Find all School-related info fast with the new School-Specific MBA Forum

It is currently 21 Sep 2014, 08:30

Close

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
Your Progress

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Joanna bought only $0.15 stamps and $0.29 stamps. How many

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 15 Apr 2012
Posts: 95
Location: Bangladesh
Concentration: Technology, Entrepreneurship
GMAT 1: 460 Q38 V17
GMAT 2: Q V
GPA: 3.56
Followers: 0

Kudos [?]: 10 [0], given: 134

Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 09 Feb 2013, 19:33
for statement 1, we get 15x + 29y = 440
as we need the value of x,so 15x = 440 -29y
x = (440 - 29y)/15
It's clear from the equation that the value of y should be 5 or 10 to be divided by 15 since x is an integer. if y = 5,you cannot divide it with 15. Now try y =10, so x = 10. Statement 1 is sufficient.

Also you can see from the 2nd statement that x=y .Statement 2 itself is not sufficient as you don't know the total price.
Intern
Intern
avatar
Joined: 23 Dec 2012
Posts: 2
GMAT Date: 04-02-2013
Followers: 0

Kudos [?]: 0 [0], given: 17

Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 22 Mar 2013, 06:37
I'd also like to add the following -

(1) She bought $4.40 worth of stamps. - We notice that the ending digit is 0, and one type of stamps costs 0.29. The only way we can have a "0" at the end with $0.29 is to multiply 0.29 with 10,20,30, etc. So if we have 10x0.29, its OK, meaning we have 2.9 worth of 0.29 stamps, and the rest worth $0.15. If we are to have 20, then we would surpass the total value of 4.40$ ( 20*0.29 = 5.8 ). So clearly she bought 10 stamps worth $0.29, and subsequently we can calculate the rest. That's why A is sufficient.

(2) She bought an equal number of $0.15 stamps and $0.29 stamps. - Clearly insufficient.
Expert Post
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Joined: 10 Oct 2012
Posts: 627
Followers: 41

Kudos [?]: 579 [0], given: 135

Premium Member
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 22 Mar 2013, 23:47
Expert's post
udaymathapati wrote:
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and $0.29 stamps.


Just a quick tool for Diophantine equations.

Lets assume the equation 4x+5y = 36. Rearrange it as x =\frac{(36-5y)}{4} or y = \frac{(36-4x)}{5}. Now, lets take the first re-arrangement. We can see that for y=0, we get an integral value for x as x=9 .Now all we have to do is keep adding 4 to the initial value of y=0. Thus, the next value of y which will give an integral solution for x is y=4,8,12 etc. One could also subtract 4 and get subsequent values for y = -4,-8,12 etc.

Taking the second rearrangement, we can see that x=-1, we have y=8. Thus, following the same logic, the next value of x, which will lead to an integral value for y is x= 4,9,14 etc.

Now back to the given problem:

We know that 15x+29y = 440.

Now, 29y = 440-15x

or y =\frac{440-15x}{29} .Now as because the statements on GMAT don't contradict each other, we know that one of the value of x and y for the above equation is x=y=10. Thus, for the above equation, keep adding 29 to x=10 for all the successive values for getting an integral solution for y. Thus, in accordance with the given sum, we can neglect negative integral values and state that x=y=10 is the only possible solution for both x,y>0.
_________________

All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions

1 KUDOS received
Intern
Intern
avatar
Joined: 09 Apr 2013
Posts: 2
Followers: 0

Kudos [?]: 1 [1] , given: 2

Diophantine Equations related Data Sufficiency. [#permalink] New post 16 Jun 2013, 16:32
1
This post received
KUDOS
Here is the method to never fail to answer correctly Diophantine-equations-related Data Sufficiency problems.

1. First, be sure that the 2 variables must be non-negative integers or positive integers and that each statement provides a linear equation relating the 2 variables. Furthermore, be sure that the 2 equations are not equivalent (2x+3y=20 and 6x+9y=60 are equivalent) and are reduced to the form: ax + by = c whith integral coefficients and constant term such that GCF (a,b)=1.

2. Find an initial Solution: "Take advantage" of the fact that statements never contradict each other and thus system of equations constructed with both statements have always at least one solution. So resolve the system of equations.

