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why does the variable rule not apply here?

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fangtray Really wants to Beat The GMAT! Default Avatar
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why does the variable rule not apply here? Post Mon Apr 30, 2012 5:38 am
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    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 bougth an equal number of $0.15 stamps and $0.29 stamps.

    I know the OA is A.

    In this case, 15x + 29y =440, and we can solve for X and Y.

    We can do this I assume, because x and y are integers. But is this ALWAYS the case? How can we spot on the exam when a linear equation will be solvable with 2 variables?

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    Post Mon Apr 30, 2012 6:22 am
    Interesting question.

    In equations of the form ax + by = c, if c is an integral multiple of both 'a' and 'b', then such an equation will have atleast one pair of integral solution.

    Case I: If c is the LCM of a and b, then the equation will have exactly one pair of integral solution (Unique solution). for e.g. 15x + 29y = 440
    Case II: If c is not the LCM of a and b, and still a common multiple of a and b, the equation will have more than one pair of integral solutions. for e.g. 15x + 29y = 880
    Case III: If c is not an integral multiple of both 'a' and 'b', then the equation will not have a pair of integral solution. But it will have infinite number of real number solutions. for e.g. 15x + 29y = 439

    In this question, since it's told that Joanna made a purchase, the Case III is ruled out. So, we can just check for the LCM to ascertain if Statement (1) is sufficient or not.

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    Brent@GMATPrepNow GMAT Instructor
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    Post Mon Apr 30, 2012 9:33 am
    fangtray wrote:
    In this case, 15x + 29y =440, and we can solve for X and Y.

    We can do this I assume, because x and y are integers. But is this ALWAYS the case? How can we spot on the exam when a linear equation will be solvable with 2 variables?
    Good question.
    If the question is a real life question in which the numbers are related to indivisible things, then you can be certain that the numbers are restricted to integers . . . in fact, positive integers.

    So, if we're talking about numbers of cats, people, cars, stamps, etc, then we can assume that those numbers must be positive integers.

    By the way, this is a very common trap on the GMAT. In fact, I address this in the following videos:

    http://youtu.be/GfbqINgCEqM
    http://youtu.be/TyK1-HsqkGk

    Cheers,
    Brent

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    ronnie1985 GMAT Destroyer!
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    Post Mon Apr 30, 2012 11:13 am
    Thank you experts...

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    fangtray Really wants to Beat The GMAT! Default Avatar
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    Post Mon Apr 30, 2012 7:46 pm
    aneesh.kg wrote:
    Interesting question.

    In equations of the form ax + by = c, if c is an integral multiple of both 'a' and 'b', then such an equation will have atleast one pair of integral solution.

    Case I: If c is the LCM of a and b, then the equation will have exactly one pair of integral solution (Unique solution). for e.g. 15x + 29y = 440
    Case II: If c is not the LCM of a and b, and still a common multiple of a and b, the equation will have more than one pair of integral solutions. for e.g. 15x + 29y = 880
    Case III: If c is not an integral multiple of both 'a' and 'b', then the equation will not have a pair of integral solution. But it will have infinite number of real number solutions. for e.g. 15x + 29y = 439

    In this question, since it's told that Joanna made a purchase, the Case III is ruled out. So, we can just check for the LCM to ascertain if Statement (1) is sufficient or not.
    wonderful explanation! thanks.

    but how do you know if C is a LCM (case 1) or just a multiple (case 2)?

    You state in case 1, C (440) is the LCM of A (15) and B (29), but neither 15 or 29 are factors of 440. what do you mean by this?

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    Post Mon Apr 30, 2012 9:54 pm
    aneesh.kg wrote:
    Interesting question.

    In equations of the form ax + by = c, if c is an integral multiple of both 'a' and 'b', then such an equation will have atleast one pair of integral solution.

    Case I: If c is the LCM of a and b, then the equation will have exactly one pair of integral solution (Unique solution). for e.g. 15x + 29y = 440
    Case II: If c is not the LCM of a and b, and still a common multiple of a and b, the equation will have more than one pair of integral solutions. for e.g. 15x + 29y = 880
    Case III: If c is not an integral multiple of both 'a' and 'b', then the equation will not have a pair of integral solution. But it will have infinite number of real number solutions. for e.g. 15x + 29y = 439

    In this question, since it's told that Joanna made a purchase, the Case III is ruled out. So, we can just check for the LCM to ascertain if Statement (1) is sufficient or not.
    Hi aneesh,

    I think even though the answer is correct, you explanation is fundamentally flawed.
    One can have several counter examples to prove it-

    Case 1 and 2-
    Let 6x + 10y = 30 (30 is the LCM of 6 & 10)

    This has 2 positive integral solutions-
    (5,0) and (0,3)

    Case 3-
    The question itself is a counter example!
    440 is neither a multiple of 15 nor 29.

