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

 It is currently 19 Jun 2013, 11:35

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

Author Message
TAGS:
Manager
Joined: 15 Apr 2012
Posts: 94
Concentration: Technology, Entrepreneurship
GMAT 1: 460 Q38 V17
GMAT 2: Q V
GPA: 3.56
Followers: 0

Kudos [?]: 7 [0], given: 134

Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many [#permalink]  09 Feb 2013, 20:33
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
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]  22 Mar 2013, 07:37
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. Senior Manager Joined: 10 Oct 2012 Posts: 332 Followers: 5 Kudos [?]: 130 [0], given: 25 Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many [#permalink] 23 Mar 2013, 00:47 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 --> inequalities-basics-154285.html Intern Joined: 09 Apr 2013 Posts: 2 Followers: 0 Kudos [?]: 1 [1] , given: 0 Diophantine Equations related Data Sufficiency. [#permalink] 16 Jun 2013, 17:32 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.

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

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

Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many [#permalink]  18 Jun 2013, 18:45
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
Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many   [#permalink] 18 Jun 2013, 18:45
Similar topics Replies Last post
Similar
Topics:
Joanna bought only $0.15 stamps and$0.29 stamps. How many 3 17 Aug 2004, 08:27
Joanna bought only $0.15 stamps and$0.29 stamps. How many 17 05 Feb 2005, 21:14
Joanna bought only $0.15 stamps and$0.29 stamps. How many 9 15 Sep 2007, 15:45
Joanna bought only $0.15 stamps and$0.29 stamps. How many 2 24 Feb 2008, 18:15
4 Joanna bought only $0.15 stamps and$0.29 stamps. How many 3 30 Nov 2009, 14:08
Display posts from previous: Sort by