3. Unicity: Once you arrive to a solution, say (x0, y0), go back to the first statement alone, for example, and check the unicity of the solution using only that statement by applying the test below. In case the solution is unique, statement 2 is superfluous and statement 1 is sufficient. The answer is A or D. In case the solution is not unique the answer is B, C or E.

Apply the test on statement (2). And update your answer.

If there is more than one solution using each statement alone then the answer is C.

-------------------------------------------------------------------------------------------------------------------------------------------------
Now here is the rule that indicates whether or not a non-negative integer solution is unique to an equation:
Suppose the equation be: ax+by=c (reduced with a, b, c positive integers. i.e. GCF(a,b)=1)
If (x0-b)<0 AND (y0-a)<0 then there is no other non-negative integer solution than (x0, y0) and the corresponding statement is sufficent.
If (x0-b)>=0 OR (y0-a)>=0 then other non-negative integers solutions exist and the statement is not sufficient.

If the variables must be positive the test is:
If (x0 - b)<=0 AND (y0 - a)<=0 then there is no other positive integer solution than (x0, y0) and the corresponding statement is sufficent.
If (x0 - b)>0 OR (y0 - a)>0 then other positive integers solutions exist and the statement is not sufficient.

Note: The test is to subtract each coefficient from the solution found for the opposite variable.
--------------------------------------------------------------------------------------------------------------------------------------------------
Let's apply this to a real GMAT problem:
A man buys some juice boxes. The boxes are from two different brands, A and B. How many boxes of brand A did the man buy if he bought $5.29 worth of boxes?
(1) The price of brand A box is $0.81 and the price of brand B box is $0.31
(2) The total amount of boxes is 9

Variables here must be positive integers -number of juice boxes- since it is suggested that some juice boxes are from brand A and the rest from brand B.
1. The equations provided are:
(1) 0.81A + 0.31B = 5.29.
(2) A + B = 9
Which are reduced to:
(1) 81A + 31B = 529
(2) A + B = 9,
which is a system of reduced, linear, non-equivalent equations.

2. Find an initial Solution:
(1) 81A + 31B = 529. GCF(31, 81)=1.
(2) A + B = 9

Mutliplying (2) by 31 and subtracting it from (1) we get:
50A=250 so A=5 and B=4.
An initial solution is (5, 4)

3. Unicity:
Unicity for statement (1):
81(5) + 31(4) = 529
Since
(5 - 31) <=0 AND (4 - 81)<=0 then there no other positive solution than (5, 4) so statement (1) is sufficient.

Unicity for statement (2):
It is obvious that statement (2) alone is not sufficient but the test is still applicable.
1(5) + 1(4) = 9
Since
(5 - 1) >0 OR (4 - 1) > 0 then there are other positive solutions than (5, 4) so statement (2) is not sufficient.
Answer A.

Hope this helps.
Intern
Intern
avatar
Joined: 09 Apr 2013
Posts: 2
Followers: 0

Kudos [?]: 1 [0], given: 2

Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 18 Jun 2013, 17:45
I'd like to provide a demonstration for the rule above.

Let's ax+by=c...................................(1)
an equation with a, b, c positive integers and GCD(a, b)=1.
If (x0, y0) satisfies (1) then
a(x0) + b(y0) = c
Any other solution can be arrived at by varying in opposite directions x0 and y0 so the ax + by keep adding to c.
a(x0 + k) + b(y0 - m) = c
a(x0) + ak + b(y0) - bm = c
a(x0) + b(y0) + ak - bm = c
c + ak - bm = c
ak = bm
it follows that ak is multiple of b (and of course of a) so it is multiple of LCM(a, b)=ab since GCD(a, b) = 1.
In other words ak can be written as ab*z for some integer z.
The same applies to bm. So,
ak = bm =ab*z
So k = b*z and m = a*z.
Returning back to our equation and reemplazing k and m by their expressions in terms of b and a we get:
a(x0 + b*z) + b(y0 - a*z) = c
Which means that the generating formula for all the solutions for (1) is:
[(x0 + b*z), (y0 - a*z)] z is an integer.
the first solutions generated are:
etc
z=-2-->(x0-2*b, y0+2*a)
z=-1-->(x0-b, y0+a)
z=0-->(x0, y0)
z=1-->(x0+b, y0-a)
z=2-->(x0+2*b, y0-2*a)
etc
So the minimum variation from (x0, y0) is:
(x0+b, y0-a) and (x0-b, y0+a)
In the case that solutions must be non-negative, then if both of the above 2 solutions is not non-negative then (x0, y0) is the only non-nergative one.