    I think the correct reasoning is in a eqn of type-
    ax + by = c
    may have a unique positive integral solution only if a and b have no common factors other that 1.

    Note that this condition is a pre-requisite but not sufficient.
    Since 15 and 29 don't have a common factor other than 1 we check for integral solution.. if there exists one we are sure that its unique.

    Please let me know if you think I missed something in this analysis.

    Regards,
    Shantanu

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    aneesh.kg GMAT Destroyer!
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    Post Tue May 01, 2012 1:25 am
    Ah, yes. It's absolutely flawed. I can't even say that I was testing you. Smile
    I was unnecessarily trying too hard to form a rule for such questions.

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    Post Tue May 01, 2012 4:18 pm
    Brent@GMATPrepNow wrote:
    fangtray wrote:
    In this case, 15x + 29y =440, and we can solve for X and Y.

    We can do this I assume, because x and y are integers. But is this ALWAYS the case? How can we spot on the exam when a linear equation will be solvable with 2 variables?
    Good question.
    If the question is a real life question in which the numbers are related to indivisible things, then you can be certain that the numbers are restricted to integers . . . in fact, positive integers.

    So, if we're talking about numbers of cats, people, cars, stamps, etc, then we can assume that those numbers must be positive integers.

    By the way, this is a very common trap on the GMAT. In fact, I address this in the following videos:

    http://youtu.be/GfbqINgCEqM
    http://youtu.be/TyK1-HsqkGk

    Cheers,
    Brent
    ughh.. wish i had youtube in China. government blocked it Sad

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    Post Tue May 01, 2012 5:06 pm
    fangtray wrote:
    ughh.. wish i had youtube in China. government blocked it Sad
    You can also watch the videos for free at: http://www.gmatprepnow.com/module/gmat-data-sufficiency

    They are videos #11 and #12

    Cheers,
    Brent

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    Post Wed May 02, 2012 4:49 pm
    Brent@GMATPrepNow wrote:
    fangtray wrote:
    ughh.. wish i had youtube in China. government blocked it Sad
    You can also watch the videos for free at: http://www.gmatprepnow.com/module/gmat-data-sufficiency

    They are videos #11 and #12

    Cheers,
    Brent
    Brent thanks! I did not even know about GMAT Prep until i went to that website. But my question is how do we KNOW there is only 1 solution for x and Y? for example if the question i started this thread with read:

    Joanna bought only 15 cent and 29 cent stamps, how many 15 stamps did she buy?

    1. she bought $8.80 worth of stamps.
    2. she bought an equal number of 15 cent stamps and 29 stamps.


    immediately B is wrong. but what about A? In this case A is wrong too! whereas when she bought $4.40, we had 1 solution for X and Y.

    But we are taught not to solve for X and Y, and in a timed exam, when you see $4.40, how can we recognize there is 1 solution or 2 or 3?

    All we get is 15x + 29y = 440 and we shouldn't waste time solving for this, but only to recognize there is either 1 or more possible answers.

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    Post Thu May 03, 2012 7:03 am
    fangtray wrote:
    Joanna bought only 15 cent and 29 cent stamps, how many 15 stamps did she buy?
    1. she bought $8.80 worth of stamps.
    2. she bought an equal number of 15 cent stamps and 29 stamps.


    immediately B is wrong. but what about A? In this case A is wrong too! whereas when she bought $4.40, we had 1 solution for X and Y.

    But we are taught not to solve for X and Y, and in a timed exam, when you see $4.40, how can we recognize there is 1 solution or 2 or 3?

    All we get is 15x + 29y = 440 and we shouldn't waste time solving for this, but only to recognize there is either 1 or more possible answers.
    If there are no restrictions on the variables (x and y), then you shouldn't solve for them in a DS question. However, if there are restrictions (such as x and y are positive integers), then you'll have to determine how many solutions there are (or whether solutions are even possible).

    Since 440 is a multiple of 5 and since 15x will give us a multiple of 5 for any integer x, we can see that y must be a multiple of 5. This is useful, since it limits our y-values to 0, 5, 10, etc.

    From here, we can create a table to test different pairs of values for x and y.

    - Try y=0: Is it possible for 15x + 29(0) = 440 for some positive integer x? No.
    - Try y=5: Is it possible for 15x + 29(5) = 440 for some positive integer x? No.
    - Try y=10: Is it possible for 15x + 29(10) = 440 for some positive integer x? Yes. Everything works when x=10 and y=10. This is one solution.
    - Try y=15: Is it possible for 15x + 29(15) = 440 for some positive integer x? No.

