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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
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.

Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
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.

Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
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. _________________

Diophantine Equations related Data Sufficiency. [#permalink]
16 Jun 2013, 16:32

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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.

Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
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.

Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
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.

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?

Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink]
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?

gmatclubot

Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many
[#permalink]
13 Aug 2014, 19:15

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