Joined: 20 Apr 2020
Posts: 44
Own Kudos [?]: 11 [1]
Given Kudos: 2
Location: India
Re: At the bakery, Lew spent a total of \$6.00 for one kind of [#permalink]
1
Kudos
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…
Intern
Joined: 22 Oct 2022
Posts: 12
Own Kudos [?]: 2 [0]
Given Kudos: 88
Location: India
Re: At the bakery, Lew spent a total of \$6.00 for one kind of [#permalink]
sujoykrdatta wrote:
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

how did you get c=0.5 and d = 20?
Math Expert
Joined: 02 Sep 2009
Posts: 94778
Own Kudos [?]: 646406 [0]
Given Kudos: 86853
Re: At the bakery, Lew spent a total of \$6.00 for one kind of [#permalink]
deep1624 wrote:
sujoykrdatta wrote:
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

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.
Senior Manager
Joined: 08 Dec 2023
Posts: 284
Own Kudos [?]: 55 [0]
Given Kudos: 1230
Location: India
Concentration: Strategy, Operations
GPA: 4
WE:Engineering (Tech)
Re: At the bakery, Lew spent a total of \$6.00 for one kind of [#permalink]
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.

­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).
Re: At the bakery, Lew spent a total of \$6.00 for one kind of [#permalink]
Moderator:
Math Expert
94778 posts