    At this point, any value of y greater than 15 will bring the total price of the stamps over $4.40, so we can stop here and conclude that there is only one possible solution: Joanna bought 10 $0.15 stamps.

    Cheers,
    Brent

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    Post Thu May 03, 2012 7:52 am
    fangtray 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.

    I know the OA is A.

    In this case, 15x + 29y =440, and we can solve for X and Y.

    We can do this I assume, because x and y are integers. But is this ALWAYS the case? How can we spot on the exam when a linear equation will be solvable with 2 variables?
    It's important to understand how the GMAT lays TRAPS.

    Quote:
    1. She bought $4.40 worth of stamps.
    2. She bought an equal number of $0.15 stamps and $0.29 stamps.
    Clearly, the two statements combined provide sufficient information to determine how many 15-cent stamps were purchased.
    Eliminate E.

    Clearly, statement 2 does not by itself provide sufficient information.
    If all we know is that Joanna purchased an equal number of each type of stamp, she could have purchased 1 of each type, 2 of each type, 3 of each type, etc.
    Eliminate B and D.

    The relationships described by the two statements are extremely straightforward.
    If the OA here is C, then virtually EVERYONE will answer the question correctly, rendering the question pretty much useless.

    Thus, the correct answer almost certainly is A.

    The best way to avoid TRAPS is to understand how the GMAT lays them.

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    Post Fri May 04, 2012 3:42 pm
    Brent@GMATPrepNow wrote:
    fangtray wrote:
    Joanna bought only 15 cent and 29 cent stamps, how many 15 stamps did she buy?
    1. she bought $8.80 worth of stamps.
    2. she bought an equal number of 15 cent stamps and 29 stamps.


    immediately B is wrong. but what about A? In this case A is wrong too! whereas when she bought $4.40, we had 1 solution for X and Y.

    But we are taught not to solve for X and Y, and in a timed exam, when you see $4.40, how can we recognize there is 1 solution or 2 or 3?

    All we get is 15x + 29y = 440 and we shouldn't waste time solving for this, but only to recognize there is either 1 or more possible answers.
    If there are no restrictions on the variables (x and y), then you shouldn't solve for them in a DS question. However, if there are restrictions (such as x and y are positive integers), then you'll have to determine how many solutions there are (or whether solutions are even possible).

    Since 440 is a multiple of 5 and since 15x will give us a multiple of 5 for any integer x, we can see that y must be a multiple of 5. This is useful, since it limits our y-values to 0, 5, 10, etc.

    From here, we can create a table to test different pairs of values for x and y.

    - Try y=0: Is it possible for 15x + 29(0) = 440 for some positive integer x? No.
    - Try y=5: Is it possible for 15x + 29(5) = 440 for some positive integer x? No.
    - Try y=10: Is it possible for 15x + 29(10) = 440 for some positive integer x? Yes. Everything works when x=10 and y=10. This is one solution.
    - Try y=15: Is it possible for 15x + 29(15) = 440 for some positive integer x? No.

    At this point, any value of y greater than 15 will bring the total price of the stamps over $4.40, so we can stop here and conclude that there is only one possible solution: Joanna bought 10 $0.15 stamps.

    Cheers,
    Brent
    HI Brent,

    How do we know that statement?

    I can see 29y=5(88-3x) so y = 5(88-3x)/29, but does that tell me Y is a multiple of 5?

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    Post Fri May 04, 2012 3:59 pm
    fangtray wrote:
    HI Brent,

    How do we know that statement?

    I can see 29y=5(88-3x) so y = 5(88-3x)/29, but does that tell me Y is a multiple of 5?
    I'll answer that question in two different ways:

    #1
    Notice that no matter how many 15-cent stamps you buy, the total value of those 15-cent stamps will always be a multiple of 5. In other words, we could use nickels only to pay for those 15-stamps.

    If Joanna spent a total of 440 cents, and she spent (some multiple of 5) cents on 15-cent stamps, then the amount she spent on 29-cent stamps must be a multiple of 5. In other words, if we let y equal the total value of the 29-cent stamps, we know that 29y must be a multiple of 5.
    Since 29 is not a multiple of 5, we know that y must be a multiple of 5.

    #2
    From your calculations, we know that y = 5(88-3x)/29
    Since x is an integer, we know that 88-3x must be an integer.
    Since y is an integer, we know that 5(88-3x) must be divisible by 29
    Since 5 is not divisible by 29, it must true be that (88-3x) is divisible by 29. In other words, (88-3x)/29 equals some integer.

    If y = 5(88-3x)/29, then y = 5(some integer), in which case y must be a multiple of 5

    I hope that helps.

    Cheers,
    Brent

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