In the case that solutions must be positive, then if both of the above 2 solutions is not positive then (x0, y0) is the only positive one.

CQFD
Manager
Manager
User avatar
Joined: 12 Jan 2013
Posts: 249
Followers: 0

Kudos [?]: 20 [0], given: 47

GMAT ToolKit User
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 10 Jan 2014, 08:46
udaymathapati wrote:
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and $0.29 stamps.



We are given: 0.15*x + 0.29*y = TC

1) Gives us TC..... And we assume this is insufficient, but that's not true. Look closely: the least common multiple between these two numbers has to be a multiple of 5, so we basically have a pretty heavy restriction for the possible combinations of the two given TC = 4.40... Also, if we take one of each, we are given: 0.15 + 0.29 = 0.44... Multiply this by 10 (which is a multiple of 5), and thus we have 4.40.. If we multiply by 5, we only get 2.20, if we multiply by 15, we get 6.60, by 20 we get 8.80.. Etc, etc... So the ONLY combination of x and y that works is x = y = 10. We have 10 of each.

2) Clearly insufficient, this only tells us that x = y, but without TC that could be any multiple of 5.

So we go with A.
Manager
Manager
avatar
Joined: 30 May 2013
Posts: 193
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
Followers: 0

Kudos [?]: 27 [0], given: 72

GMAT ToolKit User
Re: Stamps [#permalink] New post 31 Mar 2014, 07:15
Bunuel wrote:
udaymathapati wrote:
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and $0.29 stamps.


Let x be the # of $0.15 stamps and y the # of $0.29 stamps. Note that x and y must be an integers. Q: x=?

(1) She bought $4.40 worth of stamps --> 15x+29y=440. Only one integer combination of x and y is possible to satisfy 15x+29y=440: x=10 and y=10. Sufficient.

(2) She bought an equal number of $0.15 stamps and $0.29 stamps --> x=y. Not sufficient.

Answer: A.

So when we have equation of a type ax+by=c and we know that x and y are non-negative integers, there can be multiple solutions possible for x and y (eg 5x+6y=60) OR just one combination (eg 15x+29y=440). Hence in some cases ax+by=c is NOT sufficient and in some cases it is sufficient.



Hope it helps.

Hi Bunel,

How to find the equation has more number of solutions or has single solution?

Thanks in advance,
Rrsnathan.
Senior Manager
Senior Manager
avatar
Joined: 15 Aug 2013
Posts: 282
Followers: 0

Kudos [?]: 11 [0], given: 23

Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 13 Aug 2014, 19:15
Hi, It looks like we can use the property that a (multiple of x) + (multiple of x) = multiple of x. And from that, we can see that since 4.40 and .15 are multiples of 5, .29 has to be a multiplied by a multiple of 5 as well.

Can someone comment as to whether this property holds true for subtraction/division/multiplication?
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many   [#permalink] 13 Aug 2014, 19:15
    Similar topics Author Replies Last post
Similar
Topics:
8 Experts publish their posts in the topic Joanna bought only $0.15 stamps and $0.29 stamps. How many JimmyWorld 7 30 Nov 2009, 13:08
Joanna bought only $0.15 stamps and $0.29 stamps. How many evlanderwin 2 20 Jun 2009, 07:19
Joanna bought only $0.15 stamps and $0.29 stamps. How many sam76 2 24 Feb 2008, 17:15
Joanna bought only $0.15 stamps and $0.29 stamps. How many gluon 9 15 Sep 2007, 14:45
Joanna bought only $0.15 stamps and $0.29 stamps. How many DLMD 17 05 Feb 2005, 20:14
Display posts from previous: Sort by

Joanna bought only $0.15 stamps and $0.29 stamps. How many

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page   Previous    1   2   [ 28 posts ] 



cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.